SOLUTIONS MANUAL for 21st Century Astronomy FIFTH EDITION By Laura Kay, Stacy Palen, George Blumenth by StudyGuide

CONTENTS

Part 1 | Instructor’s Manual Chapter 1 | Thinking Like an Astronomer

Chapter 2 | Patterns in the Sky—Motions of the Earth and Moon

Chapter 3 | Motion of Astronomical Bodies

Chapter 4 | Gravity and Orbits

Chapter 5 | Light

Chapter 6 | The Tools of the Astronomer

Chapter 7 | The Birth and Evolution of Planetary Systems

Chapter 8 | The Terrestrial Planets and Earth’s Moon

Chapter 9 | Atmospheres of the Terrestrial Planets

Chapter 10 | Worlds of Gas and Liquid—the Giant Planets

Chapter 11 | Planetary Moons and Rings

Chapter 12 | Dwarf Planets and Small Solar System Bodies

Chapter 13 | Taking the Measure of Stars

Chapter 14 | Our Star—The Sun

105

Chapter 15 | The Interstellar Medium and Star Formation

113

Chapter 16 | Evolution of Low-Mass Stars

121

Chapter 17 | Evolution of High-Mass Stars

129

Chapter 18 | Relativity and Black Holes

137

Chapter 19 | Galaxies

143

Chapter 20 | The Milky Way: A Normal Spiral Galaxy

149

Chapter 21 | The Expanding Universe

157

Chapter 22 | Cosmology

163

Chapter 23 | Large-Scale Structure in the Universe

169

Chapter 24 | Life

175

Part 2 | Answers to Starry Night Workbook Exercises Exercise 1 | The Celestial Sphere

185

Exercise 2 | Earth’s Rotation Period

185

Exercise 3 | Motion of the Sun Along the Ecliptic

186

Exercise 4 | Motion of the Moon

186

Exercise 5 | Earth and Moon Phases

187

Exercise 6 | Sunrise on Mars

189

Exercise 7 | Precession

191

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vi ◆ Contents Exercise 8 | Kepler’s Laws

191

Exercise 9 | Flying to Mars

192

Exercise 10 | The Moons of Jupiter

192

Exercise 11 | The Rings of Saturn

193

Exercise 12 | Pluto and Kuiper Belt Objects

194

Exercise 13 | Asteroids

194

Exercise 14 | The Magnitude Scale and Distances

194

Exercise 15 | Stars and the H-R Diagram

196

Exercise 16 | Nebulae: The Birth and Death of Stars

197

Exercise 17 | Pulsars and Supernova Remnants

197

Exercise 18 | Galaxy Classification

198

Exercise 19 | Quasars and Active Galaxies

198

Exercise 20 | Views of the Milky Way

199

Exercise 21 | Globular Clusters

200

Exercise 22 | The Neighborhood of the Sun

200

Exercise 23 | Beyond the Milky Way

201

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CHAPTER 1

Thinking Like an Astronomer INSTRUCTOR’S NOTES Chapter 1 is an introduction to the measures and methods of astronomy. Major topics include

▶ our cosmic address; that is, the hierarchy of structures from solar systems to superclusters

▶ an intuitive scale of the universe ▶ relevant and relative distance scales, including the lightyear and light-travel time

▶ the scientific method and relevant vocabulary; that is,

distinguishing theory from idea ▶ reading graphs and using trends or patterns to under stand data Greetings, fellow professors! Whether you are using this textbook to teach a one semester or a yearlong astronomy course, to teach nonscience majors for general education credit, or to teach prospective physics and astronomy majors, you have an auspicious and audacious task ahead of you: to teach the whole universe, from the unimaginably small to the incomprehensibly large. As Dr. Mike Seeds of Franklin & Marshall College explains, “The Universe is very big, but it is described by a small set of rules and . . . we have found a way to figure out the rules—a method called science.” It is my hope that I can share in these “Instructor’s Notes” some successes and failures, tips and traps, in teaching this material to a diverse audience. At the beginning of every course, I provide to students an anonymous survey in which I ask them to rate their comfort and previous experiences with math and science, and a majority usually report that they consider themselves to be bad at math and afraid of science. Over nearly two decades of interacting with introductory astronomy students, I have found that they report a few common themes. First, they think that physics and astronomy are only about doing math problems, and second, much of their discomfort comes from previous experiences in which they were assumed already to be well versed in the vocabulary of science. Much as I do in my first lessons, this chapter aims to ease students into the astronomy curriculum by addressing both of those issues. Astronomy deals with numbers that span the gamut from the subatomic to the whole universe. I may be

quite comfortable discussing wavelengths in nanometers, particle densities in atoms per cubic centimeters, masses in 1030 kilograms, and distances in gigaparsecs, but I find it useful to conduct exercises with Figures 1.1 and 1.3 or show a version of the “Powers of Ten” montages to provide students with some visual context for the ranges of size, mass, speed, and time that are discussed in this course. Much of the quantitative problem solving in this course can be achieved through proportional reasoning (that is, how does brightness change if distance triples?), so in addition to asking questions about scientific notation and unit conversion, I introduce some basic ideas of how area or volume changes with size. For many of my students, this is the last formal science course they will ever take, so one of my learning outcomes is that they learn the process of science, gain scientific literacy, and understand the difference between science and pseudoscience. The seeds of these outcomes are sown in this first chapter through discussion of the scientific method and of the various logical fallacies presented in the Exploration section. Although science is ideally independent of culture or creed, it has often collided with religious or other strongly held beliefs. Therefore, because science is a human activity carried out by individuals who may hold nonscientific beliefs, I emphasize that we must construct safeguards within our work to counteract any personal bias that might taint their results. Thus, science is all about searching for objective truths that lead to conclusions that are repeatedly found to be unfalsifiable.

DISCUSSION POINTS

▶ Have students look at the sketches shown in Figures 1.1

and 1.3. Ask them if they are familiar with any of the shapes and structures shown. Where have they encountered them before? ▶ Have students think about the times given in Figure 1.3. Discuss the distances and times between our planet and nearby stars, and relate that to the likelihood that we will communicate with extraterrestrials in our lifetime 1

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2 ◆ Chapter 1 Thinking Like an Astronomer (remind students that we have only been broadcasting and listening for 60 to 70 years). ▶ Astronomers need to keep collecting data from the objects in the universe to find unexpected trends and to test new and old hypotheses. Discuss how this process has analogies in students’ own experiences. Have they ever had to collect data to learn something or to explore the unknown? One possible exercise is to have students compare their course grades with the amount of time they spend using a professor’s office hours as a gentle but realistic way to compare their actual and desired performance. ▶ Why do scientists adhere to the principle known as Occam’s razor? Is that principle an objective truth? Discuss examples of applications of Occam’s razor and examples of objective truths. ▶ Ask students if they are familiar with any scientific equations. Discuss differences and similarities between a well-known scientific equation and a world-renowned work of art. ▶ Discuss how the reclassification of Pluto as a dwarf planet rather than as a major planet makes sense in light of current scientific evidence and our understanding of the Solar System. Why did the case of Pluto create so much emotional turmoil among astronomers and the public? Is the final result of the voting at the meeting of the International Astronomical Union (IAU) in Prague representative of the majority of the astronomical community? It may be useful to have students investigate the biological reclassification of the duck-billed platypus as a parallel example that did not stir such emotional responses.

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. Radius of Earth–light-minute–distance from Earth to Sun–light-hour–radius of Solar System–light-year. Use Figure 1.3 2. (b) Theories must be testable, and a theory is valid up until a test fails. 3. (c) Patterns and order are indicative of a physical process at play. Reading Astronomy News

Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts presented in Chapter 1. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive.

1. Enceladus is 310 miles across, according to the article. The moon’s diameter is about 2160 miles, or roughly seven times bigger. According to Google Maps, the distance from Chicago to St. Louis is about 260 miles, or about 50 miles less than the size of Enceladus. The distance from Santa Cruz to Los Angeles is about 290 miles, or about 20 miles less than this moon’s diameter. The ocean is hypothesized to be about 6 miles deep and about the size of Lake Superior (about 160 miles wide, according to Wikipedia). 2. The original discovery of geysers on Enceladus was from an observation (image). 3. Scientists made this new discovery by observing both geysers and subtle Doppler shifts in the radio signals from the satellite. They did not directly observe water but concluded the presence of an ocean based on the terrain of the moon, the presence of geysers, and the possible sources of inhomogeneity in the moon’s internal structure. 4. Two key parts of the scientific process are that the hypothesis be testable and that tests be repeatable. Implicit in these is the need for multiple and independent tests and confirmations. 5. Life, as we know it, is dependent on the presence of a solvent such as water. Also, the presence of liquid water suggests warm temperatures for life-forms to grow and thrive.

Look-Back Time Simulator

Test Your Understanding

The Look-Back Time Simulator shows the finite speed of light and how the great distances to most astronomical objects cause us to observe things as they were in the past. This simulation is very useful when developing both the scale of the universe and an intuitive understanding that distant objects allow us to look back in time. Text reference: Section 1.1

1. f, e, c, b, a, g, d is the correct order from smallest to largest size. 2. (a) See Figure 1.3. 3. (b) Note that one must accumulate facts (a) to consider how they are related, and that science makes predictions based on these relationships (d), but “understanding” is the development of these relationships.

NEBRASKA SIMULATIONS

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Chapter 1

4. (a) The universe is understood to be homogeneous and isotropic on its largest scales. 5. (d) The Sun is the center of our Solar System, just one of the billions of stars in our galaxy, and one of the billions of galaxies in the universe. 6. (a) It is the distance that light travels in one year. 7. (c) Occam’s razor suggests that nature relies on the simplest (or most straightforward) processes. 8. (d) Distance units in terms of light speed are very convenient but sometimes odd to think about at first because we seem to be using time to refer to distance. This problem shows us two ways of considering the meaning of light distance. 9. (d) A reading of the plot 1.12c shows that at time step 10, the number is about 1000 whereas four orange dots to the left, it is under 100; therefore, it went up by a factor of more than 10 times. 10. (d) As the answer indicates, science relates only to the natural world. 11. (c) Our understanding must be tested and, at any time, a test could show that it is wrong. Note that this is not an issue of being worthless or incomplete, but merely reflects the fact that we are constantly testing our theories and hypotheses. 12. (c) Light travels a light-year in one year, so a star that is 10 light-years away emitted its light 10 years ago for us to see it today. 13. (d) Carbon is made inside stars. 14. (b) Except for hydrogen and helium (and a tiny bit of lithium), all the elements found on Earth were produced in stars. Note that the beryllium produced in the Big Bang was unstable and decayed long before the Earth formed. 15. b, d, a, c, e. The material for the Sun had to come before the Sun could be formed. Gas came first (b), then formed stars to make heavier elements (d), then the stars blew up to spread those elements around (a), then the gas had to collect (c) before it could form the Sun and the planets (e). Thinking about the Concepts 16. Tau Ceti e, Tau, Ceti Milky Way Galaxy, Local Group, Laniakea Supercluster. 17. The cosmological principle essentially states that every observer in the universe should find that the natural laws governing his or her local region are representative of the natural laws governing the universe as a whole. Consequently, he or she should derive the same natural laws that an observer on Earth derives.

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Thinking Like an Astronomer ◆ 3

18. 8.5 minutes. 19. Andromeda is about 2.3 million light-years away, so it would take 2.3 million years. 20. Answers will vary. An example is General Relativity superseding Newtonian mechanics, which began at the first step of the Process of Science when Einstein thought about gravity and spacetime. 21. This is a pseudoscientific theory because it is not falsifiable. Although it is possible that we may someday stumble upon irrefutable evidence that aliens visited Earth in the remote past, the absence of evidence today cannot be used to refute the hypothesis. In fact, proponents of the theory will simply argue that we just have not found any evidence yet. Evidence that could support the theory would include finding advanced technology in ancient archaeological sites or buried in old geological layers. The only tests I can think of to refute the hypothesis are: (1) to demonstrate that every piece of technology and archaeological monument could have been reasonably constructed with human knowledge of the time; or (2) to invent a time machine and return to the most likely times for aliens to have visited Earth. Because option 2 is utterly implausible, and because option 1 does not preclude alien visitations, it is impossible to falsify the hypothesis. 22. Falsifiable means that something can be tested and shown to be false/incorrect through an experiment or observation. Some examples of nonfalsifiable ideas might include emotional statements (such as “love conquers all”), and opinions (such as “coffee is better than tea”). Students may have a wide variety of these and other ideas, but all sacred cows are usually considered to be nonfalsifiable by the people holding those beliefs. Falsifiable ideas include cause and effect (coffee puts hair on your chest) and logic (if I drink coffee, I will not sleep tonight). 23. A “theory” is generally understood to mean an idea a person has, whether there is any proof, evidence, or way to test it. A “scientific theory” is an explanation for an occurrence in nature; it must be based on observations and data and make testable predictions. 24. A hypothesis is an idea that might explain some physical occurrence. A theory is a hypothesis that has been rigorously tested. 25. (a) Yes, this is falsifiable. (b) Find a sample of a few hundred children born during different moon phases who come from similar backgrounds and go to similar schools, and follow their progress for a number of years.

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4 ◆ Chapter 1 Thinking Like an Astronomer 26. In 1945, our distance-measuring methods were not correctly calibrated and, as a result, our distance to Andromeda was wrong. As we improved that calibration, we found different and more reliable measurements of its distance. In science, statements of “fact” reflect our current best understanding of the natural universe. A scientific “fact” does not imply that science has determined absolute truth; rather, it is simply a statement that this is the best understanding of nature that our current knowledge and technology supports. Over time, all scientific “facts” evolve as our knowledge base and technology grow. 27. Answers will vary. Depending upon the generality of the horoscopes, students may provide a wide array of answers for this question. For general statements, students might find that several, if not all, of the horoscopes on a given day could describe their experience. For a very specific horoscope, we expect that it should match approximately one-twelfth of the students regardless of their astrological sign. In any event, if astrology accurately reflected some natural truth, we would expect nearly everyone to find one and only one horoscope each day that describes his or her experi ence, that the horoscope would match the person’s astrological sign, and that the daily horoscope would be accurate for each person for the entire week of record keeping. Students should perform this experiment and be honest with themselves about the results. 28. Taken at face value, this is a ridiculous statement, but there are several items to consider critically before we apply a label of “nonreputable.” First, was this statement a sound byte taken out of context? Did the scientist simply misspeak when he or she might have been trying to say that we have not yet found extraterrestrial life? If, in fact, the statement can be taken at face value, then the credibility of the scientist might be called into question because he or she has forgotten that absolute truth is not falsifiable (and therefore not scientific) by definition. 29. Some scientific fields rely heavily on math, whereas others hardly use it at all. The use of mathematics is not the hallmark of good science. Rather, it is following the scientific method, which astrology does not employ. 30. Only hydrogen and helium (with perhaps a trace amount of lithium) were created in the Big Bang. Heavier elements such as carbon, oxygen, nitrogen, and iron are manufactured in the interiors of massive stars. At least one generation (and more likely several generations) of stars must die in massive supernova explosions to make heavy elements available to construct planets and the building blocks for life. Therefore,

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because all the heavy elements in our bodies were originally manufactured in stars, it is fair to claim that we are truly made of stardust. Applying the Concepts 31. Setup: Remember that to convert to scientific notation, count up all the digits to the right of the first one if the number is greater than 1, or the number of digits between the decimal point and the first nonzero digit if it is less than 1. Solve: (a) 7 3 109. (b) 3.46 3 10–3. (c) 1.238 3 103. Review: A good way to check is to use a scientific calculator, where “times 10 to the” is usually the “EE” key. 32. Setup: To convert scientific to standard notation, move the decimal point the number of digits indicated in the exponent, to the right if the number is positive, and left if negative. Solve: (a) 534,000,000. (b) 4,100. (c) 0.0000624. Review: Again, you can test this by using your calculator. 33. Setup: Distance is given in terms of speed and time by d 5 vt, where v is speed and t is time. If speed is in km/h, then use time in hours, for which we may have to convert. Remember there are 60 minutes in an hour. km 3 1 h 5 35 km. Solve: (a) d 5 vt 5 35 h km 1 3 h 5 17. 5 km. (b) d 5 vt 5 35 h 2 km 1 3 h 5 0. 58 km. h 60 Review: A good sanity check is to make sure the distance traveled is smaller if the time traveled decreases. 34. Setup: We are given the relationship surface area A is proportional to radius r squared or A  r2. To work a proportional-reasoning problem, insert the factor by which one variable changes into the formula to see how the result changes. Solve: (a) If r doubles, then r → 2r, or r changes by a factor of 2. Putting this in our formula shows A  22 5 4, or the area changes by a factor of 4. (b) If r triples,

then it changes by a factor of 3, or A  32 5 9. (c) If r 1 1 is halved, then r → r, or r changes by a factor of 2 2 2 1  1 therefore A    5 . (d) If r is divided by 3, then  2 4 2

1  1 A  5 .  3 9

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Chapter 1

Review: Note first that the change in area is much larger than the change in radius, which reflects the dependence on size squared. Note also how easy it is to do proportional reasoning rather than using the full surface-area formula (A 5 4pr2), when all we need to know is how much the result changes. 35. Setup: In this problem, we convert among distance, rate, and time with d 5 vt, or solving for time, t 5 d/v. The problem is straightforward because the units of distance are already the same. d 384 , 000 km 5 480 hr. There are Solve: t 5 5 v 800 km/hr 24 hours in a day, so this would take day 480 hr 3 3 20 day, or about two-thirds of a 24 hr month (a typical month is 30 days). Review: A typical flight from New York to London takes about 7 hours and covers a distance of about 6,000 km. The moon is 64 times farther away so it would take about 64 3 7 5 448 hours to reach the Moon using these estimates. This is about the same amount of time as we found by exactly solving the problem. 36. Setup: In this problem, we will convert between distance, rate, and time with d 5 vt or, solving for speed, v 5 d/t. The problem is straightforward because the units of distance are already the same. d 384,000 km day 3 5 5, 333 km/h. Solve: v 5 5 t 3 days 24 h 5, 333  6. 7 times faster than a jet plane. 800 Review: Using the result from problem 35, we have to travel 120/3  6.7 times faster than a jet plane, which agrees with our solution. 37. Setup: We are given the problem in relative units, so we don’t need to use our speed equation or use the actual speed of light. Instead, we will use ratios. Solve: (a) If light takes 8 minutes to reach Earth, then it takes 8 3 2 5 16 minutes to go twice as far. Neptune is 30 times farther than the Sun, so light takes 8 3 30 5 240 minutes, or 240/60 5 4 hours. (b) This means that sharing two sentences will take half a day, so it would take a few days just to say hello and talk about the weather. Review: If you watch 2001: A Space Odyssey, you will note that the televised interview between Earth and David Bowman had to be conducted over many hours and then edited for time delays. This was factually correct. Because Pluto is much farther than Jupiter, it This is about

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Thinking Like an Astronomer ◆ 5

stands to reason that it would take light and communication a lot longer still! 38. Setup: Light travels at 3 3 105 m/s. d To find the travel time, use d 5 vt or t 5 . v 6 d 56 3 10 km 5 187s. Solve: t 5 5 v 3 3 105 km/s Likewise using 400 3 106 km, t 5 1330 s. Review: Light takes about 8.3 min to travel from the sun to Earth, or about 500 sec. Our numbers are consistent with this duration. 39. Setup: We are given the relationship surface area A  r2. To work a proportional-reasoning problem, insert the factor by which one variable changes into the formula to see how the result changes. Solve: If the Moon’s radius is one-fourth that of 2 1  1 the area Earth, then its surface area is A    5  4 16 of Earth. Review: We saw this same behavior in problem 34. 40. Setup: Note that 3.6 3 104 km is 3.6 3 107 m. We will use the equation d 5 vt, where the distance is 2 · 3.6 3 107 m and light travels at v 5 c 5 3 3 108 m/s. 7

d 2  3. 6 3 10 m 1 5 0. 24 sec or about Solve: t 5 5 8 c 3 3 10 m/s 4 a second. Review: If we are receiving information by Internet satellite on a regular basis, we almost never notice a lag so the time has to be short, on the order of what we found (much less than 1 second). 41. Setup: Let the horizontal axis be time and vertical be population. If we choose to plot a graph in linear space, then a constant population will be a horizontal line, whereas an exponential growth will look similar to Figure 1.7, and a crash will be almost vertical. Solve: Answers will vary. Here is one example in which the baseline population is “1” unit. Review: Note how the growth starts out very slowly, jumps up very rapidly, and takes a nosedive down. This is what the text described. 42. Setup: We need our assumptions of speed. Assume a car goes 70 miles per hour if we include filling up with gas, eating, and restroom breaks. On foot, a person walks about 2 miles per hour with these same stops. We also need to relate distance, rate, and time with the formula distance equals rate times time, or d 5 vt. Solve: Solving for time, t 5 d/v, so by car, t 5 2,444 miles/70 mph 5 34.9 hours. Because there are 24 hours in a day, the car takes 34.9/24 5 1.45 days.

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6 ◆ Chapter 1 Thinking Like an Astronomer Note these assume you travel around the clock, which we do not usually do! Review: If you drive “almost” non-stop, you can go from NY to LA in 3 days. This is consistent with our value, because that assumed no stops at all. There are 30 days in a month, so this is 51/30 5 1.70 foot-months. There are 12 months in a year, so this would take 1.70/12 5 0.14 foot-years. 43. Setup: For water to freeze, it has to cool down to 0°C; then the liquid has to become solid. Solve: (a) This theory makes no sense to me because hot water will have to cool down much more (and therefore take much more time) than cold water once in the freezer. (b) Yes, this is easily testable. Simply try it in your dorm or room fridge (they all have little ice cube areas at the top). (c) I tried it, and it took about five times longer for the hot water to freeze, confirming my hypothesis. Review: Going back to our original physical reasoning, we see that this theory could be easily refuted without experimentation. Sometimes it is not as straightforward, and the experiments must be performed. 44. Setup: On the surface, it seems that the two pizzas cost the same number of dollars per inch; but remember that each pizza is a circle so we eat the volume, not the diameter. Solve: If both pizzas have the same thickness, then we only need to worry about area A 5 pr2; so this means area goes as size squared. That is, the 18-inch pizza is four times larger than the 9-inch one. But the 18-inch pizza costs only twice as much, so it is more economical to buy the larger one. Review: Often, larger items cost less per unit than smaller ones because almost the same amount of labor went into making them, and labor is generally the highest part of the cost. This is why you should always check the unit price when buying things. 45. Setup: For part (a), use the formula given. For part (b), we need to relate distance, speed, and time by d 5 vt, where we will solve for time. We use the formula again for part (c), where we must remember there are 24 hours in one day. C 5 2r 5 2 3 1.5 3 108 km 5 9.4 3 108 km. Now, solve: v 5 d / t 5

9. 4 3 108 km 5 1. 075 3 105 km/hr, 8 , 766 hr

or 107,500 km/hr. (c) Because d 5 vt, and there are 24 hours in one day, the Earth moves about 258,000 km per day.

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Review: It is amazing that we are hurtling around the Sun at more than 100,000 km/hr and do not even realize it! Why? Because everything else (planets, the Sun, stars) is so far away that we have no reference point to observe this breakneck speed. Using the Web 46. Pluto is about the size of the United States. So is the Moon. Venus is a little smaller than Earth, and Sirius B is a little larger. Many stars are larger than the SunEarth distance, including Enif, Deneb, R Doradus, Pistol Star, Antares, etc. Voyager 1 is about 0.002 light-year or about 3/4 of a light-day away. The Cat’s Eye nebula, Gomez’s Hamburger, the width of the Hourglass Nebula, and the height of the Blinking Nebula are about the size of the distance to the nearest star. The Milky Way is about 1021 meters across, and the Solar System is about 1014 meters (Sun to Sedna), so the Milky Way is about 10 million times larger; if a student only measures the Neptune or the Kuiper Belt, the Milky Way is about 100 million times larger. The Local Group is about 100 times larger than the Milky Way. The observable universe is about 10,000 times larger than the Local Group. 47. (a) Answers will vary. You should discuss whether the video was effective for showing the size and scale of the universe. I found this video useful but it starts a little large because I still have to think about my size in terms of the mountains. (b) Answers will vary. You should discuss whether the video was effective for understanding the size and scale of the universe. I found this video quite useful for understanding the extreme variations in scale between the atomic and entirety of the visible cosmos. 48. Answers will vary. A response will include where the image was taken, how it was made, what it shows, where that object is located, whether the explanation given makes sense to you, and whether you feel this website is useful to those who are interested in astronomy. 49. Answers will vary. You should present an article about space or astronomy, and note whether the news site or paper has a separate science section. You should also note whether this is a press release, interview, or report on a recent article. Report on how widespread the coverage is, including whether other papers picked up the news nationally and internationally, and in blogs. Comment on whether you think the story was interesting enough to cover.

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Chapter 1

50. Answers will vary. Report your reading of a science or astronomy blog. Present who the blogger is and his or her background, the topic of interest, whether it is controversial, what kinds of feedback or reader comments are present, and whether the post was interesting or engaging enough for you to read further posts on this blog. Be warned: many blogs sound authoritative but are not written by experts. So be sure to verify the credentials of the author. Exploration

Thinking Like an Astronomer ◆ 7

2. This is a slippery slope, because I am assuming that my performance on the first event must influence the next. 3. This is a biased sample, or small-number statistics, because I assume that my small circle of friends represents everyone. 4. This is an appeal to belief in which I argue that because most people believe it, it must be true. 5. By attacking the professor rather than the theory, I am committing an ad hominem fallacy. 6. This is an example of begging the question (a bit of a syllogism, too) in which the proof of my assertion comes from another of my own assertions.

1. This is an example of post hoc ergo propter hoc, in which we assume that the chain mail caused the car accident.

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CHAPTER 2

Patterns in the Sky—Motions of the Earth and Moon INSTRUCTOR’S NOTES Chapter 2 covers the causes and effects of the (apparent) motions of the stars, Sun, and Moon in our sky. Major topics include

▶ the celestial sphere ▶ the daily and yearly paths and motions of stars in our sky

▶ Earth’s axial tilt as the cause of the seasons ▶ Moon phases ▶ solar and lunar eclipses Many students come into our classes believing that Earth’s shadows cause Moon phases and that summer happens because Earth is closer to the Sun. There is a video from the late 1980s called “A Private Universe” that shows school children trying to learn the real causes of the seasons and the phases of the Moon. After a full lesson, most of the students have incorporated these original misconceptions into the actual reasons to make new, but still incorrect, explanations. The video then goes to Harvard University and asks graduating seniors and faculty the same questions, and only a physics professor answers both correctly. The point is not to humiliate any of them but to highlight that: (1) it is very hard to unravel a misconception and replace it with a correct model; and (2) a very large number of people carry around incorrect explanations of the regular rhythms of our world. Two of my own measures for the success of my introductory astronomy course are, first, whether students can correctly explain the causes for the daily and annual changes that occur around them, and second, whether they can use the processes of scientific thought to assess critically the likelihood of their own reasoning or model to explain a phenomenon. With that in mind, I spend about 2 weeks on this chapter because I find it so critical in addressing these two goals. In particular, I like to start by asking students to give me their explanations for these two phenomena, so that I can also focus on guiding them to examine why initially incorrect reasoning is flawed. I often tell my students that understanding why the wrong explanations are wrong

is an important (if not the most important) part of the process of science. There is, of course, much more in this chapter than just the causes of the Moon phases and the seasons. During these lessons, many of my students discover that they never consciously noticed that days are longer or that the Sun is higher in the sky during summer than winter; that stars move across our sky on angled paths rather than going from due east to the zenith to due west; or that the Moon can be visible during the day. To flush all this out, I spend the first full class having them just learn to describe the positions of objects in the sky. The next class is devoted to how objects at different positions in the sky move across it. We then spend one class on the paths the Sun takes through the sky at different times of year. The next class is spent unraveling all the wrong reasons commonly assumed as causing the seasons. I spend two classes on Moon phases—first on the causes and then on relating phase, position in the sky, and time of day. This slower pace allows students the time to ensure that they can explain to one another both why the right models are correct and why the wrong models are incorrect. Plus, out of everything that students will learn in the whole course that can be of practical use for the rest of their lives, I find the contents of this chapter are most relevant. I cover later chapters at a faster pace (roughly one class period once we are done with laws of physics) because by then, students have gained experience thinking “scientifically.” As one final piece of advice, I encourage students to take full advantage of the animations and simulations in this chapter. They are excellent examples of supplementary material to help students master these confusing but critical concepts outside of class.

DISCUSSION POINTS

▶ During winter in the Northern Hemisphere, Earth is closer to the Sun than at any other time of the year. Have students explain why the changing Earth2Sun distance has no bearing on Earth’s seasons.

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10 ◆ Chapter 2 Patterns in the Sky—Motions of the Earth and Moon

▶ Use Figure 2.20 to show why Earth’s shadow does not

cause the phases of the Moon. Use the same figure to explain why we can sometimes see the Moon during the day. ▶ Apply the dependence of the perspective on the sky shown in Figure 2.7 to the location of the classroom. Engage students coming from other countries or states to discuss the perspective on the sky from their birthplaces. Make them realize that these perspectives are similar for places with similar latitudes even though they may be very far apart because the perspective does not change with longitude. What are the consequences of similar perspectives of the sky on the perception of the seasons in different places? What are the differences in regions with different latitudes? Discuss how it all depends on the altitude of the Sun. ▶ To stress the cultural aspect of astronomy over the course of human history, ask students to think about why the constellations have the names they do. How did Egyptian cultures view the sky compared to Greek and Roman cultures? ▶ To help students become more comfortable with the mathematical ideas involved in astronomy, investigate Working It Out 2.1 with them. The geometric and algebraic practice will be beneficial. Have them convert Earth’s radius to miles, and compare that to other lengths such as the distance between London and New York City or between New York City and Los Angeles. ▶ To familiarize students with the night sky, discuss how to identify Polaris (the North Star) from the position of the classroom. Many students are surprised when they realize that Polaris is not that bright in comparison to other stars—its position is its most important attribute. Utilize the fact that the altitude of Polaris is equal to the latitude of the classroom. Have students identify a way to locate the South Celestial Pole using a celestial globe or map. ▶ Ask students to think about how the alignments of the Sun, Moon, and Earth would change the frequency or nature of solar and lunar eclipses. What if the Moon were twice as far away from Earth? Twice as close? What if the Sun were moved in those different positions, with or without the Moon being moved as well? What if the Moon’s orbit were inclined more or less to the ecliptic? How would things change if the Moon moved more slowly in its orbit?

integrated into assignable Smartwork5 online homework exercises. Earth Spins and Revolves This animation shows Earth as it is positioned with respect to the Sun, including motion along its orbit, spin axis tilt, and discussion of the causes for the seasonal variation in climate in terms of latitude and angle of incident sunlight. Text reference: Sections 2.1 The View from the Poles This animation shows how Earth’s rotation corresponds to the movement of the stars in the sky and the rotation of the stars around Polaris, which is very nearly at the north celestial pole. It also explores the precession of Earth’s rotation, including how the movement of the stars will look when Polaris is no longer the North Star. Text Reference: Section 2.1 The Celestial Sphere and the Ecliptic This animation shows side-by-side perspectives of: (1) the view of the sky (day and night) from a backyard; and (2) an “outside” view of Earth embedded in a concentric celestial sphere. The point of view moves with Earth as it rotates on its axis. The content of the animation focuses on the ecliptic, showing the motion of the Sun, Moon, and constellations in relation to one another. Text Reference: Section 2.1 The Moon’s Orbit: Eclipses and Phases This interactive animation explores the Earth-Moon-Sun system, building on the elements of two previous animations (“The Earth Spins and Revolves” and “The Celestial Sphere and the Ecliptic”). It shows a changing point of view from the size scale of Earth’s orbit down to the size scale of the Moon’s orbit, followed by emphasis on the Moon’s orbit to distinguish the concept of an eclipse versus a phase, and shows the relative configurations of Earth, Moon, and Sun. Text reference: Section 2.3

ASTROTOUR ANIMATIONS

NEBRASKA SIMULATIONS

The following AstroTour animations are referenced in Chapter 2 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also

Developed at the University of Nebraska2Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts

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presented in Chapter 2. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive.

Patterns in the Sky—Motions of the Earth and Moon ◆ 11

Ecliptic (Zodiac) Simulator This simulation shows the position of the Sun in the zodiac in different months of the year. Text reference: Section 2.2

Celestial and Horizon Systems Comparison This simulation demonstrates how the celestial sphere and horizon diagram are related. Text reference: Section 2.1

Seasons and Ecliptic Simulator This simulation uses the geometry of the Earth as it goes around the Sun to demonstrate why seasons occur. Text reference: Section 2.2

Rotating Sky Explorer This simulation demonstrates how rotation of the Earth leads to apparent rotation of the sky and how the celestial sphere and horizon are related. Text reference: Section 2.1

Sun’s Rays Simulator

Meridional Altitude Simulator

Paths of the Sun

This simulation shows helpful diagrams for finding the meridional (or maximum) altitude of an object. Text reference: Section 2.1

This simulation shows how the declination of the Sun varies over the course of a year using a horizon diagram. Text reference: Section 2.2

Declination Ranges Simulator

Sun Motions Overview

This simulation shows how an observer’s latitude determines the circumpolar, rise-and-set, and never-rise regions of the sky. Text reference: Section 2.1

This simulation shows the paths of the Sun on the celestial sphere. Text reference: Section 2.2

Big Dipper Clock This simulation shows how stars rotate about the North Star over time. Both daily and seasonal motions are shown. Text reference: Section 2.1 Big Dipper 3D

This simulation shows the angles at which the Sun’s rays hit different parts of Earth at different times of the year. Text reference: Section 2.2

Daylight Hours Explorer This simulation shows the hours of daylight received during the year for an observer at a given latitude. This is an important factor contributing to the seasons. Text reference: Section 2.2 Union Seasons Demonstrator

This simulation demonstrates how the stars of the Big Dipper, which are at various distances from the Earth, project onto the celestial sphere to give the familiar shape. Text reference: Section 2.1

This simulation demonstrates the changing declination of the Sun with a time-lapse movie, which shows how the shadow of a building changes over the course of the year. Text reference: Section 2.2

Coordinate Systems Comparison

Daylight Simulator

This simulation shows how the rotation of the Earth leads to the apparent rotation of the sky and associated phenomena during the day. Text reference: Section 2.1

This simulation shows daylight and nighttime regions on a flat map of the Earth. Daily and yearly motions of the sunlight pattern can be shown. Text reference: Section 2.2

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12 ◆ Chapter 2 Patterns in the Sky—Motions of the Earth and Moon Lunar Phase Vocabulary

Lunar Phase Quizzer

This simulation shows the appearance of the Moon at each of the named phases. Text reference: Section 2.3

This simulation shows the standard orbital view of the moon but with the option to hide the moon’s phase, the moon’s position, or the sun’s direction. Text reference: Section 2.3

Basketball Phases Simulator This simulation shows an illuminated basketball that can be viewed from multiple directions, providing an analogy to the Moon phases. Text reference: Section 2.3 Three views simulator this simulation shows how the phase of the Moon depends on the viewing geometry by allowing the Moon to be viewed from the Earth, the Sun, and an arbitrary point in space. Text reference: Section 2.3 Lunar Phases Simulator This simulation demonstrates the correspondence between the Moon’s position in its orbit and its position in an observer’s sky at different times of day. Text reference: Section 2.3 Moon Phases And The Horizon Diagram

Synodic Lag This simulation demonstrates the difference between a sidereal and synodic (solar) day, which arises from the Earth’s revolution around the sun. Text reference: Section 2.4 Moon Inclinations This simulation demonstrates how the inclination of the moon’s orbit precludes eclipses most of the time, leading to distinct eclipse seasons. Text reference: Section 2.5 Eclipse Shadow Simulator This simulation shows the shadows cast by the Moon and Earth due to the Sun and how they can produce the visual effect of an eclipse. Text reference: Section 2.5 Eclipse Table

This simulation correlates the phases of the Moon with its positions in the sky. Text reference: Section 2.3

This image consists of a table of solar and lunar eclipses, showing the banding that represents eclipse seasons that occur about twice a year. Text reference: Section 2.5

Moon Phases with Bisectors

Obliquity Simulator

This simulation demonstrates a method for determining Moon phases using planes that bisect the Earth and Moon. Text reference: Section 2.3

This simulation shows how obliquity, or tilt of the planet’s axis, is defined. Text reference: Section 2.5

Phase Positions Demonstrator This simulation demonstrates how planet and moon phases depend on orbital geometry. Users can drag two bodies around to see how the observed appearances change. Text reference: Section 2.3

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Patterns in the Sky—Motions of the Earth and Moon ◆ 13

The Celestial Sphere

Reading Astronomy News

This video presents the vocabulary of the celestial sphere. The instruction includes how to use an orange and basketball to visualize many of the important features of the celestial sphere. Text reference: Section 2.1

1. Solar eclipses travel over only a small swatch of the Earth when they occur; it is fairly rare for them to pass any given location, so the fact that this eclipse is the first total eclipse to hit the continental United States in about 30 years is very exciting. 2. As seen from Hopkinsville, the total eclipse will last 2 minutes, 40 seconds. This is the region of Earth where the eclipse lasts the longest because of the very particular geometry needed to produce a total solar eclipse; anywhere north or south of this region will see much shorter durations of totality. 3. There will be a lunar eclipse either 2 weeks before or after. 4. Answers will vary. 5. For all practical purposes, 0.1 seconds is not a significant enough change to be important except for those who care about making the claim of longest duration. More important factors will be weather that day, travel, and lodging.

The Cause of the Earth’s Seasons This video explains the causes for the seasons. Dr. Palen uses a bundle of spaghetti noodles (uncooked, of course) to show the difference between direct and indirect sunlight. Text reference: Section 2.2

The Earth-Moon-Sun System Three students act out the relevant motions of the EarthSun-Moon system in space. Text reference: Section 2.2

The Phases of the Moon Dr. Palen uses an orange and lightbulb to demonstrate the phases of the Moon. Students can easily reproduce this in their rooms (or your lab) using a desk lamp and any round object. Text reference: Section 2.3

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (b) The Earth rotates about the north and south poles, and the celestial poles are just an extension of this axis on the sky. 2. (d) If we are at the North Pole, Polaris is overhead or at 90° latitude, so it follows that its altitude in our sky and our latitude are the same. 3. (d) The higher the tilt, the more severe the seasons. 4. Full. See Figure 2.20. 5. The holidays would still wander because these calendars are based on a 29.5-day lunar cycle, but the wandering would be much slower than it actually is. 6. (b) If the Moon were farther away, its angular size would be smaller, and it could no longer cover the Sun to make a total eclipse. However, Earth’s shadow would still cover the Moon, so lunar eclipses would happen— (c) says they would not.

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Test Your Understanding 1. (c) Constellations are groupings of stars that appear close together on the sky and generally form a familiar pattern. However, in reality the stars are far from one another. 2. (c) Use Figure 2.2. 3. (b) The rotation of the Earth on its axis brings the Sun into and out of our sky, causing day and night. 4. (b) Currently Polaris is located very close to the north celestial pole, meaning it does not appear to the naked eye to move as the sky rotates. 5. (a) Refer to Figure 2.16, which shows how the Sun’s path compares to our orbital plane and the celestial equator. 6. (e) Seasons happen because both (b) days are longer in the summer and (c) light is more direct in the summer. 7. (d) On an equinox, the ecliptic crosses the celestial equator, the Sun rises due east, and one has exactly 12 hours of daylight. 8. (b) The Moon is in a “tidal lock” with the Earth so it spins at the same rate as it orbits. 9. (d) Using Figure 2.20, the Moon must be in the third quarter phase. 10. (a) A lunar eclipse happens when the Earth’s shadow falls on the Moon. 11. (b) Different definitions of calendar parts lead to different calendars.

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14 ◆ Chapter 2 Patterns in the Sky—Motions of the Earth and Moon 12. (d) The stars that we see at night depend on all of the options listed. 13. (c) The Tropic of Cancer is the northern tropic, and so in summer, the Sun is above the Tropic of Cancer, whereas it is daylight around the clock in the Arctic (northern) Circle. 14. (b) If you read carefully, all other answers are incorrect. 15. (a) “On the meridian” means the highest the Moon will be in the sky. If the Moon is in the first quarter phase and at this position, then using Figure 2.20, it is sunset, or on the western horizon. Thinking about the Concepts 16. Magellan could not use the North Star (Polaris) for navigation because he was in the Southern Hemisphere, thus Polaris was never above the horizon. Rather, he might have discovered that the Southern Cross constellation points approximately south. 17. Because the north celestial pole is an extension of the North Pole on Earth, if you are standing on the North Pole, you will see the north celestial pole right overhead, that is, at your zenith. 18. If Gemini is high in the night sky in the winter, then it is high in the daytime sky in the summer, which we cannot see. Thus, during the night it is behind the Earth. This is why we cannot see Gemini in the summer. 19. If I am flying in a jetliner, (a) I can tell that I am moving by watching stationary objects go past me. (b) backwards. 20. Our first question as an expert witness is whether the full Moon casts very pronounced shadows or illuminates things quite brightly and, in this reader’s opinion, that only happens if one is in an area that is otherwise extremely dark, which does not really happen in cities. That being said, the next question is whether the full Moon can cast a long shadow at midnight. To cast a long shadow, the object (Sun or Moon) must be very low in the sky, but at midnight the Moon will be at the meridian. For most observers, this is relatively high in the sky, which negates the defendant’s claim. However, if one were living around the Arctic Circle, then this argument might have some credibility because the Moon would never rise to be very high in the sky. 21. The observation is “seasons happen.” In Take 1, the hypothesis is that “seasons are due to distance from the Sun.” The prediction is that “both hemispheres will be hot at the same time.” The test is that “the seasons are

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opposite.” Because this is a failure, we return to a new hypothesis in Take 2. Here, the hypothesis is that “one hemisphere is closer to the Sun than the other because of our tilted axis.” The prediction is that “the closer hemisphere will be hotter.” The test is that “the difference in distance is too small to account for such temperature differences.” Our hypothesis failed, so on to a new hypothesis in Take 3. Here, we propose that “the tilt of our axis changes the distribution of energy on the surface.” The prediction is that “more energy will hit the Earth in summer,” which we test and confirm. With a confirmed test, we make a new prediction: “summer days are longer.” This is true, and we make another prediction, and so on. 22. The average temperatures on Earth lag a bit behind the formal change in seasons because it takes time for the Earth to heat up or cool down. Thus, although the winter starts officially in December, it takes 1 to 2 months for the Earth to cool down, making the coldest months January and February. 23. (a) The Earth takes 24 hours to complete one rotation (about its axis) with respect to the Sun. (b) The Earth takes 26,000 years to complete one “wobble.” 24. The full Moon crosses the meridian around midnight, and the first quarter Moon rises (i.e., on the eastern horizon) around noon. To answer these, use a figure such as Figure 2.20. 25. (a) Over the course of one orbit, the Earth will stay in a fixed position in the observer’s sky, because the same side of the Moon always faces the Earth. (b) The phases of the Earth as viewed from the Moon will be the opposite of those of the Moon as viewed on Earth, that is, if on Earth we see a full Moon, then on the Moon we would see a new Earth. 26. We would see a solar eclipse from the Moon. See Figure 2.29. 27. A total eclipse of the Sun casts a very small shadow on Earth and thus can only be seen from very narrow strips of the Earth, whereas the partial shadow covers a much larger area and can thus be seen by many more observers. 28. To see an eclipse at each full or new Moon requires that the Moon’s orbit be in the same plane as the Earth’s orbit around the Sun. Because this is not the case, we see eclipses only on those occasions when the two planes line up, about twice a year. 29. If the Earth’s tilt axis was close to 90°, the seasons would be very long and very severe. 30. A cyclic change in the tilt would vary the height of the sun in the sky, thereby changing the seasonal temperatures from their current pattern.

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Applying the Concepts 31. Setup: We know that it takes 24 hours for the Earth to make one revolution, so using d 5 vt we can find the circumference of the Earth. To find the diameter, we will use C 5 2πr, where r is the radius and the diameter is twice that value. Solve: If the speed v 5 1,674 km/h at the equator, then the total distance travelled in 24 hours is km 1, 674 3 24 h 5 40 , 176 km. The Earth’s radius h C 40 ,176 km is then r 5 5 5 6 , 397 km , so the 2 3 3. 14 2 diameter is twice that, or 12,790 km. Review: I can think of two ways to check this answer. First, look in Appendix 2 of the book. Second, it is about 3,000 miles from New York City to Los Angeles, and there are three time zones, which means there are about 1,000 miles or 1,600 kilometers per time zone. There are 24 time zones so the circumference of the Earth is about 24,000 miles or 38,000 kilometers. This is close to what we found so it is a reasonable sanity check. 32. Setup: For this problem, use Figure 2.6 (which shows observers on Earth at typical North American latitudes), Figure 2.7 (showing how stars move across our celestial sphere), and Figure 2.16b (which shows how the Sun moves in our sky compared to the celestial equator). Solve: (a) Answers will vary with latitude but should look like Figure 2.6. (b) Combining Figures 2.7 and 2.16b, one finds the maximum and minimum altitude of the Sun at noon on the solstices will be 23.5° above or below the celestial equator, and the celestial equator appears L degrees below the zenith or 90° 2 L degrees above the horizon, where L is your latitude. Thus, the Sun reaches (90° 2 L) ± 23.5° on the solstices. Review: If a student has access to a plastic celestial sphere with a movable Sun inside, using this tool is the best way to review the motion of the Sun in our sky and the relative positions of the zenith, celestial equator, and north celestial pole. It is worth investing in at least one of these for every introductory class. 33. Setup: For this problem, use Figures 2.6 and 2.7, which show in graphics and time-lapse photography how stars move around the north celestial pole. Solve: If Polaris is D degrees from the zenith, then your latitude is L 5 90° 2 D. So in this problem, Polaris is 40° from the zenith, therefore I am at latitude 50° which is in the southernmost part of Canada. Review: If a student has access to a plastic celestial sphere, using this tool is the best way to review the

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Patterns in the Sky—Motions of the Earth and Moon ◆ 15

answer. It is worth investing in at least one of these for every introductory class. 34. Setup: For this problem, use Figure 2.7, which shows how the stars move across our celestial sphere. Solve: If I am living in the United States (the Northern Hemisphere), then as shown in Figure 2.7, I can see a star in the southern part of the celestial sphere if it is more than L degrees from the southern celestial pole. So if I want to see a star 65° from the celestial equator, that means it is within 90° 2 65° 5 25° of the south celestial pole. To see it in the Northern Hemisphere, I need to be at this latitude or below. The only states that reach this low latitude are Florida and Hawaii. Review: If a student has access to a plastic celestial sphere, using this tool is the best way to review the answer. It is worth investing in at least one of these for every introductory class. 35. Setup: For this problem, use Figure 2.14 and note that panel (b) corresponds to the summer solstice in the Southern Hemisphere. Solve: As the Earth rotates, an observer on the South Pole will not move with respect to the Sun. In other words, the Sun will stay at the same height in the sky all day long. Using the same argument as in problem 32, the maximum height the Sun will reach in your sky is (90° 2 L) 1 23.5°, or 23.5° above the horizon. This is the height of the Sun (a) at noon and (b) at midnight. Review: One can also visualize this by combining Figures 2.7 and 2.16b, which shows the Sun 23.5° above the celestial equator and, hence, your horizon. 36. Setup: For this problem, use Figures 2.15 (how the Sun illuminates Earth at different times of year) and Figure 2.17 (how the Sun changes its height in our sky over different times of year). Solve: If the tilt of the Earth changed to D degrees, then the maximum height of the Sun above the celestial equator would also be D degrees. Because the tropics are D degrees above our equator, and the (Ant)Arctic circles are D degrees below the poles, we see that for this problem the tropics would be at latitudes ±10° and the circles at ±80°. Review: Imagine the Earth had no tilt; then where would the tropics and circles be? Now tilt the Earth by a tiny amount, and answer the question. One can easily derive the logic used in the solution in this way to verify that it is correct. 37. Setup: For this problem, use Figure 2.16, which shows how the Sun’s path compares to our orbital plane and the celestial equator. Solve: On the equinox, the highest the Sun in your sky will be L degrees below the zenith or 90° 2 L degrees

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16 ◆ Chapter 2 Patterns in the Sky—Motions of the Earth and Moon above the horizon, where L is your latitude. So the highest the Sun will be on the summer solstice will be (90° 2 L) 1 23.5°, and if the Moon can be up to 5° above that, then in Philadelphia the Moon can be as high as (90° 2 40°) 1 23.5° 1 5° 5 78.5° above the horizon. Review: This figure shows this situation as described. 38. Setup: For this problem, use Figure 2.14a (how the Sun illuminates the Earth when it is summer in North America). Solve: According to this figure, (a) if we travel to latitudes of 66.5° or higher and (b) make this trip as close to the summer solstice as possible, then the Sun will never set. Review: Knowing that it is winter in Australia when it is summer in New York City, or that in the (Ant)Arctic circles it can be daylight or night for 24 hours straight, we see that this is correct. 39. Setup: Follow the worked example in Working It Out 2.1 but replace 185 meters with 157.5 meters. Or, in the final answer, divide out 185 and multiply by 157.5 to remove one unit and replace the other. Solve: Following the first method, we use 1 3 C 5 5, 000 stadia 3 157. 5 meters/stadion to 50 find C = 50 3 787.5 km 5 39,375 km. The modern value is 40,075 km, so this is only 1.7% off. Review: Check with the second method. Using 180 meters per stadion, the example found a circum157. 5 5 39 , 375, ference of 46,250 km, and 46 , 250 3 185 as expected. 40. Setup: (a) We need to know how much time it takes for the vernal equinox to move from one constellation to another. Let us assume all constellations are equally distant from the next, then how much time is spent in each constellation for a total circuit of 26,000 years (the period of the Earth’s wobble)? (b) Use Figure 2.18b. Solve: (a) There are 12 constellations, and if they are distributed roughly uniformly around in the zodiac 26 , 000 yr 2 , 200 yr (Figure 2.13), it takes roughly 12 for the equinox to move from one constellation to the next. As extra credit, one might ask how long it will take to move from Pisces to Aquarius, because the student must figure out whether the equinox is moving from Pisces toward or away from Aquarius. The section “Earth’s Axis Wobbles” tells us that 2,000 years ago, the Sun was in Cancer on the first day of summer, whereas today it is in Taurus. Looking back at Figure 2.18b, we see that this is a change in the direction of Pisces

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toward Aquarius; therefore, the equinox need only move by one constellation or 2,200 years. (b) We find that Stonehenge was built when the equinox was exiting Taurus and entering Aries, or two constellations behind where we are now. Review: (a) Note that the question asked how long the equinox spends in a constellation, not how much time it takes to move between the two. If we assume that they are all about the same size and equally spaced apart, then the two questions are really the same! (b) Given that we determined it takes about 2,000 years to move one constellation, we need only move two constellations back. 41. Setup: Looking at Figure 2.18b, Vega is around 13000 to 14000 CE. The question is, how long will it take to move from Polaris (today) to Vega? We have all we need, as long as we remember that today is 2000 CE. Solve: 13,500 2 2,000 5 11,500 yrs. Review: We could also estimate this from Figure 2.20b as an angle. It looks like the angle between Polaris and Vega (as measured from the center of the image) is almost 180°. That means it will take about half of the 1 total period, or 3 26 , 000 5 13, 000 years, which is 2 close to what we found. 42. Setup: To solve this problem, remember there are 360° in a circle, and it takes about 29 days for the Moon to complete one orbit around the Earth, that is, to make one complete path through the fixed stars on the sky. We must find out how long it takes to move 1°. Solve: If the Moon moves 360° in 29 days, then 29 days 5 0.08 days to move 1°, and it takes 360° 24 hr 1.9 hr . We want the Moon 0.08 day 3 day to move half a degree, which will take half as long or about 1 hour, roughly. Review: There are 720 half-degrees in one revolution, day 3 30 days, which so 720 3 1 hr 5 720 hr 3 24 hr is about 1 lunar month. Check! 43. Setup: The problem gives us the formula that diameter , so compare the ratio of apparent size  distance the values for the Sun and Moon. 696 , 000 5 Solve: For the Sun, the ratio is 1. 469 3 108 1, 737 5 3. 780 3 105 4.59 3 10–3, so we see that the two are indeed about the same size. 4.73 3 10–3 and for the Moon,

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Chapter 2

Review: If we did not find that the two were roughly the same size, then the Moon would not seem to almost perfectly cover the face of the Sun during a solar eclipse! 44. Setup: Using the small-angle formula from the previsize ous problem, we know that the size  5 . In distance the case of our problem, the Earth-Moon distance is the same and so the ratio between the angular size Moon of the Moon as seen on Earth, and Earth of the Earth as seen  R on the Moon, will be the ratio Earth 5 Earth , where  Moon R Moon R indicates the actual (or physical) diameter of each object. 6371 km  5 Solve: Given the values in the text, Earth 5  Moon 1737 km 3.67, so the Earth appears 3.67 times larger to the observer on the Moon. Review: Try setting out a tennis and basketball some distance apart, and then observe each object from the other’s location. It will become clear that the change in angular size varies with the size of each object. 45. Setup: Refer to Figure 2.31, which shows why eclipses only happen at certain times of year. Solve: If the inclination of the Moon’s orbit drops, then there is a longer period of time during which the Moon can pass through the Earth’s shadow. Hence, the lunar eclipse seasons would become longer. The solar eclipse still requires a very accurate alignment of the Sun-Earth-Moon system, so this season would probably not change. Review: You can try this for yourself by going into a dark room with a single lightbulb and holding a tennis ball at arm’s length. The bulb represents the Sun, your head is the Earth, and the ball is the Moon. By turning around, you can cause solar and lunar eclipses. Using the Web 46. I am writing this in summer, and the days are becoming shorter and nights longer as I approach autumn. The shortest day occurs on December 21 and the longest on June 21. Everything is opposite for the Southern Hemisphere. 47. The answers will vary on the day of the year and time chosen. For mid-September around 3 pm, it is daytime, regardless of hemisphere. However, if one goes to the other side of the Earth, it is night. It is night directly on the North Pole, and day on the South Pole.

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Patterns in the Sky—Motions of the Earth and Moon ◆ 17

Twelve hours earlier or later, it is night where we are located and day in the Southern Hemisphere, night at the North Pole but day on the South Pole. 48. Answers will vary depending on the time of year. As one example, if the Moon is currently a waning gibbous, it will rise at 8 pm and set at 10:30 am. Over the next 4 weeks, it will become a third quarter, then waning crescent, then new, then waxing crescent, then first quarter, then waning gibbous, then full, and become waning gibbous again. A first quarter Moon rises around noon. One can observe waxing crescent through waxing gibbous during the day. 49. Answers will vary with the initial phase of the Moon. The reported phases will follow one-quarter of the full phase pattern, that is, if one starts at new Moon, then it becomes more illuminated over 7 days until it is a first quarter Moon. The Moon will change its distance from the Sun, or the stars seen behind it, every day. 50. Answers will vary according to the date in question. For example, as of October 2015, the next lunar eclipse is a penumbral (partial) eclipse on March 23, 2016, and will be just visible from the western seaboard of the United States. The next solar eclipse will be a total eclipse 2 weeks earlier on March 9, 2016. It will be visible from Sumatra, Borneo, and Sulawesi. Solar eclipses can only be seen from a tiny fraction of the Earth because the Moon’s shadow is very small and it only strikes a tiny portion of the Earth. Exploration 1. At startup, the time is noon, because the Sun is going to be at the meridian (i.e., the highest in our sky that it will reach). 2. The Moon is also at the meridian because it is located at the same position in the sky. 3. The Moon is new, as shown on the top right panel. Also, none of the illuminated face is toward us. 4. I would see a “full” Earth, that is, fully illuminated. 5. The Moon orbits counterclockwise. 6. The right side of the Moon is illuminated first. 7. If the horns of the crescent Moon point right, it must be a waning Moon. 8. At midnight, the first quarter Moon is setting. 9. The full Moon crosses the meridian at midnight. 10. At noon, the third quarter Moon is on the western horizon. 11. At 6 P.M., the full Moon is rising.

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CHAPTER 3

Motion of Astronomical Bodies INSTRUCTOR’S NOTES Chapter 3 covers force and Kepler’s (empirical) laws. Major topics include

▶ history of the geocentric and heliocentric models ▶ Kepler’s laws of planetary motion ▶ Galileo’s contributions to astronomy and physics ▶ Newton’s laws of motion I like to ask my students to pause for a few moments and try to image Galileo, on January 7, 1610, turning his telescope to Jupiter and seeing three “stars,” invisible to the naked eye, in a perfect line on either side of the bright central dot. The next day, these three stars had moved a little with respect to Jupiter, leading Galileo to believe they were just ordinary background stars. But over the next few days, his observations revealed that these “stars” were actually traveling around Jupiter. He wrote a few months later in Sidereus Nuncius that on the night of January 11, “Statutum ideo, omnique procol dubio a me decretum fuit, tres in coelis adesse Stellas vagantes circa louem, instar Veneris, atque Mercurij circa Solem” (“I therefore concluded, and decided unhesitatingly, that there are three stars in the heavens moving about Jupiter, as Venus and Mercury round the Sun”). I think we sometimes lose sight of these singularly amazing moments that took place a few hundred years ago, propelling us into the era of modern science. Part of my motivation for teaching an introductory astronomy course is to teach the process of science, and this chapter is replete with valuable lessons, such as the dangers of dogma or belief in science; scientific revolutions; evidence-based modeling; and testing theories. If you delve into the role the Catholic Church played in these early days of modern astronomy, I suggest you make it very clear to your students that you are not taking sides on religion, just reporting the historical facts. I often approach this topic with the clear statement that science has nothing to say about religion. Many students are often surprised, and I’ve found that quite often, they have grown up considering science to disprove religion. I explain that science has to do testable experiments, and for science to make any meaningful statement about the existence of God, for example, one has to conduct an experiment in

which we can guarantee God is not present. As that is not possible, any comments scientists make on this subject are personal ones, and not scientific. It may also help to tell students that the Church has been a staunch proponent of astronomy since the late 1700s, and it has been funding professional astronomy research for decades. In terms of organizing material pedagogically, the introduction of Newton’s Laws of motion to a discussion of the history of orbits may not seem entirely clear to many students. One option is to bundle this in with Chapter 4 as a prelude to Newton’s Law of Universal Gravitation. One technique I have found very helpful in motivating the understanding of these three laws is a real life coffee-cup reference in Figure 3.21. If you do not mind making a little mess on the floor, spilling some water is a very memorable demonstration.

DISCUSSION POINTS

▶ What made Galileo so sure that the Copernican model

of the Solar System was correct and the geocentric model was wrong? Did he have enough evidence to convince others? What did he do that irritated the Catholic Church, and have there been any lasting consequences? ▶ Figures 3.12 through 3.16 illustrate Kepler’s three laws of planetary motion. Demonstrate that Kepler’s laws are universal by applying them to other solar systems in which several exoplanets have been found. Use the data from http://exoplanet.eu. ▶ Use the equations in Working It Out 3.3 and Figure 3.22 through 3.23 to discuss the relationship between fuel efficiency and engine power in car design. What are the pros and cons of using lighter materials in the car industry? ▶ Think in terms of the equations in Working It Out 3.3 and Newton’s third law to discuss the physics behind the role of different players on a football team.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 3 and are available from the free Student Site 19

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20 ◆ Chapter 3 Motion of Astronomical Bodies (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises.

solar system. It shows the planets as observed from space and on our own sky. Text reference: Section 3.1

Kepler’s Laws

Synodic Period Calculator

This animation of Kepler’s three laws contrasts Earth’s orbit with that of other bodies in the Solar System. Using slightly eccentric (Mars) and extremely eccentric (Comet Halley) orbits for comparison, the three laws are illustrated showing: (1) the difference between an elliptical and near-circular orbit; (2) changes in speed as an object moves along its orbit; and (3) how period changes with larger semimajor axes. Text reference: Section 3.2

This applet calculates the sidereal period of a planet from its synodic period, or vice-versa. Text reference: Section 3.1

Velocity, Acceleration, and Inertia This animation uses the movement of an automobile in both a straight line and a circle (familiar to most students) to address the ideas of velocity and acceleration. It also explores the idea of inertia, with the addition of a trailer to the automobile to increase its mass. Text reference: Section 3.3 and 3.4

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts presented in Chapter 3. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Ptolemaic Orbit of Mars This animation shows the Ptolemaic model of orbits of Earth and Mars. The epicycle and deferent show retrograde motion. Text reference: Section 3.1 Retrograde Motion This simulation demonstrates the phenomenon of retrograde planetary motion. Text reference: Section 3.1 Planetary Configuration Simulator This simulation models the movements of the planets around the Sun in a simplified Copernican model of the

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Eccentricity Demonstrator This simulation demonstrates the parameters that define the eccentricity of an ellipse. Text reference: Section 3.2 Planetary Orbit Simulator This simulation demonstrates the behavior of planetary orbits and Kepler’s laws. Text reference: Section 3.3 Phases of Venus This simulation shows Earth and Venus going around the Sun, and what Venus will look like through a telescope from Earth. Compare these images with the Ptolemaic Phases of Venus simulation and also to Figure 3.19, which shows actual observations of Venus from Earth. Text reference: Section 3.3 Ptolemaic Phases of Venus This simulation shows Venus traveling around the Earth on its epicycle and what Venus would look like from Earth. Compare this to the Phases of Venus simulation and also to Figure 3.19, which shows actual observations of Venus from Earth. Text reference: Section 3.3

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Chapter 3

Velocity, Force, and Acceleration This simulation discusses velocity, acceleration, and force by holding and throwing a ball. Text reference: Section 3.4

Motion of Astronomical Bodies ◆

4. The video made sense to me. 5. Answers will vary. NASA could send some satellites into Mars’ orbit but spaced 90°, 180°, and 270° away, such that astronauts on Mars could bounce their signals off of these satellites and back to earth.

END-OF-CHAPTER SOLUTIONS

Test Your Understanding

Check Your Understanding

1. (d) Empirical means based only observation and not on a theoretical foundation. 2. (a) This is a description of retrograde motion that makes the other planet appear to move backward for a while and then continue forward again on the sky. 3. (b) The heliocentric model is much simpler because it does not rely on epicycles; however, it is a superior model because it makes much more accurate predictions. 4. (c) See Figure 3.7. 5. (a) Synodic positions refer to those made with respect to the sun. 6. (a) Kepler’s second law tells us that when the planet is close to the star, it moves faster, but when it is far away it moves slower. 7. (d) See Working It Out 3.1. 8. (d) All the answers listed are examples of an unbalanced force that leads to acceleration. 9. (a) Kepler’s third law tells us P2 5 A3 for all planets orbiting the Sun. 10. (b) The phases of Venus are possible only if it changes its angle with respect to the Earth and Sun by 360°, as the Moon does. Such is not possible in a geocentric model. See image below. 11. (d) Kepler’s second law says that planets move fastest when they are closest to the Sun. 12. (c) This claim violates Kepler’s second law because the planet is said to be moving slowest when it is closest to the Sun. 13. (d) The other three answers have no logical relationship to the fact that moons orbit Jupiter. 14. (b) See Working It Out 3.3. 15. (c) We assume that life will develop in relatively stable environments, which are unlikely if the planet spends considerable amounts of time at radically different distances from its central star.

1. Ancient civilizations knew planets were different from stars because they wandered through the sky. Hence, the name planet, from the Greek planets (), meaning wanderer. 2. (c) Retrograde motion requires one planet to overtake the other. Because Uranus is further from the sun, the Earth moves faster and will always overtake Uranus. 3. (b) See Figure 3.15 4. c, b, d a. The semimajor axis, or average distance from the Sun, is larger if the period is larger. 5. c, d. Phases of Venus are best explained with a heliocentric model, and for moons to orbit Jupiter, a geocentric model fails because it proposes that objects only orbit Earth. 6. (b) Acceleration is a change of velocity, which can be speed and/or direction. Reading Astronomy News 1. Conjunctions occur every 26 months. This is the synodic period. 2. During a conjunction, Mars is behind the sun; so it could only be observed during the day, and the sun is blocking it. 3. Earth will complete two full revolutions and then move an additional 2 months, or two-twelfths of a revolution. This means Earth and Mars will move the angular equivalent of 2 hours on a clock. The figure below is a copy of Figure 3.25 with the new positions noted.

Thinking about the Concepts 16. During normal motion, Mars moves toward the east. Retrograde motion for Mars is motion toward the west. In this image, Mars was in retrograde motion for about 2.5 months, or about 75 days. Mars observers would have seen Earth in retrograde motion as well.

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22 ◆ Chapter 3 Motion of Astronomical Bodies 17. Empirical means basing conclusions on what one has found in observations and data, rather than basing them on calculations and mathematical theory. 18. Saturn hardly moves through the sky in one Earth year, so its synodic period is very close to the sidereal period of Earth. 19. One possibility for the geocentric and heliocentric phases of Venus is shown below.

In the geocentric model, Venus will always appear about the same size and in a nearly full phase. In the heliocentric model, Venus will go through all the phases from “new” to “full,” and will appear largest when new and smallest when full. Because the size and phase of Venus is only observable through a telescope (i.e., it is not visible to the unaided eye), these models can not be distinguished without a telescope. 20. Two heavy objects will always hit the ground at the same time. Tissue paper will not, because the tissue is much more sensitive to air resistance. 21. By Kepler’s second law a planet is always moving the fastest when it is closest to the Sun. By the same law a planet moves the slowest when it is farthest from the Sun. 22. The eccentricity of the Earth’s orbit is currently about 0.017, which is much less than that of the moon. The moon’s eccentric orbit brings it closer or further from the Earth over time, which changes its apparent size on the sky. This leads to total versus annular solar eclipses. 23. Inertia is a description of how much an object wants to oppose a change in motion. An object with a lot of inertia is not very easy to move, and it is not very easy to change its motion once it is moving. 24. Kepler would have deduced exactly the same three laws of planetary motion. The only difference is that the actual value of an AU would be different. But because a Martian year is different than on Earth, the relationship P2  A3 would still be true. 25. Velocity is the speed of an object plus its direction. Acceleration is how quickly the velocity of an object is changing. However, an object can travel the same speed but change direction (i.e., turn), and thus,

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the velocity is changing, which means the object is accelerating. 26. Newton’s first law tells us the object in motion stays in motion unless some force/action is there to change it. 27. We wear seatbelts to protect us from acceleration. We travel at the same speed as the car or airplane, but because we have inertia of our own, we will keep moving forward (Newton’s first law) until something stops us (i.e., the seatbelt) if the car or plane suddenly decelerates. 28. In the space station, all objects will have nearly zero weight so they can all be maneuvered; however, a more massive object has more inertia, which means it is more difficult to change that object’s state of motion (Newton’s second law F 5 ma). Thus, a “heavy” piece of equipment accelerates more slowly than a “light” one because even though the astronaut is applying the same force F, the masses are different as are the objects’ accelerations. 29. Answers will vary. Scientific theories must be testable, and tests are not only confirmations of hypotheses but also refutations of challenges. Thus, challenges are, by their nature, part of the scientific process. 30. If the Earth’s eccentricity were 0.17, our planet would be substantially closer or further from the sun at the points of closest and furthest approach. This radical change in distance would dominate our seasons, causing the Earth to be scorching during the summer and glacial during winter. Applying the Concepts 31. Setup: The difference between linear and logarithmic graphs is that linear graphs have axes that increase by single counting numbers (1, 2, 3. . .), whereas logarithmic graphs increase by powers of 10 (10, 100, 1,000. . .). To read off any graph, draw a line from the point in question to each axis and find the number (or estimate the closest number) on that axis. Solve: This is a log-log graph (both axes are logarithmic). The authors have already drawn in lines to the axes. But note that the axes are P2 and A3. I estimate that the P2 is halfway between 102 and 104, so p 2  10 3 or p  1000  32 yrs . I estimate the A3 is halfway between 102 and 104 as well, so A3  103 or A 5 3 1000 510 AU. Review: We can compare these with our Appendix; also, note that because P2 5 A3 , we can test the numbers. Indeed, 322 5 103. 32. Setup: The reader can use a ruler to measure the vertical size of the two moons. An object’s apparent

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Chapter 3

(or angular) size scales with distance, so two times smaller means twice as far away. Solve: Using a ruler, I find that the right-most crescent is about 2.8 times larger than the left-most image. This implies Venus is about 2.8 times farther at far left than far right. Review: This is a very hard problem to check mathematically because we do not know at what point in the orbit each image was taken. As a rough check, use the figure provided in the solution for problem 19, and assume the gibbous Venus is at the “half ” position and the crescent Venus is “new.” This makes a right triangle with the Sun, with sides 1 AU and 0.7 AU for a total Earth-Venus distance of 12 1 0. 7 2 51. 2 AU. At “new” phase, Venus is 1 2 0.7 5 0.3 AU, so under these assumptions, Venus would appear 1.2/0.3 5 4 times larger. However, the crescent Venus we are given is not yet new, so the actual change in distance is likely to be smaller than we measured. 33. Setup: To relate period to semimajor axis for a planet orbiting the Sun, use Kepler’s third law: P2 5 A3 or P 5 A3/2. Solve: P 5 503/2 5 354 years. Review: We can check this by comparing the period of a known object of roughly the same distance. Pluto’s average distance is 40 AU with period 248 years, whereas Eris’s distance is 68 AU and period 557 years. Our hypothetical planet is between these, as expected. 34. Setup: For the semimajor axis, use Kepler’s third law P2 5 A3, solving for A. Oppositions happen on the synodic period, so use Working It Out 3.1 to find S given P is the period of Neptune and E is 1 year. Solve: For the semimajor axis, A 5 P2/3 5 164.82/3 5 1 1 1 30.06 AU. The synodic period 5 2 or S E P 1  −1  S 5  12 5 1. 006 years or 367.5 days.  164. 8  Review: We can check the semimajor axis in the textbook’s appendices, and because Neptune hardly moves in its orbit during one year, we expected a number just over one year. 35. Setup: From Working It Out 3.1, we use 1 5 1 2 1 , S E P whereas Kepler’s Law is P2 5 A3. In both cases, we are solving for P. Solve: (a) Using a synodic period of 1.278 years, we find 1  21  P 5  12 5 4. 597 years. (b) By Kepler,  1. 278  P 5 A3/2 5 2.773.2 5 4.61 years. (c) The two are very close, certainly considering significant digits.

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Motion of Astronomical Bodies ◆

Review: We expect the periods to match because the two methods must give the same results to within rounding errors from the values of each variable used. 36. Setup: Kepler’s third law tells us that P2 5 A3. To test if this assertion is correct, we figure out what distance corresponds to a period of 3 years. Solve: P2 5A3 implies that A3 5 32 or A 5 3 9 5 2. 08 AU. That is not 2 AU, although it is very close. Review: What period corresponds to a planet at a distance of 2 AU? Using Kepler’s third law, we find P2 5 23 or P 5 8 5 2. 8 years. Well, that is not 3 years as claimed. 37. Setup: Here are the data I use: Io has P 5 1.77 d, A 5 0.421 Gm (giga-meters). Europa has P 5 3.55d, A 5 0.671 Gm. Ganymede has P 5 7.14 d, A 5 1.07 Gm. Callisto has P 5 16.7 d, A 5 1.88 Gm. As long as we use the same units, we can compute the ratio P2 5 A3 with the numbers given above. 1. 77 2 5 42. 0 . Europa has Solve: We find: Io has 0. 4213 3. 552 7. 152 5 41 . 7 . Ganymede has 5 41. 7 . Cal0. 6713 1. 07 3 16. 7 2 listo has 5 42. 0 . 1. 883 Review: These are all very close, especially because I rounded. Obviously, we expected to find nearly the same ratio because all four moons, and indeed all orbiting satellites, will obey Kepler’s third law. 38. Setup: To find the speed of the planet, we could divide distance by time, but be sure to convert months to seconds. Or we could notice that we are talking about distances traveled during the same 3-months’ time period; thus, a larger distance means a correspondingly higher speed. For the second part, recall Kepler’s second law, which tells us that planets move faster when they are closer to the sun. Solve: Because the planet travels 65,000 km in 3 months rather than 30,000 km, we know the aver65, 000 5 2. 17 times higher than age speed must be 30 , 000 the first time period. Thus, the planet travels 2.17 3 10.5 km/s 5 22.8 km/s. The planet is closer to the sun during this second period by Kepler’s second law. Review: To check this problem directly compute the speed needed to travel the given distance in the given time. 39. Setup: No calculations needed for this problem. Because we are asked to compare what Mars sees of Earth to what Earth sees of Mars, we know the same things happen at the same time.

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24 ◆ Chapter 3 Motion of Astronomical Bodies Solve: Mars undergoes retrograde motion to an earthbased observer every 25.6 months. So Earth also enters into retrograde motion to a Mars-based observer every 25.6 months. Review: Recall that two observers moving with respect to each other must agree on what they see each other doing. 40. Setup: Remember the semimajor axis is half the major axis. The time to go around the sun is the period, which we find using Kepler’s third law P2 5 A3. The eccentricity can be computed from using the distance from the center of the ellipse to the sun. Solve: A sample image of this situation is shown below.

(a) The comet is moving fastest when it is closest (at left) and slowest when furthest (at right). (b) The semimajor axis is half of the full axis, or (1.26 1 5.68)/2 5 3.47 AU. The period P 5 A2/3 5 3.473/2 5 6.46 year. (c) The distance from the sun to the center is the semimajor axis less 1.26 AU, or 3.47 2 1.26 5 2.21 AU. The eccentricity e 3 A 5 2.21 or 2. 21 2. 21 e5 5 5 0. 637 . 3. 47 A Review: Compare to Figure 3.13. 41 Setup: This is a question of relativity so review that material as well as Newton’s first law. Solve: (a) I see the other car approaching me at 90 1 110 5 200 km/h, while (b) the other driver sees me approaching at 110 1 90 5 200 km/h. In other words, both (a) and (b) see the same speed of 200 km/h. Review: Remember that the total speed of two approaching objects is equal to the sum of the individual speeds, and because a 1 b 5 b 1 a, both drivers have to see the same total speed. 42. Setup: To compute the relative orbital periods of Vulcan and Mercury, we use Kepler’s third law, normal2

Review: Remember that closer to the star means faster speed and shorter period, which we have confirmed. 43. Setup: This problem asks for acceleration and gives me mass and force. Use Newton’s second law: F 5 ma. For part (b), remember that speed equals acceleration times time or v 5 at. F 100 N 5 2 m/s2 (b) Let’s find Solve: (a) a 5 5 m 50 kg the time until the refrigerator travels at 10 m/s: v 10 m/s 5 5s . t5 5 a 2 m/s 2 Review: If the mass and force given are reasonable, then I can see it taking 4 to 5 seconds until I can no longer keep up with the refrigerator. 44. Setup: To compute the acceleration of the astronaut, use Newton’s second and third law. That is, force (ma)satellite 5 (ma)astronaut , and both the astronaut and satellite experience the same force. We could calculate the force on the satellite and then the acceleration of the astronaut directly, or notice that because they both experience the same force, (ma)satellite 5 (ma)astronaut , the astronaut’s acceleration will be greater by the ratio of the masses. Solve: The astronaut’s acceleration is m 5000 aastronaut 5 satellite a satellite 5 0. 1m /s2 5 5 m /s2 . mastronaut 100 Review: To confirm, calculate the acceleration directly. The force on the satellite is F 5 ma 5 (5000 kg) 3 (0.1 m/s2) 5 500 N. The astronaut also feels 500 N, so F 500 rmN 5 5 m /s2 . her acceleration is a 5 5 m 100 kg 45. Setup: See problem 3.40. Solve: The figure will look similar to:

 P   A  izing to Mercury’s orbit:  Vulcan  5  Vulcan  .  PMercury   A Mercury  Doing some algebra, we solve for Vulcan’s orbital pe3

 A  riod and find PVulcan 5  Vulcan  PMercury .  A Mercury  Solve: Because Vulcan’s orbit is one-fourth the radius of Mercury’s orbit, we can write the ratio between the two orbits and solve PVulcan 5 or 11 days.

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13 1 PMercury 5 PMercury 4 8

(a) The major axis is 1.38 1 1.67 5 3.05 AU, and the semimajor axis is 3.05/2 5 1.53 AU. (b) The distance from the center of the ellipse to the sun is A less the 1.38 AU, or 1.53 2 1.38 5 0.15 AU. The eccentricity 0. 15 0. 15 5 5 0. 095. The Earth’s orbit has is e 5 A 1.53 an eccentricity of 0.017 so Mars’ orbit is much more eccentric.

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Chapter 3

Review: The Origins section notes that Mars has an eccentricity of 9 percent, which we confirmed in this problem. Using the Web 46. Answers will vary. An answer will include the planets visible this week. Venus and Mercury are visible only around sunrise or sunset because they remain close to the Sun in the sky (they are interior planets). Before telescopes, people knew planets were different because they moved (or wandered) through the sky. 47. Answers will vary. The answer will include opposition dates for Mars, Jupiter, or Saturn. If you take pictures of the planet during the opposition period, it is fairly unlikely that you will observe retrograde motion because it takes more than a few weeks for a planet to move appreciably with respect to the background. 48. Answers will vary. The answer will include: (a) which planets are in or near to conjunction, opposition, or greatest elongation; (b) which planets are visible in the morning or evening; and (c) a sketch of the Solar System from above. 49. Galileo’s telescope is a very small refractor. He also made protractors, sextants, and compasses. Galileo looked through the telescope with one eye and at a small “ruler” attached to the telescope with his other. This way he used the right eye to see Jupiter and its moons, and the left eye to see the scale by which he made measurements. People like to keep relics of famous historical figures; hence the reliquary of Galileo’s finger. It is a way to connect in the present to the person from the past. His tooth is also in a reliquary. 50. Answers will vary. The answer will include a planet whose orbital period is around 1 year, including the semimajor axis, the mass of the central star, and the eccentricity. The answer will also include a star with

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Motion of Astronomical Bodies ◆

multiple planets and will compute the values of P2 5 A3 to verify Kepler’s third law. Exploration 1. Mercury’s orbit is definitely an ellipse, with the Sun noticeably far from the center of the orbit. 2. Using the grid provided, I estimate the semimajor axis is about 7.8 units and the semiminor axis is about 7.6, 7.6 5 0. 974 . for a ratio of 7.8 3. Using the eccentricity formula, e 5 12r 2 5 120. 974 2 5 0 . 23 . Appendix 2 lists e 5 0.24, so we are very close! 4. When I sweep the first time (i.e., closest to the Sun) the shade region is yellow. When I sweep the second time (i.e., farthest from the Sun), it is magenta. In the yellow region, I count about 22.5 squares, and in the magenta region, I also count about 22.5 squares. When I hit fractional squares, I tried to find another one that was shaded with the amount needed to make a full square. I find about the same total area per region, as I should because Kepler’s second law tells us that we should find equal area in equal time. 5. As I change the eccentricity, the orbit becomes very elongated if e is close to 1 and very circular when e 5 0, as expected. 6. The position of the planet on the graph of axis versus period does not change, which confirms that these orbital characteristics do not depend on the eccentricity (or shape) of the orbit. 7. As I make the semimajor axis smaller, the period decreases. 8. As I make the semimajor axis larger, the period increases. 9. This confirms that the larger the semimajor axis, the longer the period of a planet.

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CHAPTER 4

Gravity and Orbits INSTRUCTOR’S NOTES Chapter 4 covers the force of gravity and the physics of orbits. Major topics include

▶ Newton’s Universal Law of Gravitation ▶ orbital motion ▶ Kepler’s third law as a law of physics ▶ why astronauts in Earth orbit experience zero gravity ▶ tidal forces The material in this chapter can pose considerable confusion to students, who find many applications of the physics to be quite counterintuitive. Why does the mass of the planet not affect the period of its orbit? How can two objects of unequal mass (that is, Earth and the Sun) exert the same gravitational force on each other? How is a bound orbit (that is, going around something) a perpetual state of free fall (that is, going toward something)? Also, the nomenclature can be somewhat confusing, as Kepler’s laws were first empirical discoveries limited to our Solar System, but thanks to Newton’s work, they are considered universal laws of physics. As with the previous chapter, I have found that students need extra time to work through these concepts. You can easily devote an entire class to Section 4.1 on gravity, a full class period on the physics of orbits in Section 4.2, and another full class on the physical (as opposed to empirical) form of Kepler’s third law. Although one can present Newton’s Universal Law of Gravitation and do a sample problem in about 5 minutes, there are a large number of nuances that we might take for granted but which students need time and guidance to master, such as: how does surface gravity change with planet mass and size; how to rank the strength of gravity between bodies of different size and mass; how does the net strength of gravity on one object change with distance as you move it from one body to another (e.g., a space probe as it travels from the Earth to the moon). When it comes to orbits, most students believe orbits must close on themselves and have a hard time visualizing a crash or a parabolic trajectory as an orbit. Again, circular and escape velocity can be presented

in 3 to 5 minutes, but discussion of relevant examples and comparison of values for different astrophysical objects is very important to student understanding. One can even start to motivate a black hole here by asking what mass/ size constraints are needed for the escape velocity to reach the speed of light. The influence of mass on orbits can be, as I noted above, quite perplexing. For example, how could a small telecommunications satellite and a giant space station at the same altitude have the same orbital period around the Earth? The textbook has chosen to circumvent this question by presenting Kepler’s third law as P 2 5 A 3 / M sun but this is only true for situations in which one mass is much larger than the other. This can cause some confusion later in the textbook when discussing binary stars for which both masses must be considered. For that reason, I prefer to teach Kepler’s law as P 2 5 A 3 / M where M 5 M 1m or M 5 M1 1 M 2 is the sum of the masses. Tidal forces can be covered in 5 minutes or an entire hour, depending on the level of detail you wish to present. Personally, I think a quantitative exploration of tidal forces is more appropriate to a student body destined to be physics or astronomy majors. There are many supplemental explorations and simulations that are helpful with these topics, as well as exercises to be found in the complementary Learning Astronomy by Doing workbook. A great way to engage students in the physics of orbits is to have them play the game Super Planet Crash at http://www.stefanom.org/spc/. I offer a few points of extra credit for students who can earn more than 10 to 20 million points over the full 500 years. Or if they want an extra challenge, have them beat my all-time high score of 202,816,975.

DISCUSSION POINTS

▶ Discuss the validity of the statement, “All objects on

Earth fall with the same acceleration, g,” in light of the universal law of gravitation. Use scaling relations to calculate the difference in acceleration between the top of

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28 ◆ Chapter 4 Gravity and Orbits Mount Everest and the deepest regions of the oceans. How significant is the change in gravity? ▶ Use the relations among weight, mass, and gravity to estimate how much of a gravitational field a human body could tolerate and still be able to stand and walk around. How much more massive would Earth need to be for us not to be able to do so? ▶ If Earth were at a different place in the Solar System, how would the gravitational force it would experience compare to our place now? How does the inverse square law of gravity affect the gravitational environment of the Solar System? ▶ Discuss the differences between tidal effects leading to ocean bulges and those that stretch the solid parts of a planet. What happens to a planet because the solid parts are not as free to move as the liquid? What does this mean for an object that experiences a large tidal force but not one large enough to break it apart? ▶ How could astronomers use their knowledge of orbits to navigate the Solar System? Discuss this in the context of bound and unbound orbits, as well as an acceleration given by the gravitational force? ▶ Discuss how the laws of physics presented in this chapter could be used to learn about the masses and/or sizes of objects other than the Sun, such as moons of planets or other stars.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 4 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises.

Tides and the Moon This section is an explanation of self gravity, why there is no gravitational force at the center of a planet, and how objects of real physical size experience tidal forces due to the inverse-square dependence of gravitational force with distance. Text reference: Section 4.3

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts presented in Chapter 4. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Gravity Algebra This simulation shows the proportional behavior of gravity on masses and distance in Newton’s Universal Law of Gravitation. Text reference: Section 4.1 Earth Orbit Plot This simulation shows the orbital period as a function of distance for satellites around the Earth. Text reference: Section 4.2 Tidal Bulge Simulation This simulation shows how the Moon and the Sun create the tidal bulges on Earth. Text reference: Section 4.3

Newton’s Laws and Universal Gravitation

ASTRONOMY IN ACTION VIDEOS

This interactive animation uses a classic thought experiment to show how force (an explosion of gunpowder) is used to move an object (a cannonball). It briefly explores the connection between this thought experiment and the real world actions of satellites and space probes. Text reference: Section 4.2

These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what they have learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5) and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack.

Elliptical Orbits

Center of Mass

This animation shows how increasing the speed of a planet in a circular orbit results in an elliptical orbit. Higher speeds result in more elongated elliptical orbits. Text reference: Section 4.2

Dr. Palen shows how we each use our own center of mass to balance, and how two people can spin around a common center of mass by linking hands. Text reference: Section 4.2

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Chapter 4

Tides

Gravity and Orbits ◆

For the Earth-Sun system,

Dr. Palen shows a kinesthetic model of how the moon causes a tidal force on Earth. Text reference: Section 4.3

1 . 99 31030 kg M Sun 5 6. 39 3105 k g/km 3 . 5 3 (1 . 49 3108 km)3 dSun Thus, we see the Moon is more important to our tides.

END-OF-CHAPTER SOLUTIONS

Test Your Understanding

Check Your Understanding

1. (a) Newton’s law of gravity states that the force is proportional to the product of the two masses and inversely proportional to the square of the distance. 2. c 2 a 2 d 2 b. As shown in Working It Out 4.2, circular velocity does not depend on the mass of the 1 ball, but does depend on . r 3. (c) This is the definition of bound and unbound orbits. 4. (d) Mass is independent of the gravitational field. It is an intrinsic property of an object. 5. (c) Plugging the mass and radius of Mars into Newton’s Universal Law of Gravitation shows that the gravitational force on Mars is smaller than that on Earth. M 6. (d) The scaling of gravity depends on Fg  2 so if d Venus has a mass 0.8 times that on Earth, and a radius 0.95 times that of Earth, your weight will be 0.8 5 0. 89 that on Earth, or about 10% less. 0. 952

1. (d) Recall that gravity scales as d22 (inverse-square law). 2. (c) The force of gravity between two objects does not depend on the size of the object. (a) If the earth’s radius were smaller, the acceleration of gravity would increase, making everything weigh more. 3. (a) Increasing the orbital speed would make the telescope travel further for a given amount of time, resulting in a larger orbit. 4. c, a, d, b. The Earth is closest to the Sun in January so the Sun’s strongest tidal force will be in that month. Thus, the neap tide in July is weaker than in January, and the spring tide in July is weaker than in January. 5. (a) Tidal locking occurred as a result of tidal forces between the Earth and Moon.

Reading Astronomy News 1. Light travels at a finite speed, so the light we receive today from a distant supernova was created long, long ago in a galaxy far, far away. 2. The scientific method requires testing, even of universal constants. Additionally, we must test whether the constant is actually “universal” as Newton proposed. 3. If G changes over time, then all forces of gravity would as well; i.e., orbital periods and distances, tidal forces, sizes of self-gravitating objects (from asteroids to superclusters). 4. Independent tests are needed to verify a theory; otherwise one can only state that the theory worked under narrow circumstances. 5. If the Sun were closer, tidal forces would be stronger. M Tidal force depends on 3 so to know how much d of an effect a changing Sun-Earth distance has compared with a changing Moon-Earth distance, consider the ratio of these. For the Earth-Moon system, 7 . 34 31022 kg M 51 . 30 3106 kg/km 3 . 3Moon 5 dMoon (384 , 000 km)3

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7. (c) Gravity depends on mass; thus, probing the effects of gravity from an object (i.e., orbits) allows one to calculate the mass of an object. 8. (a) Tidal force depends on the mass of the object creating the tide. This will make high tides higher, removing more water from the low tide areas. 9. (c) Tidal force depends on the size of the object feeling the tide, so if Earth became smaller, high tides would be lower, meaning low tides would be higher. 10. (b) The strength of gravity depends on 1/d2, so 2 times closer is 22 5 4 times stronger force. 11. (c) The strength of tidal forces depends on 1/d3, so 2 times closer is 23 5 8 times stronger tidal force. 12. (d) If both objects are tidally locked then they orbit and rotate about their axes with the same period, then (a) and (b) are both correct. 13. (b) Spring tides are the strongest and require the Sun and Moon to be aligned. 14. (d) The Roche limit is the point at which an object’s self-gravity can no longer support an object’s tidal pull of whatever it is orbiting. 15. (c) Self-gravity is the total gravitational force felt by an object because of all of its constituent parts.

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30 ◆ Chapter 4 Gravity and Orbits Thinking about the Concepts 16. The fundamental difference is that Kepler’s laws are empirical laws. They were deduced purely from observation of the motions in the Solar System. Newton’s laws are theoretical laws that describe the general behavior of all motions in the universe. A large measure of our confidence in Newton’s general laws is their ability to reduce to the more specific cases of Kepler’s laws. 17. Circular velocity is the speed needed to orbit a body in a circular path. Escape velocity is the speed needed to escape the gravitational force of a body. Escape velocity will always be larger because more energy is needed to escape an object’s gravitational force. 18. Mass is a measurement of an object’s inertia. Weight is how strongly gravity acts on the object. Mass is an internal property of an object and never changes. Weight depends on the strength of the gravitational field and, in some cases, can be zero. 19. The constant of proportionality is the gravitational acceleration. The gravitational accelerations are different because the Moon’s size and mass are different from Earth’s size and mass. Earth’s gravitational acceleration at its surface is 9.80 m/s2, whereas the Moon’s gravitational acceleration at its surface is 1.67 m/s2. 20. The comet in an elliptical orbit will eventually return to the immediate vicinity of the Sun (though we cannot say when without more information). The comet in a hyperbolic orbit will escape the Sun entirely and head out into interstellar space never to return. 21. The Earth rotates the fastest at the equator; thus, launching satellites from here gives them the highest initial speed. Because the Earth rotates to the east, launching toward the west works against this natural slingshot effect. 22. The period and distance of an orbiting object yield, via Newton’s formulation of Kepler’s third law, the sum of the masses of the object being orbited and the object that is orbiting. In the case in which the central object is much more massive than the orbiting object (i.e., a planet around a star), the smaller object can be ignored; thus, the orbital information yields the mass of the central object. 23. If an object is approaching the Sun in an unbound orbit, then either (a) the object came from outside the Solar System or (b) the object experienced some very intense gravitational “slingshot” acceleration from very close encounters with more massive objects. 24. As you move out from the center of the Earth, there is more mass interior to your radius. Because this amount of mass is what causes gravity, the strength of gravity increases.

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25. The ocean reaches its maximum recession from the beach during the low spring tides that occur at the new and full Moon phases. High spring tides occur near noon and midnight when the beach lies directly on the Sun-Earth-Moon line. Therefore, low spring tides occur approximately 6 hours earlier or later, near sunrise and sunset. The best time to go digging for clams is near sunrise or sunset during the new and full Moon phases. 26. If the Moon is in a quarter phase, then the net tidal force on the oceans is small, and there will not be a high tide. 27. The Moon attracts mass on the near side of Earth more strongly than it attracts mass at the center of Earth because of the 1/r2 relationship for gravity. For the same reason, the Moon also gravitationally attracts mass at the center of Earth more strongly than it attracts mass on the far side. As the near side of Earth is pulled away from the center, we intuitively see the near side bulge. However, because the center of Earth is also being pulled away from the far side, we get a less intuitive second bulge along the Earth-Moon line as well. Given that the tidal bulges are farther from Earth’s center than average, they represent low spots in Earth’s gravitational field. Therefore, water preferentially flows toward the centers of both bulges, raising sea levels on both the near and far sides of the planet simultaneously. 28. Solid bodies can be tidally distorted, and there is tidal distortion on the Earth. This distortion of the planet’s land masses is dissipated in frictional heating of rock. But, through conservation of energy, that dissipation comes at the expense of the Earth’s period of rotation and the Moon’s distance: the Earth is slowing down its rotational period, and the Moon is slowly drifting away. 29. Oceans cover 70 percent of Earth’s surface area, so a 1-meter increase in sea level in the ocean represents a tremendous volume of water. If that large volume of water is squeezed into a smaller area along a continental shoreline, the subsequent rise in the local sea level will be substantially higher than 1 meter. Combine that situation with receding depth as water approaches shorelines and resonant effects like those seen in the Bay of Fundy, and some places on Earth can experience 5- to 10-meter increases in sea level at high tide. 30. The Roche limit is the distance from a planet at which objects with little to no mechanical strength are ripped apart by tidal stresses. Small objects like communications satellites have a high intrinsic mechanical strength because of the chemical bonds that hold them

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Chapter 4

together. However, larger objects like planetesimals tend to be held together only by self-gravity. Furthermore, the tidal stresses inside a small communications satellite are minimal because the near and far sides of the satellite are practically the same distance from the center of Earth. Meanwhile, the tidal stresses inside planetesimals can be enormous because the near and far sides of a planetesimal may be separated by tens or even hundreds of kilometers. Applying the Concepts

Gravity and Orbits ◆

so we found the right value! Also, it makes sense that Neptune “plods” along because Kepler’s second law tells us that Neptune must travel slowly because it is so much farther from the Sun. 33. Setup: We can write the circular velocity for any object in orbit as v circ 5

GM primary r

GM Sun rVenus

r v2 M Sun 5 Venus . G Solve: Inserting values from the text and solving for mass, M Sun 5

1. 082 3108 km(35.03 km/s)2 51 . 99 31030 kg 6. 67 310220 km 3 kg 21s22

31. Setup: Newton’s law of gravity tells us that the acceleration of gravity (g) scales with the mass (M) of what 8 310 km(35.03 km/s)2 1.it; 082 makes the gravity and how far (R) you M are from that 5 51. 99 31030 kg . Sun 3 220 21 22 2 6. 67 310 km kg s is, g  M/R . We can compute the ratio of acceleration Review: This is exactly the mass of the Sun listed in g Mars M Mars R 2 Earth 5 on Mars and Earth as Note this text. M Earth R 2 Mars g Earth 34. Setup: The escape velocity from any planetary that weight is your mass times gravitational acceleration. 2GM planet . Because surface is expressed as vesc 5 Solve: Mars has 0.1 times the mass of Earth and half rplanet the Earth’s radius, that is, the Earth is twice as large. we are looking to compare the asteroid’s escape Using these values in our equation above, the ratio 2 of acceleration is (0.1) 3 (2 ) 5 0.4, that is, gravitavelocity to that of Earth’s, we will take a ratio. This tional acceleration on Mars is about 40 percent that on will also have the nice result of cancelling common Earth. This means you will weigh 40 percent less than on Earth, but note that your mass does not change be2GM asteroid cause it is an intrinsic property. If I weigh less but have rasteroid vesc,asteroid 5 . Simplifying yields factors: the same strength, I would think that I could jump vesc,Earth 2GM Earth twice as high and lift objects twice as massive. So I rEarth think Hollywood misses the mark in their movies. Review: Note that although Mars is smaller, it is sigvesc,asteroid M asteroid rEarth . 5 nificantly less massive than Earth, which is why gravity vesc,Earth M Earth rasteroid is so much less on Mars. Solve: Substituting numbers into the equation (re32. Setup: As hinted in the problem, distance equals rate membering that they must be in Earth units!), times time or d 5 vt, where the distance is the circumference 2πr of the orbit. 212 we obtain vesc,asteroid 5 10 5 1028 5 1024 . The 9 d 2 4. 5310 km vesc,Earth 1024 . Solve: Solving for time, v 5 5 164. 8 yr t escape velocity from the asteroid is only 1024 times Notice that the problem gives the Earth’s speed Earth’s escape velocity, that is, (1024)(11.2 km/s) (1,000 m/km) 5 1.12 m/s. in kilometers per second (km/s), so we really Review: This converts to about 2.5 mph. Even a good should convert years to seconds. A nifty trick to jump will launch you off the asteroid. This stands to remember is that there are roughly  3107 secreason considering how small this asteroid is. 35. Setup: Although this question involves an orbit, we onds in one year! So use this as a conversion factor: can solve it using our familiar formula that distance 9 yr 2 4 . 5310 km km/s. 5 5 . v5 3 5 equals rate times time, or d 5 vt. We are looking for 164. 8 yr  3107 s time, we know the speed, and just need the distance, which is the circumference of the Earth: 2πr 5 2 3 Review: We can actually look this up in the Appen3.14 3 6,400 km 3 40,200 km. dices and find the reported orbital speed is 5.4 km/s,

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32 ◆ Chapter 4 Gravity and Orbits d v

Solve: t 5 5

40, 200 km 5 5, 087 s or about 7 . 9 km/s

85 minutes. Review: It takes most satellites about 90 minutes to orbit the Earth, so our value is consistent. 36. Setup: The circular velocity of an orbit is and the escape velocity is vcirc 5 GM / r vesc 5 2 vcirc . We must travel above the escape speed

to leave Earth’s orbit and travel to Mars. Solve: (a) If the spacecraft is in an orbit at its circular velocity, then it is orbiting Earth in a circular orbit. (b) The spacecraft must be sped up by vesc 2 vcirc 5 (1 2

2 ) vesc 0 . 3vesc . 2

GM and M 5 5.97 3 r 1024 kg, r 5 6,370 1 350 km 5 6.72 3 106 m. Solve: Plugging these into our formula, we find 7.7 km/s. Review: Google lists the orbital speed of the ISS at 28,000 km/h or 7.77 km/s, as we found. M 40. Setup: Start with g 5 G 2Earth and solve for M. R Earth Then enter relevant values, g 59. 8 m/s2 , G 5 6.673

39. Setup: Circular velocity v circ 5

10211 m3/kg s2, and R Earth 56 , 371 km . Solve: M Earth 5 g

2 R Earth (6. 37 3106 )2 5 5. 96 31024 kg 5 9.8 G 6. 67 310211

Review: In many cases, we launch space probes in mulR2 (6. 37 3106 )2 M Earth g Earth 5 9. 8 5 5. 96 31024 kg . tiple steps, so that we can conserve fuel and use5gravity 211 G 6 . 67 3 10 to help the ship travel faster. Review: According to the text, M Earth 5 5. 97 31024 kg, 37. Setup: To solve this problem, set the gravitational so we are spot on. force from Newton’s universal law of gravitation, 41. Setup: Tidal force is directly proportional to mass and inversely proportional to the cube of the distance Mm F 5 G 2 , equal to the force you would feel standing between the objects. Writing this mathematically, we r M on Earth, that is, F 5 ma. Do not forget to use the right have Ftidal  3 . units, that is, mass in kilograms and distance in meters. r Solve: Substituting numbers into this equation, we find Solve: Setting the two forces equal, we find 3 27 FJupiter  1. 9 310 kg   Mm 1m  211 F 5 G 2 5 ma or 5   5 5310 r Fdoctor  80 kg   7 . 8 31011 m  . M m 3 5 . 97 310 24 kg a 5 G 2 5 6 . 67 310211 Review: The tidal force from Jupiter is incredibly tiny r kgs 2 (6 . 370 31 0 6 m)2 compared with the tidal force from the doctor. In fact, inverting this ratio shows that the tidal force from the 59.81m/s 2 , as expected. doctor is approximately 20 billion times stronger than Review: Notice that the acceleration of gravity dethat from Jupiter! pends on the mass and size of the planet. Indeed, a 42. Setup: We need to use Kepler’s third law in its full very small planet with high mass could have gravity 4 2 A 3 higher than the Earth. . expression: P 2 5 GM 38. Setup: We could compute the actual force of gravity Solve: The orbital separation of Ida and Dactyl is acting on the rocket at both locations or just recognize A 5 90 km 5 9 3 104 m. Plugging into our formula, 1 that gravity scales with distance as Fg  2 , so the ratio 4 2 (9 3104 )3 d P 5 51. 013105 s, or about 16 211 2 6. 67 310 4. 2 310   F on the ground to in orbit will be ground 5 dorbit . 28 hours.   Forbit  dground  Review: We expect a very short period considering the Solve: Using numbers given in the text, low mass and close orbit. 43. Setup: This problem is asking about gravitational 2 Fground  6370 1350  acceleration, so we will need Newton’s second law: 5  51. 11 , that is, gravity on Forbit  6370  F 5 ma. the ground is about 11 percent larger than when the Solve: (a) Remember that the acceleration of gravity rocket is in orbit. acts on all objects the same way, so we feel the accelReview: We know that gravity drops with altitude, eration g of gravity on us. (b) We will feel exactly the and 350 is about 5 percent of 6,370, so we expect the same acceleration g whether one instructor is strapped 2 change in gravity to be 1.05 5 1.10, as we found. to us or not. (c) Using Newton’s second law, the force

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Chapter 4

of gravity on me will be F 5 mg 5 70 kg 3 9.8 m/s2 5 686 N. (d) If I am strapped to an instructor, our total mass is doubled so the total force will be as well. Review: This problem highlights the very important fact that the acceleration of gravity on different objects does not change, but the force that they feel does because it depends on their mass. It is very important not to confuse these! 44. Setup: This question is asking about orbits, so chances are we will need Kepler’s third law. However, it mentions a star other than the Sun, so we need it in its fully expanded form P2 5 A3/M, where M is the mass of the star as measured in solar masses. Solve: (a) Using Kepler’s law, where the distance A is still 1 AU and M 5 2 solar masses, we find the period is P 5 A 3 / M 5 13 / 2 5 1 / 2 5 0. 71 year . (b) For this planet to complete an orbit with the same circumference as Earth but in 0.71 times the time, it must be traveling that much faster than Earth. 0.71 is a factor of 1/0.71 5 1.4 times smaller than Earth’s period, so the planet travels 1.4 times faster than Earth. Earth travels at 29.8 km/s, so this planet travels 1.4 times that, or 41.7 km/s. Review: Notice that the solution above, using proportional reasoning, saved us from a lot of calculation. Do the numbers make sense? Vega is more massive, so the planet must orbit faster because the star’s gravity is higher, and thus, the circular velocity must be higher as well. 45. Setup: Tidal force scales as r23. Solve: If the moon were 0.8 its current distance then the strength of the tidal force would be 0.823 5 1.95 times stronger. With a stronger gravitational pull, the spring tide would be significantly higher than it is today, and the neap tide would probably be a little higher. Review: One could always use the equation of tidal fore in Working It Out 4.4 to calculate the strength of tidal force with the moon at its current distance and again at 0.8 times that value; however, noting that none of the other parameters change except d, it seems like more work than needed. Using the Web 46. The feather and hammer video shows that in the absence of air resistance, all objects fall at the same rate regardless of mass. If I try this here on Earth, the feather will fall much more slowly than just about any object because of air resistance. Things fall on the Moon slowly compared with how things fall on Earth

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Gravity and Orbits ◆

because the acceleration of gravity on the Moon is much lower, about one-sixth. 47. I would weigh significantly more on Jupiter, about the same on Venus, Saturn, Uranus, a little more on Neptune, and a lot less on Mercury, the Moon, and Mars. It would be very difficult to get out of bed or walk on Jupiter because my muscles are not used to holding up more than double my current weight. In lower gravity, our muscles might atrophy, and it would be very hard to return to Earth. Low gravity may also cause circulatory issues as well as bone loss over time. I expect anything that lives in water and is used to little effective gravity would do fine in lower gravity fields. Similarly, life forms that are sedentary might do fine in stronger gravity. 48. a) The tennis ball seems to float at the tip of the coaster because freefall removes the effects of gravity. Newton was able to imagine a satellite orbiting Earth because he thought about the moon falling around the earth. The speed of the cannon ball determines its orbit because the ball must go further than it falls; this is dependent on speed only. Launching a satellite is difficult because it requires enormous energy to escape Earth’s orbit. b) The airplane freefalls, providing a ten-gravity experience. This is short-lived because the plane must resume flight so it doesn’t crash. The plane is almost identical to the roller coaster because both are in freefall. 49. Tidal forces make two bulges on Earth, and it takes a full day for both of these to revolve around—hence, two tides per day. In the Washington, DC, area, the maximum difference in tidal heights are about 2 feet. The time of high tide follows the moon reaching the meridian. High and low tides happen at new/full and quarter moons, respectively. 50. Tidal forces rip apart the constituent particles in a galaxy, making tails of stars. This doesn’t happen on Earth because we are a solid, bound object. Exploration 1. Mercury is farthest from the Sun when its acceleration is smallest. 2. Mercury is closest to the Sun when its acceleration is largest. 3. The largest and smallest accelerations are 0.0628 and 0.0272 m/s2, respectively. 4. The acceleration vector always points toward the Sun. 5. The force on the planet is always toward the Sun because Newton’s second law says that F 5 ma (i.e., acceleration and force always point in the same direction).

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34 ◆ Chapter 4 Gravity and Orbits 6. Mercury moves the slowest when it is farthest from the Sun. 7. Mercury moves the fastest when it is closest to the Sun. 8. The largest and smallest speeds are 59.0 and 38.9 m/s, respectively. 9. The angle between the acceleration and velocity vectors is not always 90°. 10. If the orbit becomes a perfect circle, then that angle becomes 90°. 11. When A is changed to 0.8 AU but the eccentricity is left alone, the orbit becomes larger and the period increases, but the shape of the orbital path remains the same. The planet’s highest speed is about 41 km/s.

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12. When A changes to 0.1 AU, the opposite happens from question 11, namely, the planet is zipping around the Sun now at a maximum speed of 116 km/s. 13. As I play with the value of A, I confirm that when planets are closer, they move faster, and when they are farther, they move slower. When the planet is 0.1 AU from the Sun on a circular orbit, its speed is 94.2 km/s, at 0.2 AU it is 66.6 km/s, at 0.4 AU it is 47.1 km/s, and at 0.8 AU it is 33.3 km/s. Using my scientific calcu A  lator, I find the best fit is v 5 29. 8 km/s 3  AU  which shows the dependence on A as expected.

21/2

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CHAPTER 5

Light INSTRUCTOR’S NOTES Chapter 5 covers the nature of light and matter. Major topics include

▶ the particle and wave nature of light ▶ the electromagnetic spectrum ▶ the quantum-mechanical model of atoms (including emission and absorption)

▶ Doppler shifts ▶ energy and temperature ▶ blackbody radiation ▶ the inverse-square law brightness relationship

and luminosity-distance-

Although Rømer, Fizeau and Foucault, and Maxwell typically come to mind when discussing light, students may find it interesting to learn that both Galileo and Newton dabbled with the nature of light. Galileo attempted to measure the speed of light in 1638 by using lanterns at night. Galileo and an assistant stood about a mile apart with their lanterns covered. Galileo uncovered his lantern, and when the assistant saw that light, he uncovered his. Galileo measured the elapsed time between uncovering his lantern and observing the light from his assistant. He concluded that light traveled too fast to be measured by this technique. Rather than try to measure its speed, Newton developed a theory of optics based on the “corpuscular” theory; that is, treating light as composed of tiny particles. In teaching any topic, I try to create “teachable moments” in which students become invested in learning an answer because I helped motivate them to ask the question. One example is the particle wave duality of light, which one can do in the span of 8 to 10 minutes of class time. I start by showing the first 1 minute 49 seconds of the 5-minute long video “Dr. Quantum—Double Slit Experiment” (you can readily find it on YouTube), which is an animated explanation of what happens if you pass particles and waves through two slits. I then shoot a laser through a double slit (available in most physics class optics kits) and have the students determine for themselves what light is. A batteryoperated ultraviolet (UV) sanitizing wand, soda can, and tinsel are enough to demonstrate the photoelectric effect

(https://www.youtube.com/watch?v5W038qVDGgqw). I qualitatively explain that as waves, the UV light would require days to dislodge the electrons (think about moving a bowling ball by yelling at it). Students quickly figure out the rest. As an interesting teaser for quantum mechanics and further study in physics, I show the rest of the Dr. Quantum video, in which he describes electron interference via double slits. Some of this material is quite intuitive, such as the inverse square law of brightness, as everyone knows that lights appear dimmer when they are farther away. The relationships of color and temperature or among luminosity, temperature, and size, on the other hand, can be much more problematic. Students often associate that “more” always requires “bigger,” so it can be quite counterintuitive that a hotter object is bluer (meaning shorter wavelength) or that a cooler object can be more luminous than a hotter one if the sizes are correct. Students generally benefit from a full class on blackbody radiation, including working with different Planck functions to distinguish brightness from temperature. A few other topics have consistently been challenging for all of the students I have taught. First, the notion that all wavelengths of light travel at the same speed in a vacuum is quite counterintuitive, especially when we add the fact that bluer light is more energetic. I think that students are trying to equate higher energy with faster speed. Second, color and temperature can be very challenging: Our water faucets teach us that red is hotter than blue, and we generally talk about metal as being “red hot” or “white hot” rather than “blue hot.” The term “flux” is almost a four-letter word to many students, because its meaning is not always obvious. In Working It Out 5.3 and elsewhere, it helps to spend some time differentiating between flux and brightness and give some useful analogies, such as water flowing through a hose. Also, most spectra in books show hydrogen, which can lead students to think that all atomic spectra look like hydrogen. It is useful to point out that each atom has its own complicated series of energy levels, or, spectral fingerprints. There are a number of build-your-own or preassembled handheld spectrometers (for example, http://starlab.com/ shop/Cardboard-Spectrometer-Kit.html or http://www. amazon.com/dp/B00B84DGDA) that allow students to 35

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36 ◆ Chapter 5 Light measure actual spectra. I also have found the RSpec digital spectrometer (http://www.rspec-astro.com/) to be of enormous utility, but it requires a dark room or some baffling to limit contamination from projector light or sunlight. Introduction of the UBVRI ( Johnson photographic) filters can add an unnecessary extra level of confusion, but I like to spend a few minutes talking about the practical task of measuring temperatures of stars. Many students find it fascinating that in the time needed to measure the temperature of one star with a spectrum, we can measure the temperature of thousands of stars with two images in different filters. It is always an interesting conversation to debate the merits of photometry versus spectroscopy, in terms of best using limited resources (observing time) and maximizing the information returned. Almost invariably, one student asks if it is possible to take the spectra of hundreds of stars at once, which is a great segue into modern observing techniques of multifiber spectrographs, grism imaging, and so forth.

DISCUSSION POINTS

▶ Light behaves as an electromagnetic wave that has vari-

able amplitude and frequency, but always travels at a constant speed, c, which is a universal constant. Discuss how this fact constrains our ability to obtain information from distant astronomical objects and how it limits the possibilities of interstellar travel. ▶ Use the examples in Section 5.1 and Working It Out 5.1 to discuss applications of light of different wavelengths. Have students use those equations to calculate the wavelength of x-ray light and discuss how it compares with the size of common molecules. ▶ Using the relation between the frequency of the electromagnetic wave and the energy carried by each photon (Section 5.1), have students calculate the energy of different types of light (for example, ultraviolet, visible, infrared, microwave, radio, and so forth). Discuss what types of natural processes could emit those different types of energies. ▶ Have students compare the energy levels of an atom with the bookshelf analogy. Compare the structure of an atom with that of the solar system. Discuss the differences and similarities. ▶ Think of different situations in which the Doppler shift occurs and what we have taught ourselves to learn from these (for example, we know when a motorcycle or emergency vehicle is approaching or receding based on its pitch). ▶ Compare and contrast the physical situations that lead to blackbody radiation versus emission lines and the

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physical situations that lead to emission versus absorption lines. ▶ Use a sink and variable faucet to demonstrate equilibrium, heating, or cooling of a planet based on energy inflow/outflow.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 5 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises. Light as a Wave, Light as a Particle This interactive animation defines the basic elements of an electromagnetic wave and a photon, linking the concepts of wavelength and energy through the use of a wave whose wavelength and frequency can be adjusted. Text reference: Section 5.1 Atomic Energy Levels and the Bohr Model Discussion of the atomic nature of matter, the constituent parts of an atom, the Bohr and probabilistic models of electron levels, quantized energy levels, and how emission and absorption occur. This animation helps better understand Figures 5.9 through 5.13. Text reference: Section 5.2 Atomic Energy Levels and Light Emission and Absorption Discussion and simulation of the relationship among atomic energy levels, emission and absorption, and the corresponding wavelengths/colors of light that are emitted and absorbed. This animation helps better understand Figures 5.9 through 5.13. Text reference: Section 5.3 Doppler Shift This simulation demonstrates the Doppler shift using sound from “the world’s most annoying sound on a string.” Text reference: Section 5.3

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate

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Chapter 5 Light ◆

variables and work toward understanding physical concepts presented in Chapter 5. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive.

they have learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5) and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack.

EM Spectrum Module This module surveys the electromagnetic spectrum, showing a typical astronomical image for different wavelengths of light and the kind of instrument that would take such an image. Text reference: Section 5.1

Emission and Absorption Dr. Palen explains the quantized processes of emission and absorption using a stairway (energy levels) and balls (photons of different color/energy). Text reference: Section 5.2

Three Views Spectrum Demonstrator

Changing Equilibrium

This module demonstrates how emission, absorption, and continuum (blackbody) spectra arise and what they look like. Text reference: Section 5.2

This simulation demonstrates equilibrium using water bottles to flow liquid (representing energy) into and out of a vessel. This is an analogy to energy flowing into and out of a planet to change its temperature. Text reference: Section 5.4

Hydrogen Atom Simulator This simulation models a hydrogen atom and its interactions with light, demonstrating the quantum nature of emission and absorption. The user selects different wavelengths of light and fires photons at the electron, to better understand how and why the electron moves between energy levels. Text reference: Section 5.2 Doppler Shift Demonstrator This simulation shows circular waves expanding from a source. Movement of the source or observer affects the frequency of the waves seen by the observer, demonstrating Doppler shift. Text reference: Section 5.3 Blackbody Curves This simulation demonstrates how the blackbody spectrum varies with temperature. It also shows how the relative intensities observed through different filters (including the “color index,” that is, B through V) can give an estimate of temperature. Text reference: Section 5.4

ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what

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Wein’s Law This simulation demonstrates Wien’s law by varying current through a light bulb. This also shows the Stefan-Boltzmann relationship between temperature and flux. Text reference: Section 5.4 Inverse Square Law This simulation discusses the inverse square law and provides a demonstration by inflating a balloon. Text reference: Section 5.5

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. e, c, b, d, a See Figure 5.6. 2. (d) Energy and frequency are inversely proportional to wavelength. 3. The spectral features that are present correspond to the composition of elements present in the sun, because those elements must be present to create those features. 4. (d) A blueshift is a negative shift, and the larger the value, the greater the speed. 5. (e) Bluer equals hotter for blackbodies. 6. (c) Brightness drops with distance squared.

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38 ◆ Chapter 5 Light Reading Astronomy News 1. The light took roughly 7,300 years to reach Earth. 2. The young stars are blue because they are hot. 3. The stars will give absorption spectra, whereas the gas will give emission lines. 4. Red is the lowest energy transition of hydrogen in the visible; thus, hydrogen gas clouds generally appear red. 5. This author lives in the continental United States, so the Gum nebula would not be visible. Test Your Understanding 1. (b) Earth is about 7.5 light minutes away from the sun. 2. (b) A given element is defined by its number of protons. 3. a, b, c because we can have 10-7, 10-5, and 7-5. 4. (d) The Doppler effect explains why waves are compressed in the direction of motion and stretched out in the opposite direction. 5. (b) This is a statement of Stefan-Boltzmann’s law and Wien’s law. 6. (d) As surprising as it seems, size drops out of the planet’s temperature (see Working It Out 5.4). 7. (e) Star A is farther so it must appear fainter. Because brightness scales as 1/d2, it will appear 32 5 9 times fainter. 8. (a) If less energy is radiated away, the energy is trapped, making the temperature increase to a new equilibrium level. 9. (c) The fastest speed of light is that in a vacuum. In any medium, light slows down. 10. (c) A downward transition gives off energy, meaning a photon is emitted. 11. (c) Red light is low in energy and has a long wavelength. 12. (a) Brightness scales as L/d2 so if star A is twice as far away, it must have 22 5 4 times more luminosity to appear as bright as star B. 13. (d) Using Wien’s law, T 5

2. 9 3 106 nm . 290

14. (c) Equilibrium means “stuff in” equals “stuff out.” 15. (b) Temperature is a measure of the energy in an object and because the object is made of atoms, it is equivalently a measure of the speed of the atoms (because kinetic energy is related to speed). Thinking about the Concepts 16. It is possible for light to travel at a speed slower than 3 3 108 meters per second (m/s) by having the light

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travel in a medium. Light, similar to sound, travels at a certain speed in each different kind of medium. For example, sound travels faster in helium than in air, which is why people sound funny when they inhale helium and speak. The behavior of light is no different, as it travels at different speeds depending on what it travels in, that is, air versus glass versus water. 17. Light is both a wave and a particle, depending on the size scale on which it is operating. When light interacts with atomic and subatomic particles, it acts like a particle. When light interacts with “macroscopic” things (that is, larger than atoms and molecules), it acts like a wave. 18. According to the scientific method, if one test of a theory fails, then the theory must be reconsidered and revised, or it has been proven incorrect. 19. A beam of red light can have as much energy as a blue beam since the red one has more photons. See Figure 5.7. 20. According to the cosmological principle, observers should record the same spectrum for a given element or compound whether they are making their observations from Earth or from anywhere else in the cosmos. Consider the example of hydrogen. The same set of spectral lines for hydrogen that is measured in laboratories on Earth has been observed in the atmosphere of the Sun, in the emissions of interstellar nebulae, and in the spectra of distant galaxies. Given that we see the same spectrum for hydrogen (or any other element or compound) no matter where we look in the universe, we gain confidence in the cosmological principle. 21. Atomic behavior is based on probabilities, so all we can do is say what the most likely lifetime of the state will be. That being said, it is possible to calculate these lifetimes using quantum mechanics, but this is outside the scope of the course. 22. Spectral lines tell us the composition, temperature, and density of the material, as well as its velocity. 23. Redshifted objects are moving away from us; blueshifted are moving toward us. 24. The frequency is higher, so the object is moving toward you. 25. This is an example of dynamic equilibrium because the individual visitors to the museum are constantly changing. It is still an equilibrium though, because the flow of visitors into the museum equals the flow of visitors out of the museum. 26. The blue star is hotter than the yellow one. 27. If we assume that all stars have the same luminosity, then this would tell us that the stars are at a variety of

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Chapter 5 Light ◆

distances. Knowing instead that they vary in luminosity as well, it tells us that the stars cover an even greater range of distances. 28. Life on Earth evolved with the Sun as the source of light, so creatures adapted to see in its dominant wavelength range. 29. Venus has a temperature that is tremendously higher than that predicted by a simple equilibrium model. We will learn this is because it has a runaway greenhouse effect. 30. Assuming all stars are the same size as our Sun, a red star puts out less energy per second than our sun, so a planet would have to be fairly close to a red star to have roughly the same temperature as we do. Similarly, a planet around a yellow star would be at roughly the same distance as Earth, and a planet around a blue star could be much further away. Applying the Concepts 31. Setup: Frequency, wavelength, and speed are related by c 5  3 f. 8

3 3 10 m/s c 5 380 m for 790 AM, Solve:  5 5 f 7. 90 3 105 s–1 8

3 3 10 m/s c and  5 5 53. 05 m at 98.3 FM. f 9. 83 3 107 s–1 Review: Radio waves are all around 1 meter or longer in size, so we indeed confirm that these are in the radio spectrum. Also, note that shorter frequency corresponds to longer wavelength, again as expected. 32. Setup: Frequency, wavelength, and speed are related by c 5 λ 3 f. Solve: f 5 2.45 3 109 Hz, so following the solution to question 41, λ 5 0.12 m 5 12 cm. Review: Is it odd that a “microwave” emits radiation with wavelengths of centimeters rather than microns? Microwave does not actually mean “microns” but is the region between infrared and radio, so it includes centimeter wavelengths.  v 33. Setup: The Doppler formula is 5 , where ∆λ is 0 c the change in observed wavelength and λ0 is the rest wavelength. Solve: Solving for speed, we find v 5

 502. 3 – 486. 1 3 3 108 5 10 7 m/s c5 486. 1 0 510 , 000 km/s.

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Review: In the optical, a good rule of thumb is that 1 nm of Doppler shift corresponds to a speed of approximately 600 km/s. Here, we have about 16 nm of shift, or about 16 3 600 5 9,600 km/s, as found. f v 34. Setup: The Doppler formula for sound is 5 , f0 c s where ∆f is the change in observed wavelength, f0 is the rest wavelength, and cs is the speed of sound (about 340 m/s). Solve: Solving for the frequency shift 25 v f0 5 440 Hz 5 32. 4 Hz . Because the 340 cs object is approaching, the pitch goes up so the new frequency will be 440 1 32.4 5 472.4 Hz. Review: Note that the amount of Doppler shift depends on the speed of the wave, that is, we can travel at slow speeds and still produce audible Doppler shifts, whereas very high speeds are required to produce noticeable shifts in light. 35. Setup: This is a half-life problem in which each half1 1 life, of the sample decays. In 2 half-lives, of the 2 2 f 5

1 1 1 1 decays, that is, there is 3 5 , and 2 2 2 4 1 in general, after n half-lives, n remains. 2 Solve: Here, the half-life is 30 minutes, so after 2 hours, 1 1 remains. 4 half-lives have passed so only 4 5 2 16 Because brightness depends on the number of atoms, the toy will be 16 times fainter. Review: Note that the half-life decay pattern means the number of particles remaining decreases exponentially. A common mistake is to consider it to be linear, that is, 1 1 at 4 half-lives, 3 4 5 remain, which is false. 2 8 36. Setup: Remember that brightness scales as 1/d2, where d is the distance. Solve: Neptune is 30 times farther from the Sun than Earth, so the Sun will appear to be 302 5 900 times remaining

1 1 as bright. To fainter, or equivalently,   5  30  900 Voyager 1, the Sun will appear 1242 5 15,376 times fainter. Review: Although the Sun appears as a huge, blazing bright disk in our sky, it becomes the size of a pinprick in the outer solar system.

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40 ◆ Chapter 5 Light

37. Setup: Brightness scales as

L so for two objects of d2

Lbulb Lfirefly . 5 2 2 d bulb dfirefly Solve: Solving for the distance of the bulb,

equal brightness,

Lbulb 5 5 106 5 5000 m or 5 km. Lbulb Review: This may seem very far, but remember that your eye is very sensitive to light at night. If this is just a 60 watt bulb, imagine how far a star must be to have the same brightness as that firefly! L 38. Setup: Brightness scales as 2 so for two objects of d L L equal brightness, 1 5 2 . d12 d22

d bulb 5 dfirefly

Solve: Solving for luminosity, 2

d  L2 5 L1  2  5 32L1 5 9L1 d  1

or the more distant star must be 9 times more luminous. Review: Another way to think about this is just in terms of distance. Three times farther means 32 5 9 times fainter, so the star must be nine times more luminous to appear the same brightness. 39. Setup: Power radiated is given by T4, where  is the Stefan-Boltzmann constant (Working It Out 5.3). Solve: For a temperature of 500 K, this gives power radiated per square meter of P 5 5.67 3 10–8 (5004) 5 3,540 W. Review: At 3.5 kW, this is about the power consumption of a typical house. 40. Setup: The luminosity from a blackbody of size R and temperature T is L 5 4pR2T4. Solve: Using the values given in the text, (R 5 6.96 3 108 m and T 5 5780 K), we find L 5 3.85 3 1026 W. Review: The quoted solar luminosity is 3.84 3 1026 W, so we see that this blackbody approximation is quite good. 2 , 900 , 000 nm K . T Solve: (a) If T 5 100,000 K, peak 5 2.90 nm. Review: At 100,000 K, the object is essentially emitting all its energy in the far UV (almost at x-ray wavelengths)! 42. Setup: The peak radiation for a human being can be determined from Wien’s law, but first we need to convert 37°C to Kelvin. To convert degrees Celsius to Kelvin, simply add 273 degrees. Thus, human body 41. Setup: Wien’s law tells us peak 5

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temperature is approximately (37 1 273) K 5 310 K. 2 , 900 m K . For part (b), Then use Wien’s law peak 5 T the luminosity of a blackbody is L 5 AT4, where A represents the surface area of the blackbody. 2 , 900 m K 5 9. 45 m Solve: (a) peak 5 310 K (b) L 5 AT4 5 (5.67 3 10–8)(0.25)(310)4 5 130.9 Watts. Review: Thinking about the film The Matrix, this certainly shows that the human body does put out a significant amount of energy! 43. Setup: In general, the blackbody temperature of a 1

 L (1 a)  4 1 . But planet is expressed as T 5  sun   16  d given that all other factors are equal except the albedo in this problem, we can determine the change in temperature compared with zero albedo (279 K) by tak1

 Lsun (1a)  4 1 1  16  d 4  ( ) 1 a T  . ing a ratio: 5 5  1 1 T0  Lsun (1a0 )  4 1 (1a0 ) 4  16  d   If we substitute in T0 5 279 K and a0 5 0, then T 5 279 4 1  a . Solve: (a) If a 5 0.1, then T 5 271.7 K. (b) If a 5 0.9, then T 5 156.9. Review: Note that as albedo increases, less light reaches the surface so the planet is not heated as much. This is an interesting prelude to the “snowball effect” in which one argues that if too much water is put into the atmosphere by global warming, the resulting snows in winter will reflect so much sunlight that the planet actually cools off. 44. Setup: Use the planet-temperature equation 1

 L (1 a)  4 1 T 5  sun .   16  d Solve: With a 5 0.08 and d 5 97.7 AU 5 1.5 3 1013 m, we find T 5 18.5 K. Review: Brrr. Solar heating is almost nonexistent at these distances! 45. Setup: According to our text, the blackbody tempera1

 L (1 a)  4 1 . From ture of a planet is T 5  sun   16  d /4 , which is all we need for this this we see that T  L1sun problem.

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Chapter 5 Light ◆ /4 Solve: T  L1sun 5 101/4 5 1 . 78 or the Earth would be

about 1.78 times hotter. Review: It stands to reason that the planet would warm, because the sun would put out more energy to be received.

50. Answers vary on time of year. The answer will include which parts of Earth have the highest and lowest albedos, where albedo is changing the most, whether one expects ice/snow/oceans/clouds/forests/deserts to add or take away from albedo, and what regions are not shown at all.

Using the Web

Exploration

46. (a) UV light represents hotter objects. Between optical and UV, most of the galaxy disappears! (b) Answers will vary. 47. CSI members often use ultraviolet lights to search for fluids because organic molecules/compounds in these fluoresce in UV light. 48. Thermal goggles view the world in the infrared and render this on a video screen in the optical, usually as black and white or green and white. Optical instruments exactly reproduce the optical light that is observed. 49. Many TSA imaging devices use millimeter waves to make maps of the density of objects on a person’s body. Aside from privacy arguments, it seems these are fairly harmless to the body.

1. As wavelength increases, frequency gets lower. 2. As wavelength decreases, frequency gets higher. 3. Wavelength and frequency are inversely related, that is, if wavelength increases, frequency decreases. 4. If I increase the frequency, the wavelength should shorten. 5. Yes, the wavelength shortened as expected. 6. Nothing. 7. Nothing. 8. Amplitude has no relationship to frequency or wavelength. 9. This is a wave of light, so its speed never changes in a vacuum.

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CHAPTER 6

The Tools of the Astronomer INSTRUCTOR’S NOTES Chapter 6 covers the various means by which astronomers collect data. Major topics include

▶ basic optics ▶ the design and function of an optical telescope ▶ how light is recorded by cameras and spectrometers ▶ the design and function of telescopes outside the optical region of the EM spectrum

▶ space probes ▶ other tools one does not always associate with astronomy, like computers and particle accelerators

This chapter contains material that the fast talker can cover in one lecture, which can be advantageous because by this point in the class, many students are hungry for something that feels astronomical and less “physicsy,” such as learning about the planets instead of Snell’s law. That said, a thorough exposition of this chapter is important if you want to make the distinction between “here are a lot of facts” and “here is how we discovered those facts” as you teach the later chapters; after all, were it not for telescopes, detectors, and a few space probes, we would not have the rest of the book! I like to start this lecture by asking students to name the reasons why we have telescopes. Almost without fail, “to make things look bigger” is the first reason given, followed quickly by “to see things farther away.” If you probe long enough, you can come up with three main reasons: magnification, light gathering, and resolution. And if you ask the class to rank them, light gathering will rarely place first. My subsequent lecture is devoted to explaining why I think it is the most important reason, and why magnification is the least. Increased resolution used to be more complicated to explain than light gathering, but because students today are all intimately familiar with SD, HD, and now 4K television sets, it makes a lot more intuitive sense. Still, it can help to give them a sense of resolving power and why it is important. Figure 6.11 shows what happens when two light sources are marginally resolved; we are all familiar with this problem when we see headlights coming toward us at

night, and when the vehicle is far enough away, we cannot tell whether it is a car/truck or a motorcycle. It may help to tell students their eyes have a resolving limit around 2 arcmin, so in general they cannot distinguish the two until the vehicle is about 1.5 miles away (assuming the headlights are separated by about 5 feet). A typical ground-based telescope could do that for a car 200 miles away (assuming 1 arcsec resolution), whereas Hubble Space Telescope can do it for a car about 7,800 miles away. Another interesting way to drive this home is that a typical ground-based telescope can make out the face on a U.S. quarter held 3.1 miles away (1 arcsec), whereas Hubble can easily read where the coin was minted. Finally, most students are shocked to learn that a typical radio dish cannot make out any detail on the full moon—that explains why dishes are so big and why we use arrays. Although I hate to stand on soapboxes in class, I do spend a few minutes explaining why cheap, department-store telescopes are a waste of money, because of poor optics, lousy mounts, and the promise of ridiculously high magnification. This is also a good segue into why magnification is the least important reason to use a telescope—high magnification means the image becomes very faint, and atmospheric seeing smears out most of the detail. If you have access to a few lasers and acrylic lenses, you can make a much more memorable lesson on geometric optics by showing the laser tracks passing through the lenses (or bouncing off the mirrors) than by teaching ray tracing and the lens-makers formula. It may bother your physics sensibilities to skip this formula, but at the end of the day, the math tools in Working It Out 6.1 will be a lot more useful, in my opinion. For classroom exercises and lab sessions, this chapter lends itself to hands-on activities. There are some very handy and inexpensive kits that I recommend. The Starlab Refracting Telescope Kit (sold at shop.sciencefirst.com) is a set of 10 build-it-yourself refracting telescopes with which students can perform many experiments, such as measuring focal length and resolution, estimating magnification and how much more light a telescope collects than the human eye, and observing the sky roughly as well as Galileo did with his first instruments. There are a number of build-your-own 43

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44 ◆ Chapter 6 The Tools of the Astronomer or preassembled hand-held spectrometers (for example, http://starlab.com/shop/Cardboard-Spectrometer-Kit. html or http://www.amazon.com/dp/B00B84DGDA that allow students to measure actual spectra. I also have found the RSpec digital spectrometer (http://www.rspec-astro. com/) to be of enormous utility, but it requires a dark room or some baffling to limit contamination from projector light or sunlight. Because this section discusses computers and because your students probably never lived without one nearby, it can be useful to show them what research life used to be like, by downloading a few tax forms from the IRS. At the turn of the century, “computers” were rooms filled with (mostly) women who carried out computations and data analysis, often on similarly-styled forms which guided the user to do complicated calculations in a step-by-step process. This is also a very good entry point into the larger discussion of the changing role of women in science.

DISCUSSION POINTS

▶ Students with eyeglasses can quickly find the focal point

of their lenses by holding them up to a wall opposite a window and moving the lens back and forth from the wall until an image of the window appears. Ask students with different eyesight to try this with their lenses to show differing focal lengths. ▶ Have students collect information about telescopes around the world or in space that have primary apertures larger than 3.5 meters but less than 8 m (similar to Table 6.1). Are these telescopes refractive or reflective? Where are they located? What kind of light do they observe? Discuss the trends that students find. ▶ Discuss the merits and drawbacks of putting telescopes in space versus on Earth when there is an option. Is there any reason to place a radio dish in space? Because there are windows of opacity in the IR, why would we put an IR telescope in space? ▶ Have students look at the information on the spectrographs available at the National Optical Astronomy Observatory (NOAO) website (http://www.noao.edu). What part of the electromagnetic spectrum is covered by each of those spectrographs? Discuss the connection among spectrographs, detectors, and wavelength coverage. Discuss the pros and cons of coupling a spectrograph to an adaptive optics system in an astronomical telescope. ▶ Have students collect information on telescopes that are in the process of being built (or have received funding to be built). What are the qualities of these telescopes? What can they do that the current telescopes cannot?

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▶ Have students think about the many devices that are

used in modern astronomy today. The CCD that many of them possess, either in a smart phone or a digital camera, is an integral tool. If they have a smart phone, they may also have an app dedicated to astronomy. Have students think about what it means to have such astronomical tools at their fingertips.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 6 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises. Geometric Optics and Lenses This interactive animation explores the concepts of focal point and length by showing how lenses are used to bend light and magnify images. In the interactive portion of the animation, students can adjust the distance of the object and the eye from the lens. The animation also explores the importance of telescopes as light-gathering “buckets” to see dim objects. Text reference: Section 6.1

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts presented in Chapter 6. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Snell’s Law Demonstrator This module demonstrates Snell’s Law, that is, how light is refracted when it moves between different media. The user controls the angle of incidence and the two indices of refraction. Text reference: Section 6.1 Telescope Simulator This simulation demonstrates the properties of a refracting telescope and how these vary with aperture and eyepiece selection. Text reference: Section 6.1

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Chapter 6

CCD Simulator This simulation uses rain falling into buckets as an analogy to a CCD grid of pixels. The module shows both collection and read-out. Text reference: Section 6.2 EM Spectrum Module This module surveys the electromagnetic spectrum, showing a typical astronomical image for different wavelengths of light and the kind of instrument that would take such an image. Text reference: Section 6.3

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. a, b, c. Without curved glass, a reflecting telescope cannot focus light which makes choice (d) wrong. 2. (b) In ground-based telescopes, the atmospheric blurring is almost always larger than the resolution limit of the telescope. 3. (c) CCD cameras turn photons into electrons, not protons. 4. (d) In general, a satellite costs $1 million per moving part. 5. (c) A spacecraft can measure many more things about a planet than just its images. For example, even during a flyby, a satellite can still measure gravity, magnetic fields, and charged particles near the planet. None of these can be done as effectively from Earth. 6. a, b, d. Generally speaking, pointing a telescope is not a computationally-intensive activity so (c) does not apply. Reading Astronomy News 1. Astronomers want to build at Cerro Armazones because this location is dry and cloudless almost year round. 2. Larger telescopes collect more light, meaning fainter objects can be observed. Also, larger telescopes provide higher resolution, meaning more detail can be observed. 3. Adaptive optics removes the blurring effects of the atmosphere. 4. Most other wavelengths (x-ray, ultraviolet, mid-to-far infrared) are not observable from Earth, so this telescope can only observe in the optical and near-infrared. 5. Answers will vary. At present, the telescope is fully funded and has a first-light date between 2024 and

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The Tools of the Astronomer ◆

2026. All of the ESO member states have approved this project, including Austria, Belgium, Brazil, the Czech Republic, Denmark, Finland, France, Germany, Italy, Netherlands, Poland, Portugal, Spain, Sweden, Switzerland, and the United Kingdom; Chile is also a partner. Although there has been considerable controversy over other mountaintop observatories, there does not seem to be significant opposition to the building of the E-ELT at the time of writing this document. Test Your Understanding 1. (c) The primary function of a telescope is lightgathering power so a larger diameter is better. However, please note that one must also consider focal length because this controls resolution. 2. a and e Our atmosphere blocks most UV and all gamma ray and x-ray light. 3. (a) 3, (b) 4, (c) 2, (d) 6, (e) 7, (f) 1, (g) 5 4. (b) The resolution of an interferometer is roughly given by the ratio of its observing wavelength and separation of the telescopes, l/D 5 2 3 10–6/85 5 2.4 3 10–8 radians or about 0.005 arcsec. 5. (d) Interferometry allows a much higher angular resolution than with a single telescope. 6. (c) Depending on the trajectory of light, it can bend as it moves from one medium to another, as a result of changing speeds among the media. 7. (b) The light-gathering power of a telescope depends on the area of the aperture, and because area scales as size2, two times bigger size is 22 5 4 times more light-gathering power. 8. (d) High resolution allows one to examine small scale features, which also allows one to distinguish very distant objects or separate many objects that are close together. 9. (a) Retina. See Figure 6.1. 10. (b) Turbulence in the atmosphere distorts light as it passes from space to the telescope; adaptive optics attempts to remove these effects. 11. (a) Interferometry is used to increase the resolution of a telescopic observation. 12. (d) The blurring effects of our atmosphere generally limit how well we can resolve objects in ground-based telescope images. 13. (d) Constructive and destructive interference from light as it diffractions through the slits of a grating yield spectra. 14. (b) Water vapor is very effective at blocking infrared light, so to observe in these wavelengths, one must find a very dry location, such as a high mountaintop. 15. (b) The “quantum efficiency” is how well we turn light (photons) into signal.

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46 ◆ Chapter 6 The Tools of the Astronomer Thinking about the Concepts 16. Simple lenses suffer from chromatic aberration, that is, they bend different colors by different amounts, so the resulting image is not nearly as crisp as in telescopes with multiple lenses. 17. Refracting telescopes require a straight line light path from lens to focus. Building refractors with long tubes to support increasing focal lengths becomes impractical. As the tubes grow longer, telescopes are increasingly front-loaded with weight and become unwieldy. Another problem that occurs with large refractors is aperture size. If you increase the aperture of a telescope, you increase its light-gathering power. Unfortunately, increasing the aperture size also increases the size and weight of the primary lens. Glass is heavy, and it flows like a liquid under its own weight within relatively short timescales. If astronomers go to the expense of building such massive telescopes, they want the instruments to last for decades or centuries without having to replace deformed optics every few years. 18. The telephoto lens will produce a larger image in the focal plane. The longer focal length leads to greater magnification of the subject. The trade-off for magnification is image brightness. The same amount of light is received from the subject, but the light is spread over a larger area in the focal plane, so the image appears fainter. This is a definite problem with many toy store refracting telescopes that promise 500 times power! 19. The visible portion of the electromagnetic spectrum is very small compared with the spectrum as a whole. Viewing in only the visible portion of the spectrum is rarely representative of the entire spectrum. In fact, many astronomical sources that look normal at visible wavelengths may show amazing brightness or complexity in other regions of the spectrum. Examples of astronomical phenomena that are better observed at nonvisible wavelengths include the galactic center, quasars, starburst galaxies, and protoplanetary disks. 20. The sketch will look similar to Figure 6.7. 21. This question can be answered a few different ways. First, a star is a point source, so it cannot be made to look larger or smaller; thus, one answer is that a star will appear to be the same size through all telescopes. Second, another way to look at this is to argue that brighter stars appear to be bigger in images and that the larger diameter telescope will collect more light and make the star appear larger. A third argument is that magnification depends on the ratio of the focal lengths of the objective and eyepiece, so if both telescopes have the same focal lengths, then they will have the same magnifications, making both stars appear the same size.

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22. Answers will vary. The availability of photosensitive film made it possible to expose for longer than the human eye, showing fainter features as one exposed for longer. However, film has fundamental limits in sensitivity, which the development and improvement of the CCD was able to overcome, although it took many decades to build the very large and very sensitive CCDs in use today. Technology probably drove science, because science required the larger and more sensitive CCDs, but science also drove technology because the need for these improvements has always been apparent. By the way, cell phones today use a similar type of detector (usually a CMOS). 23. An adaptive optic system will observe either a real or fake (laser generated) star and use sophisticated computer programs to control a deformable mirror in an attempt to keep the star’s image steady and as small as possible. Because all the other stars or objects one is looking at have had their light pass through the same air as our reference star, our deformable mirror corrects for much of the atmospheric seeing and provides us with higher resolution images. 24. Our eye takes an “image” about 14 times per second. Thus, we are limited by how much light we can collect— in other words, how faint of an object we can see. A camera, whether using film or a CCD chip, can collect light for much longer—the amount of time it collects light is called the integration time—and taking a much longer exposure collects much more light, which allows us to distinguish faint objects from nearby bright ones or the background. Note that the definition of Quantum Efficiency is given in the glossary. 25. The nearest star is about 4 light years away, whereas low Earth orbit is about 400 kilometers above the ground. Putting a telescope in orbit gets the telescope closer to that closest star by such an infinitesimally small amount that there is no suitable analogy that immediately comes to mind. The real reasons are to avoid weather, humidity, and other atmospheric effects. 26. Flyby spacecraft can view a large number of objects during their mission because they travel over a large region of the solar system. However, there is a limit to how much they can learn about any single object because they spend very little time at one place and generally cannot fly too close to anything. An orbiter can do a few flybys on the way to its target and can then spend its entire mission studying one object at very high resolution, but it is limited in how much it can teach us about the actual internal properties of the object. For this, we need a lander, which can yield an enormous amount of information but only about one very small region of the planet or Moon.

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Chapter 6

27. The meteorites give us a very limited view of the state of Mars. We only have a few rocks, we do not know where they came from on Mars, and they have undergone change and contamination as a result of being ejected from Mars and landing on Earth. To really study the geology of Mars (or any planet), we need to do controlled experiments, such as collecting mineral samples from known and diverse locations. 28. The Moon is tidally locked to Earth, keeping the same face toward Earth at all times. It is not possible for an Earthbound observer to see the far side of the Moon directly (save for the glimpses over the limb provided by libration). Our first look at the far side did not come until 1959, when the Soviets launched a spacecraft that looped around the Moon. 29. Detectors are located deep inside mines, for example, inside mountains, so that the rocks protect the detectors from other kinds of particles that come from space. Neutrinos are neutral and interact very weakly with matter, and the only way to detect one is to watch the result of it impacting another atom. Because this is very rare, it makes detection quite tricky. 30. The leftover radiation from the Big Bang is in the microwave region of the spectrum; there are numerous sources of microwave signal on earth, from the atmosphere, cell phone use, etc. Thus, it is better to go into space, above and out of the way of all of these sources of noise. Applying the Concepts 31. Setup: The light-gathering power of a telescope depends on the area of the aperture, and recall that area A 5 pr2, where r is the radius of the telescope. However, we do not need to use the radius, as we are comparing the sizes of both telescopes (the ratio of the radii is the same as the ratio of the diameters). Note also that we must use consistent units of size. In the example, 30 meters equals 30,000 mm, and 10 meters equals 10,000 mm. Solve: Because light gathering depends on the square of the telescope’s aperture, the astronomical telescope has a collecting power. The Thirty Meter Telescope, (3 3 104 )2 5 1. 4 3 107 times more 82 light than the human eye, whereas the 10-m Keck (104 )2 telescope gathers 5 1. 6 3 106 more light. 2 8 Review: The telescopes gather 2 to 14 million times more light than our eye! No wonder big telescopes take such amazing images!

therefore, gathers

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The Tools of the Astronomer ◆

32. Setup: There is a vertical line present at the entrance and exit of the laser. Measure the angle with respect to this. Solve: As the light enters, I measure an angle of 37 degrees. As it exits, I measure 36 degrees. These are essentially the same, as is expected because light should be bent (or refracted) by the same amount on entrance and exit, given that the block is a rectangle and, thus, its sides are parallel. Review: Note that if the sides were not parallel, the light beams would come out at different angles, which explains how prisms work. 33. Setup: The light-gathering power of a telescope depends on the area of the aperture, and recall that area A 5 pr2 where r is the radius of the telescope. Solve: Because light gathering depends on the square of the telescope’s aperture, a 16-inch telescope has a size that is 4 times larger than the 4-inch one; thus, the larger telescope collects 42 5 16 times more light. Review: Note that we did not need to use the radius, because we compared the sizes of both telescopes (the ratio of the radii is the same as the ratio of the diameters). Because light-gathering scales as the square of size, we can see that even a modest increase in size yields a large gain in improvement. 34. Setup: The diffraction limit of an optical system is given by arcseconds. Be sure to use angular measurements and lengths in the same units. Solve: Solve for D 5 (2. 06 3 105 arcsec) 5 (2 . 06 3 105 arcsec)

l  21 cm arcmin 1 . 5 arcmin 60 arcsec

5 4 . 8 3 104 cm or 480 meters. Assuming that we could detect radio waves with our eyes, we see that our pupil’s tiny size would mean we would have no angular resolution at all, so we would not be able to detect any shapes or structure, even on the largest scales. Review: Note the huge size of the radio dish that would be needed just to resolve a few arcminutes. This is why we have built huge interferometers and the dish at Arecibo. 35. Setup: Not including atmospheric seeing or manufacturing defects in the optics, the resolution of a telescope l is given by the diffraction limit  5 2 . 06 3 105 D arcsec. Be sure to use wavelength and diameter in the same units.

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48 ◆ Chapter 6 The Tools of the Astronomer

Solve: In the infrared,  5 2. 06 3 105

1,000 nm 3 1m

1m 5 0. 21 arcsec. In the optical, this is 109 nm 400 nm 1m 3 9 5 0 . 08 arcsec. 10 nm 1m Obviously, the optical observations will be higher resolution. Review: Our formula tells us that the diffraction limit scales directly with wavelength, so we can confirm that infrared (longer wavelength) will have a larger diffraction limit, thereby reducing our maximum resolution. Note that from the ground, seeing generally limits us to 0.2 to 0.5 arcsecond anyway, so both of these wavelengths could be seen without a problem. 36. Setup: The diffraction limit of an optical system is l arcsec. Be sure to use given by  5 2 . 06 3 105 D wavelength and diameter in the same units. Solve: (a) For the human eye,

 5 2. 06 3 105

5.5 3 10–4 mm  5 2. 06 3 10 5 14 arcsec. 8 mm 5

(b) This limit is 5 to 10 times smaller than the stated limit of 60 to 120 arcseconds. (c) The diffraction limit of an optical system just reports the theoretic limit but does not include real effects like seeing, imperfections in optics, and the resolution limit of the detector. In the case of our eye, seeing is not much of an issue, but our lens is not perfect, and our retina does not have rods and cones placed to provide a resolution any higher than 1 to 2 arcminutes. Review: If we found a diffraction limit in (a) that was larger than 1 to 2 arcminutes, we would know we made a mistake because the resolution limit of an optical system cannot be better than the diffraction limit. This is a good check for a numeric calculation. 37. Setup: Image size is directly proportional to the focal length of the telescope. We can write this as a simple proportion with the subscript “big” representing the big telescope and the subscript “amateur” referring to the amateur telescope. Let x represent image size, and x big f big . let f represent focal length, then 5 x amateur famateur Solve: Solving for the image size of the Moon in the big telescope, we find  250 m   f big  x big 5  x amateur 5   (13 . 8 mm)   famateur   1.5 m  5 2,300 mm or 2.3 meters.

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Review: We can directly see why it is advantageous to have very large telescopes, not only for light collection but for resolution as well. 38. Setup: The number of bits produced by a detector is given by the product of the number of pixels and the levels of brightness per pixel. Solve: The old detector produced a total of 160,000 3 256 5 4.1 3 107 bits of data per image. Today’s detector produces 109 3 32,768 5 3.3 3 1013 bits, or almost a million (800,000) times more data per image. Review: Does this make sense? Think about the first digital cameras, which could use flashcards with only 16 megabytes of memory, whereas today’s cameras require cards with many gigabytes of memory. It is the same effect. 39. Setup: The higher the efficiency, the shorter the exposure time needed to make the same observation, so the ratio of efficiency will equal the inverse ratio of exposure times. Solve: The formula for the setup statement could read Q 1 t2 5 . The CCD camera has an 80-fold improveQ 2 t1 ment in efficiency, so we will need an exposure time 60 min 5 80 times shorter than the photograph, or 80 0.75 min or 45 seconds. So we save 59 minutes, 15 seconds for each image. Review: One hour versus 45 seconds. No wonder we use CCD cameras now! With some telescopes, we can now take an image of almost the entire sky every night, whereas with photographs this was a highly ambitious feat that would take many years. 40. Setup: The diffraction limit of an optical system is l arcseconds. Be sure to given by  5 2 . 06 3 105 D use angular measurements and lengths in the same units! 1.35 cm km 3 5 5 Solve: (a)  5 2. 06 3 105 8 , 000 km 10 cm 3.5 3 10–4 arcseconds or, 0.35 milliarcseconds. (b)  5 2 . 06 3 105

550 nm m 3 9 5 1.1 3 10–3 10 nm 100 m

or about 1 milliarcsecond. Although these are fairly close, we see that if we really want ultrahigh resolution, we will use VLBA imaging. Review: Note that because of atmospheric seeing, achieving 1 milliarcsecond resolution in the optical is far harder than the submilliarcsecond resolution that is regularly achieved in the radio using the VLBA.

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Chapter 6

41. Setup: The diffraction limit of an optical system is l given by  5 2 . 06 3 105 arcseconds. Be sure to use D angular measurements and lengths in the same units! 17 mm km Solve:  5 2. 06 3 105 5 3 6 5 3 . 5 3 10–5 10 km 10 mm arcseconds or 35 microarcseconds. Review: The VLBA currently gives us a resolution of about 0.35 milliarcseconds, so this would be an improvement of a factor of 10. Is it worth it? Interstellar hydrogen emits at 21 centimeters and the universe is 90 percent hydrogen, so to be able to see this material at such a high resolution is unprecedented. 42. Setup: The small angle formula tells us that the physid . cal size s 5 2. 06 3 105 arcsec Solve: The size of an object with an angular size of 280 km 3 0 . 2 arcsec 0.2 arcsec will be s 5 5 2.7 3 2. 06 3 105 arcsec 10–4 km, or 27cm. Now, in practice, we need two to three resolution elements to really detect an object, so the orbiter can detect objects about two to three times 27 cm, or roughly the size of 1 meter. Review: For a seat-of-the-pants feel to this value, the orbiter would be able to see where everyone on a football field was standing when a certain play was being made, and it would probably know where the ball was if in midair. 43. Setup: To convert AU to km, remember there are 1.5 3 108 km in one AU. To find travel times of light signals, use d 5 vt, where here, v is the speed of light. Solve: Voyager 1 is 125 AU 5 (125 AU)(1.50 3 108 km/ AU) 5 1.88 3 1010 km from the Sun. (b) At the speed of light, the one way time for a radio signal to travel be10 d 1. 88 3 10 km tween Earth and Voyager 1 is t 5 5 v 3 3 105 km/s 5 62,500 sec or about 17.4 hr. (c) Assume that Voyager 1 was on a direct line to Alpha Centauri. Alpha Centauri is currently 4.4 ly  280,000 AU from the Sun. That implies that Voyager 1 has completed 125 3 100% 5 0. 045% of its journey to the 280 , 000 nearest star. Review: This helps us to see how far away even “nearby” objects are in space. 44. Setup: All waves follow c 5 f, so solve here for wavelength  5 c/f. 3 3 108 m/s 5 3 3 107 m or 30,000 km. Solve: l 5 10 Hz

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Review: Just to make this more readily understandable, this is 18,600 miles or more than halfway (75%) around the Earth. 45. Setup: Wien’s law gives the peak of a blackbody 2. 9 3 106 nm · K lmax 5 . T Solve: For a blackbody at 2.73 K, the peak wavelength is 2. 9 3 106 nm · K lmax 5 5 1. 6 3 106 nm 5 1.06 mm. 2.73 This light is in the millimeter or microwave regime. Review: The cosmic microwave background is at 2.73 K and is so named because the spectrum peaks in the microwave region of the spectrum, as we found. Using the Web 46. Answers will vary. The answer will discuss whether two telescopes viewed are effective for public outreach (effective for the observatory and for astronomy in general). For each telescope, the answer will include the hemisphere, what wavelengths the telescope uses, and some of its key science projects. 47. Answers will vary. After choosing a telescope, provide the telescope’s “claim to fame,” and if the website has news releases, provide a recent discovery from that observatory. 48. Answers will vary. Anyone living in a large city will suffer a good deal of light pollution; however, one can generally travel within an hour to find dark skies. Dark nighttime skies are important for most forms of night life. Bright lights at night can confuse migrating animals, stop bioluminescent animals from communicating, confuse baby turtles that navigate by the moon, influence the mating habits of some frogs, and cause overfishing. 49. At present, JWST is set to launch in 2018. This telescope will have a 6.5 meter mirror (as opposed to the 2.4 meter mirror in Hubble) and is going to observe primarily in the infrared. Instruments include the Near InfraRed Camera (NIRCam), the Near InfraRed Spectrograph (NIRSpec), the Mid-InfraRed Instrument (MIRI), and the Fine Guidance Sensor (FGS). According to NASA, JWST has four main science themes: the End of the Dark Ages—First Light and Reionization, the Assembly of Galaxies, the Birth of Stars and Protoplanetary Systems, and Planetary Systems and the Origins of Life. The total cost for JWST will be around $9 billion, give or take. 50. Answers will vary. Select a mission, and report whether the craft is still active, whether it is sending images, and what new science is coming from it.

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50 ◆ Chapter 6 The Tools of the Astronomer Exploration 1. The eraser is at the bottom of the actual pencil. 2. The eye sees the pencil upside down. 3. If I need an object to move from top to center of my field of view, I need to move the telescope down. 4. I first see a “crisp” image at 3.2. 5. The pencil is upside down. 6. The eyepiece of a telescope is very small and makes a very small image. If we move away from it, seeing the object in the telescope becomes very challenging because the eye must be exactly aligned. 7. If I rotate one way and the image is blurrier, I must rotate the knob the other way. 8. Each person has his or her own particular focal length for his or her eyes to see correctly. The focus of the

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telescope made it so my eye could see clearly but if someone has a different focal length, he or she will need to adjust the telescope. 9. The bottom telescope has the longer focal length. 10. The longer focal length has larger separation of the red and blue stars. 11. A long focal length is a much larger telescope, which means that it will be heavier and take up more physical space; this can be a problem with its own structure or fitting it into the space available. Also, long focal lengths do not view as much of the sky so a small focal length is needed if one wants a very wide view. Images will be dimmer also.

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CHAPTER 7

The Birth and Evolution of Planetary Systems INSTRUCTOR’S NOTES Chapter 7 covers the formation of stars and planets. Major topics include

▶ historical development of the nebular hypothesis for solar system formation

▶ collapse and conservation of angular momentum in a nebula

▶ formation and characteristics of a protostellar accretion disk

▶ hierarchical formation of planetesimals ▶ formation of terrestrial and giant planets ▶ why terrestrial and giant planets segregate into different regions

▶ a brief history and catalog of our Solar System ▶ the detection of extrasolar planets ▶ planetary migration I have found that one of the most challenging aspects of this chapter for nonscience students is the cause for the segregation of planets into terrestrial and giant bodies. As students work to incorporate the physics they learned in the first chapters into this discussion, many of them mistakenly conclude that the accretion disk spins most of the gas to its outer region, so there is very little material to form terrestrials; that heavy elements are attracted to the center of the accretion disk because they are more massive and feel more gravity, which is why terrestrial planets are rocky; or that giant planets form purely out of gas and never have any rocky planetesimal at their core. In addition to the discussion in this text that cooler temperatures lead to elements in solid phase, I have found it useful to frame the problem in terms of escape speeds of hot and cold atoms. In the inner regions of the accretion disk, atoms had very high temperatures and were moving faster than the escape speed of the forming planetesimals, whereas in the outer accretion disk, these gases were cold and moving slowly, such that those planetesimals could gravitationally capture abundant gas and grow to very large sizes. At present, the overwhelming majority of exoplanets are giant in mass but located at terrestrial distances. Because

the preceding discussion of planet formation explicitly prohibits this, I have found this to be a very useful topic on which to discuss the scientific process. First, the fact that we have found so many giant-type planets is a perfect example of selection effects, and I like to stress that our scientific understanding is limited by our ability to observe a complete sample. Second, I like this topic because it shows the selfcorrecting nature of science. We had a model, and new observations showed it was incomplete, so we performed further investigation and realized that we neglected some important physics (i.e., planet migration). Some students have expressed to me the belief that scientific understanding changes over time primarily because we make many mistakes, so this topic very clearly shows we make the best interpretations we can with the information we have at the time. For those interested in a more in-depth discussion of planet migration, there are many good videos of the causes available at http://jila.colorado.edu/~pja/planet_migration. html. Also, this chapter can be a good teaser for galaxy formation because the processes are quite similar.

DISCUSSION POINTS

▶ Based on the amount of material near the Sun, how likely

is it that the Sun has siblings that formed from the same parental molecular cloud? Discuss with students the possibility of stellar siblings of any kind. ▶ Have students use the equation for spin angular momentum to calculate the values for the Sun, Jupiter, Saturn, Neptune, Earth, Pluto, and a typical molecular cloud. Discuss these in the context of conservation of angular momentum. ▶ Ask students, if they had to choose just one technique among the leading choices for finding extrasolar planets that might be as habitable as Earth, which one they would pick? What makes that method advantageous over the others? ▶ Ask students to search the web to find out how many Earth-like extrasolar planets have been confirmed. Discuss the significance of that number. 51

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52 ◆ Chapter 7 The Birth and Evolution of Planetary Systems

▶ Ask students to search the web to find out how many

extrasolar planets have been found in habitable zones. What cultural impact does that have? What questions does it provoke? If there were life on those planets, would we be able to communicate with it?

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 7 and are available from the free Student Site (digital. wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises. Solar System Formation This animation begins with a rotating disk of gas that evolves to form a protostar and protoplanetary disk, and then finally forms the other objects in the Solar System. The animation emphasizes the timescales involved, significant physical features that define boundaries between different eras, and the composition at different locations that will give rise to the objects in the Solar System. Traffic Circle This animation shows an increasing number of cars entering a traffic circle. Although they come in from different directions and at different entry angles, all the cars end up spinning in a uniform disk, analogous to how material in a collapsing cloud forms an accretion disk.

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding the physical concepts presented in Chapter 7. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Influence on Planets of the Sun This simulation shows the movement of the sun due to the gravitational pull of the planets. The contribution from each planet can be isolated by toggling checkboxes. This provides a hands-on understanding of how planetary mass and distance change the planet’s gravitational effect on its parent star and, in turn, how easily the planet will be detected by radial velocity measurements. Text reference: Section 7.5

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RADIAL VELOCITY GRAPH This simulation shows a star and planet in orbit around each other while tracing out the star’s radial velocity curve. This is useful for demonstrating the physical situation that makes radial velocity measurements possible. Text reference: Section 7.5 Exoplanet Transit Simulator This simulation demonstrates how a planet passing in front of its parent star can cause dips in the star’s light curve, potentially leading to the planet’s detection. This is useful for demonstrating the physical situation that makes exoplanets detectable via stellar transits. Text reference: Section 7.5 Exoplanet Radial Velocity Simulator This simulation demonstrates how the spectrum of a star is shifted as the star and its planet orbit their common center of mass. This is useful for demonstrating the physical situation that makes exoplanets detectable via their rhythmic gravitational tugs on their parent stars. Text reference: Section 7.5

ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, which enable students to prepare for class or to review what they’ve learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack. Angular Momentum A kinesthetic demonstration of how angular momentum depends on mass, distance, and rate of spin, where Dr. Stacy Palen stands on a rotating platform and moves her arms in and out.

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (c) The number or size of planetary moons is not part of the nebular hypothesis. 2. (a) Orbital angular momentum of planets is much higher than the spin angular momentum of the central

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Chapter 7

stars, especially for planets that are very massive and far from the star. 3. (b) Hydrogen and helium are the primary gases of a stellar nebula, so a planet that has mostly these gases should have formed far from its central star, where temperatures were cool enough to allow the planetesimal to gravitationally capture this material. 4. (a) A planet (meaning large enough to clear out its orbit) is unlikely to have formed far out beyond Neptune because material moved too slowly in the formation period of our Solar System to accrete into a planet. Rather, the planet probably migrated out. 5. (c) Today, transit methods generally are the most sensitive to Earth-like planets. Reading Astronomy News 1. Answers will vary. The search for habitable planets and signs of extraterrestrial life not only is one of NASA’s current missions but also is of enormous interest to just about everyone. 2. This star must be fairly cool to provide less energy input to the planet even though the planet is fairly close. 3. Kepler’s third law gives only the sum of the masses of the planet and star, and because stars are much more massive than planets, much more orbital information is needed to extract the mass of the planet from its orbital details. This could include the orbital velocities of the star and planet, and the shape of the star’s orbit. 4. The only way to measure the composition of the planet will be through spectroscopy of its atmosphere. 5. Answers will vary. “Cousin” isn’t a scientific term in this context but one used to evoke an emotional connection to the planet and the story. Test Your Understanding 1. a, b, h, c, e, d, f, g 2. (c) Because angular momentum L = mvr, we see that if the distance r is halved, the speed v must be doubled. 3. (b) Because the inner material was hotter than the outer Solar System, the inner planets were not able to capture the large amounts of gas that are present in the Jovians. 4. (b) In the inner Solar System, the temperatures were too hot for the larger planetesimals to attract a significant amount of the gas that was present in the solar nebula.

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The Birth and Evolution of Planetary Systems ◆ 53

5. (a) Large planets close to the star tug the stars by a large amount with short periods, making them the easiest to detect. 6. (c) Current theories of protoplanetary disks suggest that they dissipate in about half the time it would have taken Jupiter to form by accretion alone. 7. (b) The distance at which planets form from their central star controls their composition. 8. (b) In the early Solar System, collisions between particles and clumps cause them to stick together and grow in size. 9. (a) Observing a transit means detecting a drop in the star’s light, which favors a large planet that is close to the star, thereby blocking a significant amount of the star’s light. 10. (a) As shown in Math Tools 8.1, spin angular momentum scales as L  R2 / P, so if the radius is cut in half, the period must drop by a factor of 22 5 4. 11. (d) Angular momentum, whether spin or orbital, depends on all three factors listed. 12. (b) Volatiles are easily evaporated or broken up with heat, and it was too hot in the inner Solar System for these compounds to be present. 13. (d) Asteroids and comets are made almost exclusively of the pristine material out of which the Solar System formed. 14. (c) Conservation of angular momentum means the direction of initial rotation dictates the direction of final rotation. 15. (d) The habitable zone changes depending on the star’s type but is considered to be the region in which liquid water could exist on a planet’s surface if it were located within that zone. Thinking about the Concepts 16. The hydrogen and most helium atoms that make up our Sun and Solar System all came from the Big Bang. Everything else was formed in previous generations of stars. 17. Stellar astronomers looking at young stellar objects have noticed that many of them are located within dark, dusty disks, as shown in Figure 7.2. Planetary scientists looking at our own Solar System noticed that the planets all lie in a disk orbiting in the same direction, and by studying meteorites they found evidence of larger objects being built up from smaller ones. This suggests that our Solar System formed from a disk of gas and dust. The two findings suggest a common origin for solar systems.

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54 ◆ Chapter 7 The Birth and Evolution of Planetary Systems 18. A protoplanetary disk is the spinning disk of gas and dust out of which planets form around the central star. The inner part will be hotter because it is closer to the central star and because the inner material gained energy when it moved in from the outer regions. 19. “Conservation” of a quantity means that the sum total of that quantity within a system must keep a fixed level. Another way to think about this is that conserved quantities are like transactions at a bank: if you borrow from the bank, you have to pay it back. Consider ice-skaters, who have a fixed amount of energy in their bodies. If they skate very quickly (i.e., a lot of energy of motion), they must get that energy from somewhere; in this case, they convert some of their own biochemical energy. An object can always gain or lose mass, energy, momentum, or angular momentum, but it has to come from somewhere. 20. If scientists find planets with highly inclined orbits, they will have to propose a formation mechanism that is consistent with all other evidence for planet formation. Perhaps these planets were ejected from one solar system and were captured by this star, or underwent numerous close gravitational encounters, for example. Scientists will consider all the relevant physics and propose the most likely scenarios, then propose tests of these scenarios and search to validate or invalidate them. 21. Spin is a conserved quantity, but it is measured by the product of how fast you rotate and how far away you are from your axis of rotation. Thus, when an ice-skater spins slowly with her arms extended, she has a fixed amount of spin. As she brings her arms in toward her body, that distance gets smaller, so the rate of rotation has to increase to keep the product of the two constant. Thus, the skater spins much faster. 22. An accretion disk is the thin, rotating disk that forms as a gas cloud collapses on itself. It is out of this disk that the central star and planets form. These disks are also found in a wide variety of astrophysical environments, as we shall later learn. Broadly speaking, this disk allows material to spiral into the star by converting its orbital energy into heat. When material spins in a disk around a star, friction between the different particles heats up that material and allows it to radiate energy. As particles give off that energy, they spiral deeper down the disk toward the star, eventually landing on the star. 23. Small grains of dust moving about the Sun in similar orbits randomly collide with each other at gentle speeds. These small grains stick to each other electrostatically to form increasingly larger seeds. As the seeds grow, gravity becomes more important. Rather

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than awaiting random collisions, the largest seeds begin to gravitationally attract nearby particles, and their growth rate increases exponentially. Eventually, the largest seeds reach planetesimal sizes. When planetesimals collide, the resulting growth leaves planet-sized objects. 24. As I blow “dust bunnies” toward each other (provided they don’t fly apart), they tend to become tangled with each other and grow in size. Although on Earth they will never grow enough in size to pull on each other in a meaningful way, this could certainly happen in outer space. Earth’s gravity and friction with the ground keep the bunnies from forming self-gravitating collections. 25. Rocky materials are refractory and condense at very high temperatures. As such, we expect that rocky materials will solidify in all regions of the nebula, including near the proto-Sun. Volatiles condense only at very low temperatures. Only the regions of the solar nebula far from the proto-Sun are cool enough for these materials to solidify. This picture is consistent with modern observations of our Solar System. Furthermore, barring other issues (e.g., migration of giant planets inward due to complex gravitational interactions), we expect the same to be true of other planetary systems. 26. The giant planets had several advantages over the terrestrial planets to enhance the growth of their atmospheres. (1) Giant planet seeds formed from more abundant volatile materials, whereas terrestrial planet seeds formed from the much less abundant refractory (solid) materials. (2) Giant planets had a much larger area from which to accumulate raw materials, whereas terrestrial planets could draw material only from a narrow zone in the immediate vicinity of the forming proto-Sun. (3) Accreting in the outer nebula left giant planets less vulnerable to the powerful solar winds emanating from the proto-Sun; the terrestrial planets were scoured clean because of their proximity to the proto-Sun. (4) Because of their ability to quickly grow very large, the Jovian planets captured and retained significant amounts of the primordial hydrogen and helium from the solar nebula. Meanwhile, the terrestrial planets were too small and too exposed to the young Sun to capture or retain significant atmospheres from the solar nebula. 27. The four methods that astronomers currently use to search for extrasolar planets are (1) spectroscopic radial velocity method: we watch the periodic Doppler shifts of emission lines from the star due to an orbiting planet dragging the star around in space; (2) transit method: we watch a star regularly change brightness due to a planet passing in front of the star

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Chapter 7

and eclipsing it; (3) microlensing method: a planet passes in front of a distant star and its gravity bends some of the star light toward Earth, making the star brighten for a short period of time; and (4) direct imaging: we take an image that resolves the planet from its central star. 28. Stars are very bright and far away, but planets appear very close to their stars. Thus, it is difficult to mask out the light of the star and still see close enough to it to see the reflected light of a planet. 29. We have only detected planets around a few hundred stars, which is a very small sample of the stars in the Milky Way, and not all of them look like our Solar System. In particular, our technology for detecting extrasolar planets is biased toward detecting giant planets in close orbits about their parent stars. Only now are we finding Earth-like planets in Earthlike orbits. Based on this sample and our current technological limitations, it is too early to make any general conclusions about whether our Solar System is unusual. 30. The Kepler satellite uses planetary transits to search for exoplanets. By “Earth-like” we first mean planets that are terrestrial and have roughly the same mass. A refinement would be planets that also have thick protective atmospheres and that orbit in the habitable zones of their stars. Applying the Concepts 31. Setup: The conversion from m/s to mph is m km mi 3600s 1 5 5 2.3mph . s 1000m 1.6km hr Solve: In Figure 7.17, the maximum radial velocity is about 35 m/s, or 35 3 2.3 5 80.5 mph, which is a 67 , 000 factor of 5 832 smaller than Earth’s orbit 80. 5 around the Sun. Review: As expected, a star moves much less than the planet orbiting it. We see here that astronomers are indeed detecting very small radial motions in a star to detect the planet orbiting it. 32. Setup: The table gives us values of planetary masses in terms of Earth’s mass, so for this question we must sum up the planet masses and then compare them to Jupiter and Earth. Solve: (a)The sum of the eight planets in our Solar System is Mp 5 (0.055 1 0.815 1 1.00 1 0.107 1 317.83 1 95.16 1 14.54 1 17.15) MEarth 5 446.66 MEarth. (b) To find Jupiter’s

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The Birth and Evolution of Planetary Systems ◆ 55

percentage of that mass, divide its mass by the total: M J 317. 83Earth 5 5 0 . 7116 ⇒ 71 . 16 %. (c) The same M P 446. 66 Earth for Earth yields M Earth 1. 00 M Earth 5 5 0. 0022 ⇒ 0. 22 % . MP 446. 66 M Earth Review: It is no surprise that the Solar System’s planetary mass is dominated by Jupiter, given how large and massive it is compared to everything else. We can simply see from inspection that almost all the mass is in Jupiter and Saturn. 33. Setup: For this problem, we will use the equation of 4 mR 2 and orbital spin angular momentum Lspin 5 5P angular momentum Lorb 5 mvr. Remember to use consistent units (i.e., time in seconds, distance in meters). Solve: For Earth’s spin angular momentum, P 5 24 h 5 86,4000, so Lspin 5

4  5.97 3 1024 kg  (6 . 378 3 106 m)2 5  86 ,4400 s

5 7. 0 3 10 33 kg m 2 / s. By comparison, using 1 AU 5 1.5 3 1011 m, Lorb 5 mvr 5 5. 97 3 1024 kg  29. 8 3 103 m/s ⋅ 1. 5 3 1011 5 2. 7 3 1040 kg m 2 /s , which is about 3.8 million times more than the spin angular momentum. Review: Recall that angular momentum depends on how fast an object is spinning and how far that object is away from its spin axis. Thus, it stands to reason that all the angular momentum of a planet is in its orbit because the planet’s distance from the Sun is so much greater than its size. 34. Setup: For this problem, we will use the equation of 4 mR 2 . However, spin angular momentum Lspin 5 5P instead of computing the spin for each planet, let’s directly compare the spins of both by constructing a ratio. This has two added advantages: (1) it gets rid of the constants in our formula, and (2) we can use whatever units we want as long as we are consistent, that is, masses in terms of Earth mass and periods in days. 2

  PEarth    Solve: LVenus 5 m Venus R Venus    LEarth  mEarth   R Earth   PVenus  1 5 0 . 003 243 Review: Although Venus is practically Earth’s twin in terms of mass and size, Venus rotates very slowly 5 0 . 815 3 0 . 9492 3

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56 ◆ Chapter 7 The Birth and Evolution of Planetary Systems about its axis, which is why it has so little spin angular momentum. 35. Setup: For this problem, we will use the equation of orbital angular momentum Lorb 5 mvr. However, instead of computing L for each planet, let’s directly compare the values of both by constructing a ratio. This has two added advantages: (1) it gets rid of the constants in our formula, and (2) we can use whatever units we want as long as we are consistent, that is, masses in terms of Earth mass, speeds in km/s, and distance in astronomical units (AU). 2

 m J   v J   rJ  Solve: 5 LEarth  mEarth   v Earth   rEarth  13. 1 km/s 5 318 3 3 5. 2 5 727 29. 8 km/s Review: Given Jupiter’s huge mass and greater distance from the Sun, it is no wonder that it has so much more angular momentum around the Sun than Earth. 36. Setup: Conservation of angular momentum states that the cloud’s angular momentum just before it collapses must equal its angular momentum after it has become the Sun. Expressing this as an equation, we have LJ

2 2 4 mcloud R Sun 4 mSun R Sun 5 . 5 Pcloud 5 PSun Solve: Because a large percentage of the cloud becomes the Sun, we will assume the mass does not change. Solving above we have

Lcloud 5 LSun or

 R   1 . 4 3 109 m  6 PSun 5  Sun  Pcloud 5  10 yr  1016 m   R cloud

51. 96 3 10−8 yr 5 0. 6 s. Review: Recall that angular momentum depends on how fast an object is spinning and how far that object is away from its spin axis. Thus, it stands to reason that a huge cloud that collapses into a star would spin at an almost mind-numbing rate if there were no means by which that cloud could transfer its angular momentum away from the central object. Thank goodness for accretion disks. 37. Setup: To compute the density of Vesta, we will assume that Vesta is a sphere. The volume of a sphere is  3 Mass d . Density is  5 . Volume 6 Solve: (a) The density of Vesta is 5

2.7 3 10 20 kg 5 3, 464 kg/m 3 . (  /6)(5.3 3 10 5 m)3

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(b) Vesta’s density is significantly higher than that of water or rock. This implies that Vesta’s composition must include a significant component of metal in addition to rock. Review: This confirms what we have found in meteors that have landed on Earth, that is, the meteors are mostly composed of metals. 38. Setup: Kepler’s Laws tell us that closer planets orbit faster. Solve: (a) The red planet is the closest so it is the fastest, followed by the blue planet and then the purple one. (b) The purple planet has the longest durations because it moves the slowest, thus it takes the longest amount of time to cross the disk of its parent star. Review: On the figure, orbital speed is represented by how often the transit occurs. The red planet has the most transits so it travels the fastest, followed by blue and then purple. 39. Setup: Doppler shifting tells us that the relative shift of a line away from its center equals the relative speed of the moving object compared to the speed of the wave. In this case, the wave is light, so our formula can be D v written as 5 . 0 c Solve: Solve our Doppler shift equation for v 1 m/s D  5 0 5 575 nm 51. 92 3 106 nm. 8 c 3 310 m/s Review: This shows us that to observe Doppler shifts in visible light, large velocities are required. 40. Setup: Doppler shifting tells us that the relative shift of a line away from its center equals the relative speed of the moving object compared to the speed of the wave. In this case, the wave is light, so our formula D v can be written as 5 , and in this problem we are 0 c looking for the shift D away from a central wavelength 0 5 575 nm. Solve: Solve our Doppler shift equation for v 0. 09 m/s D  5 0 5 575 nm 5 1. 73 3 107 nm . c 3 3 108 m/s Review: This is much smaller than the size of subatomic particles (10-15 m) and thus shows how nearly hopeless it would be to ever detect such a gravitational signal with radial velocities. 41. Setup: The Sun’s brightness will drop in proportion to how much area of the Sun is blocked. Solve: The Sun’s radius is 7 3 108 m, and Jupiter’s radius is 7.2 3 107 m. Because area is proportional to radius squared, Jupiter’s area is

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Chapter 7 2

 7. 2 3 107   7 3 108  0. 01 as much as the Sun. So the Sun’s light would drop by this amount, or about 1 percent. Review: Aliens should easily be able to detect this drop with even crude instruments. However, this drop only happens once every 11.8 years, so they will have to watch the Sun very carefully and for a long time to detect it. 42. Setup: To determine the orbital radius of a body, we use Kepler’s third law. Here, the star has 1 solar mass, so we can use P2 5 a3. Mercury orbits at an average distance of 0.39 AU. Solve: (a) We require the period in years, so convert1 year 5 0. 55 year . The average ing 200 days 3 365 days orbital distance is therefore a 5 P2/3 5 0.552/3 5 0.70 AU. (b) This planet orbits further from its parent star than Mercury orbits our Sun. In fact, it is close to the orbital distance of Venus. The planet can be expected to be warmer than Earth because it is 30 percent closer to its parent star. Review: Venus has an orbital period of 225 days and a distance of 0.72 AU. Because this hypothesized planet has a period of 200 days, we know that its distance must be just under that of Venus, as found. 43. Setup: Following Math Tools 7.3, we can estimate the R2 . percentage reduction of light during a transit as planet 2 R star In this problem, we are given units in terms of a solar radius (7 3 108 m) and Earth’s radius (6.4 3 106 m). Solve: Given the values in the problem, the percentage decrease in light during the transit is (4 . 56 . 4 3 106 )2 5 1. 4 3 103 or 0.14%. (1. 1  7 3 108 )2 Review: Transits of planets should produce very small drops in brightness given how tiny the planet is compared to the star, as we have found here. Note that in Math Tools 7.3, we see that Kepler-11C produces a drop of only 0.0008, or 0.08%, which confirms that this planet is larger than Kepler-11C. π 44. Setup: The volume of a sphere is d 3 . Density is 6 Mass 5 . However, let’s directly compare the Volume values of Earth and this planet by constructing a ratio. This has two added advantages: (1) it gets rid of the constants in our formula, and (2) we can use whatever

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units we want as long as we are consistent, that is, relative units as given in this problem. Solve: (a) This planet has a volume 3 Vp  dp  3 5  5 1. 7 5 4. 9 times tha t of Earth. V d Earth

Earth

(b) If both planets have the same density, then Mp Vp 5 5 4. 9, that is, the planet is about M Earth VEarth five times as massive as Earth. Review: A planet of this mass may be considered more of a gas giant than a terrestrial planet, so its average density may, in fact, be lower. It will be very interesting to see if we can learn whether this planet is terrestrial or giant in nature. 45. Setup: We are comparing to Jupiter, so note that MJ 5 1.9 3 1027 kg and RJ 5 7.15 3 107. Solve: (a) M 5 2.33 MJ 5 2.33.1.9 3 1027 kg 5 4.43 3 1027 kg. (b) R 5 1.43RJ5 1.43.7.4 3 107 m 5 4 4 1.02 3 108 m. (c) V 5  R 3 5  (1. 02 3 108 m)3 3 3 5 4 . 48 3 1024 m 3 . (d)  5

4. 43 3 1027 kg M 5 0 . 989 k g/m 3 , 5 V 4. 48 3 1024 m 3

which is just slightly less than the density of water. This planet must be gaseous. Review: We expect that planets this large and massive will all be gas giants, and the density we found confirms this expectation. Using the Web 46. Answers will vary. The answer will include what methods are used to detect planets from the ground and from space, whether planets have been found in each of two projects, and, if any were, what types of planets they are. A future project will also be presented, including when it will begin and what method(s) it will use. 47. Answers will vary. (a) Will show a graph showing the distances of exoplanets from its central star, noting the masses and sizes of each planet, and comparing them to our Solar System. (b) Will give a recently discovered and published planet, including its minimum mass, the semimajor axis, period and eccentricity of its orbit, its density and radius, if available, and whether it is Jovian or terrestrial. 48. (a) At the time of writing this manual, Kepler has confirmed 74 planets and has 2321 candidates. Followup observations consist of direct imaging of the star

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58 ◆ Chapter 7 The Birth and Evolution of Planetary Systems to rule out eclipsing binaries, and spectroscopy to determine the spectral type of the star, to rule out eclipsing binaries, and to try to determine the planet’s orbital details through radial velocity measurements. (b) Answers will vary. 49. (a) Crowdsourcing allows a very large amount of data to be analyzed in a short amount of time because, as the adage goes, “many hands make light work.” 50. One can make a stable system with four Earth-like planets. Adding in larger planets often causes one to be ejected; however, stable solutions can be found. In general, a massive planet is added, which causes lower-mass planets to adopt highly eccentric orbits. These slowly drift out of the bounds or have close encounters with massive planets and are flung out. By the way, my high score is 202,816,975 over 500 years. Exploration 1. Earth’s view is most like the side view. 2. This orbit is circular. 3. The maximum radial velocity of the star is about 30 m/s. 4. Between the minimum and maximum radial velocity, the phase changes by 0.5, meaning half of the full period of 1 year, or 182.5 days.

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5. If the planet moves away from Earth, the radial velocity of the planet would be positive (away from Earth) and that of the star would be negative (toward Earth). 6. The orbit is now eccentric. 7. The planet moves fastest when it is closest to the star (this is Kepler’s second law). 8. When the star moves fastest, the planet is closest to the star (this is Newton’s third law). 9. The shape of the radial velocity curve is very different for a circular and elongated (or eccentric) orbit. An astronomer would model the path of the orbit given the shape of the radial velocity curve. 10. This planet is a good candidate for a transit observation because the orbital plane has the planet eclipse the star. 11. With inclination set to 0.0, the Earth view is now like the orbit view. 12. We now do not detect any radial velocity. 13. With simulated data, the shape of the radial velocity curve is well traced by the data, as the noise is low. As inclination increases, the effect of the noise lessens. 14. At an inclination of about 4°, the data trace out a convincing agreement with the theoretical curve.

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CHAPTER 8

The Terrestrial Planets and Earth’s Moon INSTRUCTOR’S NOTES Chapter 8 covers the terrestrial worlds of the inner solar system. Major topics include

▶ the four methods of resurfacing (impacts, tectonism, vol-

canism, erosion) ▶ radioactive dating ▶ evidence for resurfacing on the terrestrial planets and the Moon ▶ how resurfacing reveals the geologic history and activity of an object ▶ the heating and cooling of planetary cores ▶ plate tectonics Before I taught my first introductory course, a friend explained that he teaches this material as “comparative planetology.” He starts by telling his students there are five ways a planet’s surface can be changed (erosion can be split into wind and water, or you can include life) and asks the class to name them. Then he shows how and why each one works on Earth and, with that toolbox, guides the students into deducing the geologic activity and history of all the other terrestrial planets. I have found this broader brushstroke approach very successful, but it means deemphasizing much of the vocabulary in the chapter. For example, in plate tectonics, I never mention the lithosphere or asthenosphere, or even crust and mantle; I just teach that the plates float on convecting magma. I do not discuss the differentiated planetary interiors, just that the interior must remain hot for the planet to have magma. I also focus on logical inference and cause and effect. For example, large bodies cool off more slowly than small ones; Earth is geologically active and the Moon is not, so we expect Venus to be active and Mercury to be dead. I have to guide students a little with this, because many of them expect that they have to memorize every term and fact in the book, and they can lose themselves in the weeds and miss the big picture. As one example, students often ask if they need to know that Martian valleys are called valles and mountains are mons, which is much less important in my view than what it means for Mars to have mountains and valleys. Obviously, these are just my opinions, and I do not wish to make any value judgments on how much geology to actually

cover. My personal teaching goal is to show a few pictures of an object and have students deduce its current geologic state and its history. During study of this chapter, I want students to deduce that because Mars has no hot spot chains of volcanoes, there are no plate tectonics. Or that because Mercury has numerous infilled impact craters and lava flows, it was volcanically active; however, as these are all pockmarked with tiny craters, Mercury is inactive or dead today. I come back to all of this a few lessons later when examining the moons of giant planets, when I let students predict the geologic states and explanations of Io, Europa, and Titan.

DISCUSSION POINTS

▶ Have students look at the pictures of the lunar high-

lands and Maria and deduce the cratering rate during the history of the Solar System. ▶ Look at the sizes and masses of the bodies in Table 8.1 and the state of geologic activity of each object. How recently did the dead planets become dead? How soon are the active ones likely to become inactive? ▶ Compare and contrast images of features on Earth and the other planets, and deduce whether the processes found on Earth were ever present on these other bodies. For example, compare Figures 8.2, 8.4, and 8.6: What can we deduce about the different characteristics of each crater? ▶ Carbon-14 is a radioisotope with a half-life of 5,700 years. Discuss the ways the dating of fossils using carbon-14 is different from the dating of minerals using the radioisotopes mentioned in Working It Out 8.1. ▶ Water ice is abundant on Earth and Mars, and it is detectable on the Moon and Mercury. Why is water so scarce on Venus?

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 8 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises. 59

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60 ◆ Chapter 8 The Terrestrial Planets and Earth’s Moon Processes that Shape the Planets This interactive animation illustrates three factors affecting the shape and resurfacing of a planet or moon, including tectonics, impact cratering, and weather erosion. Text reference: Section 8.1 Continental Drift This animation shows the original “pangaea” supercontinent drifting over time to become the continents we know today. Text reference: Section 8.4 Hot Spot Creating a Chain of Islands This interactive animation shows how a stationary source of magma coupled with a moving continental plate on top of it results in the formation of volcanic islands. Text reference: Section 8.5

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (b) Younger craters always lie on top of older ones. 2. (d) 40 seconds is two half-lives, so (1/2)2 5 1/4 will remain. 3. (d) Differentiation is what happens when oil and water separate in salad dressing. 4. (c) A dynamo requires fast rotation and a liquid core. 5. a, c, d, e. Only Venus seems to lack evidence of tectonic activity. 6. (b) The distance from the Sun has no bearing on the geologic activity of Mars. 7. a, c, d, e. Venus does have water vapor in its atmosphere but no evidence for liquid on the surface, which is why we can’t include it. Reading Astronomy News 1. The Moon is small and should have retained its internal heat for a short time; thus, we expect that it cooled quickly. 2. The new features are called mare patches because they are small, smooth regions surrounded by rough terrain, much like the mare but smaller. 3. Evidence that volcanic activity ended a billion years ago comes through crater counts. 4. Crater counts only provide a rough age, based on assumed impact rates and calibrated to rocks brought back from the Moon. Bringing back new samples of

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the regions in question will allow the rocks to be dated using radiometric methods. 5. Answers will vary. At the time of writing this manual, most search results for “Moon volcanic activity” turned up the original announcements from Oct. 2014. Test Your Understanding 1. (b) Impacts, tectonism, and volcanism build up structures on the terrestrial planets, whereas erosion tears them down. 2. (b) Newer material is always laid down on top of older material; therefore, the top layers will be younger (or the same age) as material lower down. 3. (e) All of the answers given are ways to learn about the interiors of planets. 4. (c) Angular momentum will keep the core of the Earth spinning but not heat it. Tidal effects frictionally heat an object (although on Earth this is not sufficient to completely explain our warm core). Radioactive decay injects energy into the core at a continual rate, which keeps the core warm. Gravitational contraction is important in gas giants but not for our solid/liquid core. So, we see that tidal effects and radioactive decay are the best answers. 5. (d) Each half life, one-half of the population decays, so in 4 half-lives, (1/2)4 5 1/16 remains. 6. (c) The Earth’s surface is constantly moving and changing so that its surface is renewed. 7. (c) We can study to orientation of structures in rocks to determine where the field lines pointed; by dating those rocks, we can trace the changes of the Earth’s magnetic field. 8. (a) S-waves cannot travel through liquid. 9. (d) Cratering rates are relatively well known, so younger material has fewer craters. Radiometric dating can be done on any sample that is collected and analyzed by the appropriate equipment. 10. (c) Cratering rates today are very low but were very high during the early days of the Solar System. 11. (d) Earth experiences constant resurfacing through wind, water, and volcanism. This resurfaces the planet faster than on Venus. 12. (a) The Moon has gone through periods of partial resurfacing, so the fact that it is highly cratered means that these happened long ago, due to a heavy bombardment. 13. (b) In a liquid, denser material always sinks to the bottom. If the Earth had not been molten at formation, one must ask how its layers achieved the same density structure as one would get from a molten planet.

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Chapter 8

14. (c) Although tidal effects do contribute a small amount, the predominant source of internal heating is radioactive decay. 15. (a) From Working It Out 8.2, heating scales roughly with volume (V  R3), so a planet half as large as Earth will heat (1/2)3 5 1/8 as fast. Thinking about the Concepts 16. Our Moon is similar to the other terrestrial planets (especially given our theory of how it formed) and gives us a good basis for understanding the moons of other planets. 17. Igneous rocks can be dated radiometrically because all parts of the rock formed and cooled in the same place at the same time. Therefore, any parts of an igneous rock will be the same age as any other part of the rock. Sedimentary rocks are formed as an amalgamation of deposited materials that formed in different places and at different times. As a result, different parts of the same sedimentary rock may have vastly different ages. Metamorphic rocks experience physical changes that may include one or more heating events that falsely reset their chronometers. Complex physical histories make it difficult to determine reliable ages for most metamorphic rocks. 18. We can select igneous rocks from various levels of the Grand Canyon and measure their ages using radiometric dating. We then compare ages of rocks at the rim to those along the river and note that rocks increase with age as we progress deeper into the canyon. In geology, this is called the law of superposition, and it holds generally true in most geologic strata because younger sediments are deposited on top of older sediments. 19. Earth’s internal heat comes primarily from its early accretion, radioactive decay, and friction from motion in its interior. Earth’s interior contains about 50 times the volume of the Moon’s interior, and it should therefore store about 50 times more heat energy. Earth’s surface area is about 13 times greater than the Moon’s surface area, so Earth radiates heat at 13 times the rate that the Moon does. Based on their volume-to-surfacearea ratios, Earth should take approximately 4 times longer to cool than the Moon did. 20. The Moon is smaller than the Earth, so it contained less thermal energy in its core to begin with, and a smaller core cools faster than a larger one. Thus, the Moon’s core is cool and solid whereas the larger Earth’s core has still retained enough heat to remain liquid. 21. A transverse wave is one in which the direction of motion of the disturbance is perpendicular to the

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direction of motion, whereas a longitudinal wave has both in the same direction. Shaking a taught string makes a transverse wave, whereas stretching a spring and then giving it a flick makes a longitudinal wave. 22. Earthquakes generate two types of seismic waves that pass through Earth’s interior: primary and secondary waves. Primary waves pass through both solid and liquid materials, but secondary waves pass through only solid materials. Seismometers located near an earthquake source detect both primary and secondary waves. However, seismometers located 104 degrees or more from an earthquake source detect only primary waves. This shadow zone for secondary waves is a clear indication of a liquid layer in Earth’s interior. 23. The impactor theory of the Moon is supported by the compositional similarities of the Earth and Moon, the lack of volatiles on the Moon, and the inclination of the Earth’s spin axis. However, evidence of the impacting object’s own composition has yet to be observed. 24. Venus, Earth, and Mercury all show evidence of tectonic processes. There is no active tectonism on Mercury today, but the planet displays several large cliffs that appear to be the result of buckling as the planet shrank early in its history. All the tectonic features on Mercury are compressive. Venus displays large circular tectonic fractures that some planetary scientists have interpreted as the upwelling of mantle hot spots. Over geologic timescales, large chunks of the lithosphere centered over these hot spots may melt, causing Venus’s entire surface to overturn. Tectonic activity on Earth is dominated by plate motions. The plates are moved around the surface by convection in the upper mantle. It is not clear why Venus and Earth exhibit such different tectonic behavior. 25. Plate tectonics is the fractured pieces of the planet’s upper crust floating on the denser mantle. Convection in the mantle moves these pieces of the planet’s surface around, generating new seafloor through volcanic activity in some places and subducting old seafloor back into the mantle at ocean trenches. Plate tectonics have been observed only on the Earth. 26. The largest volcano in the inner Solar System is Olympus Mons on Mars, followed by Maxwell Montes on Venus. The other large-shield volcanoes on Mars— Ascraeus Mons, Arsia Mons, and Pavonis Mons—are all larger than Mauna Loa, Earth’s largest volcano. 27. Small, simple, bowl-shaped craters with no internal structure are likely to be impact craters. A crater formed by a volcanic eruption should be easily associated with a nearby large volcano. Large, complex impact craters could also display a central peak surrounded by a series of concentric rims. A pattern of smaller, secondary

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62 ◆ Chapter 8 The Terrestrial Planets and Earth’s Moon impact craters is usually found near young impact craters. If the resolution is sufficiently high, the rim should display an inverted geologic stratigraphy. 28. Venus, Earth, and Mars all have atmospheres, so there are many processes of gradation/erosion that can change each planet’s surface. In contrast, the Moon and Mercury are geologically dead and have only impacts to change their surfaces. 29. Mars shows dry river beds, canyons, and teardrop-shaped erosions around craters, indicating that water was flowing and eroding the surface. 30. Answer will vary. The Chicxulub crater is dated to the same time as the K-T boundary, and both contain anomalous amounts of iridium and other radioactive elements; both are suggestive of a major impact. There are no dinosaur fossils found above the K-T boundary, and abundant ones below it, suggestive that this boundary marks a cataclysmic event. The K-T boundary itself is a thin layer found all over the Earth, suggestive of ash, dust, or debris on a global scale. Applying the Concepts 31. Setup: Figure 8.8 is presented in linear scale on the x-axis and the y-axis is arbitrary, where higher equals more. We see that the cratering rate drops enormously up through about 3.3 billion years and then decreases very slowly thereafter. Solve: (a) The cratering rate has changed quite abruptly around 3.3 billion years ago. (b) At present, the rate is negligible compared to 4 billion years ago. (c) If there were many small masses flying around the early solar system, we would expect there to be many collisions early on. As time goes by, those collisions clean up the debris, meaning there are fewer objects for a planet or moon to hit, and the cratering rate must drop and slow at the same time. Review: Note that very young surfaces should have very little cratering, according to the figure, which agrees with what we observe on Earth. 32. Setup: When looking at axes, remember that linear ones change by integer values, whereas logarithmic ones change by powers of some number, like 2 or 10. Solve: Figure 8.7 is shown with linear spacing. After one half-life, the number of parents and daughters are the same. This must be a general feature because when we start the experiment, we have all parent and no daughter isotopes. After one half-life, half the parents have become daughters, giving us the same number of both. Review: The y-axis of this graph might appear logarithmic only because the numbers labeled each increase

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in spacing. But in a log plot, the spacing would not change, as these numbers increased by a factor of 2. 33. Setup: Exponential growth or decay means that the rate changes as a function of time, that is, it doubles or halves each time step. Solve: The daughter growth is exponential because the number of daughters doubles with each step. Review: Note that because parents become daughters, if the change of parents is exponential decay, then the change of daughters must be exponential growth. 34. Setup: When comparing ages of a surface, look for signs of erosion. Few large craters, or filled in bottoms, indicate a younger surface. The more craters seen, the older the surface. Solve: Figure 8.18 shows a relatively smooth surface with lots of small craters. Figure 8.23 also shows a very smooth surface with fewer small craters, suggesting this region is younger. Review: Without radiometric dating, it is very hard to tell the ages of regions with roughly similar cratering, so the answer given to this question could be wrong. This is part of the nature of science—one must follow the data! 35. Setup: Remember that the surface area of a sphere is expressed as 4pr2, where r represents the radius of the sphere. 2 5 Solve: (a) Surface area of Earth: A 5 4 R 4 (6 , 378 km)2 5 5. 10 3 108 km. (b) Using the same formula, the surface area of Mars is 4p (3,390)2 5 1.44 3 108 km. (c) If 72 percent of the Earth’s surface is covered with water, then the remaining 28 percent is land area. The total land area of the Earth’s surface is then Aland 5 (0.28)(5.10 3 108 km2) 5 1.43 3 108 km2. In other words, the surface area of Mars is approximately the same as the total land area of Earth. Review: Is this a surprising result? Mars’s size is about one-half that of Earth, so Mars has about one-fourth the surface area. Given that about three-quarters of the Earth’s surface area is water, that means only onefourth of our surface area is land. So, a back-of-theenvelope calculation confirms our finding. 1 36. Setup: Kinetic energy K 5 mv 2 . 2 Solve: The kinetic energy of the ice is 1 KEice 5 (0. 001 kg )(50 3 103 m/s)2 5 1. 25 3 106 J, 2 1 whereas that of the SUV is KESUV 5 (2 , 000 kg ) 2 (90 3 103 m/hr 3 1 hr/3,600 s)2 5 6 . 25 3 105 J.

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Chapter 8

Review: The SUV barrelling along at 90 km/hr has only half the kinetic energy of the 1-gram chunk of ice hitting Earth’s atmosphere at 50 km/s. This dramatic example illustrates how speed contributes much more to kinetic energy than mass does. M 4  where V is volume   r 3  . 37. Setup: Density  5 3  V Solve: The problem gives the mass and radius, so 300 3 106 kg 5 4 , 580 kg/m 3 . (4 / 3) (25 m)3 Review: Water has a density of 1,000 kg/m3, rock a few times this, and metals a few times more. So this meteor was largely composed of metals, as we have found in previous meteors and previous problems in this chapter. 38. Setup: Working It Out 8.2 explains that the rate of cooling scales as ratio of surface area to volume, or 1/R. Solve: Using data in Table 8.1, the Moon’s size is 0.272 that of Earth; therefore, the Earth will cool off 0.272 times slower than the Moon, or alternatively, the Moon cools 1/0.272 ≈ 3.7 times faster. Review: The larger an object, the less surface area it has for its volume; thus, the slower it cools. This is why Earth is cooling more slowly than the Moon did. M 4  39. Setup: Density  5 , where V is volume   r 3  . 3  V Solve: (a) The problem gives the mass and radius, so 36 3 1024 kg 5 5, 500 kg/m 3 . 5 4 (6 , 378 3 103 m)3 (b) Because crustal rock has a density of 2,600 kg/m3, Earth must have a sizable chunk of metal in its interior to bring the average density up to 5,500 kg/m3. (c) Similar calculations for the Moon (mass 5 0.012 MEarth, radius 5 0.272 REarth (from Table 8.1) giving a density of 3,292 kg/m3. (d) The Moon’s density is much lower than that of Earth so it contains more rocks and fewer metals. Review: We know from this chapter that the Earth has a highly metallic core, which this problem confirms numerically. 40. Setup: Every half-life reduces the radioactive sample size by half. In one half-life, half of the radium would have decayed to radon. In a second half-life, half of the remaining radium would have decayed to radon, meaning one-quarter of the radium would remain, and so on. Solve: In this problem, we are told that there are three atoms of radon for each radium, which means that of four original radium atoms, three-quarters have decayed. Based on the logic in the setup, this pottery

5

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has been around for two half-lives, or 2 3 1,620 y 5 3,240 y. Therefore, the pottery is about 3,240 years old. Review: First, remember that radioactive dating is not ultra-exact, so this shard could be 3,000 to 3,500 years old and still give the same result. For a sanity check, pottery was being made that long ago, so our answer is reasonable. 41. Setup: Every half-life reduces the radioactive sample size by half. In one half-life, half of the carbon-14 would have decayed. In a second half-life, half of the remaining carbon-14 would have decayed, meaning a quarter of the carbon-14 would remain, and so on. We can write this as a formula where the fraction remaining 5 (1/2)n, where n is the number of half-lives. Solve: (a) Using our formula above, 1/64 5 (1/2)6, so this sample is 6 half-lives old. (b) Because each half-life is 5,700 years, the sample is 6 3 5,700y 5 34,200 y old. (c) If daughter nuclei were present in the sample when it formed, then we would measure a parent-to-daughter ratio that is smaller than actually occurred, meaning we would calculate the sample to have gone through more half-lives and thus overestimate its age. Review: Part (c) shows us an important issue with radioactive dating, which is why we usually try to date objects using many different radioactive isotopes. 42. Setup: Remember that every half-life reduces the radioactive sample size by half. Solve: (a) The age of the Solar System is about 4.5 billion 4. 5 3 109 8 3 105 halfyears old, which is about 5, 700 lives of carbon-14. In that time, all the carbon-14 present in an initial sample will have decayed, so radioactive isotopes with a short half-life are useless when estimating ages of objects that are many times older. (b) The number of half-lives of rubidium-87 that have passed 14 billion 5 0. 29 . In in the last 14 billion years are 49 billion this many half-lives, (1⁄2)0.29 5 0.82 of the rubidium present in a sample will remain. In other words, some rubidium (about 18 percent) has indeed decayed. Review: It is important to always date an object with an isotope whose half-life is of the same order as the age of the object. So for human-made objects, carbon-14 is very useful, whereas the other isotopes listed would be used for objects that date back billions of years. 43. Setup: This problem asks us to relate distance, speed, and time, which we know is d 5 vt. 4 , 500 3 105 cm 5 3. 75 cm/yr 1.2 3 108 yr. South America and Africa would have Solve: Solving for time yields t 5

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64 ◆ Chapter 8 The Terrestrial Planets and Earth’s Moon begun separating about 120 million years ago if we can assume that the separation rate has been constant through geologic time. Review: Our first sanity check is that this timescale is much shorter than the age of the Earth, which is important. Our second check is that animal and plant fossil remnants on different continents match, which means that the supercontinent could not have broken up billions of years ago; therefore, this number has about the right value necessary. 44. Setup: The volume of a cone is expressed as 1  V 5 bh 5 r 2 h. 3 3 Solve: The ratio of Olympus Mons to Mauna Loa is VOlympus Mons VMauna Loa

2 r  h 27  275  5  OM  OM 5  3 5 63 .   60  9 r  h ML

Review: Olympus Mons has about 63 times more volume than Mauna Loa—boy is that big! 45. Setup: Using Newton’s Law of Universal Gravitation, the strength of gravitational force on the surface of a planet scales as M/R2, so we construct a ratio of this quo2

  tient for Mars and Earth, i.e. FMars 5 M Mars R Earth .  FEarth M Earth  R Mars 

Note that Table 8.1 gives values for Mars in relation to Earth, for example, M Mars 50. 107 M Earth . Using these FMars M Mars 5 . FEarth R 2 Mars Solve: Using values from Table 8.1, FMars 0. 107 5 5 0. 377. This low gravity would have FEarth 0. 5332 meant rock would have been less compressed than on Earth, perhaps explaining why it was easier for Mons on Mars to be much higher than volcanic cinder cones on Earth. Review: A quick check on your favorite search engine will show Mars’s gravitational acceleration as 3.7 m/s2, 37.7% smaller than the 9.8 m/s2 on Earth. scaled values, the ratio becomes

Using the Web 46. Answers will vary. The response will include whether there were large earthquakes in the past week and whether any locations are unexpected given the map of earthquake frequency in Figure 8.17 of the text. You should then report where the most recent significant earthquake occurred, and how much seismic activity

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there has been in the United States in the last 30 days over magnitude 2.5. 47. (a) The far side of the Moon looks more heavily cratered. (b) Mercury looks a lot like the Moon in that it is heavily cratered, but Mercury does not have the large, smooth regions (Maria) that are present on the Moon. (c) The two hemispheres of Mars appear very different, almost as if one had been an ocean and we are seeing the dry floor, whereas the other is full of mountains and canyons. 48. (b) Mercury Mappers’ goal is to have citizens map the sizes and locations of craters from the Messenger satellite. (c) Moon Mappers is quite similar to Mercury Mapper. The FAQ is at http://cosmoquest.org/x/ moon-mappers-faq/. Basic features are marking a crater and flagging something as “odd.” The angle of the Sun causes shadows to be longer or shorter, giving the craters more or less relief (that is, making them stand out more). 49. (a) The Wide Angle Camera had a very large field of view and 11 filters, whereas the Narrow-Angle Camera had a very small field of view and only shot in black and white. Color images are false color, generally showing elevation. Messenger crashed into Mercury on April 30, 2015. Recent science results will vary. (b) Recent science results from the Mars Science Laboratory Curiosity will vary. (c) Answers on the Google Lunar X Prize will vary. 50. (a) Answer will vary. As for why we went, as President Kennedy said, we chose to go to the moon not because it was easy, but because it was hard. This was the time of the Space Race against the USSR in the Cold War, and a large part of the motivation to land on the Moon was simply to do it before the Soviets did. There was also countless science to be gained from this endeavor. The Apollo program ended for a variety of reasons— the growing cost of the Vietnam war and burgeoning federal deficit meant Congress allocated less and less money to the program. The accidents with Apollo 1 and 13 did not help endear the public to the moon missions either. Currently, China, the United States, and the European Space Agency are proposing manned missions to the Moon. (b) The Selenians (or Selenites) in Le Voyage dans la Lune live underground. The humans are able to blow them up by hitting them. The contrasts between the real Moon and that of the film are myriad, starting with the man in the moon who is hit in the eye. The Moon in the film has innumerable sharp peaks and jagged protrusions, running water, plants (mushrooms), and of course an atmosphere. This is totally anathema to the real Moon as seen by astronauts.

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Exploration 1. Early on, the number of M&Ms landing “M” side up decreases quickly. As time goes on, the rate of decrease slows. 2. If we treat radioactive decay the same way, then the rate of decay starts out very quickly and slows over time. 3. As time goes by, the number of M&Ms that remain gets smaller, as the number in the pile set aside grows. 4. If we treat radioactive decay the same way, then the number of parents decreases with time as the number of daughters grows.

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5. The plot above shows the sum, not the average. We did 10 trials, so we have to divide the y-axis by 10 to find the average number remaining over time. Thus, to find the time at which 10 remain, we must look for 100 on the y-axis. This happens around 2.5 time steps. 6. If we know how many parents the sample started with, or the ratio of parents to daughters, then we can compare the number or ratio of isotopes found to the decay rate and determine how many half-lives have passed.

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CHAPTER 9

Atmospheres of the Terrestrial Planets INSTRUCTOR’S NOTES Chapter 9 covers atmospheres, with particular application to the terrestrial planets. Major topics include

▶ formation and loss of atmospheres ▶ primary, secondary, and tertiary atmospheres ▶ the detailed structure and composition of Earth’s atmosphere

▶ magnetic fields and auroras ▶ circulation and weather ▶ comparative studies of the atmospheres of Earth, Venus, and Mars

▶ the greenhouse effect and anthropogenic global warming Most semesters, I have at least one student who comes to class believing climate change is a hoax. It is a touchy subject because this can stem from deeply held religious or political beliefs. To help all my students become more scientifically literate, I teach a lab on the differences between science and pseudoscience before this chapter. When the subject of global warming comes up during lecture, I present all the relevant data and show that current climate trends are changing faster than the geologic record. If anyone objects that this is a hoax or that they do not believe in global warming, I can refer back to that lab and we can pick apart the arguments of climate-change deniers, who are generally using the tactics of pseudoscience. Also, many students do not realize that one cannot believe in a scientific theory; it is either well tested or poorly tested and therefore makes reliable or tentative predictions. If nothing else, I believe clearly elucidating that distinction is one of the most important points of an introductory science class, and I encourage you to use this chapter to ensure that your students are literate in this difference. I find that some students can easily lose themselves in the weeds of the vocabulary of different atmospheric layers or the temperature/pressure changes of atmospheres with altitude (Figure 9.6, 9.12, and 9.16). If you opt for a broader brushstroke approach, it may help to focus less on memorizing the particular terms from the text and more on understanding why there are different atmospheric layers and why the terrestrial planets have different atmospheric structure.

How do planet rotation and solar heating affect structure, for example? To refer back to the scaffolding of learning, a comparative planetology approach teaches and strengthens the higher order processes of analysis and synthesis and will allow the student to interpret the reported discovery of a new atmospheric world (e.g. a moon or exoplanet). Because I have already covered the quantum mechanical explanation for atomic emission and absorption before this chapter, I like to explain why greenhouse gasses emit and absorb infrared photons. If you draw the structure of CO2, H2O, and methane, students will see that they are asymmet ric molecules. These can vibrate and rotate in quantized states whose energetic transitions are in the IR spectrum. At this point, I show the rovibrational action by doing the chicken dance. Perhaps they laugh at me rather than with me, but the visualization helps them remember the root cause of greenhouse absorption, which in turn helps them remember the whole greenhouse process (I have compared assessments on the greenhouse effect with and without this demo, and students retain more with it). Although not specifically discussed in the text, I like to include a short lesson on why the sky is blue. I explain that the molecules in our atmosphere scatter blue light more efficiently than red and then make drawings of sunlight coming down from close to the zenith and close to the horizon. In the former, we see blue scattered from all over the sky, making the sky blue. In the latter, because the atmosphere near the horizon is thicker, mostly red sunlight reaches us, making the sunset red. This same effect makes the blue sky look like a paler white near the horizon. I then ask why sunsets almost always look more stunning than sunrises and often show pictures that I have taken from my house during winter of both sunrise and sunset to show that they are indistinguishable. There are two reasons for this phenomenon. The first is that extra pollution and water vapor builds up over the day, which does cause sunset to look redder. But the other is that as our eyes begin to adapt to the dark (a hormonal response to our diurnal sleep pattern), our nighttime vision becomes more sensitive to reds, so the sunset looks more brilliant than it really is. This is why the photographs need to come from winter, when there are fewer pollutants and low humidity in the evening sky. 67

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DISCUSSION POINTS

▶ Earth’s atmosphere is far from being primordial. Discuss

how the current main components of our atmosphere (nitrogen and oxygen) came to be dominant and why our atmospheric composition differs from those of Venus and Mars. ▶ Our theory of stellar evolution tells us that billions of years ago, the young Sun was about 70 percent less luminous than it is currently (we will learn why in Chapter 16). Using the information given in Working It Out 5.4, calculate the equilibrium temperature of Earth, Mars, and Venus when the Sun was young. What amount of greenhouse warming would they have needed in order to have liquid water on their surfaces? Discuss how plausible it is that Mars and Venus once had liquid water on their surfaces. ▶ On the surface of Venus, the atmospheric pressure is 92 times higher and the average global temperature is 2.56 times higher than on Earth. Going up through the atmosphere, the pressure and temperature drop. At an altitude of 50 km, Venus’s atmosphere has an average pressure and temperature similar to Earth’s pressure and temperature at sea level. Could humans survive if stationed high in Venus’ atmosphere? What would they need to do to create habitable conditions? What advantages and disadvantages would there be to living on Venus as opposed to Earth? ▶ Does Mars offers conditions that make it a target for a future visit by humans? Discuss the pros and cons of actually sending humans to visit the red planet. ▶ By changing the composition of Earth’s atmosphere, humans are carrying out a dangerous experiment because their survival is linked to the outcome of the atmospheric changes. Discuss what could be done to keep this experiment under control so that humans do not create an inhospitable environment.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 9 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises (www.norton.com/smartwork). Atmospheres: Formation and Escape This animation explores the formation of planetary atmospheres, first from a planet’s original formation and then from its volcanic processes. It shows how a planet can lose or retain gas molecules depending on temperature and

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surface gravity. It also discusses the differences in chemical composition of the planetary atmospheres of Earth and Jupiter. Text reference: Section 9.1 Greenhouse Effect This animation shows the mechanism and importance of the greenhouse effect for Earth, as well as its effects on Venus and Mars. Text reference: Section 9.2

ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what they have learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5) and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack. Changing Equilibrium This simulation demonstrates equilibrium using water bottles to flow liquid (representing energy) into and out of a vessel. This is an analogy to energy flowing into and out of a planet to change its temperature. Text reference: Section 9.2 Charged Particles and Magnetic Forces This simulation demonstrates the magnetic force on moving charged particles. It includes a demonstration that there is

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no magnetic force when the charges are at rest, an introduction to the right hand rule, and an argument why magnetic fields cause charged particles to have circular motion. Text reference: Section 9.3

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. a, b, c, d 2. (c) Note that methane and ammonia are greenhouse gasses but are not as abundant as carbon dioxide and/ or water. 3. From least to greatest: c, a, e, b, d 4. (a) The troposphere contains about 90 percent of our atmosphere and is both the air we breathe and the location of weather on Earth. 5. b, a, d, c 6. a, b, c Reading Astronomy News O

Semiheavy water is 5.5 percent heavier 1 H H than regular water. 2. Regular water has a lower mass, so at a given temperature, it has a higher thermal velocity than semiheavy water. 3. Scientists studied the atmosphere over the polar regions of Mars because the poles are where much of Mars’s water is stored (in polar ice caps). 4. The water evaporated and escaped the upper atmosphere. 5. Answers will vary. 1. 2

Test Your Understanding 1. d, b, c, f, a, e 2. (a) The greenhouse effect is strongest on Venus, where it has heated the planet to temperatures higher than we use to bake bread. 3. (c) Oxygen was created by plant life out of the CO2rich secondary atmosphere. 4. (c) Although Mars, Earth, and Venus are all at different distances from the Sun, their temperatures depend much more strongly on the strength of the greenhouse effect, which itself is dominated by the kinds and numbers of greenhouse gases that are present. The radical differences between these three planets are dominated by the amount of greenhouse gas, that is, the thickness (difference in weight between pressure levels) of the atmosphere.

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5. (d) As noted in Sections 9.3 and 9.5, weather is small scale and short term, and climate describes the average state of an atmosphere. 6. (b) If particles have the same kinetic energy, then the less massive ones are moving much faster, which helps them escape the gravity of the planet. 7. (b) Venus’ exceptionally thick (in the sense of density or pressure) atmosphere translates into a very strong greenhouse effect. An analogy is that Mars has a comforter with two down feathers in it, whereas Venus is covered with several down comforters of heavy winter thickness. 8. All of these are excellent reasons to study climate in a varied range of environments. 9. (d) Dust storms cause an atmosphere to take on a reddish hue, and because dust storms on Mars are quite common, the atmosphere often takes on this appearance. 10. (a) Auroras occur when charged particles from the Sun, trapped in our planet’s magnetic fields, rain down on our atmosphere. 11. (c) Ozone is a thin layer that protects the Earth from harmful UV rays from the Sun. 12. (d) The rotation of the Earth breaks global circulation into zones. Interestingly, this is called TaylorCouette flow. 13. (c) The more molecules of greenhouse gas present, the stronger the greenhouse effect. 14. (d) It stands to reason that Earth’s average temperature tracks the amount of carbon dioxide in the atmosphere because carbon dioxide is a major greenhouse gas on Earth. 15. (c) The troposphere contains about 90 percent of our atmosphere and is both the air we breathe and the location of weather on Earth. Thinking about the Concepts 16. Hydrogen (75 percent) and helium (25 percent) were by far the dominant elements in the solar nebula, and both existed as gases in the region of space where the terrestrial planets formed. Combined, all other volatile compounds represented only a tiny fraction (less than 1 percent) of a terrestrial planet’s gas content. Therefore, the primary atmospheres of the terrestrial planets were approximately three-quarters hydrogen and one-quarter helium. 17. After the primary atmospheres were stripped from the terrestrial planets, the planets outgassed volatile material (gases) from their interiors. The gas expelled from planetary interiors combined with those materials delivered by comet impacts to produce secondary atmospheres.

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70 ◆ Chapter 9 Atmospheres of the Terrestrial Planets 18. Although free nitrogen was poorly represented in the protostellar disk, ammonia was abundant. Most of the ammonia froze as ice in the outer Solar System, becoming a significant component of comets. The ammonia was probably delivered to Earth through comet impacts during the heavy bombardment period early in the history of the Solar System. Under the conditions on early Earth, ammonia was dissociated into its constituent nitrogen and hydrogen atoms. The hydrogen was light enough to escape Earth, but the much heavier nitrogen stayed behind. 19. Impacts from icy comets and outgassing from volcanoes. 20. Greenhouse gases trap heat inside the planet, like a blanket, helping the planet to become warmer than it could if it were just heated by sunlight alone. On Earth, this raised the average temperature above the freezing point of water, making the planet hospitable for terrestrial life. 21. Plants remove carbon dioxide from the atmosphere through photosynthesis and produce oxygen as a waste product. 22. The troposphere is the lowest layer of the atmosphere; this is where ozone is a pollutant. In the higher layers (i.e., stratosphere), ozone absorbs UV photons, protecting the planet from this harmful radiation. 23. Uneven solar heating produces temperature differences at the surfaces of the terrestrial planets. Temperature differences produce pressure gradients in the atmosphere, creating regions of high and low pressure. According to the second law of thermodynamics, air flows from high pressure to low pressure, and we detect this flow of air as wind. Steeper pressure gradients lead to higher wind speeds and stronger wind. 24. (a) The Arctic ice is already floating in the ocean, so it does not add any additional mass to the oceans when it melts. (b) The glaciers that cover Greenland and Antarctica currently reside on land. When those ices melt and flow to the oceans, they add to the mass of water already in the oceans, causing the average sea level to rise. 25. Venus has perpetual, opaque cloud cover due to its runaway atmosphere. 26. The three terrestrial planets with atmospheres all have surface temperatures that are much higher than can be explained through the balance of energy received by the Sun and given off by the heated surface. This means there must be some effect at play that retains heat, that is, the greenhouse effect. 27. First of all, the runaway greenhouse effect on Venus means that the planet holds on to its heat extremely efficiently, so whether the Sun is shining or not, the planet stays warm through its very effective green-

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house “blanket.” Venus’s slow rotation means that there are not strong wind currents across the planet’s longitude, only in latitude, so hot gas at the equator is constantly mixed with cooler gas at the poles to make a uniform temperature. 28. Venus must have been cooler and had lower pressure in the past, whereas Mars must have been warmer and had higher pressure in the past. Thus, both planets had periods when they could have been more hospitable. 29. Answers will vary. 30. Answers will vary. Is it easier to evacuate Venus of its excess carbon dioxide and acidic atmospheric compounds or enrich the atmosphere of Mars? It is probably easier to bring gas to Mars than take gas away from Venus. Applying the Concepts 31. Setup: Remember that linear axes change by constant additive increments, and log axes by multiplicative ones. Solve: (a) The axes are linear. (b) The variations of CO2 and temperature are very similar. (c) The highest CO2 level was about 300 PPM, but it is 386 PPM today. Review: Because the amount of greenhouse gases in the atmosphere dictates the greenhouse effect strength, we expect the temperature and gas levels to be correlated, as they indeed are. 32. Setup: This question relies primarily on panel (a). Solve: (a) The amount of greenhouse gases began rising exponentially about 100 to 150 years ago. (b) Using panel (a), it seems the behavior of the three gases varied in unrelated ways until the last 100 years or so. (c) The rise of all three gases in modern times is most likely due to industrialization. Review: This confirms the general consensus among scientists about anthropogenic (human-caused) climate change. 33. Setup: Figure 9.6b shows that commercial jetliners fly about 12 to 15 km above the ground. Solve: It seems advantageous for jetliners to fly above the bulk of atmospheric convection, for a smoother ride, and as close to the stratospheric winds as possible, so the winds help carry the planes along. This requires that they fly above 10 km or so, and indeed they do (Figure 9.6b). Review: Interestingly, the jetstream (in which jetliners fly) was discovered during World War II, when the Japanese tried to bomb the United States mainland through balloon-carried bombs. Fortunately, these were discovered by the Air Force, and there was only one casualty.

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Chapter 9

34. Setup: This question asks about the effects of pressure in different environments. Because there is a closed bag of chips, we can assume the pressure inside the bag remains constant. What happens as the pressure outside the bag changes? Solve: At higher altitudes, the air pressure drops, so the pressure inside the bag increases. (a) Inside the plane, the pressure is lower than when the plane is on the ground. (b) Airplane cabins are pressurized, which means they are at higher pressure than the outside air when they are flying. A bag outside the plane would expand as much as it could, about five times its size, based on Figure 9.6 (and in reality would probably burst because the outside pressure is so low). Review: This explains why bottles of water sometimes leak if you open them before takeoff and lay them on their sides while the plane climbs. Similarly, at cruising altitude the pressure is lower than on the ground, so a bottle that you drank while flying will crush itself as you descend because the pressure inside is lower than the pressure on the ground. 35. Setup: Force F 5 pA, where A is the area on which the pressure acts, or F 5 ma, where m is the mass being accelerated at a. Solve: (a) A head has a typical circumference of 50 cm, meaning a radius of r 5 C/2p 5 8 cm, yielding an area A 5 pr2 5 p(0.08 m)2 5 0.02 m2. Then the force of air on a typical person’s head is F 5 pA 5 105 N/m2 3 0.02 m2 5 2000 N. (b) If this force were due to a kangaroo, the kangaroo would have to have a mass of m 5 F/a 5 2000 N/9.8 m/s2  200 kg. (c) Given that a typical kangaroo has a mass around 60 kg, we would need 200/60  3 kangaroos on a person’s head. (d) We are not crushed by this force because the bones and muscles in our body hold us up. Review: First of all, why would we ever put a kangaroo on our heads? That being said, we evolved to support the atmospheric weight on top of us. This shows, however, why we could not simply “jaunt” off to another planet such as Venus because the atmosphere would crush us. 36. Setup: Venus has an atmospheric pressure about 92 times that on Earth, or 9.2 3 106 N/m2, and an acceleration of gravity about 0.9 times that on Earth, or 8.9 m/s2. Solve: (a) Using the area of a person’s head as 0.02 m2 (see problem 35), then the force of air on our head is about F 5 pA 5 9.2 3 106 N/m2 3 0.02 m2 5 1.08 3 105 N. (b) If this force were due to a kangaroo, the kangaroo would have to have a mass of m 5 F/a 5 1.8 3 105 N/8.9 m/s2  20,700 kg. (c) Given that a typical kangaroo has a mass around 60 kg, we

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would need 20,700/60  350 kangaroos on a person’s head. (d) We would certainly be crushed by this force because the bones and muscles in our body could not hold us up. Review: When we think of traveling to other planets, we often forget about the effects of pressure and only think about temperature (and of course, whether there is oxygen). 37. Setup: Mars has an atmospheric pressure about 0.006 times that on Earth, or about 600 N/m2, and an acceleration of gravity about 0.4 times that on Earth, around 3.8 m/s2. Solve: (a) Using the area of a person’s head as 0.02 m2, then the force of air on our head is about F 5 pA 5 600 N/m2 3 0.02 m2 N. (b) If this force were due to a kangaroo, the kangaroo would have to have a mass of m 5 F/a 5 12 N/3.8 m/s2  3.2 N. (c) Given that a typical kangaroo has a mass around 60 kg, we would need 3.2/60  0.05 kangaroos, or a very tiny joey, on a person’s head. (d) We would certainly not be crushed by this force because the bones and muscles in our body already hold us up so well on Earth. Review: Given that the atmospheric pressure on Mars is so low, but the pressure in our body is about that of the atmosphere on Earth, we would probably plump up like a balloon! 38. Setup: Looking at Figure 9.6, we see pressure versus altitude in the right-hand panel of part (a) and temperature versus altitude in the left-hand panel. To relate temperature to pressure we must consider both panels at once. Solve: Figure 9.6a shows in the bottom (light blue) that temperature and pressure increase as altitude decreases, meaning pressure and temperature are related. Thus, if one goes up, the other must as well. Review: This is the essence of how refrigerators work. Gas is pressurized and then cooled down with a fan, and the gas is released into a lower pressure chamber. Lower pressure means lower temperature, so the gas ends up cooling down significantly. 39. Setup: Remember that p percent 5 0.01 3 p and added mass 3100. current mass Solve: (a) The mass of carbon dioxide in the atmosphere is 0.06 percent of the total mass of the atmosphere, or (0.06/100) · 5 3 1018 kg 5 3 3 1015 kg. (b) If we add 3 3 1013 kg per year, then we are increasing the total amount by (3 3 1013)/(3 3 1015) 5 0.01, or 1 percent per year. (c) The number of molecules is (3 3 1013)/(7.3 3 10226) 5 3.9 3 1038 molecules per year. (d) Every molecule of carbon dioxide absorbs in-

percent increase 5

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72 ◆ Chapter 9 Atmospheres of the Terrestrial Planets frared energy from the surface and helps heat it. Thus, even a 1 percent increase in the total amount (which itself is only 0.06 percent of the atmosphere) means that much more energy is being stored. Review: Although the annual increase might be small, the cumulative increase is quite large. That is, if we double the amount of greenhouse gas in the atmosphere, we also double its effect, which will warm the planet. 1 40. Setup: Kinetic energy K 5 mv 2 . 2 Solve: Kinetic energy of parcels of air on Earth and 1 2 Venus are KEEarth 5 (1. 23 kg )(10 m/s) 5 61. 5 J 2 1 and KEVenus 5 (64. 8 kg )(1 m/s)2 5 32. 4 J . 2 Taking a ratio of the two numbers shows that a parcel of air on Earth has about 1.90 times the kinetic energy of a parcel of air on Venus. (b) 1 KEMars 5 (0. 015 kg )(50 m/s)2 518. 75 J , so taking 2 a ratio of the two numbers shows that a parcel of air on Earth has about 3.28 times the kinetic energy of a parcel of air on Mars. (c) Wind erosion is only one of the gradation processes that operate on Earth. We must consider water erosion, volcanism, landslides, and plate tectonics as well. Review: In many places on Earth, the other processes are more dominant than wind erosion. Wind erosion is best observed in deserts where abrasion of buildings and scouring of the ground are common sights. 41. Setup: To solve these problems, we will assume that the volume of the container does not change. Then according to the ideal gas law, the pressure in the container is directly proportional to the Kelvin temperap T ture of the gas in the container, that is, new 5 new . pold Told T Solve: (a) The pressure in the desert pdesert 5 desert pstd Tstd 323 K 5 (10 N/m 2 ) 5 1. 18 3 105 N/m 2 . 273 K (b) The pressure in the Antarctic night 203 K 5 p Antarctic 5 (10 N/m 2 ) 5 7 . 43 3 10 4 N/ m 2 . 273 K (c) If the container walls were not rigid, they would bend outward in a convex shape when placed in the desert shade. In this case, the internal pressure would greatly exceed the surrounding atmospheric pressure.

or pdesert 5

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When placed in an Antarctic night, the container walls would bend inward in a concave shape. In this case, the surrounding atmospheric pressure would greatly exceed the pressure inside the container. Review: The same can be observed in your own room by putting a balloon on top of a large jar and watching it grow or shrink as the temperature changes. 42. Setup: To calculate speed, we will assume the kinetic 1 energy of all atoms in air are the same, and K 5 mv 2 , 2 then we can solve that the average speed of a particle 1 v . m Solve: (a) If oxygen molecules are 16 times as massive as hydrogen, and carbon-dioxide molecules are 22 times as massive, then carbon dioxide is 22/16 5 1.38 times as massive as oxygen. This means they will 1 5 0. 85 times as fast. (b) Note that as travel 1 . 38 temperature increases, both molecules gain the same kinetic energy so this ratio is fixed, as long as the gas is in equilibrium. Review: A much longer method to find this is to calculate the speeds of the molecules, knowing that at room temperature, hydrogen atoms move around 2,000 m/s. The result will be the same but a lot more work. 43. Setup: The average speed of a molecule at temperature T is given in Working It Out 9.1 as v 5 3kT / m 5 0. 157 T / A km/s where A is its atomic mass. Using Table 9.1, the average gas temperature on Earth is 288 K, and on Mars is 210 K. CO2 has a molecular weight of 12 1 2 3 16 5 44. Solve: On Earth, the average speed is 0. 157 288 / 44 5 0. 402 km/s and on Mars is 0. 157 210 / 44 5 0. 343 km/s. Table 8.1 lists the escape velocities of Earth and Mars as 11.2 and 5.0 km/s, respectively. The Earth, therefore, holds onto its gas much better than Mars. Review: Working It Out 9.1 suggests that if the molecular speed is less than one-sixth that of the escape velocity, the planet will hold onto its atmosphere over its lifetime. For Earth and Mars, these are 1.9 and 0.83 km/s, respectively. These are both higher than the average molecular speed of carbon dioxide on each planet, explaining why each planet has held on to some of its gas. But because the difference is much less for Mars than for Earth, we see why Mars has lost most of it whereas the Earth has lost much less.

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44. Setup: 6.4 millibars 5 0.0064 bars. Solve: Using the figure (we are on the far left side), we see that we could have to climb about 36 km above sea level to reach such low pressures. Review: This is about four times higher than Mount Everest, which shows just how thin the atmosphere is. 45. Setup: The atmospheric pressure on Venus is 92 bars. Solve: At 10 m per bar, you would have to dive 920 m below the surface of the ocean to achieve the same pressure you would feel at Venus’s surface. Review: 920 m is about half a mile. This is about 33 percent deeper than any person has dived using a specially designed hard-shell suit. Using the Web 46. Answers will vary. Note that at the South Pole, the hole is generally largest in late September to early October. The ozone is depleted over winter by strong winds blowing pollutants and ice crystals into the upper atmosphere, and this hole becomes larger by the springtime Sun. A good answer will note whether recent holes are larger or smaller than previous ones and whether there is a general trend in the hole sizes with time. 47. (a) Planet Four data came from the HIRISE camera on the Mars Reconnaissance Orbiter. The project’s goals are to identify and measure features on the surface of Mars. This is a slow and time-consuming process, and thus, it is aptly suited to crowd sourcing because many hands make light work. (b) The MAVEN mission’s goal is to explore Mars’s upper atmosphere, ionosphere, and interaction with the sun and solar wind. This is an orbiter mission. MAVEN contains many instruments, including the Solar Wind Ion Analyzer, Solar Wind Electron Analyzer, Solar Energetic Particle, SupraThermal and Thermal Ion Composition, Langmuir Probe and Waves, and Magnetometer. The mission will give scientists a full profile of the top of the Martian atmosphere and allow them to determine the role that the loss of atmospheric gas to space played in changing the Martian climate through time. 48. (a) Answers will vary. A good answer will address the state of Arctic sea ice and how it compares with previous years (currently, it is melting more and faster than in recent history). You should also comment on the state of Antarctic ice and how changes in ice levels affect the planet’s albedo and temperature of the Earth. (b) In the last few years, the surface air temperature has bounced up and down around 0.1 degree from its

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average. The average is fairly constant. The 5-year mean smooths over the yearly “bumps and wiggles” to show average trends. On average, it is about 0.8°C warmer today than in 1880. (c) Answers will vary for much of this. Mauna Loa is a good site for measuring carbon dioxide because the site is about 3.4 km above sea level and, therefore, receives typical wind patterns but is unaffected by pollution. 49. (a) In 1896, the first calculation of global warming from human emissions of carbon dioxide was reported. It seems other gases were not considered until the mid1970s. Scientific opinion started to converge in 1977 on global warming as a climate risk. Mars and Venus provided test beds for extremes of greenhouse gases by studying where the planet freezes or boils because of the amount of such gases present. Aerosols make the air “dirty,” which reflects light back into space before it hits the Earth, thereby cooling the planet down. (b) From the report: “Anthropogenic greenhouse gas emissions have increased since the pre-industrial era, driven largely by economic and population growth, and are now higher than ever. This has led to atmospheric concentrations of carbon dioxide, methane, and nitrous oxide that are unprecedented in at least the last 800,000 years. Their effects, together with those of other anthropogenic drivers, have been detected throughout the climate system and are extremely likely to have been the dominant cause of the observed global warming.” The report continues, “In recent decades, changes in climate have caused impact on natural and human systems on all continents and across the oceans. Impacts are due to observed climate change, irrespective of its cause, indicating the sensitivity of natural and human systems to changing climate.” Polar regions have been impacted in terms of glaciers, snow, permafrost, rivers, lakes, costal erosion, rising sea level, terrestrial ecosystems, marine wildlife, and livelihood/ economics of human activity. Past and present measurements help us to understand climate activity and allows us to model climate change in the future. (c) Answers will vary. Students will do a trial run of global climate simulation software, to learn what contributes to global warming and cooling. The instructor may wish to provide a more specific assignment using the program and the Earth Exploration Tool. 50. (a) Answers will vary. In most Mars movies, people are able to breathe, gravity and pressure are equal to those of Earth, and there is water available. (b) Everything is wrong with the end scene of Total Recall. Water vapor could not cover the planet in just 2 to 3 minutes, and it would not create an oxygen-rich environment.

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74 ◆ Chapter 9 Atmospheres of the Terrestrial Planets Exploration 1. I naïvely expected the water level to rise. 2. The water level did not rise. 3. As the Artic sheet melts, the ocean levels should not rise. 4. I expected the water level to rise.

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5. The water level did indeed rise. 6. When surface ice melts, the ocean levels will rise. 7. Most population centers are near the coasts, so sea rising will dramatically affect a large amount of the global population via flooding.

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CHAPTER 10

Worlds of Gas and Liquid—the Giant Planets INSTRUCTOR’S NOTES Chapter 10 covers the giant planets. Major topics include

▶ the common characteristics of Jovian planets ▶ atmospheric circulation, turbulence, and weather ▶ interior composition and structure ▶ magnetic fields ▶ exoplanets and planetary migration Much as with the terrestrial planets (Chapter 9), I use “comparative planetology” to teach the Giant (or “Jovian”) planets. I discuss Jupiter in some detail and then use images and data for the other three planets to deduce how they differ. As an example, I use optical and infrared images of Jupiter to show that the colorful regions (belts) are warmer than the whiter ones (zones), indicating that we are seeing down deeper into the atmosphere in those belts. I then ask why we do not see similar structure on Saturn. (We do not see such obvious structure because its atmosphere is much thicker and opaque.) What does this imply about Uranus and Neptune, given their surface features? If you take this broader brushstroke approach, you may want to remind your students that the names and chemical compositions of different atmospheric layers, the names of planetary rings, and the inclinations of the magnetic fields are less important than knowing why the atmospheres are differentiated, why the planets all have rings, and how magnetic fields influence the planets and their moons. I like to perform a demonstration in class of turbulent fluid flow that mimics Jupiter’s banding. It requires a video camera hooked up to a projector, a good quality food processor with its cutting blade, and a bag of honey roasted peanuts (or peanuts that still have the thin paper skins on them). Many students do not know that they can make delicious peanut butter this way, but more on topic, a few minutes after turning on the processor, the peanuts will liquefy, and the path of the blade through the viscous fluid leaves turbulent eddies that can be seen in the motion of the darker brown specks. These look very similar to the images of turbulence in Figure 10.6. There are two good opportunities to discuss planet migration. First, one can motivate it by comparing the high methane content of Uranus and Neptune with the primordial

content of Jupiter and Saturn, and also by noting that at Neptune’s distance, planet formation timescales are longer than the age of the Solar System. The Origins section of this text gives an alternative entry point. You might also choose to wait until Chapter 12 because planet migration comes from a planet clearing out the space around it, which is the distinguishing feature between planets and dwarf planets. Because our students still remember Pluto being demoted during their childhood, they all want to know why that happened, and I find that migration works well in that story. Either way, I recommend the excellent videos of planet migration at http://jila.colorado.edu/~pja/planet_migration.html.

DISCUSSION POINTS

▶ The atmospheres of Jupiter and Saturn are closer to being

primordial than those of the other planets. Discuss the connection between the mass of a planet and the composition of its atmosphere and how this relationship can be understood. ▶ Giant planets rotate faster than terrestrial planets and have systems with rings and moons. Have students discuss the differences between giant and terrestrial planets in context of what they have learned in this chapter and what they have previously learned about planet formation and angular momentum. ▶ Ask students if they expect the interior temperatures of the giant planets to be close to the equilibrium temperatures as calculated in Working It Out 5.4? If not, why? Do the giant planets have a greenhouse effect? ▶ Based on our four giant planets, what characteristics do we expect giant extrasolar planets to have? Rings? Moons? Turbulent and banded atmospheres? Storms? How might we observe these? ▶ The temperatures and pressures of the interiors of the giant planets create interesting phenomena. For example, the metallic nature of the liquefied hydrogen is not something easily produced in our laboratories. Discuss the implications of the existence of these phenomena and the scientific creativity needed to create theories for these and many other not-before-seen phenomena in the universe. 75

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76 ◆ Chapter 10 Worlds of Gas and Liquid—the Giant Planets

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (a) Uranus and Neptune are ice giants. 2. (b) The color of an atmosphere depends on how the particular elements that make it up absorb and reflect light. 3. (b) If a planet gives off more energy than it receives, it must have a source of its own. 4. (c) The radiation belts are created from charged particles being dragged around the planet due to its rotation. 5. a, d, b, c Reading Astronomy News 1. The Great Red Spot has only been seen since the invention of the telescope, which was a little over 400 years ago. 2. The size of the GRS is measured by comparing its dimensions with that of Jupiter. 3. Jupiter’s cloud tops are opaque, making it hard to see below them. 4. Dropping a probe into the GRS could be hard for a few reasons, including the enormous pressure and very high wind speeds. 5. Answers will vary. Test Your Understanding 1. a, c, f, e, b, d. See Figure 9.7, which shows the convection cycle on Earth. 2. (b) The state of matter (solid, liquid, gas) depends not only on the temperature but on the pressure as well. For example, at low pressures, water at room temperature can boil (an easy experiment your professor may show you if he or she has a vacuum bell jar) and at very high pressures, even ultrahot materials can stay liquid or solid. 60 sec km 3. (b) 26 min 3 3 30 5 46 , 800 km . Half of min s this is about 23,000 km. 4. (a) Uranus’s motion did not follow the path predicted by Newtonian and Keplerian physics, suggesting another large mass was tugging on it. 5. (c) Uranus’s seasons are most extreme because the inclination of its rotation axis is the highest. 6. (b) The giant planets’ magnetic fields are very strong and extend far into outer space. However, at cloud tops, they are as weak as the field on Earth, and as shown in Figure 10.12, the fields do not necessarily pass through

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the planet centers nor are they necessarily aligned with the rotation axes. 7. (b) Note that passing in front of the Sun is still an occultation because the Sun is a star. However, we usually refer to this as a transit. 8. (a) The timing of an occultation gives us the size of the planet. 9. (d) Jupiter and Saturn are composed almost entirely of hydrogen and helium, similar to the Sun. 10. (d) Temperature changes with altitude in cloud layers. 11. (b) The Great Red Spot was first observed by Hooke and Cassini in 1664 to 1665 and has been raging continuously since. 12. (a) This is why they are called Ice Giants. 13. c and d. Resonant orbits and conservation of angular momentum through accretion can cause planetary migration. 14. (b) Giant planets rotate in about the same number of hours as an Earth day, but they are much larger, meaning they have much higher rotational velocities, which yield much faster zonal winds. 15. (d) “Hot Jupiters” are giant planets found very close to their central stars. Thinking about the Concepts 16. Several differences between terrestrial and giant planets are summarized in the following table: Terrestrial

Giant

• Small worlds with solid surfaces

• Large worlds with no solid surfaces

• Tightly spaced within 2 AU of the Sun

• Spread out beyond 5 AU from the Sun

• Dense rock and metal composition

• Lightweight gaseous composition

• Thin secondary atmospheres

• Thick primary atmospheres

• Weak or no magnetic fields

• Powerful magnetic fields

• Few moons and no ring systems

• Numerous moons and rings

17. The composition of the solar nebula was approximately 98 percent hydrogen and helium with only 2 percent heavy elements. Jupiter’s protoplanetary seed grew very quickly (less than10 million years) beyond the “snowline,” allowing it to capture and hold the hydrogen and helium that prevailed in the solar nebula. Earth, however, was too small to retain hydrogen and helium. As such, Earth contains mostly the heavier elements that were present in the solar nebula within the “snowline.”

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18. The first answer is the size of a planet. We know how fast the planet moves and its distance from us, so the duration of an occultation gives us its physical size. However, we can also watch the variation of starlight as the occultation takes place, and study the presence and properties of the planet’s atmosphere. 19. Rapid rotation, which also produces very strong Coriolis forces. 20. Uranus is discovered → Hypothesis: We can predict its orbit → Prediction: Based on its distance from the Sun, we predict Uranus’s orbit → Test: Uranus’s orbit does not match predictions → FAIL → New hypothesis: There might be something pulling on Uranus → New prediction: There might be another planet → New test: Find the planet → Success: We found Neptune. 21. Giant planets rotate very rapidly, and rapidly rotating spheres tend to expand along the equator and contract along the poles, making the planet oblate, or “squished.” 22. Variations in color are because we are seeing different depths into the atmosphere. For example, the white clouds are high in the atmosphere and highly reflective; thus, we are basically seeing reflected sunlight. Colorful regions are those for which we see deeper into the atmosphere and effectively see “smog,” that is, chemical pollutants (e.g., sulfur, organic compounds) that have caused the gases and ices to take on reddish colors. Methane gas, abundant in the atmospheres of Uranus and Neptune, strongly absorbs the longer red wavelengths of visible light. Therefore, we see only the shorter blue and green wavelengths of light reflected by these planets. The absence of clouds in the atmospheres also prevents the strong color from being washed out by scattering effects. 23. All the giant planets have rotation axes that are inclined, and a planet with an inclined axis will experience seasons. Of these, Jupiter’s inclination is very small, so its seasonal effect will be very low. Saturn’s and Neptune’s axes are similar to Earth’s, so these planets’ seasons will be similar. On the other hand, Uranus’s inclination places its axis in its orbital plane, so its seasons are very severe. 24. The extraordinarily high pressure in Jupiter’s interior prevents individual atoms and molecules from moving around randomly in such a confined space. Instead, atoms and molecules are forced into such close proximity that they remain loosely bound to each other even at high temperatures. Random particle motions in a system are characteristic of a gaseous state. Loose bonding between particles in a system is characteristic of a liquid state.

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25. We measure the positions of clouds as they move across the surface of the planet, and compare those speeds with each other (e.g., in latitude) and also with our assumed rotation speed of the interior of the planet, based on its magnetic fields. 26. The Great Red Spot is a massive storm that has been raging on Jupiter for at least 400 years. 27. The extra energy comes from ongoing gravitational contraction and differentiation in the interiors of Jupiter, Saturn, and Neptune. 28. The bright radio source emanating from Jupiter is synchrotron radiation. It is emitted by the energetic motion of charged particles trapped inside Jupiter’s powerful magnetic field. 29. Auroras on the giant planets are caused by the same mechanism as on Earth. Charged particles from the solar wind become trapped in the planet’s magnetic fields and spiral back and forth between the poles. Occasionally they “spill” down over the magnetic trap and onto the planet’s upper atmosphere, where they collide with atmospheric particles, releasing energy and creating the aurora light show. In addition, Jupiter’s moon Io connects to Jupiter by a magnetic “flux tube” that creates an additional aurora. 30. Migration of outer planets could leave inner ones alone completely, or could completely disrupt their orbits and send them flying out into space. Applying the Concepts 31. Setup: Read the figure caption to learn the difference between the images. Solve: The top panel shows the planets as they appear scaled to Jupiter’s size. The bottom panel shows how they would appear to an observer on Earth in relation to the size of Jupiter on the page. Review: Note that Uranus and Neptune are so far away that from Earth they only appear like small blue dots, even though you can see great detail on Jupiter and Saturn. This is why we had to send satellites like Voyager 1 and 2 to take close up images of these ice giants. 32. Setup: Remember that Neptune is very distant and hard to observe from Earth and has only been visited one time by the Voyager 2 satellite as it flew past. Solve: It is most likely that we have little information about the winds in these regions and a lack of line indicates that we do not know the behavior. If there were zero velocity, the line would be vertical in these regions. Review: It is important to recognize that even though we know much about our Solar System, there is still much to be discovered.

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78 ◆ Chapter 10 Worlds of Gas and Liquid—the Giant Planets 33. Setup: When a gas is compressed to very high densities and pressures, the electrons are stripped from the nuclei, and the resulting plasma behaves like a liquid metal. Solve: The interiors of Jupiter and Saturn reach at least 10,000 K, hot enough to ionize the electrons off of the hydrogen atoms. Review: Note that Uranus and Neptune do not have enough internal gravity to reach these high temperatures. 34. Setup: Using a ruler, I find that the distance from the edge of Jupiter to the torus is about 2.5 times the diameter of Jupiter. Accounting for the half of Jupiter that is its own radius, this is about three Jupiter diameters or six Jupiter radii. Solve: Jupiter’s radius is about 7 3 104 km. At six times Jupiter’s radius, the torus is about 6 · 7 3 104 5 4.2 3 105 km. Review: Online, we find that the torus is about the same distance from Jupiter as the moon Io itself, which is 4.2 3 105 km. 35. Setup: The distance-luminosity relationship indicates that the intensity of light decreases with the square of 1 1 . We will use this latter the distance b  2 or d  d b equation with proportional reasoning. Solve: The question asks us how far we would have to be from the Sun for it to appear as bright as the full Moon, that is, 400,000 times fainter than it does now, which means brightness changes by a factor 1 . Inserting this into our distanceof b5 400 , 000 1 5 632 or the 1 / 400 , 000 distance is 632 times farther than we are currently. Because we are 1 AU from the Sun, we would have to move to 632 AU to see the Sun as bright as the full Moon. Review: Note that this distance is about 21 times farther than Neptune’s orbit of 30 AU. The solution may be checked by directly computing a ratio using our distance-brightness formula d 5 bsun or dsun b

brightness equation, d 

bsun 400 , 000 5 1 AU 5 632 AU. b 1 36. Setup: To find the size of Uranus, we must compute distance based on speed and time, that is, d 5 vt. However, note that we are given time in minutes, not seconds. d 5 dsun

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60 s 5 2 , 220 s, so the total occultation time min of 37 minutes, 2 seconds is 2,220 1 2 5 2,222 s. km Solve: a) d 5 vt 5 23 3 2 , 222 s 5 51,100 km. s (b) If Uranus had not passed directly in front of the star, the occultation time would be shorter than expected and would have only given us a lower limit to the planet’s actual size. Review: We are told throughout this text that Uranus is a giant planet. The size we computed, about 50,000 kilometers, certainly is much larger than Earth’s diameter of about 12,000 kilometers. 37. Setup: Oblateness is defined as O 5 (Req 2 Rpole)/Req. Solve: Given the oblateness and equatorial radius of Jupiter, we solve Rpole 5 Req(1 2 O) 5 66,852 km. Review: The percentage difference between the polar and equatorial radii is (71,500 2 66,852)/66,852 5 0.069 or 6.9 percent. So we see that the oblateness and percentage difference between the radii are nearly equal; hence a very oblate planet will have a large difference between the equatorial and polar radii, as expected from the name “oblate.” 38. Setup: Nitrogen has an atomic mass of 14, sulfur is 32, and hydrogen is 1. Solve: There is 1 N, 1 S, and 5 H, so the total molecular mass is 14 1 5 3 1 1 32 5 51 atomic mass units. Review: Ammonium hydrosulfide, as a molecule, has about the same mass as Vanadium and just a little less than an iron atom. 39. Setup: Because volume scales with r3, we can compare the volumes of the two planets and assume that this is the number of Earths that can fit inside. If we imagine stacking golf balls in a box, there will be space between each ball, so in reality the number we get will be larger than the actual number of Earths that could fit inside. M For part (b), remember density  5 . V 37 min 3

69 , 900 510. 97 times 6 , 370 larger than Earth. So, using our volume-radius relationship, the volume will be V  r3 5 10.973 5 1,320 times larger. This suggests that, at most, 1,320 Earths could fit inside Jupiter. (b) To compare the relative density of Jupiter to Earth, use the density equation, in which Jupiter’s mass is 318 times larger than Earth, and its volume is 1,320 times larger than Earth’s,

Solve: (a) Jupiter’s size is

then  5

M 318 5 5 0. 24 times that of Earth. In V 1, 320

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Chapter 10

other words, even though Jupiter is more massive and bigger than Earth, its density is about one-quarter that of Earth. Review: We can recompute the above values using direct ratios rather than proportional reasoning: V J r 3J 69 , 9003 (a) 5 5 51, 320. (b) V r 3 6 , 3703

J 

M J / VJ M / V

M J V M V J

318 M V 5 0.. 24. M 1, 320 V

40. Setup: Referring back to Chapter 2, recall that the maximum height of the Sun at a pole is equal to the planet’s inclination; for example, in the summer at the North Pole, the Sun is 23.5 degrees above the horizon. Solve: Using this logic, in the summer, the Sun on Uranus would be 98 degrees above the horizon, or 8 degrees from the zenith. Review: Eight degrees from the zenith is very high in the sky, and because the pole would point toward the Sun continuously during the planet’s rotation, that means it would indeed be very warm on the pole compared with the other side of the planet, which is in permanent night time. Hence, our argument that the seasons will be very severe. 41. Setup: The cloud has moved from 122 degrees to 118 degrees, or a total of 4 degrees in 10 hours. To compute the actual speed, we need to know what distance corresponds to 1 degree, which we can do two ways. First, we can use the small-angle formula, where the physical r , or compute size s covered by an angle  is s 5 57. 3° distance as a ratio of the circumference. To compute speed, remember distance equals rate times time, or d 5 vt. Solve: Because the cloud has moved from 122 degrees W to 118 degrees W, it has traveled eastward. Winds blowing from the west to the east are westerly winds. Using the small-angle formula, the physical distance covered by 4 degrees on the surface of Jupiter is r 71, 500 km 3 4° s5 5 5 4 , 990 km. This was 57. 3° 57. 3° d 4 , 990 km 5 traveled in 10 hours, so the rate v 5 5 40 h t 499 km/h. Review: Let’s check by doing a ratio based on circumferences. The circumference of Jupiter is C 5 2pr 5 2p 3 71,500 km 5 449,250 km. There are 360 degrees in a circle, so 4 degrees is 4/360 5 1/90 of a circle. Then the distance covered is one-ninetieth of C,

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449 , 250 km 5 4 , 990 km , which is the same as 90 above. 42. Setup: The power (or flux) given off by a blackbody is related to temperature by the Stefan-Boltzmann law, F  T4; so we can compute the ratio of energy absorbed to energy radiated by comparing the ratio of T4 F temperatures, that is, in 5 4in . Fout Tout Solve: Using our ratio above, we find that Fin 954 5 51. 8, or Saturn gives off 1.8 times (80 perFout 82 4 cent) more energy than it receives. Review: How is this possible? Gravitational contraction of the planet is continuing, so the planet’s energy output includes not just solar warming but the planet’s own internal energy. 43. Setup: The power (or flux) given off by a blackbody is related to temperature by the Stefan-Boltzmann law, F  T4, so we can compute the ratio of energy absorbed to energy radiated by comparing the ratio of tempera4 tures, that is, Fin 5 T in . 4 Fout Tout Solve: Using our ratio above but solving for 1/4

F  Tout 5 Tin  out  . Because Neptune gives off 2.6 times F  in

more energy than it receives, we find that Tout 5 47 K(2.6)1/4 5 60 K. Review: For the planet to give off more energy than it receives, it must be warmer than its predicted temperature based just on solar heating, as we found. Also note that Neptune’s temperature is 58 K, very close to what we found. 44. Setup: Looking at the panels in Figure 10.8, note that the rate of increase of pressure with depth is the change in pressure compared with the change in depth, that is, the ratio of the two. However, because each panel gives a different range of depths, it is important to compare “apples with apples,” that is, compute and compare the actual ratios rather than simply inspect them visually. Solve: Starting with panel (a), note that the pressure changes from 0.1 to 10 atmospheres as one travels from 0 to 2100 kilometers below the surface. So the rate of change for Jupiter is about (10 3 0. 1) atm 5 0. 1 atm/km. For Saturn (panel b), 100 km the pressure changes from 0.1 to 10 atmospheres as one travels from 0 to 240 kilometers, for a rate of

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80 ◆ Chapter 10 Worlds of Gas and Liquid—the Giant Planets (10 2 0. 1) atm 5 0. 04 atm/km, that is, 240 km pressure increases much more rapidly on Jupiter than on Saturn. Repeating the same exercise for Uranus (panel c) the pressure increases to 200 atmospheres at a depth of 150 kilometers, or a rate of change of 1.33 atmospheres per kilometer, whereas for Neptune (panel d) the pressure increases by about 500 atmospheres at a depth of about 125 kilometers, or a rate of change of about 4 atmospheres per kilometer. Thus, the pressure change is greatest for Neptune. Based on these values, Saturn has the lowest rate of change, and Neptune has the largest. Review: Are these results surprising? The ice giants are much denser than the gas giants and thus, should have a larger increase of pressure with depth, as we found. Using this same logic, Saturn is less dense (overall) than Jupiter and shows the same trend. 45. Setup: Look at Figure 10.8 and read off the temperature from the red trace at an altitude of 100 kilometers. It may help to use a ruler to draw lines from 100 kilometers to the trace and then from the trace to the temperature scale at the bottom. Solve: Jupiter: about 210 K; Saturn: about 100 K; Uranus: about 90 K; Neptune: about 100 K. Review: Are these results surprising? One might expect that because the ice giants are so far from the Sun, their temperatures would be much colder than in the closer gas giants. However, remember that for gas, temperature and pressure are related by the ideal gas law, and the ice giants’ atmospheres are under significantly higher pressure. change of

Using the Web 46. Answers will vary. The answer will include whether Cassini is still active, what was recently discovered about Saturn or its rings or moons, and why this discovery is important. 47. Answers will vary. The answer will include recent images from Cassini from http://ciclops.org, including the wavelength of observation, whether the picture is true color or not (and why), and why the pictures are significant. 48. (a) From the mission pages, Juno will (1) determine how much water is in Jupiter’s atmosphere, which helps determine which planet formation theory is correct (or if new theories are needed); (2) look deep into Jupiter’s atmosphere to measure composition, temperature, cloud motions, and other properties;

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(3) map Jupiter’s magnetic and gravity fields, revealing the planet’s deep structure; and (4) explore and study Jupiter’s magnetosphere near the planet’s poles, especially the auroras—Jupiter’s northern and southern lights—providing new insights about how the planet’s enormous magnetic force field affects its atmosphere. (b) At the end of the mission, Juno will deorbit into Jupiter. (c) By looping, Juno gathers extra speed (sometimes called a gravitational slingshot) to reach the planet Jupiter faster without having to carry a lot of heavy rocket fuel. (d) Galileo made the first observations of Jupiter and its moons, and these were pivotal discoveries in changing the notion of our place in the universe from geo to heliocentric and in ushering in the modern age of science. Certainly, Galileo deserves this kind of public (and interplanetary) acclaim. 49. (a) Both spacecraft are entering the heliopause; the actual distances will vary. The Voyager images of Uranus and Neptune provided the first close up observations of the planets and their moons. (b) Answers will vary. 50. Answers will vary. Exploration 1. Neptune is an ice giant and significantly larger than any of the terrestrial planets. For it to have completed a single orbit in 17.6 hours (about three-quarters of an Earth day) means that it is rotating very quickly! Because the Great Dark Spot and this small storm rotated different amounts within the same time period, it means that the cloud tops experience “differential rotation,” that is, the speed of clouds change with latitude. 2. To measure the approximate radius at which the small storm travels, I laid down a ruler along Neptune as shown in the figure below. On my copy of the text, the segment from the limb to the line through the center of the planet is about 24 mm. The distance to be measured is shown as the horizontal segment below. 3. The circumference C 5 2pr 5 2 3 3.14 3 24 mm 5 151 mm. 4. Using a similar method as described in question 2, I find the distance the storm has moved from the first to second image is about 13 mm. So the total distance it traveled around the planet is about 151 1 13 5 164 mm. T C or solving for the rotation period 5. 5 t D C 151 mm T 5 t 517. 6 h 516. 2 h. This is less than D 164 mm 17.6 hours, as suggested in the hint, because in one

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Chapter 10

period the storm would have moved one time around the planet. That it moved more than one revolution means that the elapsed time was longer than the actual period of rotation. 6. Repeating the above exercise, I find that the limbto-center radius is about 21 mm, and the storm moved about 7 mm. The circumference of this circle is 132 mm and, in this case, the storm moved 132 2 7 5 125 mm. Using our period equation, C 132 mm T 5 t 517. 6 h 518. 6 h. D 125 mm 7. These two periods differ by about 2.5 hours (16.2 to 18.6 hours) or about 13 percent. So they are close but certainly not the same.

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8. These findings tell us that either (a) the center of the planet completes a revolution faster than the regions nearer to the poles or (b) this method is subject to large enough errors that we cannot really trust the results. 9. We can use periodic radio bursts caused by rotation of the planet’s magnetic fields to determine the rotation rate. We could also improve the method done above by fitting circles to the orbits of the storms rather than straight lines, as shown below (Neptune is a threedimensional object, after all).

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CHAPTER 11

Planetary Moons and Rings INSTRUCTOR’S NOTES Chapter 11 covers rings and moons of planets. Major topics include

▶ the moons of our solar system ▶ formation and behavior of regular and irregular moons ▶ comparative investigation of the geologic activity and history of moons

▶ the characteristics and structure of planetary rings If you teach “comparative planetology” in Chapter 9, the discussion of moons in this chapter can be flipped by showing images of many of the moons and asking students to deduce the current geological activity and the geological history of each rather than using the text to give those answers to the students. I also like to have students rate each moon’s chance of hosting life. I used to refer to Mimas as “The Death Star” and call potentially habitable moons “possible Endors,” but I’m finding that Star Wars (episodes IV and VI) is less and less in current students’ zeitgeist, so use these references with care. Io always seems to be a fascinating surprise to students. A plethora of eye-popping images are available at http:// photojournal.jpl.nasa.gov/target/Io, showing how quickly this moon resurfaces itself. An easy demonstration of the tidal heating process of Io can be performed with a thick paperclip and glass of cold water. Open the paperclip, flex it back and forth a few dozen times, and then plunge it into the water. By flexing it, you can easily heat the metal to many hundreds of degrees, which will result in a loud hiss when the paperclip is plunged into the water. Jupiter’s tidal forces flex Io’s solid surface by up to 300 feet; by contrast, the Sun and Moon only move water around on Earth by about 60 feet. No wonder Io is so volcanically active! If you do not assign problem 35, I also recommend discussing how Io has turned itself inside out over a dozen times in its lifetime. This is a very exciting time in the field of habitable moons. Obviously Europa has been a target of interest for quite some time, and currently, an extensive flyby mission, named the Europa Multiple Flyby Mission (formerly Europa Clipper), has received extensive funding to move into development stages. Meanwhile, since the last edition of this textbook was released, astronomers have concluded

that Saturn’s moon Enceladus also has a subsurface ocean of liquid water. An interesting question for students is whether we should try to land on either moon and enter its ocean. On the one hand, what could be more exciting than turning on the lights and seeing a possible extraterrestrial ecosystem. On the other hand, bacteria are such tenacious survivors that we can be almost guaranteed that we would contaminate that environment as soon as our probe entered it. The ethical question—a real life version of Star Trek’s Prime Directive is not just an academic one because today’s students will be tomorrow’s policy makers and will have to make judgments on important scientific issues such as these.

DISCUSSION POINTS

▶ A hundred years ago, Mars was suspected to be the most

likely place to host extraterrestrial life in the solar system. In the last few decades, it was Europa, and in the last few years, Enceladus took that distinction. Why does the object of our focus keep changing? Is this a strength or weakness of science? ▶ Discuss what we know about Enceladus that has led scientists to suspect it is the most likely object in the solar system to host extraterrestrial life. ▶ Given the range of extremophiles on Earth (that is, bacteria in deep ocean geothermal vents or in ice lakes frozen for millions of years), discuss the likelihood of the presence of life on moons, such as Europa or Titan. ▶ All four giant planets sport ring systems, but none of them are alike. Compare and contrast what we know about each ring system and speculate on what kinds of rings we would expect to see in giant exoplanets. ▶ One of the great unsolved mysteries in planetary science is the diversity of satellites in each of the giant planetary systems. Why did some moons in the same system develop active geology whereas others did not? Why did some moons maintain active geology whereas others became inactive early in their histories? ▶ Each of the four Galilean moons of Jupiter display different levels of geologic activity. Discuss why each one has the level it does in context of its distance from Jupiter. 83

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84 ◆ Chapter 11 Planetary Moons and Rings

▶ Neptune and Uranus have both fewer moons and smaller moons than Jupiter and Saturn. Does this fit into our currently solar system formation scenario?

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. a, b. Note that many moons are about the size of Mercury. Answer (d) would be right if it said most known planets, but it said all. 2. a, d. 3. c, b, a, d. These have been ranked in order from most to least geologically active. 4. (a) Ring material is temporary so long lasting rings probably have a source. 5. (b) Rings are brightest when looking back toward the central star, in the shadow of the planet. Reading Astronomy News 1. This article suggests that either a moon is forming or being destroyed in the A ring. Either way, it links moons with rings because the formation/destruction of one is found inside the other. 2. A moon this small is not likely to be geologically active because of low internal energy. But . . . who knows!? 3. Although there is a general trend that the mass of a moon increases with orbital radius, this is not a rule. For example, Iapetus has less mass than Titan but is three times further away. 4. Saturn’s moons are named after Greco-Roman mythological characters and giants from other mythologies. Peggy will surely not last. 5. As of August, 2015, there is no news of Cassini’s return to Peggy. Test Your Understanding 1. (a) This is also the basis of comparative planetology. 2. (b) The proximity of Io and Europa to Jupiter means these moons experience tidal stresses that have continually heated the interiors enough for both moons to stay geologically active. 3. (d) Note that by life, we mean “life as we know it.” 4. (d) Tidal stresses in Io’s orbit around Jupiter deform the moon’s radius by about 100 meters per orbit. 5. (a) Of the choices given, the narrow gaps and oscillating edges seen in rings are due to the gravitational effects of small moons. 6. a, d. These are the best choices. We do not know how recently the rings formed.

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7. (c) Physics was able to provide an explanation of this ring even though it was not obvious at first. 8. (a) Moons are expected to form out of the same material as the host planet; thus, a retrograde moon is likely to have been captured. 9. (b) Rings are best observed when lit from behind; thus, being behind the planet looking toward the Sun. 10. (b) There was more material in the outer solar system from which the planets and their moons could form. 11. (a) According to the IAU definition, a moon is the satellite of another body. 12. (c) Although (a) and (b) are ways to date objects, one has to land on the Moon to perform them, which only leaves us the study of surface features from imaging. 13. (d) More precisely, it is the tidal stretching of Io by Jupiter’s gravity that keeps Io’s interior molten hot. 14. (c) If a moon was formerly active, it would have covered some of its older craters. This does not preclude those regions receiving a few new craters, but because most of the cratering happened early in the solar system’s history, one expects few new craters in regions that were resurfaced. 15. (c) The volcanic ejections inject new particles into the ring system. Thinking about the Concepts 16. Tidal stresses from Jupiter continuously flex Io by many tens of meters per orbit and keep the interior mantle molten. 17. Like terrestrial volcanism, cryovolcanism involves the ejection of liquefied materials at high pressures. Cryovolcanism is volcanic activity that occurs with materials that are liquid at very low temperatures. For example, Enceladus exhibits cryovolcanism with liquid water substituting for magma. Triton exhibits cryovolcanism with liquid nitrogen substituting for magma. Io exhibits it through constant volcanic eruptions of magma. 18. Europa’s surface is spectroscopically dominated by water ice. The source of this ice must be liquid water in its interior. Furthermore, we see that the surface is broken into pieces that fit together like pieces of a jigsaw puzzle, just like the tectonic plates that float on Earth’s molten mantle. The absence of impact craters also indicates that the surface must be relatively young, and the material used for resurfacing the planet could come only from its interior. Finally, the tidal stresses experienced by Io are also experienced by Europa to a lesser degree. If Europa is the rock-ice mixture it is believed it to be, tidal heating could liquefy the ice in the planet’s interior.

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Chapter 11 Planetary Moons and Rings ◆

19. Methane is not expected on Titan because UV photons from the Sun photodissociate the molecule. Although copious amounts (about 1.4%) are found, there should be little to none because of this effect. 20. Titan resembles Earth with the nitrogen composition of its atmosphere and surface features that are similar to rivers and lakes. On Titan, methane appears to play the same role that water plays in the hydrologic cycle on Earth. 21. Ganymede shows grooves and ridges on its bright terrain, clearly indicating that it had tectonic activity at some point in its past. Other moons such as Tethys, Ariel, and Miranda also show evidence of past cryovolcanic or tectonic activity. Tethys has a huge canyon system that spans most of its circumference. Ariel’s surface exhibits smooth plains and complex rifts, indicating that cryovolcanism and tectonic activity operated over portions of its surface. Finally, Miranda displays enormous fault scarps, suggesting either that its history was marred by a catastrophic breakup and accretion or that differentiation stopped prematurely when the moon cooled. 22. Inner planets lack the debris that outer planets have, and therefore, there is not a source of material to form rings today. Note that this does not preclude them from having transient rings in the past! Occultations of stars by planets are the best ground-based method used to look for rings. If the star’s brightness varies just before or after the star is occulted by the planet, then the chances are that a ring of material has caused that dimming. 23. Ring arcs, such as those seen in Neptune, are denser regions of a ring, so that when viewed, one sees only the denser material and could interpret the ring as being incomplete—that is, an arc of material rather than a complete ring. 24. Planetary ring material is likely produced by the breakup of moons, asteroids, or comets that fall inside a planet’s Roche limit. It is also produced as nearby moons shed material through volcanic processes, impact events, or (in the case of Titan) atmospheric escape. 25. Gaps can be created by small moons, which clear out a lane of material as they orbit within the ring. Larger gaps are created by orbital resonances of shepherd moons. 26. Planetary rings are fairly dense so that particles constantly collide with each other and maintain a fairly tight, thin ring. Diffuse rings are sufficiently low density so that collisions are rare, allowing the particles to have noncircular orbits. As a result, diffuse rings are often thick (puffy).

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27. The discovery of kinks and knots in Saturn’s rings meant that either our understanding of rings was completely wrong or that we had an incomplete theory. Because scientific understanding is provisional, our understanding of rings was correct until we learned more about them and required a theory that incorporated more physics. Thus, our understanding grew with more observations to become a more complete explanation. 28. Particles in rings are constantly undergoing collisions, which sometimes impart enough energy to the particle that it escapes from the ring. Thus, all rings are doomed to dissipate unless there is a source feeding ring material to make up for the particle losses. However, shepherd moons on either side of a ring herd the ring particles back into position. This shepherding process keeps the ring confined to a narrow region of space. 29. The E ring of Saturn is being fed by eruptions from Enceladus and may survive indefinitely, or at least as long as Enceladus is active. 30. Answers will vary considerably. Europa is one of the most commonly cited moons because it may harbor life as we know it in the liquid water under its surface. In such a case, a probe would have to land on the moon, bury itself into the ice until it reaches the surface ocean, and return both imaging and microscopic analyses of the water. Applying the Concepts 31. Setup: To solve this problem we will use the escape 2GM , as given. Note that R the radius of the moon is given in kilometers but must be converted to meters. Solve: (a) velocity equation, v esc 5

v esc 5

26 . 67 310211 8 . 9 31022 2GM 5 5 R 1 . 82 3106

2,500 m/s or 2.5 kilometers per second (km/s) (b) 2. 5 km/s 5 2. 5 times faster The escape velocity is 1 km/s than the typical speed of gas escaping from Io’s vents. This means most of the vented material will fall back on the moon, rather than escaping into space. Review: The book explains that Io has essentially turned itself inside out a few times, meaning that the moon has expelled almost its full mass of material from its volcanic eruptions. This is only possible if its gravity is sufficient to pull all that material back onto the moon

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86 ◆ Chapter 11 Planetary Moons and Rings rather than letting it escape into space. Thus, our finding is consistent with this necessity. 32. Setup: The first equation of Working It Out 11.1 gives 3 4 2 A 3 10 A 5 5 . 9 3 10 kg by inserting the us M 5 G P2 P2 value for G. Use P in seconds and A in km. Solve: Europa has an orbital period of 86 , 400 s P 5 3. 55 d 3 5 3. 07 3105s and a semi-major d axis A 5 6.71 3 105km. Inserting these into our 5

(6 . 71 310 km) formula yields M 5 5. 9 310 5 (3 . 07 3105 s)2 1.9 3 1027 kg. Review: The reported mass of Jupiter is 1.898 3 1027 kg so we are correct! 33. Setup: Using any moon of Saturn, calculate the A3 ratio 2 . and calculate the mass shown in Working It P Out 11.1 and problem 32. 20

Solve: Titan orbits at a distance of 1.22 3 106 km 86 , 444 s 1. 35 3 106 s. and has a period of 15. 95d 3 d Now, M 5 5 . 9 31020

(1 . 22 3106 )3 5 5 . 9 31026 kg . (1 . 35 3106 )2

Review: The mass of Saturn is 5.7 3 1016 kg. 34. Setup: Linear axes increase in integer increments, logarithmic in multiples of 10. Solve: Both diagrams have linear axes. Panel (b) has axes that are 15 times larger than in panel a. Review: Note how small the orbits from panel (a) are in (b), which confirms how much more space is being shown. 4 35. Setup: Surface area is 4pr2 and volume is  R 3. 3 The relationship between the two is that (volume) 5 (surface area) 3 (thickness). The time is takes for amount something to happen is t 5 . rate Solve: (a) Using Io’s radius, we find a surface area of 4.14 3 1013 m2 and a volume of 2.51 3 1019 m3. (b) The volume of material deposited on the surface of Io each year is Vdeposit 5 surface area 3 thickness 5 (4.140 3 1013 m2)(0.003 m) 5 1.24 3 1011 m3. (c) The time t that it takes to deposit the equivalent of Io’s entire volume onto the surface is determined by dividing Io’s volume by the deposition rate

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2. 504 31019 m 3 Volume 5 5 2.0 3 Deposition rate 1. 24 31011 m 3 / yr 108 yr. Io will essentially turn inside out every 200 million years. (d) The solar system is about 4.5 billion years old, so Io has turned “inside out” about 22 to 23 times. Review: This problem shows us just how “active” Io really is. 36. Setup: According to Working It Out 11.2, tidal force t5

M planet M moon R moon

, so use this as the guide d3 to see how changes in some factors affect the tidal force. Solve: If the size of the moon Rmoon increases but its mass stays constant, the tidal force will increase. Likewise, if the moon’s size decreases, so will the tidal force. The formula shows us that if the planet’s mass increases, tidal force will increase. Review: Note that the strength of tidal force scales like the force of gravity, multiplied by the relative size of the moon to its distance from the planet. Thus, we can see that the masses and sizes of the planet and moon directly affect the tidal force. 37. Setup: Use the full equation for tidal force: M planet M moon R moon F 52G . d3 Solve: Plugging in values from the book, we find the tidal force is scales as

26 . 67 310211

5 . 97 31024 7 . 35 31022 1 . 74 3106 5 (3 . 84 3108 )3

1.80 3 1018 N. Review: Compare this with the strength of gravity between the Earth and Moon of about 2 3 1020 N, and we can see that the tidal force is much smaller. 38. Setup: Use the tidal force formula, M planet M moon R moon , where the “moon” is now F 52G d3 replaced by the astronaut’s values for mass and size. Solve: We will use a mass of 65 kg and size of 1.7 m because that’s a typical size for a person. The tidal force of the Earth on the astronaut is 26. 67 3 10211

5 . 97 3 1024 601 .7 5 [(380  63533) 3 103 ]3

2.7 3 1024 N. Review: There are about 10 N in 1 kg, so this is a total force so miniscule the astronaut would never feel it.

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Chapter 11 Planetary Moons and Rings ◆

39. Setup: If all else is held equal, then tidal force 1 scales as 3 . d Solve: The graph of tidal force will fall off much faster than that of gravity. Here is a graph of tidal force (red) versus gravity (green) for arbitrary masses and sizes. Review: Note that there is a region in which the tidal force is stronger than gravitational attraction, but it is very close to the planet. 40. Setup: A 7:6 resonance implies that the ring particles orbit Saturn seven times for every six orbits of Janus. This means that their orbital periods are 6/7 as long as Janus’s orbital period. 6 6 Solve: Pring 5 (16 h 41m ) 5 (16.6833 hours) 5 7 7 14.3 hours. Review: The orbital period of the ring particles is 14.3 hours or 14h18m. 2GM 41. Setup: Escape velocity v esc 5 . R Solve: Janus has a mass of 1.9 3 1018 kg and a radius of 8.95 3 104 m. The escape velocity will be 2 ⋅ 6. 67 3 10211 1. 9 3 1018 5 53 m/s. v esc 5 8. 95 3 10 4 Review: This is about 120 miles per hour, that is, an airplane could escape from Janus. 42. Setup: To compute the total mass of material in the ring, we need to compute the volume of the ring material. We can approximate the ring as a disk with a smaller, disk-shaped hole in the center. The volume of a disk with radius r and thickness h is V 5 pr2h; so the volume of the ring is the volume of the whole disk minus the volume of the inner material, that is, 2 v 5  h ( rout 2 rin2 ). Finally, to compute the mass of material in the ring, we substitute the volume of the ring into the well-known relationship between m mass, density, and volume  5 . Solving for mass V 2 2 gives M 5 h ( rout 2 rin ) . Solve: Inserting values given in the text, we find kg   M 5  150 3  (10 m) (2. 35 3 108 m)2 2  m  (1. 84 3 108 m)2  5 1. 01 3 1020 kg . Review: We know from the text that the total mass in the rings is small. In this case, the total mass is 0.0017 percent that of Earth.

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43. Setup: From the previous problem, the mass of the material in B ring is 1020 kg, whereas Mimas is 3.8 3 1019 kg. Solve: The ring mass is

1020 kg 5 2. 7 times 3. 8 3 1019 kg

larger than that of Mimas. Review: This is significant because it suggests that the ring was probably formed by the tidal disruption of a medium-sized moon rather than from the slow escape of particles from nearby moons. 44. Setup: To relate period and distance, use Kepler’s law, that is, P2 5 A3. 3/2

P  184 , 000  5 0. 69 , that is, the inner Solve: in 5  Pout  235, 000  particles orbit in 69 percent of the period of the outer ones. Another way to explain this is to invert the ratio, that is, the outer ones take 1.44 times longer to complete an orbit. Review: We usually think of rings as being solid bodies, but this shows us that the ring is clearly made up of material that is doing different things at different places. 45. Setup: Escape velocity scales as M / R from the formula in Working It Out 11.3. Solve: The escape velocity formula tells us that a more massive planet will have a higher escape velocity, and a smaller one will have a higher one as well because making R smaller makes 1/R larger. Knowing the escape velocity of a planet, we still need to know the radius to know the mass. Review: Why is the escape velocity larger if the planet’s size is smaller? Because gravity is stronger the closer you are to the mass. Using the Web 46. Answers will vary. The answer should report the date, time, and location of Jupiter’s four moons. Then, moving forward in time, which moons (if any) are in transit. If possible, the student will include sketches of the moons as observed in a telescope over a few nights. 47. Answers will vary. The answer will provide a few recently discovered moons of Jupiter, and the orbital details (eccentricities, inclinations, pro/retrograde directionality). The answer should discuss where the moons fit on the graph of orbits of other moons, where the moons’ names come from, and why the moons may be labeled as “provisional” (this is because the details may still be tentative because more observations might be needed).

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88 ◆ Chapter 11 Planetary Moons and Rings 48. Answers will vary. Cassini is in its extended mission and was still active as of September 2012. The aurora hiss is caused by electrons moving along magnetic field lines. Most of these come from the moon Enceladus. 49. (a) JUICE will study Ganymede, Callisto, and Europa. Project goals include understanding the conditions for planet formation and emergence of life. Currently, the project is in development stage. (b) Destination Europa is advocating for enhanced exploration of Europa. Europa Clipper has been renamed Europa Multiple Flyby Mission, or just Europa Mission; the mission is in development stages with funding from NASA and Congress. The mission will make many close flybys of Europa to study the ice shell and confirm the presence of a subsurface ocean and to study the composition and geology of the moon. 50. Answers will vary. At present, exorings and an exomoon have been proposed around J1407b, and there

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is some question whether an exomoon exists around WASP-12b and MAO-2011-BLG-262. Exploration 1. In the copy I have, the image is 50 mm in height. 2. This is probably good to 2 mm. 3. Radius is half the diameter, or 25 mm. 4. Io has a radius of 1,821 km. 1, 821 km 5. The scale is 5 72. 8 km/mm. 25 mm 6. km per mm 7. The black ring is about 6.5 mm across. 8. 6.5 mm 3 72.8 km/mm 5 473 km. Arizona is about this big. 9. The plume is about 4 mm, or 4 mm 3 72.8 km/mm 5 290 km tall. This is about the length of Pennsylvania!

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CHAPTER 12

Dwarf Planets and Small Solar System Bodies INSTRUCTOR’S NOTES Chapter 12 rounds out the solar system with everything that is not a planet or moon. Major topics include

▶ dwarf planets ▶ asteroids ▶ comets ▶ meteors and meteorites ▶ impacts Today’s students still grew up in a world in which Pluto was a planet that was taken away from them, and many still have very emotional reactions to the subject. Every semester a new hoax is posted on the Internet claiming Pluto’s planet status has been reinstated, so the topic is still very much in the minds of the public. I believe it is critical that students in this class fully understand how Pluto’s reclassification carefully follows the scientific method and shows the power of its self-correcting nature. We hear the term “belief ” and emotional appeals used quite a lot in the media and from politicians and pundits over a broad range of scientific issues, many of which are hotbutton topics that may prove difficult to discuss in the classroom. The issue of Pluto is politically benign enough that it does serve as an excellent proxy to discuss the broader questions, such as should science try to make the public happy or report the best understanding available today? How does science deal with personal bias? How should science engage with the public when a self-correction happens? Stepping from one proverbial third rail to another, this chapter is filled with fine-grained details in which many students could easily become lost if not properly guided. As I have indicated before, many students tend to equate learning with memorizing everything in a chapter. Between all the names of dwarf planets, asteroids, and comets, and all the different types and locations of each, one can easily become inundated in terminology and miss the broader message of how all these smaller bodies are a natural outcome of our solar system formation model. Although much of introductory astronomy does not lend itself to tabletop demonstrations, this chapter does. If you have access to dry ice and well-insulated gloves, an

easy and entertaining demonstration is to build your own comet. Line a very large bowl with a heavy duty trash bag, and pour in a few cups of crushed dry ice, a few spoonfuls of dirt, a squirt of corn syrup (organic material), a dash of vinegar (amino acids), a dash of rubbing alcohol (methanol), and a cup of cold water. Stir with a big spoon for a minute. Gather the trash bag around the mixture, and squeeze the ingredients from all sides for a few minutes. Voilà, comet. If you have an old-fashioned overhead projector, place the comet on top; the heat from the projector will make it sputter and form an upward-pointing tail. Of course, a real comet is about 50 percent dirt, but with that amount in this demo, the object will never solidify. Small chunks of meteorites can be purchased online for very little money, and with just $100, you can easily amass a collection of different types of meteors so that students can literally touch and feel what they have been reading about. Impacts are also fairly easy to reproduce in the classroom using nothing more than a few large plastic bins from a big-box home store, a sack of playground sand, a 10-pound bag of flour, and a billiard ball or a dense 1-pound ball bearing. A natural topic for this chapter is the impact that destroyed the dinosaurs. There is some fascinating new research underway that suggests the story most of us learned (and indeed the one told in this textbook) is wrong. After the impact of the Baptistina asteroid that created the Chicxulub crater off the coast of Mexico, many scientists have told the story that a giant cloud of ash covered the atmosphere of the Earth, creating a centuries-long winter that froze and starved the dinosaurs to death. But if you watch slow-motion videos of impacts (like balls falling into sand), you will find two distinct features that counter this scenario. First, the impact creates an enormous amount of heat; second, there is always a tremendous rebound plume of material that shoots straight upward. New thinking is that during this impact, many cubic miles of Earth were vaporized and launched into space, which then cooled and collided again with our atmosphere. The thermal energy generated across the planet during that hypershower of meteors could have heated the atmosphere to 700° to 1000° F, wiping out all animal life on the surface in a matter of minutes. What 89

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90 ◆ Chapter 12 Dwarf Planets and Small Solar System Bodies survived were the smaller animals living underwater and underground, that is, small mammals and tiny dinosaurs destined to be the small reptiles and birds of today. It is just a hypothesis, but an intriguing one! Finally, the New Horizons space probe flew past Pluto (and its moons) while this edition of the text was being written. The few images that were released as I wrote this manual were just a small tantalizing sample of what is certain to be a treasure trove of fascinating data, both for scientists and the public alike. It is worthwhile to spend a few minutes using your favorite search engine to find the most up-to-date images and analyses of Pluto before covering this topic.

DISCUSSION POINTS

▶ The resolution made by the International Astronomical

Union in 2006 reclassifying Pluto and other solar bodies as dwarf planets has been considered controversial. Discuss whether such a controversy is based on scientific grounds or whether there are also some historical and cultural aspects. Ask your students to investigate the biological reclassification of the duckbilled platypus or the colugo, and have them discuss why these did not bring about such controversy. ▶ Asteroids and comets often collide with the major planets. Discuss what we have learned about the history of the solar system and what astronomers are doing now to help assure us that there will not be an impact with the Earth without advance notice. ▶ Given the constraints imposed by the current economic climate in the United States, there are intense discussions among politicians and scientists about what extraterrestrial object should be the next target to be visited by humans. The current candidates are a return to the Moon, a near-Earth asteroid, or Mars. Discuss your views about which should be the next object, if any, that NASA astronauts visit. ▶ Earlier in human history, the sudden appearance of a comet in the sky was often regarded as an omen, and often as a fear-inducing event. Without the knowledge that we have today, why do you think that would be the case?

Cometary Orbits This animation shows the difference in orbits between short-period comets and long-period comets. It also provides the orbits of Saturn and Jupiter for a comparison. It features both edge and face views, highlighting the difference in orbital planes of the short- and long-period comets. Text reference: Section 12.3

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. According to the IAU definitions, these objects are dwarf planets and not moons because they do not orbit another body. 2. (c) Volcanic activity implies the body was, at one time, geologically active. 3. (c) The text describes comet nuclei as similar to deep-fried ice cream “with a soft and porous interior.” We know that the interior of a comet is made of ices (water and carbon dioxide) mixed with dust and organic material. 4. a, b, c, e 5. The origins of a meteorite come from its precollision orbit, its structure, and its chemical makeup.

ASTROTOUR ANIMATIONS

Reading Astronomy News

The following AstroTour animations are referenced in Chapter 12 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises.

1. Spectral lines and isotope ratios are both atomic processes, which follow fixed laws of physics, such that observations of either spectral lines or isotope ratios tell us important information about the medium we are investigating.

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Chapter 12

2. Water in Comet 67P has a different ratio of hydrogen to deuterium than water on Earth, whereas the water on Comet Hartley 2 matches that of our water. It is, therefore, unlikely that Earth’s water came from a comet like 67P, but it could have come from a comet like Hartley 2. 3. At present, the water on Comet 67P is a closer match to that found on other comets and not like that found on Earth or Hartley 2. 4. Other sources of Earth’s water include that it was on the planet at the time of its formation, that it was “beaten out of other minerals” as the planet formed, or that “wet planetesimals” from near Jupiter collided with the early Earth. 5. Dawn arrived at Ceres on March 6, 2015. At the time of writing this manual, bright spots were found on Ceres that were hypothesized to be water ice or even eruptions from water volcanoes. Test Your Understanding 1. (c) Most planetesimals were swept up by larger bodies (like planets). 2. (b) Stony-iron meteorites come from the transition zone between the iron core and rocky mantle. Compared with the size of a planetesimal, this is a very small region. 3. (d) A comet’s ion tail always points away from the Sun. 4. (a) Near Earth objects may collide with Earth; hence, our need to try to detect and prepare for such events. 5. (c) Meteorites are not debris from comets. 6. Perihelion is the point in an orbit closest to the Sun; aphelion is the point in an orbit farthest from the Sun. 7. (b) Comet nuclei only form comae and tails when they are heated by the Sun’s radiation as they approach perihelion. 8. (a) Asteroids are small chunks of rock and metal orbiting the Sun. Most are found between Mars and Jupiter, but they are not restricted to only that location. 9. (a) Short-period comets tend to have prograde orbits within the plane of the planets, whereas long-period comets orbit in random orientations. 10. (d) Averages mean that one must consider a large sample. In this case, a long period of time will yield roughly the average number of comets found. 11. (b) This is the “asteroid belt.” 12. (b) Comets contain a large amount of carbon dioxide and water ice. The impacts would have delivered this water to the planet.

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Dwarf Planets and Small Solar System Bodies ◆ 91

13. (b) A large clump of iron is likely from an object which differentiated such that the heavier metals sank into a central region together, whereas lighter rocky material moved to the upper regions of the object. 14. (c) Comets leave behind particles of dust and ice from their tails as they pass the Sun; the Earth passes through these producing meteor showers at roughly the same time each year. 15. (d) Dwarf planets do not have enough mass to have cleared out the space around them. This means they have lower mass than planets. Thinking about the Concepts 16. Pluto differs from the other planets in a number of ways. It does not have a nearly circular orbit and does not orbit in the plane of the other planets, which suggests that its formation was different than the eight “classical” planets. Pluto also is not massive enough to have gravitationally swept out the space around it, whereas all the other planets are, which has allowed them to remain in the nearly circular and planar orbits. 17. Pluto is no longer classified as a planet because it did not clear out the orbital space around it. It was originally classified as a planet because we did not know anything about it other than it orbited the Sun; as we learned more, we discovered we had to change our formal understanding and categorization. 18. Asteroids are loose agglomerations of refractory rock and metal held together only by self-gravity. Comet nuclei are “dirty snowballs” that are composed primarily of ices and organic materials held together by a loose rocky matrix. 19. Meteoroids are particles of space debris smaller than about 100 meters in diameter. Meteors are ionized trails left by the passage of meteoroids and interplanetary dust through the upper atmosphere. Meteorites are chunks of meteoroids that survive their atmospheric passage to land on the surface. 20. Comet nuclei are kilometers across compared with the typical millimeter sizes of the meteoroids that produce meteors. Comet tails can be hundreds of millions of kilometers long compared with meteor streaks that are only a few kilometers long. Even at closest approach, most comets still remain tens of millions of kilometers away from Earth. Meteor streaks occur less than 100 kilometers above Earth’s surface. Comets remain bright for weeks as they pass through the inner solar system. Meteors remain visible for at most a few seconds.

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92 ◆ Chapter 12 Dwarf Planets and Small Solar System Bodies 21. The age of a meteorite is determined by the time at which it cooled from its original molten state. Because C-type meteorites are older than S-type and M-type meteorites, we can conclude that C-type asteroids cooled quickly and probably remained undifferentiated. On the other hand, S-type and M-type asteroids probably came from larger, differentiated planetesimals that took longer to cool. We can also conclude that because of their older ages, C-type asteroids were likely never incorporated into larger planetesimals. 22. The vast majority of asteroids lie on stable orbits in the asteroid belt, posing no threat to Earth. However, if an asteroid happens to cross Earth’s orbit at the same time Earth is passing through that region of space, the likelihood of a catastrophic collision increases. Earth-crossing asteroids are also much easier to reach with spacecraft because the Sun’s gravity brings the asteroids to us. In the future, near-Earth asteroids are probably going to be mined for their resources long before their main belt cousins. 23. Position yourselves several kilometers apart and watch the same section of sky. When you see a meteor, you should each describe the path of the meteor against the background stars. Because the meteor is an atmospheric phenomenon, you and your friend should notice a very large parallax shift. Of course, this experiment should not be conducted during a meteor shower or a meteor storm because you and your friend have to be certain that you both observed the same meteor! 24. With an origin of “only” 1 billion years, the meteorite did not come from a primitive solar system object like an asteroid or a comet, nor did it come from a world like Mercury or the Moon that was geologically inactive 1 billion years ago. Instead, the meteorite probably had to come from Mars. Note that we could not conclusively rule out Venus without further laboratory testing, but it is highly unlikely that even a large impact event could eject rock at speeds greater than the 35 km/s circular speed that Venus has in its orbit around the Sun. 25. The Kuiper Belt is a disk of cometary nuclei, whereas the Oort cloud is spherical. Also, the Kuiper Belt is very close to the Sun compared with the Oort cloud. So Kuiper Belt comets will orbit roughly in the disk of the solar system and will have short periods, whereas Oort cloud comets will orbit randomly and will have very long periods. (Note: By “short,” we do not mean a few years, but short compared with the periods of the Oort cloud comets.) 26. Comets contain a nucleus, a coma, a hydrogen halo, and several tails. The nucleus is the actual “dirty snowball” source of the material that forms the coma and

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the tail. It is the smallest part of the comet, usually only a few kilometers in diameter. It is also by far the most massive part of the comet. The coma is the bright, fuzzy ball of gas that surrounds the nucleus, typically reaching a million kilometers in diameter. The tails are the largest and most impressive parts of a comet. The ion tail is formed when the solar wind pushes the coma gases away from the Sun. The dust tail is formed from small dust particles entrained in the gas flow and pushed away from the Sun (solar radiation pressure). The neutral sodium tail is the straightest antisolar tail and is formed by sunlight pushing neutral sodium atoms away from the nucleus. (There is also a diffuse neutral sodium tail coincident with the dust tail from which sodium is released from the dust. These tails can sometimes stream hundreds of millions of kilometers away from the coma.) 27. There is very little mass in the tail of a comet. What little cyanide Earth absorbed as it passed through the comet’s tail was utterly dwarfed by the cyanide that already exists at Earth’s surface. 28. Comets have a dust and an ion tail. Ion tails are pushed directly away from the comet by the solar wind, whereas the dust tails are pushed away both by solar wind and solar radiation. The two tails need not point in the same direction because the shape of the dust tail depends on the particle makeup of the dust and how it is accelerated away from the comet by the Sun’s light. 29. Zodiacal light is sunlight that is scattering from a disk of dust that permeates the inner solar system. The source has long been debated. First thought to be from the tails of active comets and asteroid collisions, recent work has shown it is likely from nearly dormant Jupiter-family comets. 30. Comets and asteroids might have brought the chemical seeds of life to Earth, or the impacts might have created enough energy to cause chemical reactions that formed self-replicating proteins in our “primordial soup.” Applying the Concepts M 4 where V 5  R 3 . Escape V 3 2GM velocity v esc 5 . R Solve: (a) Using the numbers given in the text and noting that the radius of the comet is 2 km or 2,000 m, we find . The density of water is 1013 kg r5 5 300 kg/m 3 . The density of 4 3  (2000 m) 3

31. Setup: Density r 5

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Chapter 12

water is 1,000 kg/m3, so the comet is about 3.3 times less dense than water. (b) The escape velocity is 26 . 67 31011  1013 5 0. 82 m/s. 2000 Review: We expect the comet to be much less dense than water because it is mostly made of dry ice. We also expect the escape velocity to be very low because the comet has such a low density (that is, a large size for a small mass). 32. Setup: The rotation speed is (circumference)/ d (period) 5 , where d is the diameter. P 3. 14 3 975 km Solve: v 5 5 340 km/hr . 9 hr Review: This is about 210 mi/hr or the speed of a small aircraft. 33. Setup: Figure 12.12 shows a box 200,000 on a side, with the Oort cloud filling up most of it. It notes that Proxima Centauri is about 270,000 AU away. Solve: If we assume the Oort cloud extends all the way to the side of the box, then it is about 100,000 AU in 100 , 000 5 0. 37, or about 37 percent of radius. This is 270 , 000 the way to the nearest star. Review: Although space is mostly empty and the distances between the stars are vast, we see that much of the interstellar space can still be filled by very weakly bound objects such as comet nuclei. 34. Setup: Pluto’s semimajor axis is 39.5 AU and its eccentricity is 0.249. For Eris these are 67.7 AU and 0.441, respectively. The perihelion and aphelion are given by (1  e) A, where e is the eccentricity and A is the semimajor axis. Solve: For Pluto, the perihelia and aphelia are 29.74 and 49.3 AU, respectively. For Eris, these are 37.8 and 97.6 AU, respectively. Review: Note that the more eccentric the orbit, the larger the change in distances between nearest and closest approach. Pluto’s perihelion is actually smaller than that of Neptune (30.4 AU). 1 35. Setup: The kinetic energy of an object is mv 2 in 2 Joules. An H-bomb has a yield of 4.2 3 1015 J. Solve: For a mass of 4.6 3 1011 kg and speed of 40,000 m/s, the kinetic energy is 1  4. 6 31011 (40 , 000)2 5 3 . 68 31020 J, or 2 87,600 H-bombs. Note that this does not depend on what is hit, just what is doing the hitting. v esc 5

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Review: We clearly see how much energy is contained within a fast-moving object, and why the impacts of even smallish objects with the Earth are such a dangerous situation. 36. Setup: Using the small-angle formula s 5 d, where s is the physical size of the object, d is its distance, and  is its angular size, we find that Charon appears  5 1,186/19,600 5 0.06 radians or 3.5 degrees. Solve: Using the angular size just calculated, we find (a) Charon appears about 7 times larger in Pluto’s sky than our Moon. (b) Charon and Pluto share a locked orbit, so Charon will always appear in the same place to a Pluto observer (and is never visible to someone living on the side that faces away from its moon). Review: Note that Charon’s proximity means that it will have a much stronger tidal effect on Pluto than the Moon does on Earth. This explains the tidal lock in part (b). 37. Setup: The probability of being struck by a meteorite  A you  in your lifetime is Phit 5   3 Rate 3 Time. A Earth

Solve: Substituting appropriate numbers into the  0 . 25m 2  3 equation yields Phit 5  14 2  5 310 m   365. 25 days   3(100 yr) or  800 meteorites/day 3 1 yr Phit 51 . 5 3108. Review: These are approximately the same odds as winning the lottery. 38. Setup: (a) Kepler’s third law allows us to compute the mass of Electra. Newton’s form of Kepler’s third law 4 2 3 r can be written mathematically as P 2 5 GM Electra and rewriting this to solve for the mass of Electra gives 4 2 r 3 M Electra 5 . For part (b) we need to compute GP 2 4 M volume (V 5  r 3 ) and density r 5 . 3 V Solve: (a) Using the equation above, M Electra 5

4 2  (1. 35 3106 )3 2

51. 27 31019 kg.

 86 , 400 sec  6. 67 31011   3 . 92 day s 3  day  5 1.27 3 1019 kg. 4 3 The density of Electra will be V 5  r 5 3 15 3 3.16 3 10 m . To compute the density r, we substitute

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94 ◆ Chapter 12 Dwarf Planets and Small Solar System Bodies the results into the standard relationship for mass and M density r 5 5 4 , 020 kg/m 3 . V Review: This is quite a bit denser than typical rock, so we can conclude that Electra contains a significant amount of metal in its composition. 39. Setup: The Kirkwood Gap is in a 3:1 resonance with Jupiter, meaning asteroids in this gap complete three orbits in one orbit of Jupiter. In other words, the period of an object at this distance is one third that of 11. 86 5 3. 96 yr . We can use Kepler’s third Jupiter, or 3 law to find the distance with P2 5 A3. Solve: A 5 P2/3 5 2.5 AU. Review: A quick search on Google confirms that the gap is indeed at 2.5 AU. 40. Setup: Because the Sun is the primary mass for all three comets, we can use the simple version of Kepler’s third law to compute their semimajor axes P2 5 A3, where P is in years and A is in AU. If we assume that the perihelion distances for all three comets approach zero, then the aphelion distances will be approximately twice the semimajor axes. Solve: (a) Using Kepler’s third law for Comet Encke, the period is 3.3 years yielding a semimajor axis of 2.2 AU. For Comet Halley, the period is 76 years yielding a semimajor axis of 17.9 AU. For Comet HaleBopp, the period is 2,530 years yielding a semimajor axis of 185.7 AU. (b) Using Working It Out 12.2 for the perihelion and aphelion distances, they are 0.34 and 4.1 AU for Comet Encke, 0.59 and 35.3 AU for Comet Halley, and 0.93 and 370.4 AU for Comet Hale-Bopp. (c) Encke probably came from the Kulper belt but suffered an orbital change. Halley probably came from the Kulper B. H. and Hale-Bopp from the Oort Cloud. (d) Comet Hale-Bopp should be the most pristine because its long orbital period means it has made fewer passes around the Sun and lost less of its volatile inventory. Comet Encke should be the least pristine because its frequent trips around the Sun rapidly deplete its volatile inventory. Review: To check our distances, consider Figure 12.12, which shows many comets relative to the solar system. Encke is about at two-thirds the distance of Jupiter (about 5 AU), Halley is past Neptune (about 30 AU) and Hale-Bopp is far beyond Pluto. 41. Setup: From 230 BCE to today (2015) is a total of about 2,245 years. Solve: (a) 2,245 years divided by the period of 76.4 years yields 29 orbits. (b) 29 orbits times

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3 3 1011 kg is 9 3 1012 kg lost. (c) This represents 9 31012 kg 3100 5 4 percent of its total mass. about 2. 2 31014 kg Review: If Comet Halley lost a huge amount of its mass at each passage, then we should expect to have seen considerable changes in the comet’s brightness over the past few millennia. Furthermore, the comet would be extremely short-lived, which would raise the question “Why are we able to see it at all?” The fact that it has been losing only a small fraction of its mass during this time frame is consistent with our observations. M 42. Setup: Density r 5 . V Solve: (a) Using the numbers given in the text and noting that the radius of the comet is 5 km or 5,000 m, we 2 . 2 31014 kg 5 420 kg/m 3 . The density of 4 3  (5000 m) 3 water is 1,000 kg/m3, so the comet is about 2.4 times less dense than water. Review: We expect the comet to be much less dense than water because it is mostly made of dry ice. N 43. Setup: Just like mass density, number density n 5 , V where N is the total number of particles present, and n is the number of particles per unit volume. Solve: At 200 molecules per cubic centimeter, the comet tail would need a volume of find r 5

1019 particles N 5 5 5 31016 c m 3 to con3 n 200 particles/cm tain 1019 molecules. Review: We expect this volume to be huge given how few molecules there are per volume element in the comet’s tail. Indeed, this volume is equivalent to a cube that is 3.7 3 105 centimeters (or 3.7 kilometers) on a side. 44. Setup: The energy from an impact is kinetic, for which 1 we use K 5 mv 2 . Note that 5 km/s is 5,000 m/s. 2 Solve: Using the mass and speeds given, 1 K 5 4. 6 31011 (5, 000)2 5 5 . 75 31018 J. 2 Review: For comparison, this is about 1,370 H-bombs, using problem 35. 45. Setup: The removal rate is (total mass removed)/ (time to remove it). If the total amount of dust remains V5

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Chapter 12

approximately constant, this represents equilibrium between the dust removed from the system and the dust being produced by comets. As a shortcut, note that there are 31.6 million seconds in 1 year. Solve: In dynamic equilibrium, the rate of replacement equals that of removal, which is 1 yr 1016 kg 3 510 , 500 kg/s . 30 , 000 yr 3. 16 3106 s Review: Dust is removed and replaced at a rate of approximately 10,000 kg (or ten tons) every second. Using the Web 46. Answers will vary with time. (a) As of February 2013, Mike Brown has identified eight objects that are almost certainly dwarf planets and 394 objects that are possibly dwarf planets. It is very difficult to certify a dwarf planet because we must know each object’s size (and hence its shape), which may sound simple, but these are dim and far away, making this determination quite difficult. (b) New Horizons is past Pluto. 47. Answers will vary with time. (a) As of October 2012, Dawn is at Vesta and is traveling to Ceres as of February 2013. So far, we have learned that Vesta has had complex geology with faults and troughs and that water-bearing material was probably incorporated into the object throw of the slow impact of early solar system material. 48. (a) The goals of asteroid zoo are to use images from the Catalina Sky Survey with crowd-sourcing to look for objects moving across the sky, which could be new minor planets. (b) The goals of CosmoQuest’s asteroid (or Vesta) mapper is to use crowd-sourcing to identify craters to determine the stratigraphy (or the layered history of the surface) to reconstruct Vega’s past. 49. Answers will vary. The response will include if there is any news from NASA’s Asteroid Watch program, including a new discovery or asteroid flyby. If so, students will discuss whether the asteroid was studied by a spacecraft, orbiting telescope, or ground-based observatory, and what was learned about the object. 50. Answers will vary. The answer will include whether there are any comets observable with the naked eye, whether any near-Earth asteroids will be approaching

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in the coming months, and how close a few known near-Earth objects will be at their closest approach. The student will list the eccentricity and semimajor axis of each object and compute the aphelion and perihelion. Exploration 1. To investigate the position of the asteroid, I made a 200-dpi scan of the page, cut out the squares in an art program, saved them, and then did a quickly repeating slide show (essentially making a digital flipbook). The following image is a sum of the four images with the location of the asteroid marked with arrows showing its movement.

2. An early human eating on the savannah had very little to defend himself or herself from prey other than his or her wits, so if we detect subtle movements, then we are more likely to see approaching danger and escape before it is too late. 3. As noted in the text, a more sophisticated method is to subtract the two images, which gets rid of everything that is constant and only shows what has changed between the two images. In these negative images, we would expect a white background with two dark spots showing where the asteroid was in the first and second images.

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CHAPTER 13

Taking the Measure of Stars INSTRUCTOR’S NOTES Chapter 13 describes how we measure most stellar properties. Major topics include

▶ the relationship between luminosity, brightness, and distance. ▶ parallax and spectroscopic parallax. ▶ composition and stellar spectral types. ▶ binary stars and how they are used to determine stellar masses and size. ▶ the Hertzsprung-Russell (H-R) diagram.

In a typical introductory course, we generally have so much material to cover that we spend most of our course focusing on the five Ws (who, what, where, when, why) with very little time available to explore how we discovered all this knowledge. I am particularly fond of this chapter because it provides the observational background on which much of the rest of the book is based. However, because my students have spent the preceding two-thirds of the course focusing on those five Ws (laws of physics, planets), I have found it helpful to tell them explicitly that we are going to shift gears into the how of observational astronomy here. Some of this material is quite intuitive, such as the inverse square law of brightness, as everyone knows that lights appear dimmer when they are farther away. On the other hand, the relationships of color and temperature or among luminosity, temperature, and size can be much more problematic. Students often associate that “more” always requires “bigger,” so it can be quite counterintuitive that a hotter object is bluer (meaning shorter wavelength) or that a cooler object can be more luminous than a hotter one if the sizes are correct. This is, therefore, another chapter in which I slow down, spending at least two weeks to cover: (1) parallax and distance; (2) the inverse square law; (3) blackbodies revisited; (4) luminosity, temperature, and size; (5) binary stars, mass, and size; and (6) how we measure these through imaging and spectra. There are a few points of confusion in this chapter that I used to take for granted until faced with lines of

confused students outside my office. Here are some example questions:

▶ “How can a star have a blackbody and absorption spec-

trum at the same time?” We model stars as blackbodies (dense objects) but talk about their absorption spectra (low-density gas). The nuance is that the interior acts like a blackbody, but the cooler outer atmosphere produces the absorption. ▶ “How can a cool star be more luminous than a hot one?” Figure 13.8b is a new graphic that may help a lot, as it shows graphically how the flux changes between hot and cool objects. ▶ “Aren’t all A-type stars the same?” Not all A-type stars are main sequence stars, as there are different luminosity classes. ▶ “How can atomic spectra look different if they all have the same pattern as hydrogen?” Not all atomic spectra look like that of hydrogen; it is just that hydrogen is an easy spectrum to plot. In truth, each atom has its own complicated series of energy levels. ▶ “Why can’t we measure the masses of all binary systems?” Eclipsing and spectroscopic binary systems are only useful if they are almost edge on. Introduction of the UBVRI ( Johnson photographic) filters can add an unnecessary extra level of confusion, and indeed, the textbook largely glosses over them. However, I still like to spend a dozen minutes talking about the practical task of measuring stellar properties. Many students find it fascinating that in the time needed to measure the temperature of one star with a spectrum, we can measure the temperature of thousands of stars with two images in different filters. It is always an interesting conversation to debate the merits of photometry versus spectroscopy, in terms of best using limited resources (observing time) and maximizing the information returned. Almost invariably, one student asks if it is possible to take the spectra of hundreds of stars at once, which is a great segue into modern observing techniques of multifiber spectrographs, grism imaging, and so forth. 97

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98 ◆ Chapter 13 Taking the Measure of Stars

DISCUSSION POINTS

▶ Students will learn in Chapters 19 and 20 that galaxies are far larger than 1 kpc. Given that we can reliably measure parallax to 0.001 arcsec, have students discuss what this means for measuring distances to most objects in the universe. ▶ Discuss the Sun in terms of its relation to other stars concerning size, temperature, mass, luminosity, and its spectrum. ▶ Figure 13.5 shows that most stars in the galaxy are low mass, whereas Figure 13.16 shows that most of the closest 16,000 stars are of higher mass than that of the Sun. Are these two findings consistent? Discuss the information conveyed by those two figures and how to reconcile them. ▶ Discuss the history of the stellar classification system, including its original incarnation and its rearrangement once temperature was realized to be the important quantity. How is this an example of the scientific method at play? ▶ Discuss the utility and ambiguity of the spectral classification of stars. Have students look for information about the new spectral classes defined in the 21st century, such as the “L,” “T,” and “Y” dwarfs. Discuss their spectral properties. ▶ Is there anything about a star that can be determined if we do not know its distance? What about a cluster of stars?

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 13 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises.

found in that place, with basic information about luminosity, surface temperature (and color), and the life stage of the star. Text reference: Section 13.4

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding the physical concepts presented in Chapter 13. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Parallax Calculator This module shows how the observed parallax motion of a star changes with its distance from Earth. Text reference: Section 13.1 Stellar Luminosity Calculator This simulation shows how a star’s luminosity depends on its temperature and radius, using the blackbody nature of stars. Temperature is selected by spectral type, allowing the user to see the corresponding temperature. Results for luminosity are given in MKS and solar units. Text reference: Section 13.1 Center of Mass Simulator This applet shows how the center of mass of two objects changes as their mass changes. The user can modify both the masses and the separation. Text reference: Section 13.3

Stellar Spectrum This interactive animation explores how the spectra emitted by different atoms can be used to determine the chemical composition of a star—information that is used to infer the star’s mass and age. In the interactive portion of the animation, students can select various elements and see spectra produced. They can also see the emission spectra of various stars and try to determine their chemical compositions. Text reference: Section 13.2 H-R Diagram In this animation, students move a position marker across an interactive Hertzsprung-Russell (H-R) diagram. An offset window shows an example of a typical star that would be

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Eclipsing Binary Simulator This simulation models the motions of two stars in orbit around each other and the combined light curve they produce. The user can control multiple parameters to see how they affect the light curve, including the temperature, mass, and radius of both stars, and the orbital separation and eccentricity. The user can also change the orientation of the binary system (edge on to face on). The simulation also includes several presets for real binary systems. The applet shows both stars on an optional H-R diagram, which is also useful as it allows the user to explore the mass-radius-temperature relationship, not only on the main sequence but also along giant branches. Text reference: Section 13.3

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Chapter 13

Taking the Measure of Stars ◆

Hertzprung-Russell Diagram Explorer

Reading Astronomy News

This simulation introduces the H-R diagram, a plot showing the relationship between luminosity and temperature for stars. The user not only sees the location of the star as one changes its parameters, but also visualizes its size and color with respect to the Sun. Text reference: Section 13.4

1. One parsec is 3.26 ly, so 7,500 ly is about 2,300 pc; 10,000 ly is about 3,070 pc. 2. d 5 1/p so a star 7,500 ly away has a parallax of 1/2300 5 0.0004 arcsec. A star at 10,000 ly has a parallax of 0.0003 arcsec. 3. Exposures are taken 6 months apart to provide the largest baseline; see Figure 13.1. 4. From space, images of stars are not subject to the atmospheric turbulence on Earth; thus, their positions can be recorded much more accurately. 5. Measurements of high distances are always calibrated using nearer methods; the best calibration is trigonometric parallax, which provides the strongest foundation on which all other methods are based. However, spectroscopic parallax is fairly uncertain because all stars of a given spectral type are treated as identical, which is not true. Thus, the improvements here will be limited for the particular case of spectroscopic parallax.

Spectroscopic Parallax Simulator This module simulates how spectroscopic parallax works. The spectral type and luminosity class determine the observed spectrum of a star from which the star’s luminosity can be estimated. The simulation shows how the strength of relative absorption features change with spectral type, the location of the star on the H-R diagram, and the logarithmic relationship between distance and apparent magnitude. Text reference: Section 13.4

ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what they have learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack. Parallax This simulation discusses how to experience parallax with your own eyes and how astronomers use this same process to measure the distances to stars. Text reference: Section 13.1

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (b) Brightness follows an inverse square law, so to be twice as far away but appear equally bright, the star has to be 22 5 4times more luminous. 2. (d) Using the Working It Out 13.3, L  R2T4 so if L is the same but T is doubled then R must be quartered. 3. a, b, d. Period and distance give the sum of the two masses, whereas velocities allow the individual masses to be determined. 4. b, d. Stars at the lower right of the H-R diagram are cool, low luminosity, small, and if they are main sequence stars, they have low mass as well.

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Test Your Understanding 1. (b) Because the two stars are at the same distance, the only reason one would appear brighter is if it is more luminous. 2. (a) Because a star acts like a blackbody, we know that a bluer intrinsic color means the star’s surface is hotter. 3. (c) Because the stars are at the same distance and appear the same brightness, they must have the same luminosity. The fact that one is red means it is cooler; therefore, it must be larger to have the same luminosity as the blue star. 4. (a) For a given line to appear, the atoms must be present and in the right energetic configuration, which is highly dependent on temperature. 5. (b) Star A is moving three times faster than star B; therefore, it must be less massive. 6. (b) If stars are at the same distance, then brightness indicates luminosity. Because star A is brighter than B but at the same temperature, it must be larger; that is, A is a giant, and B is a dwarf or main sequence star. 7. (d) If stars are at the same distance, then brightness indicates luminosity. Because star A is brighter than B but the same color, then it must be larger. We can rule out either star being a red giant because of the blue color, leaving only answer (d). 8. (a) See Figure 13.20. 9. (a), (c), and (e). Because both stars are main sequence, more massive implies hotter, larger, and more luminous.

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100 ◆ Chapter 13 Taking the Measure of Stars 10. (d) Because parallax depends on the distance of the planet from the Sun, a larger orbital distance would make the pc longer and would allow us to see parallax motion of more distant stars. 11. (a) More distant stars have smaller parallax angles. 12. (a) Brightness b changes as L/d2, so the fact that star A appears twice as bright but is twice as far away means that L  b 3 d2 5 2 3 22 5 8, using proportional reasoning. 13. (b) Helium has four times the mass of hydrogen. 14. (b) Capella will be made of yellow (G-type) and red (M-type) stars, and because both pairs are close to each other, the brighter G-type stars will dominate. Therefore, the color will appear yellow. 15. (a) See Figure 13.13. Thinking about the Concepts 16. As parallax angles become smaller, they approach the precision limits of our instruments. For example, consider an instrument that can measure any parallax with a precision of ±1 milliarcsecond. For an object with a 50 milliarcsecond parallax, the uncertainty is only 1 milliarcsecond out of 50 milliarcseconds or 2 percent. On the other hand, an object with a 2 milliarcsecond parallax has an uncertainty of 1 milliarcsecond out of 2 milliarcseconds, or 50 percent! Therefore, due to the limited precision of our instruments, uncertainties in distance grow as parallax angles get smaller. 17. Size and mass are distance-dependent quantities. Size depends on distance because objects that are farther away look smaller. Mass depends on distance because we need to know the size of a binary companion’s orbit to accurately determine the star’s mass. Temperature, color, spectral type, and composition are all distanceindependent quantities that can be determined from a star’s spectrum. Because stellar spectra show the same lines whether viewed from up close or from halfway across the galaxy, the measured properties of a star derived from its spectrum are not influenced by the star’s distance. 18. (a) The color of a star is a direct measure of its temperature. Blue stars have hot surface temperatures, and red stars have cooler surface temperatures. Because the golden star lies closer to the redder side of the spectrum than the sapphire blue star, we conclude that the sapphire blue star has the higher temperature of the pair. (b) Blue stars emit more light per unit area than yellow stars. Therefore, if the blue star was the same size as

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or larger than the golden star, it would be the brighter star in the pair. Because our observations demonstrate that the golden star is actually the brighter of the two, we must conclude that it has a much larger surface area than the blue star to appear more luminous. 19. Every blackbody emits light at all wavelengths; the question is just how much light is emitted. A star with a 2,500 K blackbody emits mostly in the red and infrared but still gives off a measurable amount of blue light. 20. Betelgeuse is cooler than Rigel. Nothing can be said about their luminosity or size without knowing their distance. 21. The original classification of stars was based on the strength of hydrogen absorption lines in their spectra, with A-class stars exhibiting the strongest absorption. Hydrogen absorption lines are strongest at temperatures near 10,000 K, but stellar surface temperatures span the range from 3000 K to 30,000 K. At surface temperatures well below 10,000 K (F, G, K, and M stars), much of the hydrogen in a star’s upper atmosphere remains unexcited, so fewer electron transitions are observed in the stellar spectrum. At temperatures well above 10,000 K (O and B stars), most of the hydrogen in a star’s upper atmosphere is ionized, so no electrons remain in the hydrogen to produce spectral lines. When spectral classes are arranged as a temperature sequence, the O and B spectral classes move ahead of the A-class stars despite their weaker hydrogen lines. As a final note, the missing letters in the OBAFGKM sequence are a result of merging the original 22 spectral classes into the seven larger spectral classes that we have today. 22. We can directly measure stellar masses only by using Kepler’s third law, which requires us to observe the orbit of a companion. To accurately measure the semimajor axis of the companion’s orbit, we need to know how its orbit is inclined to our line of sight. Only eclipsing or visual binaries allow us to measure both the orbital period of a companion star and its semimajor axis directly. Note that we can use Kepler’s third law with spectroscopic binaries as well; however, because we do not know the orbital inclination of the companion star, we can determine only a minimum mass for the system, not the exact mass. 23. Stars are all governed by the same laws of physics, and their evolution depends almost exclusively on their mass (which controls their central temperature, pressure, etc.), so we assume that if two stars have similar composition and mass, they will act and evolve the

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Chapter 13

same way because there is no physical mechanism to change this. 24. Binary stars can transit each other just like planets, so any mission that is looking for regular time variability is going to discover binary stars (and pulsating variables as well). 25. Answers will vary. Scientific collaboration requires the open sharing of data, division of labor, participation of many different people in the analysis and interpretation, and corroboration of findings through many different people performing the same work. It is not just that many hands make light work, but that many hands also check each other. 26. As the distance from the Sun to the planet increases, 1 arcsec corresponds to a farther distance, so if we move to Mars we will be able to measure the distances to even farther stars. Jupiter would open up the sky to even more distant stars, and Venus would make us able to measure only closer ones. 27. Hydrogen absorption is weak in O stars because most of the hydrogen is ionized; that is, there are no electrons bound to the nucleus to absorb photons. In G stars, the hydrogen is neutral but also sufficiently cool that most of its electrons are in the ground state in which only UV photons can be absorbed. A-type stars have the right temperature to have many hydrogen atoms, both neutral and in the first excited state ready to absorb optical photons. 28. Edge-on observations are best for eclipsing and spectroscopic binaries because there is no inclination effect to change. Face-on observations are best for visual binaries. 29. (a) Star 2 has a smaller orbit because it has a higher mass; therefore it moves less about the center of mass. (b)

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Taking the Measure of Stars ◆ 101

(c)

30. If the Sun were a blue, main sequence star, its luminosity would be thousands or even tens-of-thousands times higher than it is now, which would scorch the Earth to cinders. If the Sun were a red, main sequence star, its luminosity would be hundreds or even thousands of times lower than it is now, making Earth a potentially-frozen wasteland. I would expect Earth would not be in the habitable zone for either case. Applying the Concepts 31. Setup: Parallax angle scales inversely with distance: p51/d, so if an object is twice as distant, its parallax angle is twice as small. Solve: If a star C were located four times further than star A, it would appear to move four times less each year than star A, but because it is only two times further than star B, it would move two times less than B. Review: You can try this directly using the method shown in Figure 13.1a, by changing the furthest distance of the two pencils you are holding. 32. Setup: Parallax angle scales inversely with distance: p51/d, so if an object is twice as distant, its parallax angle is twice as small. Note that one-third of a degree is two-thirds as small as one-half of a degree. Solve: An object that moves two-thirds as much will be 1.5 times further away. Review: This object is 1.5 times the distance. Does this make sense? If the parallax angle was one-fourth a degree, it would be at twice the distance. One-third is between one-half and one-fourth, so yes, this makes sense. 33. Setup: Quite often in astronomy, graphs need to cover a huge range of parameters, which is easier to visualize using logarithms as they spread out the whole range of values so that one can identify the positions of points with both small and large values equally well. To find

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102 ◆ Chapter 13 Taking the Measure of Stars the difference in luminosity between the least and most luminous stars, read off the values on the y-axis. Solve: The most luminous star is shown with about 106 LSun, whereas the faintest are about 2 3 10–3 LSun for a total difference of about 109 or a billion. Review: It is remarkable that stars that have a small range of masses and temperatures can have such a huge range of luminosities; we will learn why this happens in future chapters. 34. Setup: Remember that when reading logarithmic axes, each “tick mark” is a power of ten. Solve: (a) On the left side of the graph, the stars have 1024 solar luminosities, or are 10,000 times less luminous than the Sun. On the right, the stars are 106, or a million times more luminous than the Sun. Review: This graph shows stars spanning 1010 or a range of 10 billion in luminosity. Wow. 35. Setup: In the figure, temperature is on the bottom axis, luminosity is on the left side, mass is indicated along the main sequence, and size is shown by diagonal lines in the figure panel. Solve: A star 10 times the mass of the Sun is found at the upper left of the main sequence. It has a luminosity of around 10,000 solar luminosities, a temperature around 23,000 K, and a radius around 8 solar radii. Review: Note that this star, although only 10 times as massive as the Sun, is almost 10 times larger, 4 times hotter, and 10,000 times more luminous. This is consistent with Working It Out 13.3. 36. Setup: Parallax uses the parallax formula d 5 1/p, where d is in parsecs and p is in arcseconds. To convert parsecs to light-years, recall 1 parsec 5 3.26 ly. Solve: The distance to Sirius is d 5 1/0.379'' 5 2.64 or 3. 26 ly 2. 64 pc 3 5 8. 6 ly. pc Review: Remember that a parallax of 1 arcsecond equals a distance of 1 parsec. Sirius has a distance close to one-third of an arcsecond, so its distance should be about 3 parsecs, as we found. 37. Setup: Remember that luminosity L  R2T4 and that if two stars are the same spectral type, then they have the same temperature. Solve: Both stars are the same spectral type (A) and therefore have the same temperature. If the “Pup” appears 6,800 times fainter and is the same distance as the “Dog Star” (which they are because they are a binary pair), then the “Pup” is 6,800 times less luminous. Because luminosity L  R2T4, we can use the above information to infer that the “Pup Star” must be R  L/T 4 1/6,800/14 5 1/6,800 51/83 times the size, or 83 times smaller than the “Dog Star.”

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Review: If two stars are the same temperature and one is very faint, then it must be very small. Why? All blackbodies with a temperature T give off the same amount of energy per tiny patch of area. So if a star has small luminosity, it has small area—that is, small size! 38. Setup: For binary stars, we know the ratio of the masses is equal to the inverse ratio of the orbital ve M v locities; that is, 1 5 2 , and Kepler’s law for two M 2 v1 a3 . M11 M 2 Solve: (a) For binary stars, we know the ratio of the masses is equal to the inverse ratio of the orbital M1 v 5 2 , so if the companvelocities; that is, M2 v1 ion has 2.35 times higher velocity, it has 2.35 times lower mass. Its mass is then about equal to that of the Sun because Sirius A has a mass 2.35 times larger than the Sun. (b) Using a total mass of 1.00 1 2.35 5 3.35 solar masses and a period of 50 years, we find stars is p 2 5

a 5 3 (M1 1 M 2 ) P 2 5 3 3.35502 5 20 AU . Review: In our Solar System, a period of 50 years would place a planet between Saturn and Uranus (about 14 AU). Because the Sirius system has more mass, the period of a given object will be shorter than were it orbiting the Sun. It therefore makes sense that the companion must be more than 14 AU away to have a period of 50 years. 39. Setup: To solve this problem, we begin by noting that the apparent brightness (b) of a star is proportional to its luminosity (L) and inversely proportional to its disL tance squared (d2) as b  2 . d Solve: Taking a ratio of the brightness of Polaris to Sirius and solving for distance yields dPolaris 2,350 L 5 23 3 5 49.6. Polaris is 49.6 22L dSirius times farther from us than Sirius is. Knowing from problem 45 that Sirius is 8.6 ly from the Sun, we compute that Polaris must be 426 ly from the Sun (49.6 3 8.6 ly 5 426 ly). Review: When we talk about “nearby” stars, we are really including those up to many hundreds of light-years from Earth. 40. Setup: Parallax uses the parallax formula d 5 1/p, where d is in pc and p is in arcseconds. Solve: Using the parallax formula, the distance is 131.1 parsecs. The uncertainty means the distance could be as low as 0.00763 2 0.00164 5 0.00599 arcsec, or as

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Chapter 13

high as 0.00927 arcseconds, yielding distances of 108 and 167 parsecs. The uncertainty is therefore half this difference, or about 30 parsecs. Review: Remember that a parallax of 1 arcsecond equals a distance of 1 parsec. Betelgeuse has a parallax close to one-hundredth arcsecond, so its distance should be about 100 parsecs, as we found. The uncertainty is about four times smaller than the parallax, so the corresponding uncertainty in distance should be about one-fourth the actual value: 30 parsecs is about one-fourth of 135.9 parsecs (more or less), again confirming our calculation. 41. Setup: Parallax uses the formula d 5 1/p, where d is in pc and p is in arcsec. Luminosity, brightness, and distance are related by L  bd2. Solve: Rigel is at a distance of d 5 1/0.00412'' 5 243 pc, based on its parallax. Rigel is twice as far as Betelgeuse, so Rigel must be about four times more luminous than Betelgeuse to appear the same brightness. Betelgeuse is red and so much cooler (perhaps half the temperature of Rigel), so it must be larger than Rigel to give off so much energy. Review: To check whether Betelgeuse is larger than Rigel, using our luminosity, size, and temperature relationship that L  R2T4, we find Betelgeuse should be about R  L / T 4 (1/4)/(1/2)4 5 4 5 2 times larger. 42. Setup: We know that the brightness of an object decreases as the square of the distance according to 1 b  2 . We can form a ratio of the brightness of the d Sun to the brightness of a distant solar-mass star, yielding a distance dStar 5 dSun 3

bSun . bStar

Solve: (a) Using the numbers in the problem, we find dStar 5 1 AU 3 1.6 3 1013 5 4 3 106 AU . (b) Converting this distance to light-years yields 149.6 3 106 km 4 3 106 AU 3 3 1 AU 1 ly 5 6.32 ly. 9.46 3 1012 km Review: This is a shockingly small number, which tells us that most of the stars that we see in the sky are significantly brighter than our Sun. 43. Setup: Figure 13.9 suggests that the center of mass is moved toward the heavier mass by the factor by which it is greater than the lighter one. Solve: If the masses are equal, the center of mass will be at the middle of the beam, or directly between the two masses; that is, d1 5 d2. If one mass is twice that of the other, then d2 5 2d1.

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Taking the Measure of Stars ◆ 103

Review: The formula for center of mass is x cm 5

m1 x1 1 m2 x 2 . If we let the center of mass be the m1 1 m2

origin (xcm 5 0), then we find the above answers for both situations. 44. Setup: Wien’s law tells us the star emits most of its 2. 9 3 106 nm . radiation at  5 T Solve: At 10,000 K, Wien’s law gives 2 . 9 3 106 nm 5 5 290nm . This is in the nearT UV. This star will appear very blue/white to us. Review: Although the majority of the energy comes out in the near-UV, this star is still emitting copious amounts of radiation in the visible spectrum, which is why we can see a blackbody that peaks outside the visible range of light. 45. Setup: Using the relationship between brightness b, luminosity L, and distance d, L 5 4πd2b, we can solve for the luminosity of the Sun, knowing we are 1 AU, or 1.5 3 1011 m, away and b 5 1,470 W/m2. Solve: This yields a luminosity of 4 3 1026 W. Review: This is very close to the actual value of 3.8 3 1026 W. Using the Web 46. As of August 2015, Gaia had been operating for a year and a half. As listed on the mission page, the satellite will “precisely chart [the] positions, distances, movements, and changes in brightness” of about 1 billion stars. The L2 orbit was chosen to provide an unobstructed view of the sky. 47. As the inclination increases, the orbit goes from looking circular to linear. The simulation shows an elliptical orbit until the inclination is exactly 90 degrees. As separation increases, the dips during an eclipse become narrower, and vice versa as the stars approach closer. If star 2 grows three times larger and cools to 4000 K, eclipses disappear at an inclination around 77 degrees. This shows that highly inclined systems are the easiest to detect by looking for eclipsing light curves. 48. Answers will vary with time. The student will report the number of confirmed planets and how many of these are eclipsing binaries. From the eclipsing binary catalog, a few stars will be listed, including their inclination and the depth of their light curves. 49. Answers will vary. My favorite constellation is Orion. Two stars in particular have very different colors: Betelgeuse (red) and Rigel (bluish white). The three

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104 ◆ Chapter 13 Taking the Measure of Stars brightest stars are Rigel (T 5 12,100 K, L 5 117,000 LSun, d 5 773 ly), Betelgeuse (T 5 3100 K, L 5 120,000 LSun, d 5 640 ly), and Bellatrix (T 5 22,000 K, L 5 6,400 LSun, d 5 243 ly). 50. Answers will vary. Exploration 1. As I move to the left of the H-R diagram, the star’s radius becomes smaller. 2. As I move to the right of the H-R diagram, the star’s radius becomes larger. 3. As I move up the H-R diagram, the radius becomes larger. 4. As I move down the H-R diagram, the radius becomes smaller.

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5. To move the star into the white dwarf area of the H-R diagram, the temperature must be very high and the luminosity very low. 6. In general, the nearest stars are cooler than the Sun. 7. In general, the nearest stars are less luminous than the Sun. 8. In general, the brightest stars are hotter than the Sun, but there are still many that are cooler. 9. Except for one straggler, all the stars are more luminous than the Sun. 10. The brightest stars all have high luminosities and a range of temperatures. The nearest stars almost all have lower temperatures and luminosities than our Sun. This shows us that the nearest stars are not always the brightest stars, meaning that space is more filled with low-luminosity stars than high ones.

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CHAPTER 14

Our Star—The Sun INSTRUCTOR’S NOTES Chapter 14 is an extensive overview of the Sun. Major topics include

▶ hydrostatic equilibrium ▶ energy generation ▶ energy transport ▶ the atmosphere of the Sun ▶ probing the solar interior with neutrinos and helioseismology ▶ sunspots and solar activity

Chapter 14 can pose a bit of a chicken-and-egg conundrum for teaching stellar evolution. Did our understanding of the Sun inform our theories of stellar evolution or did we use the Sun to test models that were developed first? The truth is that both occurred together. After experimenting with different orders and different levels of detail, I have come to take the broad-brushstroke approach to Chapter 14: I use our knowledge of the Sun to illustrate explicitly our understanding of all stars. To make the chapter less about “the Sun as an isolated body” and more about “the Sun as a star,” I teach this material a little differently than as presented in the text. It may seem kitsch, but I open this section by playing The Sun Song recorded by They Might Be Giants. Most of the salient details of a star are right in the refrain: “The Sun is a mass of incandescent gas / a gigantic nuclear furnace / where hydrogen is built into helium / at a temperature of millions of degrees.” Otherwise stated, a star can be modeled like a blackbody, a star is powered by nuclear fusion of hydrogen into helium, and this only happens if the core temperature is at least many millions of degrees. If you want to be a little more accurate, you can also grab their sequel, “The Sun is a Miasma of Incandescent Plasma.” Frankly, I wish there were a fun song for just about every topic in this course because using them really does help with reaching learning outcomes. I follow the song with our observational understanding of the visible surface: absorption lines show that the Sun has a cooler, outer atmosphere; limb darkening indicates the interior is gaseous (as opposed to solid); turbulence

indicates energy is transported to the surface via convection (as I indicated earlier, I return quite often to convection because it causes stars to climb the giant branches); sunspots reveal differential rotation (further evidence that the Sun is gaseous and not solid); and sunspots and storms indicate the presence of magnetic fields, the complicated interplay of which we still do not fully understand (that is, it is okay for science not to have all the answers). Then I show how helioseismology reveals the interior structure, consisting of a nuclear core and inert envelope. I finish by showing how nuclear fusion works, and why it can power the Sun whereas chemical reactions and gravitational contraction cannot. As a side note, I save hydrostatic equilibrium for Chapter 15: as it is part of stellar evolution, hydrostatic equilibrium provides a “natural safety valve” of stability against thermal anomalies in the core. As in previous chapters, my personal feeling is that much of the vocabulary and graphs and measurements, especially in sections 14.2 and 14.3, can cause students to focus too much on memorizing small details and miss the important connections that will carry forward into other chapters. If you do prefer to cover this material in class, I urge you to spend a little extra time distinguishing that which they must commit to memory. Solar neutrinos provide an excellent opportunity to talk about the scientific method in action. When it was proposed that nuclear fusion could power the Sun, neutrino detection became the necessary test of this theory. But how to detect something whose very nature is to be inherently undetectable? Then, when we did collect data, we found a result quite different from what we expected, forcing us to face the classic conundrum: was our theory wrong or had we left out some important physics? Far too often, non-science majors approach science as a subject for which all the answers are already known, so I stress the fact that the resolution of this problem came during their own lifetimes. I highly urge you to visit the Solar Dynamics Observatory (SDO) website (http://sdo.gsfc.nasa.gov) the morning of your lecture(s) and grab the latest optical through 105

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106 ◆ Chapter 14 Our Star—The Sun x-ray movies of the past 48 hours of solar activity. The detail of the imaging is astounding, and there is something about seeing real-time sunspots, storms, and ejections (as opposed to images in a text) that inspires students. Two other easy demonstrations can be accomplished in a quick “field trip” outside. Students can point handheld spectroscopes (such as the Project Star Spectrometers at http:// www.sciencefirst.com) at clouds, buildings, or the ground and observe the Sun’s absorption lines. Poking a small hole in a piece of cardstock allows you to safely show the photosphere projected on the ground. Alternatively, you can look directly at the Sun using eclipse glasses, welding goggles rated No. 14 or higher, or inexpensive aluminized Mylar filters. Of course, repeat 27 times to your students never to look directly at the Sun, even for a second, without these special filters. If it helps, tell them that I looked directly at the Sun through cloud cover during an eclipse, sunburned my retinas, and could not see for a week.

DISCUSSION POINTS

▶ Discuss everyday examples of energy transport by con-

duction, convection, and radiation. ▶ Discuss how the solar neutrino problem exemplifies the scientific method. ▶ Real-time images and movies of the Sun are available in stunning detail from the SDO satellite at http://sdo. gsfc.nasa.gov. Display some of these and discuss whether features or events from this chapter are seen in the data. ▶ Look at the three pictures of the Sun displayed in Figure 14.22; each was taken using a different wavelength of light. Discuss the differences and similarities revealed by looking at the Sun through different wavelengths of light. ▶ An active field of research is the investigation of whether or not variations in the solar activity may induce changes in Earth’s weather. A possible link between these two phenomena is the so-called Little Ice Age that took place during the Maunder Minimum. Discuss whether or not Earth’s global warming during the past century could be related to solar activity. ▶ Compare and contrast how seismology has been used to understand the interiors of Earth and the Sun.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 14 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises.

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The Solar Core This animation emphasizes the process of fusion that generates the radiation at the core of a star (such as our Sun) whose pressure balances the self-gravitation of all the gas out of which the star is made. Text reference: Section 14.1

ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what they have learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5) and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack. Random Walk This video provides a visual demonstration of the random walk of photons out of the Sun, by having students catch and throw a small ball in a lecture hall. This video shows how slowly a ball moves from the front to the back of the room, because of the random directions it takes. Text reference: Section 14.2

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (c) One clue is the name “hydro” because fluids always deal with pressure instead of force. 2. (b) Fusion works via E 5 mc2. 3. (c) Neutrinos stream easily out of the Sun, allowing us to detect them to corroborate our understanding of nuclear fusion.

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Chapter 14

4. (b) The photosphere’s thinness causes it to look well defined instead of fuzzy. 5. (d) Sunspots are still quite hot, just not as hot as the photosphere; thus, they look dark in comparison. Reading Astronomy News 1. A “CME” is a coronal mass ejection; that is, a large mass of solar material that is blasted into space from the Sun. 2. A CME contains huge numbers of charged particles, which create electric and magnetic storms. This could disable all electronic devices, erase all hard drive data, and even set ablaze anything connected to a telephone wire. 3. The ejection was made in a direction away from the Earth; thus, it passed the Earth without incident. 4. Modern society, without electricity, computers, and telecommunications lines, would be a radical change to almost every aspect of our current way of life. One can imagine that if a storm were strong enough and knocked all of these out permanently, we would be forced to adopt a much lower-tech lifestyle, perhaps not as bad as the Middle Ages, but certainly rivaling that of the late 1800s. 5. The video is the same as this article, but with added graphics. From the video: The Carrington Event of September 1859 was a series of powerful CMEs that hit Earth head-on, sparking Northern Lights as far south as Tahiti. Intense geomagnetic storms caused global telegraph lines to spark, setting fire to some telegraph offices and disabling the “Victorian Internet.” The video does help illuminate some of the pieces of the article through graphics and animation, which makes it more helpful to a nonspecialist. Test Your Understanding 1. (a) Only neutrinos can escape from the center of the Sun unchanged, and helioseismology allows us to probe the Sun’s interior just as earthquakes on Earth allow us to probe our planet’s core. 2. e 5 a 5 g; f; h 5 b; d 5 c. See Figure 14.4. 3. (a) Magnetic activity causes these phenomena, as explained in the text. 4. (b) The first statement is hydrostatic equilibrium, and the second is energy equilibrium. Both conditions are necessary for a star. 5. (b) Fusion requires protons to overcome their mutual repulsion, which requires very high temperatures and densities.

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Our Star—The Sun ◆ 107

6. (b) Sometimes, not finding what we are looking for indicates that we do not understand that thing, not that the thing is missing. 7. (c) Sunspots are a magnetic phenomenon. 8. (d) Neutrinos stream out of the core of the Sun almost instantaneously. 9. (d) There are very few atoms emitting energy in the corona, so it can be very hot and very faint. 10. (c) As the Sun spins, Earth moves around the Sun in its own orbit. 11. b-c-d-f-g-e-a This is shown graphically in Figure 14.10. 12. (b) Charged particles are effectively trapped along magnetic fields, so when those field lines break, the particles are free to escape. 13. (c) See Figure 14.10. 14. (d) Nuclear fusion requires very high temperatures (tens of millions of degrees) and pressures, which are only found at the centers of stars. 15. (c) The solar wind is a stream of neutral and charged particles streaming out of the Sun, which carries with it the Sun’s magnetic field. It is the influence of the Sun’s charged particles with our magnetosphere that changes its shape. Thinking about the Concepts 16. If the star starts making too much energy, hydrostatic equilibrium causes the star to expand, which releases that excess and slows the production of energy in the core. Having lost the excess energy, gravity contracts the star until hydrostatic equilibrium is again achieved, and the star is back to its original size, temperature, and luminosity. As a result, the star keeps its energy generation and energy loss at a constant, self-regulated value. 17. Nuclei all have positive charges, which repel each other. To fuse two atoms, very high temperatures (millions of degrees), pressures, and densities are required so the two nuclei can overcome this amazingly strong repulsion and fuse. The oceans, and indeed every natural environment on Earth, do not satisfy these conditions. 18. Neutrinos are very hard to detect because they carry no charge, travel at nearly the speed of light, and interact very weakly with matter. To detect a particle, we have to watch the reaction of that particle striking or scattering off another one; because neutrinos hardly interact with matter, it is very rare that we detect them. 19. In the proton-proton chain, four hydrogen atoms do not come together at one time to produce a helium atom. Rather, the hydrogen atoms are slowly built up to helium via the proton-proton chain. First, two

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108 ◆ Chapter 14 Our Star—The Sun hydrogen atoms come together to form a heavy hydrogen or deuteron. Note that for this to occur, one proton must turn into a neutron, which releases a positron and a neutrino. Then the deuteron fuses with a proton to form light helium. Two of these light heliums then fuse to form a helium nucleus with two hydrogen nuclei left over. This is how the pathway looks: 1 1

H  11 H → 21 H  v  e

2 1

H  11 H → 23 He

3 2

He  23 He → 42 He  2 11 H

20. Fission requires that a large number of heavy elements already exist. The Sun is made up almost exclusively of light elements (H and He, for the most part), so there are very few nuclei available for fission within the Sun, or any other star. 21. If a huge amount of hydrogen suddenly fused in the Sun, the core would heat up, which would cause it to expand. The core would then cool down, fusing less hydrogen per second and producing less energy. The core would then contract, and hydrostatic equilibrium would return; the Sun would be stable again. Now all that energy needs to get out of the Sun; some of it will go into making the Sun expand a little, which would also make the Sun become a little brighter. Of course, we might not see this for hundreds of thousands of years. 22. Answers will vary. If neutrino oscillations had not been discovered, a variety of other neutrino-oscillation detection experiments would have to be conducted to conclusively rule out this idea. If the oscillations still were not discovered, then either our understanding of nuclear physics and/or how fusion powers the Sun would need to be revised. 23. The solar neutrino problem stems from a measurement of the solar neutrino flux that was only onethird as large as the flux predicted by stellar structure models. Initially, astronomers thought there was a problem with the stellar structure models of the Sun, but today we know that neutrinos carry a tiny mass that allows them to oscillate between three different flavors: electron, muon, and tau. Our detectors on Earth are sensitive only to electron neutrinos, but at any given time there is only a one-in-three chance that a neutrino will be an electron neutrino. Thus, our measurements of solar neutrinos missed the remaining two-thirds of the flux that contained muon or tau neutrinos. 24. The Sun is made only of gas, so it has no actual surface at all. The photosphere is as deep into the Sun as we can see because it becomes opaque at the base of

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this layer. Think of it a little like looking down into the ocean—at some depth one can no longer see any further. This is not a solid surface, just the depth at which the water becomes too opaque to see through. 25. The winds from solar flares contain huge amounts of charged particles, which can disrupt or destroy electronics. Thus, a flare could knock out any satellites (including telecommunication, spy, and weather satellites). Also, periods of increased solar activity cause Earth’s upper atmosphere to expand slightly. This increases the drag on orbiting satellites, causing their orbits to decay. Unless they are boosted back up again, this effect can cause them to crash to the ground much sooner than expected (or desired). 26. The solar corona is the outermost gas of the Sun, which has very high temperatures and low density. Because it is faint, we can see it only if we block out the light of the rest of the Sun, either during an eclipse or with a special circular disk called a coronagraph. 27. By Einstein’s E 5 mc2, the difference in mass becomes energy used in binding the new nucleus together. 28. Sunspots tell us the Sun’s overall rotation period, and more importantly, that the Sun has differential rotation, whereby the equator rotates faster than the poles. 29. Different parts of the Sun give off their light at different electromagnetic wavelengths, because of their temperatures or the particular radiative processes that happen in those regions. Anything that gives off UV or x-rays is best studied from space (the chromosphere and corona); however, because our atmosphere distorts sunlight, studying the photosphere at high resolution is useful from space. 30. The interaction between the solar wind and interstellar medium is an important region of study, as it probes our understanding of both environments. But for us in particular, the heliosphere protects us from high-energy cosmic rays; thus, when the Sun is in a period of lower activity, we are more susceptible to damage from these particles. Applying the Concepts 31. Setup: To determine whether a graph is linear or logarithmic, check whether major tick marks increase by a constant additive factor or by powers of 10. Solve: (a) The height axis increases by adding 1000 km to each tick mark; thus, it is linear. (b) The density axis increases by a multiplicative factor of 100 for each tick mark; thus, it is logarithmic. (c) The temperature axis increases by a multiplicative factor of 10 between each tick mark; thus, it is logarithmic.

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Chapter 14

Review: The range of values of the Sun’s characteristics (like temperature, density, or pressure) is very large and thus requires a log scale to show all of those values. 32. Setup: Figure 14.19 shows (b) the location and number of sunspots versus time and (c) the strength and position of magnetic fields versus time. If we are to argue that spots result from strong magnetic fields, then they should appear in roughly the same regions over time. Solve: Inspection of the figures shows that the butterfly shape seen in panel (b) is reproduced in panel (c), confirming that the two effects are correlated. Review: If sunspots were not correlated with magnetic field positions, then we would expect there to be no visual similarity between the two panels. 33. Setup: Figure 14.18 shows a small patch of the Sun’s surface with a high-resolution image of sunspots that are present in that area. Figure 14.19 shows the number of sunspots present as a function of time over the last few centuries. We can answer (a) with Figure 14.18, (b) and (c) with Figure 14.19, and (d) with Figure 14.24. Solve: (a) Looking at Figure 14.18b, it appears that about 10 percent of the square is covered with sunspots. However, this little square is only a tiny fraction of the entire Sun’s surface. The box looks to be about one-fifth of the radius of the Sun, so if its radius is 5, the area of the Sun is 4pR2 5 314, and the box has an area 1 5 0.3 of about 1 3 1 5 1; therefore, only about 314 percent is covered by these spots. (b) The average number of spots that occur at solar maximum varies between 100 and 200 spots. (c) On average, near solar maximum, there are hundreds of spots, whereas in question (a), the date was near solar minimum. This would suggest that at maximum, the fraction of the Sun covered by spots is a few hundred times higher, that is, 2 to 3 percent of its surface. (d) Compared to the graph of irradiance shown in Figure 14.24, we see that there is a direct correlation between the number of spots and the Sun’s brightness, which makes perfect sense because the more spots that are present, the more of the Sun’s hot surface is covered. Review: Note that although a small percentage of the Sun’s surface may be covered with spots, the overall energy coming out of the Sun (the right-hand panel in Figure 14.24) is essentially constant with time, that is, this is a very small effect. 34. Setup: For a gamma ray to emerge 2.3 seconds after being formed, it must travel straight out of the Sun. Is this possible? Solve: We have learned that a photon scatters around in the Sun, traveling about 1 cm before hitting another

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Our Star—The Sun ◆ 109

atom. Thus, the distance required for a photon to escape the Sun is much longer than the straight path out. Review: Note that neutrinos can stream freely out of the Sun, so they emerge in about 2.3 seconds. M M 35. Setup: Density  5  3. V R Solve: (a) Putting in the values for the Sun’s mass and size as compared to Earth, we find that the Sun’s den300 , 000 sity is 5 0. 3 times that of Earth, or about 1003 1,650 kg/m3. (b) This is about 65 percent more dense than water. Review: We know the gas giants Jupiter and Saturn have very low densities and are made mostly of hydrogen and helium, like a star, so it stands to reason that a star’s density will be low as well. 36. Setup: Einstein’s equation E 5 mc2 can be solved for mass: m 5 E/c2. A joule is 1 kg m2/s2 (note that the units match Einstein’s equation). Note that a watt is a joule/ second so the amount of energy (in joules) given off in 1 second is numerically equal to the luminosity of the star. Solve: Inserting the solar luminosity and speed of 3.85 3 1026 kg m 2 /s2 E 5 light, we find m 5 2 5 c (3 3 108 m/s)2 4.3 3 109 kg. There are 1,000 kilograms in a metric ton, so this mass equivalent is 4.2 million metric tons burned each second. Review: The mass burning rate of the Sun is about 4 billion kilograms per second, or about 9 billion pounds per second. Although this may seem huge, remember the mass of the Sun is 2 3 1030 kg, so the amount of mass burned is tiny in comparison. 37. Setup: This problem asks for the mass lost over a given amount of time. In the previous problem, we found the mass burning rate is 4.2 3 109 kg/s, so the mass lost in a given time is just the rate times the time. To calculate the percent mass lost, remember M % mass 5 lost 3 100 M total Solve: (a) If the Sun’s rate of burning has remained constant over the last 4.6 billion years, then the kg total mass lost is Mlost 5 4.2 3 109  4. 6 3 s 7 3.15 3 10 s 109 yr 5 6.0 3 1026 kg. (b) The yr fraction of the Sun’s mass that has already been converted to energy is 6 3 1026 kg % mass 5 3 100 5 0.03 % . 2 3 1030 kg

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110 ◆ Chapter 14 Our Star—The Sun Review: Note that despite converting 100 Earth masses to energy in 4.5 billion years, the Sun still retains an astounding 99.97 percent of its original mass! 38. Setup: Use Working It Out 14.1. mass so this star will Solve: Lifetime scales as Iuminosity 2 1 5 as long as the Sun. live about 12 6 Review: Another test is that lifetime scales roughly as 1 M–2.5 and 2–2.5 5 or about one-sixth. 5.7 39. Setup: Neutrinos and photons both travel at the speed of light, and the Sun is about 8.3 light-minutes away. Solve: (a) Neutrinos hardly interact with any other matter, so they essentially stream freely out of the Sun. Thus, they will arrive at Earth 8.3 minutes after they are created. (b) A photon takes about 300,000 years or longer to travel from the Sun’s core to its interior, so any physical change that occurs in the heart of the Sun will not be seen in visible light until 300,000 years after the change occurs (or longer). Review: Another good example of this is SN 1987A, which was observed in neutrinos before the visible light reached us because the neutrinos streamed right out of the collapsing star, whereas the light took much longer to escape. 40. Setup: The Sun is 149.6 million kilometers from Earth (on average). To find the time it takes for a particle in the solar wind to reach us, remember d 5 vt. 8 d 1.49 3 10 km Solve: Solving for time, t 5 5 5 v 400 km/s 3.72 3 105 s. Review: In other words, if a solar flare happens today, it will not reach Earth until 4 to 5 days later. 41. Setup: Because the photosphere and sunspot both act like blackbodies, we know that the brightness b  T4. 4  Tspot  bspot 5  . We can construct a ratio b  T  Solve: The sunspot is 70 percent as bright as the photosphere, which means in our ratio above bspot

5 5,780 4 0.7 5 5300 K. b Review: Blackbody brightness scales strongly with temperature, so the spot appears much fainter than the photosphere because its temperature is lower. 42. Setup: Solving E 5 mc2 for the mass will yield the amount of mass that is converted to energy. To figure out how much energy would come from our textbook, we will consider how many 5-megaton bombs are in the book, based on our answer to part (a). Tspot 5 T 4

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Solve: (a) The mass equivalent to one bomb is E 2 3 1016 J 5 5 1.1 3 103 kg or c2 (3 3 108 m/s)2 1.1 gram. Because the book weighs about 1 kilogram, there are about 909 bombs worth of energy in one textbook. (b) If one 5-megaton bomb uses up 1.1 gram of mass, and the Earth’s mass is 6 3 1027 g, then the Earth 1.1 lost 3 100% 5 1.8 3 1026 percent. 6 3 1027 Review: This problem shows us just how much energy there is inside a small amount of mass and why fission and fusion are such amazingly energetic processes. 43. Setup: This is an example of a Fermi problem in which one makes many assumptions to estimate a very complicated (or seemingly impossible) problem. The important thing to remember is to document all the assumptions and assume that if you overestimate some, you will underestimate others and your final answer will be within a factor of 10 (or so) of the real value. Solve: Some estimates give the yearly consumption of power on Earth at 10 GW (10 billion watts). A watt is a unit of energy consumption, that is, energy per second, so the total amount of energy in 1 year is J s J 5 3 3 1020 . A “J” or joule is 1013  3 3 10 7 s yr yr the amount of energy released by 1 watt in 1 second. Over 500,000 years, we will consume about 1.5 3 1026 J, whereas in 1 second, the Sun releases 4 3 1026 J. So we see that in 1 second, the Sun does indeed out produce all that the existing power plants could produce in over a half million years. Review: Earth-based problems like this sometimes help us to understand just how “large” a large number is, such as how much energy is produced by the Sun. 44. Setup: Let’s treat this problem by rounding numbers to powers of 10. For example, the mass of the Sun is about 1030 kg, and it is mostly hydrogen, which has an atomic mass of 10–27 kg per atom. Thus, there are about 1030 kg 5 1057 atoms. The text suggests that 1027 kg/atom each atom could produce 10–19 J of energy in a chemical reaction. Solve: The total amount of energy stored in the Sun under chemical reactions would be the product of [the energy per reaction] times [the number of possible reactions], which is the number of atoms. Thus, the total energy available is about 10–19 · 1057 5 1038 J. At an energy-output rate of 1026 W, the Sun would use up its fuel in 1012 seconds or a few hundred thousand years. m5

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Chapter 14

Review: Given that Earth is some 4 billion years old, this problem shows us that the Sun clearly cannot be powered by chemical reactions. 45. Setup: Use all values in MKS units and the lifetime will be in seconds. Remember there are 3.15 3 107 sec in one year. Solve: 11 2 30 2 GM 2 (6.67 3 10 J m/kg )(2 3 10 kg) 5 5 (7 3 106 m)(3.8 3 1026 J/s) RL 1.1 3 1015 s or about 33 million years. Review: The Sun can, and did, power itself via gravitational energy for a few million years, when it was a protostar contracting to land on the main sequence and initiate fusion. Using the Web 46. (a) From the website: “The data will allow researchers to learn about solar storms and other phenomena that can cause blackouts and harm astronauts.” This satellite is providing the highest resolution measurements ever, with unprecedented time resolution. (b) Answers will vary based on the sunspot activity present. Magnetic fields are strong in the neighborhood of any sunspots; and in the AIA images, one sees more field lines coming from these regions than elsewhere. (c) STEREO “consists of two space-based observatories—one ahead of Earth in its orbit, the other trailing behind. With this new pair of viewpoints, scientists will be able to see the structure and evolution of solar storms as they blast from the Sun and move out through space” (from the Web page). 47. (a) The science mission of IRIS is to “observe how solar material moves, gathers energy, and heats up as it travels through a little-understood region in the sun’s lower atmosphere” called the Transition Region (from the Web page). At the time of writing this manual, IRIS is just past its expected 2-year mission duration, but science results were very sparse. (b) SOHO images are much smaller and lower resolution than those from SDO. (c) Magnetic fields occur in a vast array of astrophysical processes, and their effects are critical to understanding nearly any dynamic process. The Sun is a nearby laboratory for “magnetohydrodynamics” and therefore offers a prime resource. DKIST will observe in the near UV (0.3 to 0.35 nm). The location is optimal as it offers a stable atmosphere and occurs at latitudes that maximize the number of daylight hours. The facility is scheduled to be completed in or around 2016. 48. (a) Answers will vary. The answer will include whether there were flares today, the sunspot number, how many

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Our Star—The Sun ◆ 111

sunspots would be expected (based on where we are in the solar cycle), and whether there are any coronal holes. (b) Answers will vary. Students will save a screenshot of classification of sunspots. (c) No answer needed. 49. (a) From their website: “All of the energy of NIF’s 192 beams is directed inside a gold cylinder called a hohlraum, which is about the size of a dime. A tiny capsule inside the hohlraum contains atoms of deuterium (hydrogen with one neutron) and tritium (hydrogen with two neutrons) that fuel the ignition process . . .NIF was designed to produce extraordinarily high temperatures and pressures—tens of millions of degrees and pressures many billion times greater than Earth’s atmosphere. These conditions currently exist only in the cores of stars and planets and in nuclear weapons. In a star, strong gravitational pressure sustains the fusion of hydrogen atoms. The light and warmth that we enjoy from the Sun, a star 93 million miles away, are reminders of how well the fusion process works and the immense energy it creates. Replicating the extreme conditions that foster the fusion process has been one of the most demanding scientific challenges of the last half century. Physicists have pursued a variety of approaches to achieve nuclear fusion in the laboratory and to harness this potential source of unlimited energy for future power plants.” (b) A Google search for He3 on the Moon suggests that China, Russia, Germany, and Japan may be investigating helium on the Moon. 50. (a) As of mid-August of 2015, Voyager 1 is 132.27 AU from the Sun, whereas Voyager 2 is 108.78 AU from the Sun. Voyager 1 is believed to have crossed into interstellar space whereas Voyager 2 is still inside the “heliosheath,” the boundary between the solar system and interstellar space. (b) Answers will vary. IBEX has been taking data since 2009, but results have been sparse since early 2014. As of late 2013, results included: discovery of rapid (6 month) time variations in the heliosphere-interstellar interaction; first observations of the heliotail and the influence of the interstellar magnetic field on it; discovery of our Sun’s slower motion with respect to the interstellar medium; discovery that the heliosphere is likely to have a bow wave and no bow shock; and discovery that the very local interstellar medium is rotating ahead of the heliosphere. Exploration 1. The positron carried away the positive charge. 2. A neutrino is an almost massless particle that is produced in nuclear processes involved with electrons and positrons. The neutrino is produced in this reaction.

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112 ◆ Chapter 14 Our Star—The Sun 3. Yes, the same reactions happened in both panels. 4. Deuterium and hydrogen nuclei entered the reaction, and light helium (3He) resulted. 5. Two positive charges came in and two left so charge was conserved. 6. A gamma ray is a high-energy photon. A gamma ray was produced during the reaction. 7. Two 3He entered the reaction, and an ordinary helium, 4 He, was produced. 8. Because six hydrogen nuclei are used but two are returned, it is possible to consider two of the hydrogen nuclei as catalysts.

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9. The gamma rays produced during these reactions leave the Sun as light. Note that the positron will annihilate and produce two further gamma rays. 10. The two hydrogen nuclei that are released at the production of 4He could be involved in further reactions. Also, the positrons will be used almost immediately when they annihilate with free electrons. 11. The helium nuclei that are produced will stay as they are for a long time.

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CHAPTER 15

The Interstellar Medium and Star Formation INSTRUCTOR’S NOTES Chapter 15 covers the interstellar medium and star formation. Major topics include

▶ the different components, densities, and phases of the

interstellar medium ▶ the effects of interstellar dust on starlight ▶ molecular clouds as sites of star formation. ▶ stability, collapse, and fragmentation of molecular clouds ▶ formation and evolution of a protostar ▶ hydrostatic equilibrium ▶ brown dwarfs ▶ Young-Stellar objects, such as Herbig Haro and T Tauri, and bipolar outflows With an entire universe to cover in just one or two semesters, it is all-too-easy for students in our classes to fall into the trap of treating each lesson/chapter/section/module/ week as information that can be learned independently and is treated as unrelated to any other material. Borrowing a page from Fight Club, addressing this “single serving” effect can be one of our greatest pedagogic challenges. As I have noted throughout this manual, one of my teaching goals is to show students how a small number of physical laws and principles can be used to interpret a huge variety of situations. This chapter offers the perfect opportunity to address that head-on, because here we return to old material in new ways. For example, we revisit much of Chapter 7, but focusing on the central star rather than the planets. We return to Chapter 5, to discuss waves, blackbodies, and atomic transitions. We introduce hydrostatic equilibrium, a (Section 5.4) balance of gravity (Chapter 4) with pressure. I think it is important to discretely point out these correspondences, to show students that much of science is how the same players take on very different roles. In previous chapters, I have often urged the instructor to cut down some of the content in the chapter and use broader brushstrokes. Here, I often find myself going the opposite direction, but not always. I say very little about the classification of brown dwarfs or the role of H2 as a source of opacity in protostars, for example, because they are a fairly specialized topics which require more time

than I have available to make them readily understandable. On the other hand, I spend extra time in many sections, because I think a little more is needed than is in the text itself. I like to show images of similar galaxies or stars at different distances from Earth, so students can directly see the effects of extinction and reddening. I also like to show dramatic images comparisons of visible extinction and infrared transmission, such as the dust lane in the Sombrero galaxy, as imaged by Hubble and Spitzer. On the topic of molecular clouds, the book does not entirely address the question of why stars form there as opposed to other places in the ISM. Section 15.2 can be supplemented by a discussion of the ideal gas law and hydrostatic equilibrium: if a cloud starts to collapse, the gas will heat up and provide thermal stability against that collapse. But in a cold, dense, dark molecular cloud, collisions between atoms and dust efficiently cools the gas so it can continue to collapse. This situation is not present in other phases of the ISM. Dust not only reddens and extinguishes light, but it also reflects that blue light, creating eerie blue “reflection nebulae.” This can be seen in the bottom left of Figure 15.8a, and surrounding the Pleiades cluster in Figure 15.24. It is worth a quick drawing showing how the stars must be in front of the dust for us to see the reflected blue light, and furthermore, that every dust cloud that we see extinguishing starlight from Earth could look like a blue reflection nebula to an observer on the other side of the cloud. Fragmentation and collapse happen because the clouds are larger than their Jeans’ radii. One can make a hand-wavy derivation of this by comparing thermal versus gravitational energy or by using a plastic water bottle to demonstrate dynamic stability and instability. Fill the bottle with a few inches of water, set it the correct way on a table, and blow on the bottle. Flick it. Even shove it. It will not fall over because it is dynamically stable. Turn it over and balance it on its cap and, even with a gentle breath, you can knock it over. There is a lot of detailed physics buried in Figure 15.20, which you may want to avoid. Why do low mass stars 113

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114 ◆ Chapter 15 The Interstellar Medium and Star Formation drop almost vertically, then shoot up and to the left on the H-R diagram, whereas high-mass stars move almost horizontally to the left? Why do the stars take little backand-forth jogs right before they land on the main sequence? Rather than show the evolutionary tracks, I instead focus more on the collapse time for protostars of different mass (see “The Influence of Mass” subsection at the end of section 15.4). That higher-mass stars evolve faster is a common theme throughout the next few chapters. It can shock students to learn that a 60 solar mass star can collapse to the main sequence in just a few tens-ofthousands of years whereas a star such as our Sun takes tens of millions.

DISCUSSION POINTS

▶ Discuss how stars and planets contain only the material

Star Formation This animation starts with the spiral arm of a galaxy and moves to the interstellar medium and then to giant molecular clouds. We then show turbulence and fragmentation, and finally we see the formation of protostars, protostellar disks, and jets. Text reference: Section 15.2

from the interstellar clouds from which they formed.

▶ Discuss what role stars play in the temperature of the gas

and dust clouds in which they are situated. ▶ In this chapter we learn that stars form inside molecular clouds. Why? What is special about molecular clouds, as opposed to other regions of the interstellar medium? ▶ How likely is it that the Sun has siblings that formed from the same parental molecular cloud? Discuss with students whether those solar siblings exist and, if they do, where they might be? ▶ Why are the locations of molecular clouds and H II regions generally correlated? ▶ An interesting analogy to the collapse of a molecular cloud is the collapse of a house of cards. Have students imagine or actually build a house of cards that is a few layers high, and then flick one of the bottom cards. What happens to the cards in upper layers? What phenomena in interstellar space could cause the collapse of a molecular-cloud core? ▶ Interstellar dust makes everything look fainter and redder than it really is. Discuss what we would conclude about observations of distant objects if we did not know about these effects. ▶ Compare and contrast the discussion in this chapter of collapse of a molecular cloud to a protostar to the discussion in Chapter 7 of solar system formation.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 15 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises.

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Angular Momentum This simulation is a kinesthetic demonstration of how angular momentum depends on mass, distance, and rate of spin, in which Dr. Palen stands on a rotating platform and moves her arms in and out. Text reference: Section 15.2

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (d) Interstellar dust both dims and reddens distant objects. 2. (c) Collapse takes place when regions become denser than their surroundings. 3. (a) Gravitational energy is converted to thermal energy (increasing the temperature) as the protostar compresses itself. 4. (a) Molecular clouds are dark and dusty but also the cradles of star formation, making most of the protostars hidden from view. Reading Astronomy News 1. Answers will vary. Typically scientists want to identify interstellar dust grains because that is the material out of which the solar system—and we—formed. 2. The researchers looked for chemical signs of interstellar origin and travel to distinguish interstellar dust from solar system dust. They also looked for dust entering away from the Sun in the direction of the interstellar dust stream.

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Chapter 15 The Interstellar Medium and Star Formation ◆ 115

3. A complete analysis of these particles will involve destroying them, for example in scanning-electron microscopes. Thus, they cannot be further studied. 4. Citizen scientists aided in searching for trails in the aerogel and informing the astronomers of these discoveries. This rapidly sped up the process of analyzing the gel. 5. This problem sends the user to Stardust@Home, to look for these tiny tracks of dust in the aerogel. New results will vary. Test Your Understanding 1. a, b, d. It is hard to make hot gas very dense, because the pressure is very high (note that a star is really a plasma). 2. (b) Although the effects of dust can be observed in the optical (extinction), dust can be directly observed by its thermal glow in the infrared. 3. (a) The Sun’s composition of heavy elements is fairly close to the interstellar medium because the Sun formed out of this gas and it takes a very long time to significantly change the chemical composition of a star. 4. (c) Dust grains are about the size of optical and UV wavelengths, which make them very effective at stopping those “colors” of light. 5. (c) Supernovae (and very bright, hot stars) are quite effective at heating up interstellar gas to extremely high temperatures. 6. (a) Spectra tell us the composition, density, and temperature of the interstellar medium. Note that the Sun’s composition does tell us about our local interstellar medium; although a satellite would be able to tell us about that material as well, no satellites have reached the actual interstellar medium yet. 7. (a) “Bi” means two and “polar” means “the end” so bipolar means related to both ends. 8. (c) The process of star formation is too slow to see it happen in real time, but we have observed many objects at different stages, allowing us to piece together a complete picture of the process. 9. (d) When the spins of the nucleus and electron flip with respect to each other, a 21-cm photon is emitted. 10. (c) A vertical line on the H-R diagram means changing luminosity at constant temperature (or color). 11. (a) Gravity pulls inward, and pressure pushes outward. 12. (c) A bar or lane of dust in the distant galaxy is blocking its light from reaching us, making the region look dark. 13. (d) A 21-cm photon is released when the spin state of an electron flips over, from being aligned with the nucleus to unaligned.

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14. (a) The dark bands that we see are either a conspicuous lack of material (which is unlikely) or the presence of something such as dust-blocking light. We have confirmed the presence of dust by looking at these objects in the infrared. 15. (c) The mass of a star dictates nearly all aspects of its evolution. Thinking about the Concepts 16. Dust grains are comparable in size to visible-light wavelengths, so they act as a barrier to visible light. Gas molecules are much smaller than dust grains, so they do not obstruct light at visible wavelengths. 17. Star formation occurs only where the interstellar medium is dense, cold, and dusty. Because of the dust, these regions are hidden from view in visible light. However, infrared radiation is less affected by interstellar dust, making it easier to peer through the veil at these wavelengths. In addition, the low surface temperatures of protostars mean that they emit significantly in the infrared region of the spectrum. 18. Interstellar gas clouds are dense and are distributed sporadically because only about 2 percent of the volume of our galaxy is filled with them. The rest of the galaxy is filled with hot, low-density intercloud gas. 19. Most of the material of the cloud collapses into the disk that forms the star(s) and planets of that system. If the cloud is large enough, then gas from the cloud can remain and enshroud the star. If the star is bright (O or B star), then it will ionize the gas and can disperse it through its powerful radiation. 20. Temperature measures the average kinetic energy of particles in a system, whereas heat measures the total thermal energy of the system. Even though the water is 100°C cooler than the oven, the number of hot-water molecules touching your skin in boiling water is far greater than the number of hot-air molecules touching your skin in the oven. So even though the oven is hotter, your skin experiences far more collisions each second with water molecules in the pan than it does with air molecules in the oven. As a result, the boiling water transfers heat to your skin much more rapidly than the oven does. This same principle applies in the interstellar medium. Even though temperatures reach millions of kelvins in the space between stars, there are very few particles to carry this energy. Assuming that you could float in the interstellar medium, the density of the surrounding material is so low that your body would radiate its internal heat much faster than it would absorb heat from the surrounding environment. In

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116 ◆ Chapter 15 The Interstellar Medium and Star Formation other words, despite a temperature that is measured in millions of kelvins, you would freeze rather than cook in the interstellar medium. 21. Scientists believe we are located in a large bubble of hot gas because much of the interstellar medium surrounding the Solar System is very hot and very low density, whereas outside a radius of about 650 ly, the interstellar medium shows all of its different phases. 22. Tenuous gases have very few collisions with other atoms or dust; thus, the atoms have a hard time getting rid of their energy, and as a result, they stay hot. On the other hand, dense gases have frequent collisions; thus, they can cool easily. In general, cooling scales as density squared. 23. We observe 21 cm radiation from all directions in the sky and across the universe, showing that the laws of quantum physics we discovered here on Earth are valid everywhere. 24. In interstellar clouds, no mechanism (such as gravity) exists to separate hydrogen from the carbon and oxygen, so we can safely assume that if carbon and oxygen are present in the cloud, hydrogen is present as well. When temperatures are low, carbon and oxygen combine with each other to produce carbon monoxide (CO). Under those same low-temperature conditions, atomic hydrogen will chemically bond with itself to produce molecular hydrogen (H2). Thus, by observing CO emission and making the assumption that if there is CO, there is also molecular hydrogen, we can map the distribution of molecular hydrogen in giant molecular clouds (and we have done so for the whole galaxy). 25. A giant molecular cloud (GMC) is a cradle of star formation. It is a huge, cold, dense, and dark interstellar gas cloud. In these clouds, gas is sufficiently cold and of such high density that small perturbations can cause a cloudlet to collapse and form a star. 26. As the cloud collapses, the central regions become more massive, more compressed, and hotter (all of the gas releasing its gravitational energy to collect at the center of the cloud heats up this gas to very high temperatures). When the central temperature of the cloud reaches a few million degrees, hydrogen fusion begins, and this energy source opposes collapse. Also, the gas rotates about a common center and this orbital motion (that is, angular momentum) opposes collapse. 27. As a protostar collapses, more mass implies higher gravity, but that higher gravity compresses the gas further. This compression raises its temperature and thus its pressure, so the star maintains hydrostatic equilibrium at all times.

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28. In the range from about 10 to 80 times the mass of Jupiter, it is difficult to classify a brown dwarf as either a planet or a failed star. Some brown dwarfs in this mass range can temporarily sustain low levels of nuclear fusion in their cores; others cannot. Some are hot enough to produce a visible light glow, but others are not. In the absence of a formal definition, I would be hesitant to classify anything larger than about 10 Jupiter masses as a planet. The term brown dwarf already describes these objects well. 29. H2 is a source of opacity, in that once it exists in a gas, it is able to absorb photons and use that energy to split the extra electron off of the ion, rather than allowing the photon to escape. If a protostar cools to the point that a lot of H2 can form, those ions make that gas opaque and thus trap radiation trying to escape, which heats the gas back up. On the other hand, a hot star will ionize the H2, making the gas less opaque, so that radiation can escape and the gas will cool. 30. A planet can only form out of the elements available, so a molecular cloud that has very few heavy elements will be unlikely to form any terrestrial planets. Higher levels of heavy elements help the gas cool faster, which helps the gas collapse faster and can affect the kinds of stars formed within the cloud. Applying the Concepts 31. Setup: Recall that dust grains are comparable in size to visible-light wavelengths, but they extinguish UV and blue light much more effectively than redder wavelengths. Solve: Dust “reddens,” which means it makes stars look redder than they really are. Thus, if we gauge the temperature of a star by its color, we will always estimate it as being cooler than it really is. Review: Note that dust also makes a star fainter, which makes us think it is farther away than it really is. 32. Setup: Review the material in Section 5.2 “The Quantum View of Matter” and Section 5.4 “Temperature.” Solve: (a) Proton and electron. (b) Because the kinetic energy depends on the mass and speed of a particle, if two particles have the same energy, the more massive one will travel slower. The electron is about 1,800 times less massive than the proton, so it will move much faster. Review: In general, astronomers talk about the electron temperature or density of a gas, and this is because the electrons move so much more rapidly and therefore collide more often, producing most of the pressure in a gas.

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M if we want to calculate from V first principles, but knowing the density of gas in the interstellar medium, we can estimate that of dust by noting that the mass of dust in the interstellar medium is about 100 times less than the gas component. Solve: We know the mass density of dust is about 0.01 times that of gas. Because the gas density is given around 10227 kg/m3, dust has a mass density of 10229 kg/m3, or dividing by the mass of a dust grain, 10229/10212 yields a number density of 10212 grains/m3. Review: The Earth has a volume of about 1 billion km3, which means that in a volume the size of the Earth, there would be about 1 million dust grains. That is not very much. 34. Setup: Look at Figure 15.3, which shows the spectra of a star seen with and without dust. Remember Wien’s 2 . 9 3106 nm  K , which we solve to find law,  max 5 T 2 . 9 3106 nm  K . T5  max Solve: We need a peak of about 460 nm for the star without dust and 720 nm with dust. These yield temperatures of about 6,300 K without dust, and 4,000 K with it. Obviously, the differences are significant. Review: Note that the star not only appears far too cool but its spectrum faded from a peak around 1 to 0.08; that is, the star became about 12 times fainter because of the dust. 2. 9 3 106 nm  K 35. Setup: Remember Wien’s law,  max 5 . T Solve: Inserting a temperature of 8000 K yields 363 nm. Review: Note that the gas will not actually radiate a blackbody spectrum peaking around 360 nm because this material is low density and thus only produces emission/absorption lines. 36. Setup: Wien’s law solved for temperature gives 2 . 9 3 106 nm K T5 , and remember that there are max

33. Setup: Density  5

1,000 nm in 1 mm. 2. 9 3 106 nm K 5 10 , 000 K . 1000 nm 0. 29 m 3 1 m Review: This implies that there are stars with a surface temperature around 10,000 K in the nebula. 37. Setup: To ionize a hydrogen atom, the electron must be stripped away from the nucleus, that is, given more Solve: T 5

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than its binding energy. If the atom is in the ground state, this is 13.6 eV, whereas in a typical excited state in the interstellar medium, this would require 3.4 eV. Now ask which stars produce more high-energy photons that would contain at least this much energy. Solve: (a) A B8 star is very hot, meaning that it produces many blue and UV (high-energy) photons that can ionize the hydrogen, whereas a K0 star is very cool and produces mostly red (low-energy) photons. (b) It is clear that the temperature of the star is most important if one wishes to ionize the interstellar medium. Review: Most bright nebulae glow in emission lines because there are hot stars (such as O and B types) that are heating the gas so that the atoms can produce emission lines in the optical. 1 38. Setup: Remember that kinetic energy K 5 mv 2 . 2 Solve: If a proton and an electron have the same 1 1 kinetic energy, we can write me v e2 5 m p v 2p , or re2 2 writing the equation to solve for the ratio of electron mp v . Finally, subspeed to proton speed gives e 5 vp me v stituting numbers gives e 5 1, 850 5 43. vp Review: The electron travels over 40 times faster than the proton, which explains why pressure and temperature are usually measured by the electron properties because these are the particles that interact the most. 39. Setup: This is a simple d 5 vt problem with conversions. 1. 5 3 1012 km 5 33,000 years. Solve: t 5 1. 5 km/s 3 (3 3 107 s/yr) Review: Of course, the actual time will differ because: (a) the particle is accelerating as it travels; and (b) no particles fall straight into the center, but rather they have to orbit in a spiral that very slowly reaches the center. This is why star formation takes much longer than just a few tens of thousands of years. 40. Setup: Because carbon represents 3/7 of the mass of a CO molecule, the mass of carbon in the molecular 3 cloud is MC in cloud 5 100 M 5 42. 8 M . Next, we 7 consider the mass of hydrogen in the Sun relative to the mass of carbon. The Sun contains 2,400 hydrogen atoms for every carbon atom. However, each carbon atom has a molecular weight of 12, whereas each hydrogen atom has a molecular weight of only 1. In other words, it requires 12 hydrogen atoms to equal the mass

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118 ◆ Chapter 15 The Interstellar Medium and Star Formation of a single carbon atom. Therefore, the Sun contains only 200 times as much hydrogen as carbon by mass. Solve: If we assume that the ratio of hydrogen mass to carbon mass in the Sun is representative of that in a molecular cloud, the molecular cloud of interest conM tains M H in cloud 5 H in Sun MC in cloud 52000 3 42. 8 M MC in Sun 5 8,572 M. Review: Note that the actual fraction of carbon, by mass, is 42.8/8,572 5 0.5 percent, which reinforces the fact that most of a cloud’s mass is hydrogen. 41. Setup: If it takes N seconds for something to happen on average, then the chances of it happening in 1 sec1 . The rate at which photons are emitted ond are just N is the probability of emission (p) times the number of objects (n) that can emit. Solve: (a) The probability that any atom makes this transition in 1 second is simply 1 p5 5 2. 86 3 10215. (b) As described in 3. 5 3 1014 the setup, the rate at which photons are emitted will be p 3 n, where n 5 6 3 1059 atoms. The rate of photon emission will be about 2.86 3 10215· 6 3 1059 5 1.7 3 1045 photons/sec. (c) Comparing this rate to the rate of photon emission from a solar-type star (1.8 3 1045), we find the rates are almost identical. Review: The difference between the 21-cm emission and that of the star is that the solar-type star primarily emits more energetic, visible-light photons, whereas the 21-cm neutral hydrogen radiation is found at the less energetic radio wavelengths. 42. Setup: There are 86,400 seconds in 1 day, so if we are compressing time, we must determine the relative fraction of a certain phase out of the Sun’s whole lifetime and then multiply that by the length of a day to find out the scaled time. Solve: (a) The collapse of the Sun takes only 0.3 percent of the Sun’s total lifetime: 30 3 106 yr 3 100 5 0 .33 percent 10 3 109 yr If we shrink the Sun’s lifetime down to a single day, the collapse time is tcollapse 5 (0.003)(86,400 s) 5 259 s or a little over 4 minutes. (b) The Sun spends one-third of its protostellar collapse on the Hayashi track, or 1 4. 32 min 5 1 . 44 min. 3 Review: If the Sun’s lifetime is compressed into a single day with its birth occurring at midnight, the

percent collapse 5

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entire protostar collapse period is finished by about 12:04 a.m. 43. Setup: L 5 4πR2T4, but we can write this in terms of the Sun’s values as L( L  ) 5 R(R  )2 T (T )4 , which can be solved for radius as R(R  ) 5

L( L  )

. T (T )2 3500 Solve: The protostar is 3,500 K, or about 5 0. 61 5770 the temperature of the Sun, with a luminosity 200 times higher. Inserting these in our formula, we find 200 R( R  ) 5 5 38 , that is, the protostar was (0. 61)2 about 38 times larger than the Sun. Review: We see clearly here how large protostars were, and also how a cool, but bright object can truly be a giant compared to the size of the Sun. 44. Setup: Brightness scales as 1/d2; 30 Doradus is 160 , 000 5 123 times farther than Orion. 1, 300 Solve: Orion is about 123 times closer than 30 Doradus, so the stars will be 1232 5 15,150 times brighter than they were. If they were 1/6 as bright as the faintest visible stars (to the naked eye), they would now become about 2,500 times brighter than those faintest stars. Review: The 30 Doradus region is full of the most luminous O- and B-type stars; this shows just how luminous those stars really are. 45. Setup: L 5 4πR2T4, but we can write this in terms of the Sun’s values as L( L) 5 R(R  )2 T (T )4 . Jupiter’s is about 10 times smaller than the Sun, and this brown dwarf will have a temperature that is 5.8 times smaller than the Sun’s (which we approximate here to be 5800 K). Solve: Using our relationship above, the brown dwarf 2

1  1  1  5 will have a luminosity L 5     10   5. 8  113, 000 or about 113,000 times lower luminosity than the Sun. 1 5 8. 84 3 1026. A student might also answer 113, 000 As such, we would need 113,000 brown dwarfs to equal the luminosity of the Sun. Review: A brown dwarf is indeed very faint, which shows why they will be so hard to detect directly around bright stars without specialized imaging techniques.

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Using the Web

Exploration

46. Answers will vary. The response will describe a few pictures of molecular clouds, where the pictures were taken, at which wavelengths, with which telescopes, which wavelengths the colors represent, and whether the images are true color or false color. 47. (a) Answers will vary. The response will describe a recent story from the Spitzer Space Telescope about star formation, including what was observed, which wavelengths are shown in the images, how such images (even in false color) help astronomers to analyze the images, and a discussion of why it is better to study star formation in the infrared than the optical. 48. SOFIA flies on an airplane to rise above most of the water vapor and turbulent atmosphere that hampers high-resolution images in the infrared. Answers will vary about what has been detected by the instrument. 49. Students will search for stardust as part of the citizen science project and submit a screenshot of their results. This project is to help identify dust particles that may have been captured with the Stardust spacecraft. 50. Answers will vary. The response will include whether the story on brown dwarfs is from an observatory, a NASA mission, or a press release. What is new and why is it interesting?

1. There are 20 coins. 2. I removed about 14 coins (or ions) to hit the 5,000 K mark. That leaves six in the star. 3. With ions removed, heat can escape and the star cools down. 4. I put about 10 coins (or ions) back, for a total of 16 to reach 5,000 K. 5. Now that there are more ions, the temperature of the star will rise. 6. Based on this pattern, the presence of ions heats up the star, at which point the ions are eventually destroyed, and then the star will cool down again. When it is cool enough, ions reform and the star heats back up. This is just a heating and cooling cycle. 7. The spreadsheet will contain this: star heats → ions are destroyed → star loses energy and cools → ions reform → star heats. Note: Obviously, inside the star, there are not 20 coins but huge numbers of hydrogen ions, which the coins represent. This Exploration was designed to show how the relative presence or absence of such ions can have a significant effect on the heating and cooling, and hence the size, of the star.

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CHAPTER 16

Evolution of Low-Mass Stars INSTRUCTOR’S NOTES Chapter 16 covers the stellar evolution, primarily focusing on low-mass stars. Major topics include

▶ how stellar mass determines main-sequence lifetime ▶ stages of single low-mass stellar evolution: the main sequence, subgiant, red giant, horizontal, and asymptotic branches, mass loss, planetary nebula, and white dwarf ▶ changes in the core and envelope, and location on the H-R diagram, as single low-mass stars evolve from the main sequence to become white dwarfs ▶ degeneracy and its effects in stellar cores ▶ evolution in close binary systems, including mass transfer, novae, and Type Ia supernovae

On the first day of class, I ask my students how many of them have an uneasy feeling about taking a science class, and after a few moments of hesitation, most students raise their hands. Once we start talking, a common theme is that they felt previous science classes had too much jargon and rote memorization. I truly believe that this introductory astronomy class can be taught with very little memorization of facts and figures, and I make it a common theme throughout all my lectures. Unfortunately, there is no way to avoid the fact that stellar evolution requires a lot of it. For example, what are the names of each branch or evolutionary phase? In relative terms, how long does each phase last? When will a star leave one branch and enter another? What elements are present in the core? Is each element burning or inert? Are any elements degenerate? What is the envelope doing? The list goes on. With all these facts and figures, it can be hard for students to move into the higher stages of learning (analysis and synthesis, to use Bloom’s taxonomy), such as why the star has its current structure or how a star will evolve/react given a certain internal state or change. My first solution is to be upfront with the class, that Chapters 16 and 17 are fairly memory intensive. The second solution, which might offend some purists, is that I brush more of the nuances under the proverbial rug to streamline the story and give a few less details to commit to memory. I do not discuss the small changes on the Hertzsprung-Russell (H-R) diagram that stars make while

on the main sequence. Similarly, I do not mention degeneracy in the RGB (Red Giant Branch) core nor the helium flash. The helium flash is dramatic, but it is not observable. A 5-M. star with a nondegenerate core does not experience a flash but still becomes a horizontal branch star. Also, high-mass stars still climb supergiant branches, but they have nondegenerate cores. So in my mind, the nuances of degeneracy and flashes can serve to confuse the story more than illuminate it. Another simplification is that I teach the horizontal and asymptotic giant branches as “the exact same thing that happened from the main sequence to red-giant branch, except now helium is burning to carbon in the core.” What I do focus on are the differences between core and shell burning, and between burning and inert gas; the snowball effect (Figure 16.5) that causes giant branches; and why these different processes move the star along the H-R diagram as we observe. My feeling is that by restricting the material to these topics, we can dive deeper into the physics and apply them more generally, not just to more readily understand horizontal branch and AGB stars, but to tie this chapter more concretely to Chapter 17 when we discuss high-mass stellar evolution. Because zombies are heavy in our zeitgeist these days, I like to end stellar evolution with binary stars and the story about Type Ia supernovae, if for no other reason than to keep students intrigued by dropping teasers like, “and Friday, we will learn how stars return from the dead.” For that reason, I often insert section 16.5 between sections 17.3 and 17.4. A wonderful resource for stellar evolution is http:// rainman.astro.illinois.edu/ddr/stellar/beginner.html, which contains a number of movies showing stars of different mass evolving across the H-R diagram. If you are adventurous, use the “Advanced Page” at the bottom of the screen on which you can create customized movies. You control the masses, metallicities, and duration of the movie and can add in starbursts and control the star-formation rate. I have made movies of 50 stars (from 0.1 to 50 M.) evolving over periods from 15 Myr to 15 Gyr. Not only do these show each star going through each of its evolutionary phases, but they also clearly show how H-R diagrams reveal the age of 121

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122 ◆ Chapter 16 Evolution of Low-Mass Stars the cluster. These can be interwoven between Chapters 16 and 17, to show the relative timescales of each branch of evolution and to help students visualize how the stars evolve through the H-R diagram. Whether we like to admit it or not, our students have grown up as a video generation, so it often helps to provide video and animation content whenever possible, even if the same things are clearly shown in static images in their textbook.

DISCUSSION POINTS

▶ According to the main-sequence lifetimes given in

Table 16.1, many generations of high-mass and intermediate-mass stars preceded the formation of the Solar System. Discuss what properties of the Solar System can be identified as coming from those previous generations of stars. ▶ Have students look for examples of degenerate matter and be prepared to discuss them in class. Discuss astronomical objects in which degeneracy pressure can be important. ▶ Compare and contrast the evolution from mainsequence to RGB with that from horizontal branch to AGB. If a star could start fusing carbon, predict how the star would subsequently evolve. ▶ How will the evolution of the Sun from its beginning as a protostar to its end as a white dwarf affect Earth? In particular, have students estimate or investigate whether the Sun will swallow Earth when it becomes an AGB star. Also, have students use the equations in Working It Out 5.4 to estimate the equilibrium temperature of Earth at various points of the Sun’s evolutionary track. ▶ Planetary nebulae have nothing to do with planets. Discuss why they bear such a potentially misleading name, and find other examples in which names are not in good agreement with the nature of the objects that they describe. ▶ Planetary nebulae are formed from stellar winds, and since stars are spherical, one might naïvely expect the nebulae to be as well. Have students investigate the morphology of planetary nebulae to test this hypothesis. Discuss what physical mechanisms could be at work to explain the diversity of planetary nebulae seen in those images. ▶ Compare and contrast novae and Type Ia supernovae. Will one always occur before the other? Are all binary stars doomed to experience one or both?

NEBRASKA SIMULATIONS Developed at the University of Nebraska2Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts

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presented in Chapter 16. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. H-R Diagram Explorer This simulation introduces the H-R diagram, a plot showing the relationship between luminosity and temperature for stars. The user not only sees the location of the star as one changes its parameters, but also visualizes its size and color with respect to the Sun. Text reference: Exploration

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (a) More massive stars have higher core temperatures and therefore burn their fuel faster. 2. (c) When the Sun is no longer able to burn any fuel in its core, it will be because the core is full of inert helium. 3. (a) The gravitational collapse of inert cores eventually allows them to heat up to the ignition temperature of helium. 4. (a) The “planetary nebula” is a misnomer and has nothing to do with planets; it is the outer layers of the star that have been blown into space and heated up by the exposed, hot white dwarf. 5. (d) To explode, the white dwarf must reach the critical mass of about 1.4 solar masses, which it does by accreting mass from its overgrown companion. Reading Astronomy News 1. In this context, “young” means a white dwarf that was recently formed (or more precisely, exposed because the outer layers of its parent star were blown off). 2. The spectra from the white dwarfs show the emission or absorption patterns corresponding to various chemical elements; see Chapter 5.2. 3. The atmospheres of stars are mostly made of H and He, so one would expect that to make up most of any gas left over on a white dwarf. The presence of significant amounts of metals, especially iron or silicon suggest they were present before the star formed and may have been accreted onto the star’s core if it engulfed its own planets during its giant phases. 4. Far-UV wavelengths do not penetrate the Earth’s atmosphere; a telescope must therefore travel above the atmosphere to observe at these wavelengths.

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Chapter 16 Evolution of Low-Mass Stars ◆ 123

5. In this study, white dwarfs with significant amounts of metals in their atmospheres constitute around onethird of all white dwarfs. This is of the same order of magnitude as the number of Sun-like stars with planets (22%). If a white dwarf needs planets to become enriched with these metals, then it makes sense that the two numbers should be close. Test Your Understanding 1. From shortest to longest: c-e-f-b-d-a. The higher the mass, the shorter the lifetime. 2. a-e-d-c-f-b-g. This can be confirmed using Figure 16.13, which shows the evolution of a low-mass star like the Sun. 3. (c) A horizontal path means the star changes surface temperature at a constant luminosity. 4. (a) Degenerate matter does not expand when you heat it, and as you increase the mass, the density increases as well so a clump becomes smaller as it grows more massive. 5. (a) Higher mass means more gravitational compression (pressure) in the core, which yields higher temperatures that cause the fuel to be burned at a faster rate. 6. (b) This situation is shown in Figure 16.5 as the red giant branch. 7. (d) Because stars take so long to evolve, we piece together their evolution by looking at many different objects, each of which is at a different point in its evolutionary path. A very short phase, such as the helium flash, will therefore be almost impossible to observe. 8. (d) A giant star is so large that its surface gravity is very low, allowing the radiation pressure to accelerate material away from the surface. 9. (d) The white dwarf at the center of a planetary nebula excites the electrons in the surrounding gas, creating emission lines. 10. (c) As an AGB star loses its mass, it peels away the outer layers, exposing the hotter layers underneath. 11. (b) Gravitational collapse converts gravitational energy into thermal energy, even for degenerate material. 12. (c) A Type Ia supernova is believed to result from a white dwarf that grew to 1.4 M. 13. (c) A nova (or supernova) makes a star grow much brighter, and in general, this appears as a new star in the sky in which the pre-nova star was too faint to see. 14. (a) Shell burning causes the core temperature to increase, which in turn makes the H-burning shell fuse more rapidly, cause the star to become larger and more luminous.

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15. (c) The “lower left” of the H-R diagram is for stars that are very hot and have very low luminosity. Thinking about the Concepts 16. To burn helium, a star must have a massive amount of it in its core and the core must be hotter than about 100 million K. With today’s chemical makeup of stars, the first is unlikely. More importantly, if a star did have a central temperature hot enough to burn helium, it would still burn all the hydrogen that was present because H burning starts at much lower temperatures. 17. If a main-sequence star started burning H at a faster rate, the energy generated would increase. That energy would be used to lift the atmosphere of the star, making it grow larger and cool down. This would drop the core temperature, returning the star to its original size and luminosity. 18. When a star starts burning a new element, it raises its temperature but expands to reach hydrostatic equilibrium; thus, its density drops. 19. The birth mass of the small star in a close binary pair does not determine that star’s ultimate destiny. As its higher-mass companion evolves into a red giant (or supergiant), it will overfill its Roche lobe and transfer mass to the smaller companion. The addition of mass to the smaller companion shifts it higher on the main sequence and changes its evolutionary track. 20. If all Type Ia supernovae are shown to be more luminous than previously thought, then our distance estimates to them would increase. 21. Yes. Stars change structure on the main sequence. Fusion turns four H into one He, which takes up less space, so the core slowly contracts and heats over the main-sequence lifetime. 22. (a) Depending on where it happens to be in its orbit, Jupiter is anywhere from 4.2 to 6.2 AU from Earth at any given time. If Jupiter were a G5 star that was 80 percent as luminous as the Sun, it would appear 22 times fainter than the Sun at closest approach and 48 times fainter than the Sun at farthest approach. During Jupiter’s opposition, we would experience daylight for all 24 hours of the day, but Jupiter’s light would provide only 4 percent of the illumination that the Sun provides. Thus, the half-day when light was coming from Jupiter would be equivalent to a deep twilight. (b) As the Sun evolved into a red giant, it would overfill its Roche lobe and the G5 star would strip some of its outer layers (frying the inner planets in the process). Eventually, the Sun would settle into a white dwarf. Later, as the G5 star evolved into a

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124 ◆ Chapter 16 Evolution of Low-Mass Stars red giant, the white dwarf Sun would steal back some hydrogen from the G5 star and form an accretion disk. When hydrogen from the accretion disk landed on the white dwarf ’s hot surface, it would cause the Sun to explode as a nova. 23. Both follow the Hayashi track, but in different directions. 24. The red giant star is a giant, hence has much more surface area and can thus put out much more energy per second than a smaller horizontal branch star. Also, hydrogen fusion produces much more energy per fusion than helium, so more energy is produced during the red giant phase. These combine to yield a larger luminosity for the giant. 25. If a star can start burning oxygen, it will go on to burn heavier elements up to iron, and then will undergo a Type II supernova. On the H-R diagram, it will make small loops between a horizontal branch area and a giant branch area, just as it did during the red giant to horizontal giant phases. These loops will be small because the time during which the star burns elements heavier than helium is increasingly short. 26. A white dwarf is just a cooling blackbody of constant size, so as it emits radiation, it cools and hence its luminosity drops as well. 27. A degenerate gas does not expand when you heat it, so there is no safety valve to slow down the nuclear reactions. 28. If two binary stars merge, the result could form a single star with a more massive H-burning core. It would thus appear as a higher-mass, main-sequence star. This is one theory for “blue stragglers.” 29. (a) T CrB (pronounced “Tee Cor Bor”) is a binary system. We know of no other mechanism for generating novae except through mass transfer in a binary system. (b) Hydrogen from the accretion disk falls onto the hot surface of the white dwarf, causing periodic outbursts of uncontained nuclear fusion. (c) As long as there is hydrogen streaming from the red giant companion to the accretion disk, and as long as the mass of the infalling hydrogen does not cause the white dwarf to exceed the Chandrasekhar limit and detonate, then novae can recur. 30. Low-mass stars are also low luminosity. The low mass means it is easier for a planet to tug the star, yielding larger Doppler shifts in the star’s spectra that would be easier to detect than for a high-mass star. The lower luminosity means that one could more easily block the star’s light to look for orbiting planets around the star in imaging. These are two key reasons why low-mass stars are more attractive when searching for planets.

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Applying the Concepts 31. Setup: On linear axes, a straight line appears straight, whereas an exponential appears curved. On logarithmic axes, however, a straight line is curved and an exponential is straight. This is because the logarithm undoes an exponential. Solve: In Figure 16.1, the axes are logarithmic, so the line that is drawn is a straight line; however, it does not represent a linear fit, but rather an exponential function. Review: Note that our approximation predicts that some stars will be a little brighter or fainter than they really are. These errors may be a factor of two or more, so although they are good for instructive purposes, one must be careful when using them in calculations that require high precision. 32. Setup: Figure 16.2 shows the percentage of the mass present that is hydrogen or helium as a function of position in the Sun. The amount of the Sun that is turned from H into He is the area of He in the last panel above 30 percent. We will approximate that by a triangle. Solve: The area of a triangle is base 3 width, or 0.2 3 70 5 14 percent. Review: The Sun uses up its core hydrogen during its main-sequence lifetime, so we expect the amount of hydrogen inside it to drop over time, as we found. 33. Setup: Figure 16.13 shows the evolution of a 1M star. The axes are logarithmic in luminosity, with the current star having a luminosity of 1L. To compare the brightness (that is, luminosity) of the star between two new periods, simply measure the luminosities on the y-axis. Note that the question asks us to compare phases to the top of the giant branch, which occurs around 7,000 L. Solve: (a) On the main sequence, the star has a luminosity of 1 L, so the giant star is about 7,000/1 5 7,000 times brighter than the main-sequence star. (b) During the horizontal branch, the star is around 100 L, so the giant star is about 7,000/100 5 70 times brighter than the horizontal branch star. Review: A giant star is not only physically huge but incredibly luminous. 34. Setup: We need temperatures and timescales for each phase shown in the H-R diagram in Figure 16.13. The main sequence lasts about 10 billion years at 6,000 K. The red giant branch is about 200 million years at 4,000 K. The horizontal branch is about 100 million years at 5,000 K. The AGB is about 1 million years at 3,500 K. The planetary nebula is about 50,000 years up to 200,000 K. The white dwarf cools for about 10 million years to reach the dot on the white dwarf path around 100,000 K.

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Chapter 16 Evolution of Low-Mass Stars ◆ 125

Surface Temperature (K)

Solve: 300 000

30 000

3 000 10.00

10.10

10.20

10.30 Billions

Time since birth of star (billions of years)

Review: Note that the planetary nebula phase is just a blink of an eye in a star’s evolution. 35. Setup: The mass-luminosity relationship can be written as L(L) 5 M(M)3.5 because our Sun has one solar mass and one solar luminosity. Solve: We find (a) 0.1 L, (b) 530 L, and (c) 1.6 3 106 L. For comparison, Table 16.1 gives approximate luminosities around (a) 0.08 L, (b) 480 L, and (c) 500,000 L. Review: It is hard to directly compare the luminosities in Table 16.1, because those exact masses are not listed. However, we see that the luminosities we found are on the same order as actual stars. 36. Setup: The stellar-lifetime formula can be normalized 1010 yr to the Sun’s life as lifetime 5 2.5 . M Solve: We find (a) 56 billion years, (b) 110 million years, and (c) 360,000 years. For comparison, Table 16.1 gives approximate lifetimes around (a) 54 billion years, (b) 120 million years, and (c) 360,000 years. Review: It is hard to directly compare the lifetimes in Table 16.1 because those exact masses are not listed. However, we see that the ages we found are on the same order as actual stars. 2GM , but to com37. Setup: The escape speed is v esc 5 r pare speeds between two similar objects, we can construct a simple ratio of the two:

v RGB M RGB R now 5 . v now M now R RGB

Note that vnow 5 617 km/s as shown in Working It Out 16.2. Solve: Using the same formula as above, the escape speed for the Sun as an AGB will be v RGB 0.7 1 5 5 0. 059 of today’s value, or 0.059 3 1 200 v now 617 km/s 5 37 km/s. As escape velocity drops, it is easier for mass to leave the Sun, meaning that the Sun’s rate of mass loss will increase.

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Review: Remember that the force of gravity scales inversely with size squared (F  1/r2); thus, it makes sense that the strength of gravity at the surface of the Sun will drop considerably as it becomes a red and asymptotic giant, and with it the escape velocity. 38. Setup: If the rate scales as T n, then an increase of x percent in temperature results in a change of (1 1 x)n in energy generation. Solve: (a) If the temperature increases by 10 percent, the rate of hydrogen fusion increases by (1.1)4 5 1.46 times. (b) If the temperature increases by 10 percent, the rate of helium fusion increases by (1.1)40 5 45.2 times. Review: The lifetime of a phase decreases as energy generation increases because the fuel that is available is used up that much faster. 39. Setup: This is a simple d 5 vt problem where we must convert units. It turns out 1 km/s is about 1 pc/Myr. Therefore, 20 km/s is about 20 pc/Myr. pc Myr 3 50 , 000 yr 3 6 5 1pc . Solve: 20 Myr 10 yr Review: Note that a planetary nebula grows about as large as the space between stars! 40. Setup: The star starts at 0.6 M and must grow to 1.4 M, so it has to gain 0.8 M. We must therefore find out how long it takes to gain this much mass at a rate of 1029 solar masses per year. Similar to distance, mass gained equals rate times time or M 5 rt. Solve: The time it takes to transfer 0.8 M is 0. 8 M t 5 29 5 8 3 108 yr or 800 million years. A 10 M  / yr typical low-mass star generally lives on the order of billions of years. Review: The fact that mass transfer happens faster than the typical lifetime of a low-mass star is an important confirmation of our theory of how Type Ia supernovae happen. 41. Setup: Remember Kepler’s law in general form: A3 P 2 5 . However, this does not give us speed. For M that, we remember that v 5 2pr/P for a circular orbit, where r is the distance of the orbit, that is, A. Solve: A white dwarf has a mass of 0.6 M and a typical radius of 5,500 kilometers or 3.3 3 1025 AU. Then the period of the material orbiting just above the white dwarf is: P 5

(3. 3 3 1025 )3 A3 5 5 8 s. The orbital M 0.8

speed is therefore v 5

2 5, 500 km 5 4320 km/s. 8s

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126 ◆ Chapter 16 Evolution of Low-Mass Stars Review: A white dwarf is very small in astronomical terms, and therefore, we expect that matter will be orbiting around it very rapidly, which this problem confirms. 42. Setup: Density is mass over volume, so it is inversely proportional to the cube of a sphere’s diameter. If we take the ratio of densities between Earth and the white  dwarf, we find that Dwd 5 3  D .  WD Solve: Substituting in numbers yields: Dwd 5 3

5, 500 kg/m 3 (12 , 700 km) 5 224 k m. 1 3 109 kg/m 3

Review: 224 kilometers is about 140 miles, or about the length of New Jersey. That is a pretty small size to pack the Earth into! M ; so 43. Setup: Density is mass/volume or  5 4 / 3 r 3 3M . 4 Solve: The text says that degenerate material can have a density of 1,000 kg/cm3 or more. If the Sun’s mass had this density, its size would be solving for size, we find R 5 3

3  2 3 1030 kg 5 8 3 108 cm 4  1, 000 kg/cm 3 or 8,000 kilometers. Review: Earth’s radius is about 6,000 kilometers, so the Sun’s size would be around that of Earth, if the Sun were degenerate. 44. Setup: Write the luminosity law in terms of solar units L(L) 5 R(R)2T(T)4. In this case, the white R dwarf ’s radius of 107 m 3 5 0 . 014 R , or 7 3 108 m

main-sequence lifetime is closer to 13 billion years will not be significantly less massive. Using the Web 46. The “KAIT” supernova search project is an automated telescope that searches for supernovae in distant galaxies by regularly imaging the galaxies and using sophisticated software to detect whether any objects within those galaxies have changed brightness. The rest of the answer to this problem will include how many supernovae were found in a given year and how bright they looked compared to their host galaxies. 47. The American Association of Variable Star Observers (AAVSO) collects data from amateur astronomers on the brightness of variable stars. Their instructions will help you to follow and report the brightness of a variable star. 48. Here are responses from the “Properties of Planetary Nebulae” lab. Part A 1. and 2. will look like this sketch:

Brightness

R53

Distance

Solve: Plugging in temperature, we find (a) 1.8 3 1013 L; (b) 1.8 3 105 L; (c) 1.8 3 1023 L; and (d) 1.8 3 10211 L. Review: We see that a white dwarf is amazingly bright when it is first exposed but becomes almost unobservably dim when it cools. 45. Setup: For a star to have reached the red giant today but be as old as possible, the star would need to have a main-sequence lifetime just short of the age of the universe. Use Table 16.1 to answer this. Solve: Looking at the values in the table, the star should be between G5 and K0 type or 0.8 to 0.55 solar masses. Review: Remembering that our Sun has a mainsequence lifetime around 10 billion years, a star whose

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3. This model does seem spherical with “limb brightening,” as the lab suggests. 4. A sketch of the mass-loss history might look like this:

Mass loss

 T  L( L ) 5 2 3 1024   5 1 . 8 3 10219 T 4 .  5780 K 

Time

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Chapter 16 Evolution of Low-Mass Stars ◆ 127

5. The fact that there is not a hard outside edge suggests that there was mass loss that ramped up and then became steady with time. Part B 1. If the image height is 2 arcmin, then the nebula’s edge is about 1.2 arcmin, or 71 arcsec. 2. The radius is about 35 arcsec. 3. R 5

3. 4 3 1016 km 3 35 5 5. 8 3 1012 km 206,265 12

4. Because d 5 vt, t 5

5. 8 3 10 km 5 2 . 9 3 1011 s 20km/s

or 9,000 yrs. 5. 5.8 3 1012 km is 5.8 3 1012/1.5 3 108 5 38,400 AU. That is huge compared to the Solar System. Part C 1. The volume of the entire sphere is 4 (5. 8 3 1012 km)3 5 8. 2 3 1038 km 3 . The inner 3 sphere has a radius of about half that of the full nebula, yielding a volume of one-eighth of the entire sphere, 1.0 3 1038 km3. 2. The total volume is about 7.2 3 1038 km3. 3. The mass M 5 V 5 1.7 3 10210  7.2 3 1038 5 1.2 3 1029 kg. 4. In solar masses, this is 1.2 3 1029/2 3 1030 5 0.6 M. 5. If there are 700 nebulae spilling 0.6 M into the interstellar medium every 9,000 years, then there are about 700 3 0.6/9,000 5 0.05 M/yr. 6. Given that there are about one full solar mass per year forming but only about 5 percent of that coming from planetary nebulae, there must be other sources of enriched gas for the interstellar medium. 49. Answers will vary. You will find an example each of spherically symmetric, bipolar, and point-symmetric

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nebulae. You will label the asymmetry, the location of the central star, show the symmetry axis if there is one, or identify the symmetric features. 50. Answers will vary. You will find a press release for a planetary nebula and for a white dwarf, present the observations that are reported and why they are important. Exploration 1. 5,800 K 2. One solar luminosity 3. One solar radius 4. As a star moves up and to the right in the H-R diagram, the test star becomes larger, brighter, and redder. 5. As I moved the star, it changed to 4,900 K. I know the temperature changed from the figure because the star became redder (for a blackbody, redder means cooler). 6. The star is now at 3.5 solar luminosities. 7. The star is now at 2.6 solar radii. 8. When a blackbody grows cooler but larger, it is possible for the change in size to outpace the change in temperature. As such, each “patch” of the star may give off a little less energy than before, but there is so much more area that the total amount of energy given off is higher, that is, the star has higher luminosity. 9. The changes in the star’s size and color are even more dramatic than what happened when I played with the star for question 4 because I moved it only a small amount the first time. 10. The size of the star is the most noticeable change to me. 11. As I drag the star horizontally across the H-R diagram, it shrinks in size and becomes blue. 12. By the time I am on the far left of the diagram, the star is very small. 13. As I drag the star down and to the right, the star becomes redder but changes size by only a small amount. 14. The star is still very small compared to the Sun.

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CHAPTER 17

Evolution of High-Mass Stars INSTRUCTOR’S NOTES Chapter 17 covers the life and death of high-mass stars. Major topics include

▶ changes in the core and envelope as single high-mass

stars evolve from the main sequence to become Type II supernovae ▶ pulsating variables ▶ core-collapse supernovae ▶ chemical enrichment from supernovae ▶ neutron stars and pulsars ▶ measuring the ages of star clusters from their H-R diagrams Similar to the previous chapter, this material involves a reasonable amount of information that must be committed to memory. To that end, I try to use real-world objects to provide a sense of scale or appropriate analogy. In my department, we have some 1-pound steel balls, which happen to come from a kinetic-energy demo, in which knocking them on either side of a piece of paper will light the paper on fire. I like to carry one of the balls during my lessons on high-mass stellar evolution, as it represents the equivalent size of the Washington Monument compressed to the density of an inert iron core or the size of Mount Everest compressed to the density of a neutron star. When I talk about the implosion of the core and rebound off of the stiffened neutron star, I also like to drop the ball from above my head onto my lectern. It makes a very loud bang, which wakes up the drowsy, but also demonstrates the power of a shock wave—if this 1-pound ball dropped 3 feet onto a table hurt your ears, imagine 3 solar masses crashing down thousands of kilometers. Today’s generation of students has grown up learning much of what they need from videos, so I try to include as many as I can, and this chapter lends itself nicely to the inclusion of both simulations and real time-lapse data. For example, there is little that elicits more gasps from the class than the pulsar-wind nebula movies of the Crab Nebula available at http://chandra.harvard.edu/photo/2002/0052/ animations.html. An artistic rendering of how the nebula was formed can be found at https://www.spacetelescope. org/videos/heic0515a/. There are lots of animations of

the expansion of the SN 1987A ejecta and hydrodynamic simulations of the explosion available on YouTube. A wonderful resource for stellar evolution is http:// rainman.astro.illinois.edu/ddr/stellar/beginner.html, which contains a number of movies showing stars of different mass evolving across the H-R diagram. If you are adventurous, use the “Advanced Page” at the bottom of the screen, on which you can create customized movies. You control the masses, metallicities, and duration of the movie and can add in starbursts and control the star-formation rate. I have made movies of 50 stars (from 0.1 to 50 M.) evolving over periods from 15 Myr to 15 Gyr. Not only do these show each star going through each of its evolutionary phases, but they also clearly show how H-R diagrams reveal the age of the cluster.

DISCUSSION POINTS

▶ We are taught that chemical elements heavier than iron

cannot be fused as a source of energy, yet supernovae produce them. Discuss how this is possible. ▶ Compare and contrast Type Ia and Type II (core collapse) supernovae. Under what conditions will they occur? How common are they? What remnant (if any) is left behind? Are heavy elements created? Which is likely to be responsible for the heavy elements out of which we formed? ▶ Cepheids and RR Lyrae variables are standard candles. Discuss their importance for understanding the universe. ▶ Discuss what you would do if you suddenly heard that Eta Carinae has exploded as a supernova. How likely would it be that you could witness this event? What problems could there be to study this event?

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130 ◆ Chapter 17 Evolution of High-Mass Stars the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. CNO Cycle Animation This simulation shows how carbon is used as a catalyst in more massive main-sequence stars to fuse hydrogen into helium. Text reference: Section 17.1 H-R Diagram Explorer This simulation introduces the H-R diagram, allowing the user to choose the temperature and luminosity of a star, and then showing a star’s size and color compared to the Sun, and the star’s location in the H-R diagram. The user can also overplot many different values, such as luminosity classes or nearby stars. Text reference: Section 17.1 H-R Diagram Star Cluster Fitting Explorer This simulation plots H-R diagrams for different star clusters and allows the user to perform a zero-age, main-sequence fitting to find the main-sequence turn-off. The user can also determine the distance using the distance modulus. Text reference: Section 17.4

ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what they’ve learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5) and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack. Type II Supernovae Dr. Palen shows how core collapse and rebound causes the ejection of the star’s outer layers, with the classic demonstration of dropping two balls on top of each other, making the top ball fly off with enormous energy after the pair hits the ground. Text reference: Section 17.2 Pulsar Rotation Dr. Palen uses a foam ball and a few flashlights embedded within it to demonstrate the lighthouse effect of pulsars. Text reference: Section 17.3

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END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. b, d. High mass stars have higher internal temperatures, so they burn fuel faster and use the CNO cycle to burn H to He. 2. (b) Iron can’t fuse and give off energy, which causes the star to implode. 3. (d) Pulsars are often found inside supernova remnants, which tie the two together. 4. (b) Stars at the top of the main sequence live for a very short amount of time, so if the main-sequence turn off for a cluster is at this location, the cluster must be very young. Reading Astronomy News 1. The stars in a cluster are all born at about the same time; therefore, the most massive ones can be expected to explode at about the same time. 2. X-rays cannot be observed from the ground because our atmosphere blocks them; we must travel above the atmosphere to observe in this wavelength region. 3. The DXL instrument was able to distinguish between x-rays produced within and outside our solar system and found that 60% of the observed x-rays came from outside it. 4. The x-ray gas is heated to at least a million degrees Kelvin; however it is safe to put your hand into that gas because it is so diffuse that no atom would ever hit it and transfer that energy (conduction). 5. Answers will vary. I found the video helped visualize the local bubble and the method of charge transfer. Multiple Choice 1. (c) High-mass stars burn successively heavier elements in their cores while continuing to burn the lighter elements in a shell. This creates an “onion skin” of burning layers. 2. e-a-f-b-c-d. See Figure 17.4. 3. (d) Elements heavier than iron cannot be fused inside a star because fusion of these atoms requires energy. Thus, they are created only in explosive events such as supernovae. 4. (d) A pulsar is a spinning neutron star whose magnetic axis is different from its spin axis; the magnetic axis acts as a lighthouse. 5. (a) As the star moves to the right in the H-R diagram, it cools off and becomes redder.

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Chapter 17

6. (a) The star moves to the right at roughly constant luminosity, so it becomes red and giant. 7. (e) The Sun is 5 billion years old, so it must lie between panels (e) and (f), which show clusters at 1 and 10 billion years, respectively. 8. (b) High-mass stars fuse hydrogen via the CNO cycle, low-mass stars by the proton-proton chain. 9. (a) The layers of high-mass stars correspond to which elements are being fused, and heavier elements (that is, higher atomic number) are fused at higher temperatures, meaning closer to the core. 10. (a) Eta Carinae is very high mass, and as a result is also driving off a huge amount of its material into space, yet another reason it is an extreme example. 11. (b) To fuse two iron nuclei, energy must be absorbed because elements heavier than iron are also more massive than their constituents. 12. (a) All core collapses produce supernovae. Most produce neutron stars, whereas the most massive progenitors produce black holes. 13. (a) Supernova remnants emit across the electromagnetic spectrum, from the radio to x-ray. 14. (b) Neutron degeneracy pressure holds up a neutron star. 15. (b) Very young stars that are turning off the main sequence will have very short lifetimes, meaning they are very high mass. These are located at the upper left of the H-R diagram. Thinking about the Concepts 16. In low-mass stars, the proton-proton chain involves only protons, deuterium, and helium-3. With a lower repulsive (that is, Coulomb) barrier to overcome, this process operates most efficiently at the temperatures and pressures in the cores of low-mass (that is, low core temperature) stars. The CNO cycle uses 12C as a catalyst to fuse hydrogen into helium. The carbon nucleus collides with a proton to become nitrogen. After several stages of fusion and radioactive decay, 4 He splits from 16O, leaving a 12C nucleus to start the process again. The CNO cycle generates more energy to support a star, but it requires the high temperatures and pressures in the cores of high-mass stars to overcome the large Coulomb barrier between carbon and hydrogen. Low-mass stars have difficulty achieving these conditions. 17. When it begins to burn helium, a low-mass star has a degenerate helium core. As such, the core begins burning helium in a sudden thermonuclear runaway that causes a thermal flash. In the high-mass star, the core is not degenerate, so the high-mass star eases into helium burning.

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Both burn helium by the triple-alpha process; however, the way this process begins is dramatically different. 18. The core of a high-mass star is convective and well mixed, plus temperatures are higher, which all oppose degeneracy because the atoms do not become forced too close together. 19. Post-helium burning for high-mass stars becomes shorter for each cycle because in each successive burning cycle, there are fewer and fewer nuclei to be fused (numerous lighter nuclei are combined to make a single heavier nucleus). Second, fusion of heavier nuclei requires much higher energy (or temperature) meaning the collisions happen far more often. 20. Cepheid variables are supergiant stars that are visible over intergalactic distances. Because the Cepheid has a regular period, it is easy to detect even at large distances by watching the brightness variations of stars in a galaxy. The period of a Cepheid’s variability is directly related to its luminosity, so all we need to do is measure the period over which its brightness fluctuates. Once the luminosity of the Cepheid is known, we can use the distance-luminosity relation to accurately determine the star’s distance. Over intergalactic scales, the distance to the Cepheid is essentially the same as the distance to its host galaxy. 21. Supernovae influence new stars by synthesizing nuclei that are heavier than iron and seeding the clouds in the interstellar medium with these heavy elements. When new generations of stars are formed, they incorporate the heavy elements into their compositions. Supernovae also provide new star systems with heavy material out of which to build planets. Furthermore, the shock wave from the expanding supernova bubble can compress interstellar clouds, triggering their collapse into new stellar systems. 22. A Type II supernova is both an implosion and explosion. First the core implodes; then the resulting release of energy drives a shock wave through the star that blows it up. 23. Eta Carinae will probably appear similar to what was observed with SN 1006, which was recorded by astronomers as being so bright, one could read by its light at night. 24. The neutrinos from SN 1987A escaped directly from the core, whereas the shock wave from core collapse took a few hours to break out of the envelope. 25. The farther material falls, the higher the kinetic energy grows. Because neutron stars are much smaller than white dwarfs, inward-falling material has more distance to fall and gain speed before reaching the accretion disk. The kinetic energy of the inwardfalling material varies as the square of its speed, so even

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132 ◆ Chapter 17 Evolution of High-Mass Stars a modest increase in speed translates to a tremendous increase in kinetic energy. 26. The ejecta from a supernova may start out traveling at 10 percent the speed of light, but that material loses energy as it blows into low-density interstellar gas and dust, and thus the ejecta slow down. 27. Astronomers assume that all stars in a cluster formed at approximately the same time. Based on our observations of star-forming regions, this appears to be a reasonable assumption. We know that high-mass stars evolve much more quickly than low-mass stars; so when all the stars in a cluster are plotted on an H-R diagram, there will be some “turnoff point” on the main sequence at which stars are beginning to evolve into giants. A young cluster will still have its high-mass stars residing on the main sequence, but an old cluster will not. In a very old cluster, even some of the lower-mass stars will have evolved into subgiants and red giants. Thus, by noting the point at which its stars are turning off the main sequence, astronomers can estimate the age of a cluster. 28. The Big Bang produced only H and He, yet even the oldest stars that we see show evidence of heavier elements in their atmospheres, which must have come from a previous generation of stars. 29. Binding energy refers to the amount of energy needed to break a nucleus into its components. One can also think of it as the ability of the strong nuclear force to overcome the electrostatic repulsion between atomic nuclei. Higher binding energies require more effort to break nuclear bonds, but those bonds also release more energy when broken. Therefore, when lighter elements are fused into heavier elements (up to iron), the lost mass is converted to energy via Einstein’s equation: E 5 mc2. The difference in binding energies between the beginning and end products of a reaction is the energy given off in that reaction. 30. All of the elements heavier than helium were made inside stars. Although AGB stars create some C and O, most of the “metals” in the universe were made in supernovae and distributed into space in their ejecta. Thus, most every atom on Earth that is heavier than helium came from a stellar explosion. Applying the Concepts 31. Setup: Figure 17.2 shows how much of the star is composed of H or He at different radii. The core is demarcated by an arrow out to about 33 percent of the total radius on the x-axis. The fraction of the core that is He can be read off of the y-axis. Solve: At zero years, the core is 30 percent He. At 7 million years, the core is 100 percent He.

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Review: The main sequence ends when the core has run out of hydrogen to burn; therefore, this figure shows us that the main-sequence lifetime of a 25 M star is about 7 million years. This is staggeringly short compared to the 10 billion years for our Sun. 32. Setup: To gauge how a star compares to the Sun, remember that the Sun has values of one solar mass or solar radius and a surface temperature around 5,800 K. The uppermost main-sequence star is the little blue dot at top left. Solve: The uppermost main-sequence star has a luminosity of about 2 3 105 (or 200,000) times more than the Sun. Its radius is about 10 R. Its temperature is about 40,000 K, which is 40,000 4 5,800 5 6.9 times hotter than the Sun. Review: Note how this star is about a factor of 10 (or less) hotter and larger, but hundreds of thousands times more luminous. It is no surprise that these stars die so quickly, because to give off this much energy per second they have to burn through their fuel at breakneck speeds. 33. Setup: A typical star’s core takes up about 25 percent to 33 percent of the star’s interior. To judge whether or not each star is to scale, compare the size of the core to the whole star, and read the figure caption. Solve: For the first three stars shown, the core takes up about 25 percent to 33 percent of the star’s volume, which is about what we expect, and thus these are to scale. In the last figure the core takes up about half of the star’s volume, and we also see that the scales to the edges of the nonburning envelopes change dramatically, which shows that this is not to scale. Review: An onion-skin model such as that shown in the bottom star in Figure 17.4 is hard to show to scale to the star’s actual size, which is why it was exaggerated in this figure. 34. Setup: Use d 5 vt where v 5 1500 km/s and t 5 2014 – 1054 5 960 yr. Be sure to convert t to seconds. Solve: t 5 960 yr 3

3 . 15 3 10 7 5 3.02 3 1010s, so the yr

size of the nebula is d 5 vt 5 1500 km/s 3 3.02 3 1010s 5 4.5 3 1013 km. Review: This is about 1.4 pc, roughly the distance between two stars in our galaxy. 35. Setup: Figure 17.6b shows the period-luminosity relationship for a classical Cepheid. To determine the luminosity, simply find the point on the top line that matches the period. Solve: For a period of 10 days, the luminosity is approximately 3,000 L. However, a distance cannot be

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Chapter 17

determined to this star until we also know the star’s observed (or apparent) brightness. Review: It is easy to measure the average brightness of a Cepheid by observing it regularly over its cycle, which then lets us determine its distance using L  bd2, where b is the brightness and d is the distance. 36. Setup: Recall that linear plots increase by single units whereas log plots increase by powers of 10. Solve: Figure 17.13 is a logarithmic plot. Oxygen is found at about 1023, which indicates that for every 1 atom of hydrogen, there are this many oxygen. Another way to look at this is that there is 1 oxygen for every 1000 hydrogen. Review: We expect every element to have abundances less than hydrogen and helium because they were made in stars after the first two were created. This expectation is borne out in the data. GM 37. Setup: Math Tools 17.2 tells us that gravity g 5 2 , R so we need only plug in relevant values rather than the 15 km and 2 M used in that sample problem. Solve: 6. 67 3102112. 82 31030 5 3 . 7 31012 m/ss2 g5 3 2 (10 310 ) Review: This is about 370 trillion times stronger than on Earth. Wow! 38. Setup: Math Tools 17.2 tells us that escape velocity v esc 5 2GM / R , so we need only plug in relevant values rather than the 15 km and 2 M used in that sample problem. Solve: v esc 5 2. 667 310211 ⋅ 2. 8 ⋅ 2 31030 /(10 3103 ) 5 2 . 7 3108 m/s. Review: This is 90 percent the speed of light, so this neutron star is almost a black hole! 39. Setup: This problem gives us all the information we need in terms of solar masses and solar luminosities, so we need only construct the ratios necessary to answer the questions. Solve: (a) The total luminosity of the 50,000 average-mass stars is the number of stars (50,000) times the brightness per star (0.08 L), which yields 4,000 L. This is still 2.5 times less luminosity than a single 20 M star, which puts out 10,000 L. (b) The total mass in average-mass stars is the number of stars (50,000) times the average mass of each star (0.5 M), which yields 25,000 M. This is 1,250 more mass than a single 20 M star. (c) Based on the calculations above, average-mass stars make up most of the stellar mass in the galaxy; however, the high-mass stars make up most of the light.

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Review: We know that low-mass stars are long-lived and much more common, so it makes sense that they would make up most of the stellar mass of a galaxy. Given that luminosity scales so strongly with mass (M3.5), it also makes sense that massive stars will make up most of the light output. 40. Setup: This problem is an exercise in unit conversion. Remember there are 2 3 1030 kg in one solar mass, and 5.25 3 105 min in one year. 30 yr M 2 310 kg 3 3 5 5. 25 3105 min yr M 3.8 3 1023 kg (b) In one minute, Eta Carinae loses

Solve: (a) 0.1

3. 8 3 1023 kg 5 5. 25 times more mass than the 7. 35 31022 kg Moon. Review: The mass-loss rate of Eta Carinae is truly huge compared to most stars. Most stars lose mass at rates 1,000 to 1 million times slower. 41. Setup: Mass-loss rate is given by (mass lost)/(time). Solve: A 25M that loses 20% of its mass loses 5M. If it loses this over 7 million years, the mass loss rate is 5 M 5 7 . 1 31027 M  /yr . 7 3106 yr Review: Typical mass-loss rates range between 10210 21026 M  /yr . 42. Setup: Using the period-luminosity relationship L(L) 5 335 P, we find the luminosity of Delta Cephei is 1,809 L and the more distant one is 18,090 L. Using our equation for parallax-distance, Delta 1 1 5 303 pc. To find the Cephei is at d 5 5 p 0. 0033 distance to the more distant Cepheid, we use the L inverse-square law for brightness: b 5 2 . d Solve: (a) In this case, the more distant star is 10 times more luminous and 0.001 times as bright, so solving for distance, we find d 5 L / b 5 10 / 0 . 001 5100 times as far away, or 30,300 parsecs. (b) The more distant star is more than 1,000 parsecs away, and thus we cannot use parallax to measure its distance. Review: In this problem, we computed two rungs of the distance ladder: first we found the distance and luminosity of a star one way (parallax), and then used the same star to calculate the distance to a more distant object. This is how we extrapolate to very distant objects. For example, we might calibrate a supernova in the same galaxy as the distant Cepheid and use that calibration to measure distances to galaxies even farther way.

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134 ◆ Chapter 17 Evolution of High-Mass Stars 43. Setup: This problem refers to orbits, which we usually solve using Kepler’s third law. However, note that the co-rotating satellite must complete 30 orbits of the neutron star every second. In other words, it travels the circumference of its orbit 30 times each second, where circumference is C 5 2pr and we know the speed is c (the speed of light). Solve: If it is traveling at the speed of light, the circum300 , 000 km/s 5 ference of its orbit is C 5 c /  5 30 rev/s 10,000 km. Now, because the circumference C 5 2pr, we find r 5 1,560 kilometers. This is much larger than the neutron star’s radius, about 10 kilometers. Review: This radius is important, for the following reason. The magnetic field of the pulsar rotates at the same rate as the pulsar itself, and particles travel on those field lines. Beyond 1,560 kilometers, the particle would start rotating faster than light, which cannot happen. This region is where much of the gamma and radio emission originates. 44. Setup: Density  5 M / V and the volume of a sphere 4 is  r 3 , where r is the radius. Solving for radius, 3 r 5 3 3 M / 4 . Solve: The mass of the Earth is roughly 6 3 1024 kg, and the density of a neutron star is listed in the text as 1018 kg/m3. Using our formula, the equivalent size of the Earth would be r 5 3 3 ⋅ 6 3 1024 / 4 3 1018 5 133 meters. A meter is about a yard, so this is just larger than a football field (100 yards). Review: Indeed, a neutron star is amazingly dense! 45. Setup: We are given the density of a neutron star is about 1019 kg/m2 in section 17.3. Use the setup from Problem 44 and M 5 1 M to find the radius of the object. 1030 3 1018 5 3300 m or 3.3 km. 4 Review: A neutron star with the mass of the sun will be smaller than the island of Manhattan! Solve: r 5 8 32 3

Using the Web 46. You will complete lab exercises on Cepheid variables. The first lab is “Stellar Heartbeats.” The star’s brightness varies with Julian Day as recorded at the top of the screen. The second lab is “A Variable Star in Cygnus.” This is a fairly hard lab because the finder chart is fairly dim and challenging to use. The final result will be a plot of magnitude versus time that should show the

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star fade and rebrighten. It may be hard to tell whether the star is periodic or not. W Cyg is in fact a Cepheid with a period about 131 days. A longer-term light curve from the AAVSO (www.aavso.org/lcg) will show that periodicity. 47. Answers will vary. The response will include a few recent supernovae found, including their type, how bright they were, why they are fainter than the nova examined in Chapter 16, and whether Type Ia or Type II supernovae are more common. 48. The Palomar Transient Factory (PTF) survey finds supernovae by looking for “transients,” or objects that change in brightness over the course of the night or between nights. As of September, 2015, it had found 1692 SNe Ia, 113 SNe Ib and Ic, and 620 SNe II. The replacement camera will cover a much larger field of view, allowing for the entire night sky to be monitored daily. It is not scheduled to start collecting data until 2017. 49. Gaia is an astrometry mission that will map out the exact positions (and determine parallax distances) to a billion stars. This will be accomplished by monitoring the positions of these stars over time; any stars that are changing brightness during this time will naturally be recorded; hence, an ancillary science result will be light curves of known and new variable stars, novae, and supernovae. 50. Answers will vary. You will report on any news from Einstein@Home, look for pulsars using their BOINC applet, and report your results. Exploration 1. 12C and hydrogen (1H) are involved in the first reaction. 2. The proton (hydrogen) is represented by blue. 3. The blue wiggle is a gamma ray emitted from the reaction. 4. This first reaction creates the nitrogen isotope, 13N. 5. The resulting nucleus has gained a proton, and because the number of protons in a nucleus defines the atom, this nucleus has changed from carbon (six protons) to nitrogen (seven protons). 6. The yellow ball appearing is a spontaneous decay. 7. The yellow ball represents a positron. 8. The dashed line represents a neutrino. 9. The resulting nucleus has had a proton inside it converted (by positron decay) into a neutron. Thus, the 13 N becomes a 13C. 10. As each proton comes into the nucleus, it gains one proton, raising the mass number (the number of nucleons) by one and turning it into the next higher element on the periodic table.

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Chapter 17

11. During these collisions, gamma rays are emitted. 12. The transition from 15O to 15N is a spontaneous decay. 13. This mimics the decay process when 13N decayed to 13C. 14. The nucleus that began this reaction was 12C. 15. The top red ball has 12 nucleons while four came into the process, so the bottom nucleus has four nucleons.

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16. The bottom red ball has two protons. 17. The bottom red ball has two neutrons. 18. The bottom red ball is 4He. 19. The net reaction is 41H → 4He. 20. In this process, the 12C simply acts as the catalyst, or intermediary, because it is returned back to its original state.

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CHAPTER 18

Relativity and Black Holes INSTRUCTOR’S NOTES Chapter 18 covers Special and General Relativity. Major topics include

▶ relative motion and inertial frames of reference ▶ special relativity: length contraction, time dilation, and relativistic mass

▶ general relativity: mass distorts spacetime, orbital pre-

cession, gravitational lensing, gravitational redshift, and gravitational time dilation ▶ black holes: formation, event horizon, spaghettification, and Hawking radiation This chapter is, by far, my favorite. Black holes are invariably popular, consistently ranking highly among the topics in which students express interest on the first day of class. The weird and wacky world of relativity is also quite awe inspiring. If you are interested in recruiting for your major, I have had a few students who came to the college to pursue a nonscience major but decided to become physics majors or minors after we covered this material. There are compendia of animations and videos to demonstrate the consequences of special and general relativity; however, classroom demonstrations are quite hard. If you have the time, a longer discussion of muon detection is a good way to start, as well as talking about results from atomic clock experiments, to show that Special Relativity has been extensively tested and proven accurate. If you have the physical space and a little time, a sheet stretched over a large hoop can make a great visualization of spacetime, in which large balls warp the space and small marbles show orbital motion. A good example can be found at https://www.youtube.com/watch?v5 MTY1Kje0yLg. You can also purchase a spiral well (such as the kind at science museums, but smaller) which does a very good job of replicating warped space and singularities. I purchased a 2-foot well from spiralwishingwells.com (contact them for an educational discount) which I use with an overhead document camera. With this setup tuned to the right brightness, the projection does not show the warped shape so one only sees the ball going in a circle—a

perfect demonstration of how warped space can be understood as a gravitational force. A smaller demonstration of mass bending light can be achieved by cutting the base off of an inexpensive wine glass and moving it slowly across a black dot on a piece of paper. The shape of the glass base closely resembles that of a lens with a caustic, thereby demonstrating weak and strong gravitational lensing and Einstein rings. You may want to warn your students not to try this with grandma’s crystal, however. A decade ago, there was always one student in my class who had relished the reading of a number of popular science books and bounded forward asking about a huge host of what I consider to be speculative science, such as wormholes, white holes, warp drives, etc. Today, that number can reach as high as half the students, because YouTube and blogs allows anyone to propose anything they want. I love student enthusiasm and try hard to not curb it, but use these as teaching moments to underscore the importance of being skeptical and of understanding the differences between testable and speculative science. This generation often responds very well to “Remember what Abraham Lincoln said: Don’t believe everything you read on the Internet.”

DISCUSSION POINTS

▶ Discuss why we never experience relativistic effects in our daily lives.

▶ Do relativistic effects make intuitive sense or do they run

counter to our expectations? Is that an acceptable part of science? ▶ Relativistic effects are rarely observed on Earth; is it possible to learn everything just by studying what happens here? ▶ Does relativity mean that most of the physics we learned in previous chapters is wrong? ▶ A common misconception about black holes is that they are giant vacuum cleaners, relentlessly sucking inside everything around them. Discuss whether this view is realistic and the possible origins of this idea. 137

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138 ◆ Chapter 18 Relativity and Black Holes

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (d) The speed of light is constant for all observers. 2. (a) Moving clocks run slow, so the moving ship will measure a smaller amount of time than the ship at rest. 3. (d) Mass bends space. 4. (c) The event horizon of a black hole scales directly with mass. Reading Astronomy News 1. When joined, the mass of two neutron stars is higher than neutron degeneracy pressure can support, so it collapses into a black hole. 2. The stars approach because their orbital energy is bled off with gravity waves and lost through tidal distortion. 3. Highly warped and very rapidly changing. 4. Neutrons are captured into the nucleus and then decay into protons, building up heavier elements. 5. We are not able to observe these mergers in real life, because we do not know where they will happen, they are too far away to observe, and they happen too fast to observe meaningfully even if we could take care of the first two problems. Test Your Understanding 1. c-a-b. A white dwarf comes from a star from under 8 M, a neutron star from a massive star, and a black hole from the most massive stars. 2. (c) You see the fly moving at 50 2 7 5 43 km/h. 3. (d) The speed of light is the same for all observers. 4. (c) A moving clock runs slow. 5. (c) Moving objects appear shorter than they really are in their direction of motion. 6. (d) When you orbit around another object, you feel no gravity because you are in constant free fall. 7. (d) General relativity explains gravity on all mass scales and predicts Newton’s laws when the mass is low. Thus, it does not replace Newtonian physics, but it does explain it within a larger context. Thus, we also learn why Newton’s laws work, not just what they are. 2GM 8. (b) The Schwarzschild radius R s 5 2 increases c linearly with mass. 9. (c) A neutron star is held up by neutron degeneracy, which can be overcome by the inward force of gravity if the mass of the neutron star is sufficiently high. 10. (a) Whether one is moving at classical or relativistic speeds, relative motion can be detected, provided the detector is sufficiently sensitive.

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11. (c) The speed of light is the same for all observers. 12. (c) The force of gravity does not depend on the speed at which particles are moving, only their masses and distance. 13. (d) Special relativity tells us that no massive object can travel at or faster than the speed of light. For the interested reader, the “velocity addition formula” from a modern physics course shows us the ships will actually see each other approaching at 0.8c. 14. (b) The period at which light flickers limits the size of the region that is emitting the light, and because the flickers are very fast, we deduce that the region is sufficiently small to be consistent with a black hole. 15. c, d, f. We see jets or jet-like streams of material from pulsars, protostars, and planetary nebulae. Thinking about the Concepts 16. Absent other information about the object of interest, the astronomer would not be able to tell the difference between a gravitational redshift and a recessional redshift. Both produce the same Doppler redshift in the object’s spectrum. 17. Both observers (you looking out the back of the ship, and a person on a slowly moving planet) will see the light traveling at exactly c, because the speed of light is the same for all observers. 18. A photon is a massless particle, and all massless particles travel at exactly the speed of light. 19. Twin A returns younger than twin B but not because of time travel. Rather, twin A aged at a slower rate than twin B because twin A accelerated and decelerated from moving close to the speed of light. Note that with physics as we know it, time travel is not possible. 20. It is possible for temporal order to be changed or even reversed, depending on the frames of reference. Consider a person standing in the middle of a train car with two flashbulbs on either end of the car. If the bulbs flash at the same time, the light reaches the observer at the same time. However, if the train is moving away from an observer on the ground, the further pulse will reach the middle of the car first, and the closer pulse will reach there at the end. 21. Moving clocks run slowly, so the person’s pulse rate will appear extremely slow to you. 22. A person’s mass does not change with his or her speed, but his or her volume will. Because moving objects are shorter, the person’s volume will appear to decrease. Density  5 M/V, so if volume appears to drop, the density will appear to go up. 23. No, the cosmonaut is moving with the ship, so he will see no length contraction. Length contraction is

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Chapter 18

observed only by an outside viewer who is not moving at that high speed. 24. Moving objects appear shorter to someone at rest. If the ball appears round to you (who is at rest), then it must be elongated (that is, an oval or rugby-shaped ball) on the ship. 25. Moving objects appear shorter in their direction of motion. If the meter stick still appears to be 1 meter long, then it must be standing straight up rather than lying down in the direction of motion. 26. There is absolutely no reason for anyone to be remotely concerned about a black hole located a few light-years from Earth. It will not affect us any differently than the stars of similar mass that are already at that distance. It is a common misconception that black holes are giant “vacuum cleaners” that actively suck up everything around them and will instantly destroy us all. 27. If a star fell into a black hole, it would become increasingly redshifted by gravity. We would see it appear redder and redder as it approached the event horizon. 28. The effects of relativity show themselves only in extreme circumstances, such as traveling close to the speed of light or being right next to a black hole or neutron star. In our common lives, special and general relativity are equivalent to the laws of physics that we encounter on a daily basis. 29. It is exceedingly unlikely that we will ever have fasterthan-light travel because special relativity tells us that nothing can travel faster than the speed of light. Also, we have never found evidence for anything that travels faster than light, which supports this claim. Alas. 30. If a gamma-ray burst were nearby and directed toward us, the burst of high-energy photons could wreak genuine havoc on Earth. The energy would destroy much of the ozone layer, irradiate the Earth with deadly radiation, and possibly sputter off much of the atmosphere. Applying the Concepts 31. Setup: Figure 18.1b shows that when the car is moving, rain seems to approach the car, but when the car is still, it falls vertically. If we are to measure the speed of the rain, then we have to determine whether this angle depends only on the speed of the car, or also on the speed of the rain. A thought experiment would be, what would the situation look like if the rain were stopped? In the right-hand panel, it would move horizontally toward us at the speed on our speedometer. As the rain falls down faster, but we stay at constant speed, it will appear to fall more horizontally. So yes, there must be a way to determine the speed of the rain.

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Solve: Suppose the rain falls down at speed v and we move to the right at speed u, then the angle  at which v it appears to fall will be given by tan  5 . u Review: If the rain is not falling, the rain will appear to come right at me at  5 0°. Mathematically, for this situation, v 5 0°, so tan  5 0°, which verifies that  5 0° Now, if instead we are stopped, we expect  5 90°; being stopped means u 5 0° or tan  5 , which verifies  5 90°. 32. Setup: Figure 18.1 shows how rainfall appears aberrated to drivers, whereas Figure 18.2 shows the same effect for starlight. Follow the arguments in the solution to question 31. Solve: Light travels at speed c, and we move to the right at speed u, thus the angle  at which it appears to fall c will be given by tan  5 . u Review: See question 31. 33. Setup: Einstein’s formula says E 5 mc2 such that energy can be turned into mass. Solve: Because higher energy can translate into higher mass, a hot cup of coffee has more thermal energy and thus more energy that could be transformed into mass than, say, a cup of iced coffee. However, these differences are infinitesimal in everyday terrestrial examples. Review: Any object with more energy has the potential to gain mass, such as a wound-up watch (the kind that uses springs). 34. Setup: In Figure 18.5, time is shown on the bottom and speed on the left as a fraction of c, so 0.6 means 0.6c. The red line shows Newtonian (that is, incorrect) speed, whereas the blue line accounts for relativity. To convert from c to km/h, recall the speed of light is 1.08 3 109 km/h. Solve: (a) The two curves, and thus Newtonian and relativistic effects, begin to significantly differ at about 0.33 to 0.5c. (b) 0.33c is 0.33 3 1.08 3 109 km/h 5 3.6 3 108 km/h. Typical human speeds of travel are in the hundreds of km/h (100 km/h on the highway, a few hundred in an airplane). We do not experience the effects of relativity because we never travel fast enough. Review: In general, one can ignore the effects of rela1 tivity if the speed is under c , as shown in this figure. 2 35. Setup: Figure 18.6 shows the Lorentz factor  as a function of the fractional speed of light. Solve: One can quibble about the exact number, but  does not deviate significantly from one until the speed 1 1 is about to c . 3 2

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140 ◆ Chapter 18 Relativity and Black Holes Review: In general, one can ignore the effects of 1 relativity if the speed is under c , as shown in this 2 figure. 36. Setup: As noted in the problem, there are 60 arcsec in an arcmin, and 60 arcmin in a degree, or 3,600 arcsec in one degree. There are also 100 years in a century. So, we want to convert 2 degrees per century to arcsec per year. Solve: degrees 3600 arcsec century arcsec 2 3 3 5 72 . century degree 100 y ears year This is certainly large enough to measure because telescopes can measure positions smaller than 0.01 arcsec. Review: The measurement of the perihelion of Mercury is the easy part, and its precession has been known for a long time. Explaining it was the hard part, which has been one of the great successes of general relativity. 37. Setup: Figure 18.12 shows how the gravitational bending of light by a massive object makes the star appear to be at a different position than it really is. This deflection is caused by the massive object warping space, so that light takes a curved path, as shown in Figure 18.10a. Solve: Because the warp of space increases linearly with the mass of the object warping space, doubling the mass of the Sun will double the angular deflection of the star, that is, the distance between the stars would increase. Review: Measuring the angular deflection of light around objects is another way of measuring the mass of the object, although it can be challenging because we have to know where the background star was supposed to appear. 38. Setup: Working It Out 18.1 shows us that the elapsed time when moving at high speeds changes by the Lorentz factor , which equals 1.15 for a speed v 5 0.5c, as stated in the problem. Solve: tEarth 5 tmoving 5 1.15 3 4 yr 5 4.6 yr. Review: The twins would not experience a humongous change in age from this problem, but the Earthbased twin would still be about seven months older. 39. Setup: Working It Out 18.1 tells us that tEarth 5 tmoving, where for v 5 0.999c,  5 22.37. Solve: For the twin problem, tEarth 5 tmoving 5 22.37 3 4 yr 5 89.48 yr. Review: In this case, the Earth-based twin will probably be deceased by the time the space traveler returns.

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40. Setup: The Schwarzschild radius is given by 2GM R s 5 2 . Note that c 2 5 9 3 1016 m 2 /s2 . c Solve: A 70-kg person has a Schwarzschild radius of Rs 5

2  6. 67 3 10211  70 kg 5 10225 m. 9 3 1016 m 2 /s2

Review: This is about 10 billion times smaller than the size of a proton! 41. Setup: The Schwarzschild radius is given by R c2 2GM R s 5 2 , so solving for mass, we find M 5 s . c 2G 2 16 2 2 Note that c 5 9 3 10 m /s . Solve: A black hole with RS 5 1.5 km has a mass of 1, 500 m  9 3 1016 m 2 / s2 51030 kg , which is 211 2  6. 67 3 10 0.5 M. Review: This tells us that the Schwarzschild radius for the Sun is about 3 km, which is a very useful number to remember. 42. Setup: To find the Schwarzschild radius, use 2GM R s 5 2 . Note that c2 5 9 3 1016 m2/s2. Tidal force c 2GMmd Ftidal 5 , where d is the size of the Earth and r r3 is the Earth-Sun distance. Solve: (a) The mass of the moon is 7.3 3 1022 kg, so its Schwarzschild radius is M5

Rs 5

2  6. 67 3 10211  7. 3 3 1022 kg 5 1023 m 9 3 1016 m 2 / s2

or 1 mm. (b) If the Moon turns into a black hole, none of the variables in the tidal force equation are changed, thus its tidal force on Earth will not change. (c) Any collapse of an object into a black hole is expected to produce gravitational waves. Review: Notice that a black hole has a very small radius, which allows objects to be much closer to the mass (that is, R can get very small), which can make the tidal force huge. But at ordinary distances, the effects of gravity and tidal forces remain the same, even around a black hole. 43. Setup: One critical principle of relativity is that the speed of light is the same for all observers (that is, in all reference frames). Solve: Because the speed of light is constant for all observers, everyone sees the laser travel at 3 3 108 meters per second (m/s).

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Chapter 18

Review: Always remember, light travels at the same speed no matter who is emitting it, watching it, or receiving it. 44. Setup: Working It Out 18.1 tells us that t Earth 5 t moving d and dmoving 5 rest , where for v 5 0.995c,  5 10.  Solve: (a) On the spaceship, the star would appear 25 5 2 . 5 ly away. Traveling close to c, it would to be 10 therefore take about two and a half years for this ship to reach the star. A person on Earth will see this take just over 25 years because one travels 1 ly in 1 yr traveling at c. The aliens on that planet will also see the ship cross 25 ly at a speed of almost c so it will take just over 25 years for them as well. (b) A return will happen 50 years later for the people on the ship, so many people they knew might be dead. Review: Relativistic space travel certainly makes it easy for a space explorer to visit a planet, but hard for him or her to tell his or her family and friends about it. 45. Setup: Black holes still follow Kepler’s third law, A3 p2 5 , where the period is in years, distance M1  M 2 in AU, and masses in M. Solve: Just as a planet’s mass is so small compared to its star that we can ignore its mass in Kepler’s third law, so too can we ignore the mass of the spaceship. As such, the mass of the black hole becomes A3 13 5 5 4 M . P 2 0. 52 Review: Note that this answer to this problem would be same if the black hole were a star. M BH 5

Using the Web 46. On the “relativistic roller coaster,” colors look different at first because the speed of light is very slow, and we are moving fairly quickly; therefore, we see exaggerated effects of the Doppler effect. On the desert road, we see a rainbow from extreme Doppler shifting, in which far away we are moving toward the light so it is blueshifted, whereas light from our side/rear is moving away from us so it is redshifted. When you approach an object like the Sun at relativistic speed, we see the effects of beaming and aberration in which all the light seems to emanate from one point; we also see the extreme Doppler shifting in which all colors blueshift.

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47. Answers will vary. (a) Report on a recent supernova, gamma-ray burst, or stellar black hole result from the Swift telescope. Two neutron stars that merge will have a higher mass than neutron degeneracy can support, so the result would have to be a black hole. (b) Report on a recent news story on black holes from the Fermi telescope. This observatory is looking for the high-energy signatures of relativistic jets being accelerated away from black holes. 48. Answers will vary. Report on recent news from the LIGO project. The four listed sources of gravitational waves are continuous, in spiral, burst, and stochastic. 49. Answers will vary. Report on what has been observed and learned from the NuSTAR mission. This observatory is observing in high-energy x-rays. 50. As you travel into the Schwarzschild black hole, light from objects in the background becomes increasingly distorted, but as the black hole fills your field of view, you see very little, time slows down, and eventually everything becomes black. The distortion is called gravitational lensing. Inside a black hole, all light goes straight to the center, so there will be nothing to see inside the object. The Reissner-Nordström black hole has a “wormhole,” which is a mathematical result of the charged sphere that could, in principle, take one to a new place in space. However, you will still die as you pass through the black hole due to “spaghettification.” Moving toward the Sun at close to the speed of light, you see the effects of aberration, in which angles change so the Sun first appears to move away from you, then Doppler shift and the Sun changes color, followed by beaming where the Sun appears brighter than it really is and time dilation makes time slow down as you approach the Sun. Exploration 1. The ball goes straight. 2. The ball goes fairly straight, provided the sheet is big enough; otherwise, they are deflected by small amounts. 3. The ball is deflected by a large amount. 4. The ball hits the bowling ball. 5. Numbers 2 and 3 change depending on speed, whereas number 4 does not. The faster the balls travel for 2 and 3, the less their paths are deflected. 6. As more golf balls fall into the pit, the total mass in the pit increases, so the pit will become deeper and larger. 7. Light is like a superfast moving golf ball. Its path will not be deflected as much as our golf balls, but will still be deflected.

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142 ◆ Chapter 18 Relativity and Black Holes 8. The other golf ball will make a huge dip in the sheet, and both balls will probably start rolling around each other. Certainly, the golf balls and sand would move differently if rolled anew, and in the end, all the balls and sand will probably end up together in a single, very deep and wide pit.

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9. Close to a black hole, time and space are all highly warped/distorted, so nothing would appear as we would see it if there were no black hole.

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CHAPTER 19

Galaxies INSTRUCTOR’S NOTES Chapter 19 is an introduction to galaxies. If you used the fourth edition of this text, most of the material in this chapter was found in Chapter 20 of that edition. Major topics include

▶ galaxy morphology ▶ structure, kinematics, and composition of spiral and elliptical galaxies

▶ the Cosmological Distance Ladder ▶ the discovery of Hubble’s law ▶ rotation curves and dark matter ▶ AGNs and supermassive black holes One of the best pieces of advice I received when I started teaching is that we are not really ready to remember an answer until we have asked ourselves the question. To that end, I like to introduce galaxies by having students create their own classification scheme before I say a single word about galaxies. I print out a page or two with color thumbnails of 10 to 20 spirals, ellipticals, and irregulars, including at least one edge-on spiral that shows a dust lane and at least one merger. I give the students 10 to 20 minutes to create and use a classification system. Once we start the lesson on galaxies, we can refer back to their work, noting how they might have missed certain attributes or might need to reclassify some galaxies or merge certain classifications. This is also a great time to remind them why Pluto was reclassified. Once we understand Hubble’s classification system and have correctly identified their galaxies, we discuss his tuning fork diagram as an evolutionary sequence. Referring back to Chapters 16 and 17, I ask leading questions and guide students to conclude that spirals have blue stars and dust, which signals active star formation, whereas ellipticals only contain red, that is older, stars. From this, they generally make the connection that ellipticals cannot evolve into spirals and the testable prediction that spirals should have abundant gas whereas ellipticals do not. At this point, they have essentially taught themselves Sections 19.1. With all the recent news about precision cosmology and the Higgs boson, I am finding that many people, not just students, carry around the misconception that physics and

astronomy have all the answers and we now understand everything. The subject of dark matter is the perfect opportunity to stress that we only understand what makes up about 10 percent of a galaxy, and as a teaser for later chapters, it turns out we only understand what makes up about 5 percent of the universe. At this point, at least one hand shoots up to ask whether it is “okay” for so much of the universe to be a mystery—a perfect opportunity to talk about the process of science. One of the most common misconceptions I have encountered is that black holes are the great Dyson vacuum cleaners of the universe, perpetually sucking up everything around them. For example, the most common answer to my question “What would happen to Earth if the Sun suddenly became a black hole” is that it would be immediately sucked inside. It does not help that we often show renderings of black holes inside accretion disks. If that misconception persists after your coverage of relativity in Chapter 18, you may need to provide your students the reassurance that the supermassive black holes of galaxies (especially our own) are never going to suck us inside of them. However, you can really ramp up their fascination with the subject by showing them images that point to the binary black holes in M31, or the behemoths like the 6.4 billion solar-mass black hole and giant jet in M87, or the 21 billion solar-mass monster in NGC 4889 in the Coma Cluster.

DISCUSSION POINTS

▶ Have students look for differences between ellipticals

and spirals that yield clues about their composition and star-formation activity. ▶ Ask students to look for examples of spiral patterns in nature, such as those found in some types of shells. Discuss whether there is any common underlying physics between spiral galaxies and that occurring on Earth. ▶ Discuss the evidence for dark matter. Foster a debate among students about whether they consider massive compact halo objects (MACHOs) or weakly interacting massive particles (WIMPs) to be more likely the stuff that composes dark matter. 143

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▶ Does every galaxy have a potential AGN inside it? Discuss

what would be the observable consequences if the center of the Milky Way were to light up as an active galactic nucleus. Do students believe it was ever active in the past? What evidence could we look for to support or refute this? ▶ Astronomers regularly measure the redshift for objects in the universe because, as shown in Working It Out 19.2, redshift yields the distance to objects at cosmological distances. Have students use Figure 19.9 to place the distances estimated from cosmological redshift in the context of the distance ladder. ▶ Compare and contrast the accuracy of measuring distances via redshift against that of other distance-measurement techniques discussed in this and previous chapters.

variables and work toward understanding physical concepts presented in Chapter 19. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. NAAP Lab: Spectroscopic Parallax Simulator This simulation shows how the technique of spectroscopic parallax works. The user chooses a star’s spectral type and luminosity class, as well as the star’s apparent magnitude, and the stars distance and position on the H-R diagram are shown. Text reference: Section 19.2 NAAP Lab: Supernova Light Curve Fitting Explorer

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 19 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises. Hubble’s Law This animation uses Hubble’s observed expansion of galaxies as a starting point for an exploration of galaxy expansion, expansion of the universe, the cosmic microwave background radiation as resulting from redshift, and the Big Bang. Text reference: Section 19.2 Dark Matter This animation presents the idea that dark matter is made of some alien stuff that has not been found on Earth, but which makes up about 90 percent of all matter in the universe. Text reference: Section 19.3

The user compares an observed supernova spectrum to a template, which allows measurement of the distance modulus and the distance to the supernova. Text reference: Section 19.2 Galaxy Redshift Simulator This simulation demonstrates the redshift of a galaxy’s spectrum due to expansion of the universe and the effect this shift has on the galaxy’s brightness as observed through various filters. Text reference: Section 19.2

Active Galactic Nuclei This still visualization shows an active galactic nucleus in which emission from a very compact region in the galaxy’s center outshines the brightness of billions of stars. Text reference: Section 19.4

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate

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Dr. Palen uses a few household items to show how orientation changes our interpretation of a galaxy’s shape. Text reference: Section 19.1 Size of Active Galactic Nuclei Dr. Palen and a group of students show how the time variability of a signal cannot be any faster than the physical size of the region that emits that light. Text reference: Section 19.4

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Chapter 19

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (d) Another term for this is “galaxy morphology.” 2. (c) Spiral galaxies have abundant gas and dust, the fuel for star formation. 3. (c) Hubble’s law is also called Hubble’s redshiftdistance law. 4. (a) Dark matter is matter that exerts gravity but does not show up in the budget of luminous matter. 5. (c) A supermassive black hole has been found at the center of nearly every galaxy. Reading Astronomy News 1. M60-UCD1 is 15.3 million pc, or 15.3 Mpc, away. 2. There are 140 million stars in a galaxy 500 times smaller than the Milky Way, with a diameter around 300 ly. If we let the galaxy be a sphere of radius 150 ly, it has a volume of 14 million ly3, or about 10 stars per ly3. 3. The black hole mass was estimated using the distances and speeds of stars within the galaxy. 4. In this merger, most of the original galaxy was stripped away during the encounter, unlike the images shown in Figure 19.23. 5. This video suggests the galaxy formed by multiple passes near the center of M60. Test Your Understanding 1. (a) Up to 90 percent of a galaxy’s mass is dark matter. 2. (c) Kepler’s laws use orbital characteristics of stars to determine the mass interior to that star’s orbit. 3. (c) An object cannot vary in time any faster than the time it takes for light to cross it. This gives an upper limit to the size of any variable object.  756 . 3  656. 3 5 5 0. 15. 4. (d) Redshift z 5 656 . 3 rest 5. (a) We are limited to using Cepheid variables by their faintness and the sensitivity of our best telescopes. 6. a and b. All spirals have a bulge and disk. 7. E5. A rugby ball is ellipsoidal, and thus would have a large “E number.” 8. (a) Morphology is a description of the appearance, shape, and size. 9. (a) An E0 is spherical and thus results from perfectly randomized orbits. 10. (c) There must be extra mass at large distances to provide the gravity to keep stars rotating at roughly constant velocity with radius. 11. (c) Hubble’s law indicates a direct relationship between recessional velocity and distance.

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12. (a) This is the definition of the Hubble constant. 13. (c) This is a restatement of our observations of flat rotation curves. 14. (d) In the Unified Model of AGN, the main difference between different types depends on the viewing angle, with a smaller contribution from the amount of accretion feeding the black hole. 15. (b) One AU is roughly 8.3 light minutes. Ten hours is 600 70 AU. 600 minutes, so the size is 8.3 Thinking about the Concepts 16. The Great Debate was the meeting of the National Academy of Sciences to decide whether “spiral nebulae” were small clouds that were part of the Milky Way, or large “island universes,” as Kant described them. This single question determined the scale of the entire universe because if the spiral nebulae were just small parts of our own galaxy, it implied that the Milky Way was the entire universe, as opposed to the vast expanse that we now know it to be. 17. Cepheids found in the Andromeda Galaxy (M31) by Hubble showed that M31 was far too large and too far away to be part of our own Milky Way. 18. Standard candles are not exact but are fraught with uncertainty, so it is very important to make multiple measurements to a distant galaxy so that one can have an answer with confidence. 19. Distance-measurement techniques are generally only applicable to a certain distance range. To calibrate a given technique, we need to independently know the distance to a number of objects first; so, for example, we calibrate the Cepheid distance formula using parallax or spectroscopic parallax. Then we calibrate Type Ia supernovae using Cepheid variables found in the same galaxy. In such a way, we build up each “rung” of a distance ladder that lets us probe out to larger and larger distances. 20. A Type Ia supernova is a standard candle only if we know exactly how intrinsically bright each one is. For that, we have to know its progenitor. 21. The luminosity of the blue regions in spirals is dominated by young blue stars. This implies that they are regions of active star formation that contain significant gas and dust. Meanwhile, the luminosity of the red regions is dominated by cool red stars. This implies that they are regions lacking active star formation, though gas and dust may or may not be prevalent. Note that this does not imply that regions of space that are red are only full of old stars. For example, HII regions are also red. 22. The bulge of a spiral has an older population of stars and the stars are on randomized orbits, which mimic on a small scale the structure of an elliptical galaxy.

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146 ◆ Chapter 19 Galaxies 23. A quasar is an enormously powerful engine at the center of a galaxy, powered by huge amounts of mass falling into a supergiant black hole, all stuffed into the size of our Solar System. 24. Quasars appear to be associated with the early stages of galaxy formation, and because distant galaxies are younger than nearby ones, we have to look to distant sources to find sufficiently young objects. 25. An active galaxy exhibits strong x-ray and radio emissions from its nucleus. Bipolar jets of hot gas are also emitted along the magnetic axis of the galaxy. These emissions indicate that the supermassive black hole at the galaxy’s center is feeding on infalling matter. In contrast, the nucleus of a normal galaxy is quiescent, suggesting that the supermassive black hole is dormant. This model follows Occam’s razor because all AGN can be explained by a single physical situation. 26. AGNs are about the size of our Solar System. We know because the time variability of an object constrains its size, that is, an object cannot be much larger than the timescales of variability; otherwise, that signal would be washed out. 27. Stars in the heart of a normal galaxy may occasionally wander too close to the supermassive black hole, where they are captured into elliptical orbits. As these stars approach the black hole, they pass inside the Roche limit and are shredded by tidal forces. As the stellar remnants fall toward the black hole at a large fraction of the speed of light, they flatten into an accretion disk. Eventually, all of the kinetic energy of the stellar remnants is converted into radiation that is emitted across the entire electromagnetic spectrum. Charged particles are trapped by the black hole’s magnetic field and funneled along its rotation axis to create radio jets. The normal galaxy has been converted into an active galaxy. Alternatively, a large galaxy may absorb one of its satellites. When material from the satellite galaxy falls into the supermassive black hole, it is converted to radiation that lights up the galactic nucleus. Something similar (on a much larger scale) occurs during galaxy mergers, and it is not surprising to see that many AGNs are associated with galaxy collisions. 28. To detect supermassive black holes at the center of a galaxy, we need to either see their gravitational influence on their immediate surroundings or catch them in the act of “feeding.” Because the centers of galaxies are very small, we cannot easily observe stellar motions in the centers of distant galaxies, and thus we can infer the presence of a supermassive black hole only if a galaxy is currently active. 29. (a) A close-by rung is needed to calibrate a more distant one; therefore, there must be overlap so that we

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can reliably use the more distant rungs. (b) At the far right of the distance ladder, we find nothing past Type Ia supernovae because at these distances we are running out of standard candles that are bright enough to be seen that far away. 30. Near the centers of galaxies, there can be abundant star formation, which means a great deal of ultraviolet energy which would “cook” planets. The close proximity of many stars means a number of tidal encounters which would also disrupt planetary orbits. Applying the Concepts 31. Setup: Figure 19.14 shows a typical rotation curve of a spiral galaxy using the black curve and left-hand scale on the g-axis. On the right-hand scale, higher in g means there is more mass in the galaxy within that radius, shown for normal matter (red) and dark matter (blue). Solve: (a) 25,000 pc is where the rotation curve peaks. The data values are about 150 km/s and the errors are about 10 km/s in length. So, the measurement error is 10 km/s. (b) The percentage error is 10 3 100% 5 7%. (c) Note that the error bars still 150 trace out the same shape, so varying the black curve within the error bars will not change the conclusion at all. Review: The evidence for flat rotation curves in galaxies has been extensively studied with a variety of techniques, and the fact that the error bars are as small as they are shows that this is in fact a very reliable finding. number , so 32. Setup: Density is, in this problem, n 5 volume the number of galaxies in a given volume is N 5 nV, 4 where V is the volume of a sphere,  r 3 . Note that 3 we call n the “number density” to distinguish it from mass density . Solve: 3   10 68 gala x ies 4 16 m    10 pc  3. 08 3 10 N 5 10 pc  3  m3 51012 galaxies. Review: One trillion galaxies is a good back-of-theenvelope number to have handy, just like a galaxy typically contains 100 billion stars.    rest 33. Setup: The redshift of a galaxy is z 5 obs .  rest The recessional velocity (for redshifts less than one) is v 5 cz, and the distance is given by the Hubble relationship d 5 v/H0.

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750 nm  656. 28 nm 5 0. 143. 656. 28 nm (b) Using this redshift, we find a recessional velocity of v 5 0.143c 5 4.29 3 107 m/s. (c) Using the Hubble relationship with this velocity, Solve: (a) z 5

4. 29 3 104 km/s 5 613 Mpc. 70 km/s/Mpc Review: A galaxy with high redshift has high recessional velocity and high distance. We see in this problem that this galaxy (at about 600 megaparsecs) certainly follows this trend. Now imagine how far a galaxy will be at redshifts greater than one! 34. Setup: Hubble’s law is v 5 H0d, where H0 5 72 km/s/ Mpc and redshift relates to speed by v 5 cz. Solve: The galaxy redshift gives a recession velocity of v 5 zc 5 0.158 3 3 3 105 km/s 5 4.74 3 104 km/s, which with Hubble’s law yields d 5 v/H0 5 (4.74 3 104)/72 5 658 Mpc. Review: This shows us that just because the redshift is small (for example, around 0.1 or 0.2), that does not mean the object is nearby. It is still very far away. 35. Setup: Ten billion times the luminosity of the Sun is 1010L Solve: 3C 273 has a luminosity of 1012L , so the hypothetical galaxy is 100 times fainter. Review: Quasars are extremely bright objects, as shown in this problem, where 3C 273 is 100 times brighter than a typical, local galaxy. 36. Setup: The distance-luminosity relationship d5

Lquasar d

2 quasar

Lgalaxy d

2 galaxy

yields dquasar 5 dgalaxy

Lquasar Lgalaxy

Solve: This quasar is 1 million times brighter than our galaxy, and is 2 megaparsecs away, so dquasar 5 2 Mpc

106 5 2 , 000 Mpc 1

away (or

1,000 times farther). Review: Quasars are all located at very large distances, as this problem demonstrates. 2GM 37. Setup: The Schwarzschild radius R s 5 2 , or we c can use the scaling relation that a black hole with l M has a Schwarzschild radius of about 3 km. Solve: Using our formula, 26 . 67 3 1011 26 3 10 9 2 3 1030 5 Rs 5 9 3 1016 7.7 3 1013 m or about 513 AU.

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Review: Using our scaling relation, the Schwarzschild radius is 26 3 109  3 km 5 7.8 3 1010 km 5 7.8 3 1013 m, as found above. Notice that this is about as large as the distance to our Kuiper Belt. 38. Setup: An object can be no larger than the distance light travels in the time during which the brightness varies. Solve: Because the object varies with an 83-minute period, it can be no larger than 83 light-minutes across. Remember that it takes sunlight about 8.3 minutes to reach Earth; so, 1 AU is 8.3 lightminutes, and therefore the object is 83 light-minutes. 1 AU 5 10 AU across. 8. 3 light-mi n utes Review: This is about the distance to Saturn. Given that the newspaper describes the object as “cosmological” (which I interpret to mean at a very large distance), I conclude it is a quasar. 39. Setup: We use E 5 mc2 to convert mass into energy because mass feeds the quasar. The lifetime of an object is (the amount of fuel available) divided by (how quickly the fuel is used). Solve: The total amount of energy present is 20 percent kg 20  108 M  2 3 1030 5 of 108 solar masses, or 100 M 4 3 1037 kg, which yields a mass energy of E 5 mc2 5 4 3 1037 3 (3 3 108)2 5 3.6 3 1054 J. Because the power output is 1041 W, the lifetime is 3. 6 3 1054 J yr 5 3 . 6 3 1013 s  5 1041 J/s 3 . 15 3 10 7 s 1.14 3 106 yr, or about 1 million years. Review: In cosmological time spans, a million years is very short, which shows that the engine of a quasar is very short-lived, about as long as the most massive stars. 40. Setup: We use E 5 mc2 to convert mass into energy. Solve: Converting half of the solar-mass star into energy produces E 5 1030 kg  (3 3 108 m/s)2 5 9 3 1046 J. Review: The Sun puts out 4 3 1026 J in 1 second, or over its 10 billion year lifetime (3.15 3 1016 s) about 1.2 3 1043 J. Thus, we see that the energy released during infall into a black hole is significantly more than via normal stellar processes. 41. Setup: Distance, luminosity, and brightness are related L , where the typical 4 B luminosity of a Type Ia supernova is 3.7 3 1036 W (see Working It Out 19.1). by L 5 4  d 2 B or d 5

Solve: d 5

3. 7 3 1036 5 1. 7 3 1026 m or 5,540 Mpc. 4 1017

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148 ◆ Chapter 19 Galaxies Review: The most distant Type Ia supernova observed is at z 5 1.55, or about 4,500 Mpc, so we see that this hypothetical supernova is even further and fainter! 42. Setup: We use E 5 mc2 to convert mass into energy. Solve: (a) Converting one-tenth of the Earth mass into energy produces E 5 0.1 · 5.97 3 1024 kg  (3 3 108 m/s)2 5 5.4 3 1040 J. (b) This is 5. 4 3 1040 5 1. 4 3 1014 or 140 trillion times more 3. 8 3 1026 energy. Review: A black hole can give off more light per second than even an entire galaxy of stars. 43. Setup: There are 3.15 3 107 sec in 1 year and one solar mass has an equivalent energy of E 5 mc2 5 2 3 1030 kg  (3 3 108 m/s)2 5 1.8 3 1047 J. 3 . 15 3 10 7 s Solve: 2 3 10 41 J/s 3 5 6 . 3 3 10 48 J/yr. yr Dividing this by the energy equivalent of the Sun, we find the quasar consumes 35 solar masses per year. Review: This may seem like a lot, but consider that a supermassive black hole has a mass of millions to billions of solar masses; thus, it only needs to accrete a miniscule fraction of its total mass. d converts 44. Setup: The small-angle formula D 5 57. 3  from angular size  to physical size D given the distance d. Solve: Solve for angle 57. 3 D 57. 3 3 30 , 000 5 0. 1. 5 5 d 17 3 106 (b) This is about 4 times smaller than the full moon (which is about half a degree in size). Review: If this jet were bright enough, it would be visible to the naked eye. 45. Setup: The speed of light is 1 light-year per year (a very convenient number) so let’s work in these units. Solve: If the distance is 5,000 light-years, and the gas is moving at the speed of light, it was ejected 5,000 years ago. Here, we are moving at 99 percent the speed of light, so it takes 101 percent of the time to travel, or 5,050 years. Review: Another approach is to use 5, 000 ly d 5 5 5, 050 yr. d 5 vt or t 5 0 . 9 9c 0. 99 ly Using the Web

the luminosity, and then compare this with the average brightness of the star to find its distance. Notice that we also have to worry about the effects of extinction by dust between the galaxy and us, and we have to assume that all Cepheids of a given period have the same luminosity. 47. Answers will vary. The response will include recent pictures of galaxies, where the pictures were taken (that is, what size telescope, and from the ground or space); whether the galaxies are face-on, edge-on, or inclined; what wavelengths were imaged; are the images false colors; and what the colors indicate. 48. (a) Complete a citizen science project on the Galaxy Zoo and submit the copy of the completed project. 49. Answers will vary. Report on (a) recent news from the Fermi Telescope about dark matter, and (b) whether the Alpha Magnetic Spectrometer on the International Space Station has discovered anything promising about dark matter. 50. (a) Answers will vary. Report on the Swift observatory about supermassive black holes. (b) NuSTAR is a space-based x-ray telescope specializing in observations of very high-energy x-rays. The response will include recent discoveries as well as a description of the design of the telescope. Exploration 1. Very round ellipticals tend to be the easiest to identify (like the E3 or E4 labeled A in the image to the left). 2. Very “lenticular” galaxies are hard to identify, such as galaxy B to the left, or very small galaxies, such as galaxies at C, which could be just about anything. 3. Inclination makes it very hard to tell if the galaxy is a spiral and its subtype, or if it is a very eccentric elliptical. Size (that is, very small galaxies) makes it very hard to tell if a galaxy is elliptical or spiral at all. 4. I tend to agree on ellipticals most often with my partners. 5. I tend to disagree most often on the subtype of spirals with my partners. 6. It is easier to classify galaxies if one has (a) much larger- or higher-resolution images and (b) an example catalog of many different types to which one can directly compare.

46. The website shows an animation of a pulsating Cepheid variable. Knowing the period, we can calculate

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The Milky Way: A Normal Spiral Galaxy INSTRUCTOR’S NOTES Chapter 20 is an exploration of the Milky Way. If you used the fourth edition of this text, this material was found in Chapter 21 of that edition. Major topics include

▶ the size and morphology of the Milky Way ▶ the nature of spiral arms ▶ the process of chemical enrichment ▶ the structure of the disk and halo ▶ evidence for dark matter ▶ evidence for a supermassive black hole ▶ the Local Group and eventual collision with Andromeda This chapter can be looked at either as an introduction to the Milky Way or as an examination of how our galaxy supports the topics put forth in the previous chapter. I used to teach the Milky Way first and then use those findings to interpret other galaxies, but I prefer the structure of this textbook, in which we first learn about galaxies in general and then home in on the specific observations and insights that our home galaxy has to offer. In particular, I believe this latter approach better follows the scientific method because in Chapter 19 we present our theories of galaxies, and in this chapter we test them. One thing that I like to add to my lesson about the Milky Way is the history of mapping the galaxy (or universe, as it was often known), such as the various attempts made by William Herschel in the late 18th century. It is a useful lesson in how dust is literally in the way of everything we look at, one important but often overlooked reason why there are so many uncertainties in astronomical measurements. Also, if you are interested in including songs in your class, Eric Idle’s Galaxy Song is a perfect way to open this chapter, although the last stanza is a little irreverent. The careful listener will note that many of his values are different from today, which opens up a good discussion about the process of science and the challenges of discovering the structure of something from the inside looking out. To teach an astronomy course necessarily implies that we will talk about tremendously large quantities as if they

are quite commonplace. I can easily say that the galaxy is 100,000 light-years across (Figure 20.8), but how easily will students develop an instinctive appreciation for what this means? One useful exercise is to have students calculate the scale of the face-on view of the galaxy (Figure 20.2), and then locate some of our nearest neighbors. For example, point out the first letters of each labeled object (Sun, Perseus, Scutum, Orion) and ask students to guess where Alpha Centauri (4.4 light-years), the Pleiades (525 light-years), and Betelgeuse (643 lightyears) are. Chances are good that they will indicate Alpha Centauri is at the “S” in Sun, the Pleiades are at the “S” in Scutum or “P” in Perseus, and Betelgeuse is at the “O” in Orion. Figure 20.2 has a scale of roughly 1,300 lightyears per millimeter, meaning the dot marking our Sun is more than 1,500 light-years across. They will probably be shocked to learn that all three objects are located within the Sun’s yellow dot. Another excellent exercise for students is to use Kepler’s third law to compute the mass of our supermassive black hole Sagittarius A* (Sgr A*) using Figure 20.18 and Working It Out 20.2 but for a different star than S0-2. In case any students ask, the name “Sgr A*” was inspired by standard naming conventions in radio astronomy. The “A” refers to the first radio source discovered in the constellation. As co-discoverer Bob Brown explained, “Scratching on a yellow pad one morning I tried a lot of possible names. When I began thinking of the radio source as the ‘exciting source’ for the cluster of H II regions seen in the VLA maps, the name Sgr A* occurred to me by analogy brought to mind by my PhD dissertation, which is in atomic physics and where the nomenclature for excited state atoms is He*, or Fe*, etc.” (Astron. Cachr./AN 324, No. S1, 1–8, 2003). Andromeda and the Milky Way are likely to collide in the next 6 to 8 Gyr, and with an abundance of images of mergers available, we have a good sense of what that would look like from the outside. There have been some interesting simulations/movies made of what this would look like from space and from Earth. You can find dozens of these on YouTube.

149

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150 ◆ Chapter 20 The Milky Way: A Normal Spiral Galaxy If you want to include a little “challenge” section in this chapter, you can introduce the notion of mean-free path, which we alluded to in Chapter 14 as a random walk. With an estimate of the number of stars in the galaxy, space between them, and average speed, your students will discover just how few stellar collisions occur inside a galaxy. It drives home the fact that, although galaxies contain hundreds of millions of stars, they are mostly empty space.

DISCUSSION POINTS

▶ Have students make a step-by-step storyboard of the

formation and evolution of the Milky Way by taking into account the components that we see today and including the chemical evolution of the galaxy. Is this formation story also applicable to other galaxies? If so, which ones? ▶ Discuss whether or not the Milky Way Galaxy would be likely to have the same appearance as today if it had formed in a dense cluster of galaxies rather than in a sparsely populated group. ▶ Is the Sun a typical star in the Milky Way? Have students try to find any properties that make the Sun special. ▶ As the Sun moves in its orbit around the galactic center, what objects could it encounter during its lifetime? Discuss the possible influence of those objects on Earth. ▶ Consider the challenges of measuring the structure of your college or university campus by sitting at one single location armed only with a telescope. Discuss why it is both expected and acceptable that our understanding of our galaxy has changed over time.

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Milky Way Rotational Velocity This simulation shows a plot of the rotational velocity of stars at varying distances from the center of the Milky Way. A draggable cursor allows determination of the contained mass implied by the curve. Text reference: Section 20.3 Milky Way Habitability Explorer This module allows the user to estimate the trade-off between abundance of heavy elements and likelihood of a catastrophic event (supernova, stellar collision, or near-stellar encounter) at different distances from the center of our galaxy. Text reference: Chapter 20 “Origins”

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. Answers will vary: The Milky Way contains gas and dust as well as ongoing star formation. Observations of 21-cm radiation emitted from neutral hydrogen clouds show the existence of spiral arms. The galaxy has a well-defined rotation curve that flattens toward the visible edge. Finally, most stars in our galactic neighborhood share the same relative motion around the galactic center. 2. The disk contains old and young stars; the halo contains old stars. 3. (c) Dark matter is only detectable by its gravity. 4. Andromeda is moving toward us because of our mutual gravitational attraction. Reading Astronomy News 1. The stars were 5 million trillion km, or 162 kpc away. 2. On Figure 20.16, this measurement would be off the page to the right. 3. This mass estimate (8 3 1011 M) is about 8.6 times the mass of our galaxy inside the Sun’s radius (Working It Out 20.1). 4. This additional dark matter is distributed throughout the galaxy and its halo. 5. The new estimate of the galaxy’s mass is lower than previously thought, making the escape velocity lower as well because there is less gravitational mass from which to escape. Test Your Understanding 1. (c) RR Lyrae variables in globular clusters allow us to reliably determine the distances to these clusters,

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which are distributed throughout the galaxy and its halo. 2. (a) It is difficult to determine the structure of something when you are embedded inside it and when dust obscures much of what you can see. 3. (b) Older stars have been “kicked out” of the midplane by gravitational interactions with other stars and massive molecular clouds. 4. (a) The glow of synchrotron radiation from cosmic rays spiraling in the galactic magnetic field is a direct confirmation of the presence of large-scale fields. 5. (a) Observations of stellar orbits lead to the conclusion that there is a very compact and massive object at the galactic center. 6. (a) We already see tidal streams of stars from the Magellanic Clouds being stripped off by the Milky Way’s gravity. 7. (c) The distribution of globular clusters shows that we are not located at its center. 8. (b) Shapley used RR Lyrae (variable stars) to estimate the distance to each globular cluster. Without those stars, this distance estimate would not have been possible. 9. (a) The flat rotation curve can only be explained by the presence of a significant amount of matter that we cannot see. 10. (b) The name is a bit of a misnomer because these are not rays of energy; at first, scientists did not know what they were, so they were called “rays.” 11. (c) All indications suggest that we are living in a spiral with a small bar. 12. (c) The disk, and specifically the spiral arms, is where the youngest stars are located because this is where star formation is the most active. 13. (d) Both disk and halo stars can have the same temperatures. 14. (d) It is very hard to detect many galaxies because they have “low surface brightness,” which means that their light is spread out (diffuse). 15. (d) The galactic habitable zone considers where stars could form that may host life-bearing planets, but does not discuss the particulars of the solar systems themselves. Thinking about the Concepts 16. Science is about providing the best explanation possible at the time, but its self-correcting nature allows for those explanations to change, without judgment or repercussions, as more is learned. Shapley measured the wrong distances because he did not know about

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the dust, but the conclusions he drew were nonetheless critical because he made the first data-driven measurements of our galaxy’s size and our location within it. Further refinements later on have not changed those basic conclusions. But without those first measurements, subsequent advancements in our understanding might not have occurred. 17. Globulars are distributed roughly spheroidally, showing that there is a large, spheroidal halo that encloses the disk of the galaxy. The center of that spheroid is roughly 8,000 pc from the Sun, showing that we are roughly that distance from the center of the galaxy. 18. Open clusters are young and therefore will generally be found near spiral arms and regions of recent star formation. 19. If the gas in the inner part of the galaxy has undergone much more nuclear processing, then it is more metal rich than the gas in the outer galaxy; hence, stars in the inner part of the galaxy will be more metal rich. One can only make the argument that “young stars are always more metal rich” if one assumes that the stars being compared all came from the exact same source of gas or if one knows the detailed chemical enrichment history of the gas. 20. Doppler shifts of 21-cm radiation emitted by neutral hydrogen clouds show material moving around the center of the galaxy. This is shown in Figure 20.15. Note that dust obscures our view of the galaxy in the visible and ultraviolet regions of the electromagnetic spectrum. However, it is transparent to the longer 21-cm radiation emitted by neutral hydrogen clouds. This makes 21-cm radiation a useful tool for peering through the dust to see the galaxy’s structure. 21. Similar to most other spiral galaxies, ours has a flat rotation curve, which implies that up to 90 percent of the mass of our galaxy must be dark matter. 22. Halo stars have a much lower metal content than disk stars, and they are not moving with the disk; instead, halo stars are moving up and down through the disk on spheroidal orbits around the galactic center. 23. Synchrotron radiation comes from cosmic rays spiraling through the galaxy’s magnetic field. 24. There is far too much dust along the path to the galactic center to see optical light from that region. Whereas dust is opaque to optical and UV light, it is transparent to radio, x-ray, and infrared; hence, we can (and do) observe this region in these wavelengths.

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152 ◆ Chapter 20 The Milky Way: A Normal Spiral Galaxy 25. Sgr A* is believed to be the massive object at the very center of the galaxy. It was first detected as a strong radio source and later in x-rays. 26. The Doppler shifts of stars near the galactic center are very large. This implies that there must be a tremendous mass packed into a small volume in the galactic center. Gamma ray, x-ray, and radio emissions (particularly synchrotron radiation) emanating from the galactic center are also black hole signatures. 27. The Large and Small Magellanic clouds are irregular satellite galaxies of the Milky Way. They are the remnants of dwarf galaxies that crashed through the galactic disk, stripping some of our galaxy’s gas and dust in the process. So besides being at a comparable distance from our galactic center (which is why they look like small clouds), they also look spectroscopically similar to the Milky Way because of the gas and dust they “stole” during previous passages through the galactic disk. 28. Galactic satellites are the remnants of the small protogalaxies that first formed in our dark matter halo. Most of them coalesced to form the Milky Way; however, there are a number of them that remain in orbit around our galaxy. Many satellite galaxies have low surface brightness, so they are hard to detect against the foreground stars of our galaxy itself. Many are also located on the other side of the galaxy, and because of the same sources of opacity that make it impossible to see the galactic center, these dwarf galaxies cannot be detected because the stars can’t be seen in the first place. 29. (a) If the Sun were located near the galactic center, the night sky would be extremely bright because of the very high density of stars surrounding the Sun. (b) If the Sun were located near the center of a large globular cluster, both the day and night skies would be filled with thousands of stars. However, most of these stars would be Sun-sized and smaller. As a result, we would not be bathed with the intense radiation that we would face at the galactic center. (c) Inside a molecular cloud, no stars would be visible in the background. We would see the Sun dimmed by the thick dust, and even the planets might be hidden by the dense fog. Meanwhile, the sky would glow a dull red because the cloud would likely be heated by nearby stars and protostars. 30. A galactic habitable zone is the region of a galaxy in which we believe stars capable of hosting life would be found. This excludes regions of rapid star formation (both because of the UV light and frequent supernovae), close stellar encounters, and regions that have enough heavy elements to form rocky planets.

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Applying the Concepts 31. Setup: Figure 20.8 shows the disk and halo of our galaxy. Note that the halo is listed as about 90 kpc in diameter, and the disk of the galaxy is shown at 30 kpc. Solve: The disk is about one-third the size of the halo. Review: In general, the disk of a galaxy is much smaller than its halo. 32. Setup: Figure 20.16 shows the rotation curve for our galaxy, where the y-axis shows orbital speeds in km/s, and distances (on the x-axis) in kpc. The period of a circular orbit is equal to the distance traveled divided 2 r . by the orbital speed, or P 5 v Solve: (a) At 6 kpc, a star is orbiting at about 3 . 08 3 1013 km 230 km/s. (b) 6 , 000 pc is 6 , 000 pc 5 pc 1.85 3 1017 km. The time to orbit once is the period, 2 1. 85 3 1017 km p5 5 5 . 04 3 1015 s and because 230 km/s there are 3.15 3 107 sec in 1 year, this is 160 million years. Review: Our Sun takes 230 million years to orbit, so it makes sense that a star that is closer to the galactic center will complete one orbit in less time. 33. Setup: Figure 20.16 shows the rotation curve for our galaxy. At a distance of 50 kpc, the Magellanic Cloud would follow a circular orbit at a speed of about 170 km/s. The time to complete one orbit is the 2 r . period P 5 v Solve: 50 kpc is 3. 08 3 1013 km 50 , 000 pc 5 1.54 3 1018 km, pc 2  1. 54 3 1018 km 5 5.7 3 1016 s. Because 170 km/s there are 3.15 3 107 sec in 1 year, this is 1.8 billion years. Review: At this rate, the Magellanic Cloud has completed just a few orbits since the Milky Way was formed. 34. Setup: To compute the number of orbits of the Sun, we note that if one trip takes x years, then n trips takes n times as long. The number of trips made in a given time T is just the ratio of the two. Solve: The number of trips the Sun has made is 4. 6 3 109 yr 5 20 trips. 230 3 106 yr Review: Although the Sun’s orbital period seems very slow, we have circled the galaxy many times. so P 5

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Chapter 20

35. Setup: To solve this problem, we could use the information given to determine the orbital velocity of the Sun, and then apply this to the proposed globular cluster to find its period. Or we could consider using ratios because both objects will have the same orbital speed (after all, the rotation curve is flat). The period of orbit is equal to the distance traveled divided by the orbital 2 r , so because both objects have the speed, or P 5 v rglobular r . same speed,  5 P Pglobular Solve: (a) Solving our equation for the period of the r globular, its period will be. Pglobular 5 P globular 5 r 8 , 300  9 , 000 pc 5 480 Myr . (b) The total 230 Myr 8 , 300 pc 13 3 109 yr number of orbits will be 27 orbits. 480 3 106 yr Review: In problem 34, we found the Sun has orbited the galaxy about 20 times in the last 5 billion years, whereas here we see the globular has made about the same number of orbits in about three times the time. 36. Setup: Use the distance-luminosity relation L L1 L2 5 2 or dcluster 5 dstar cluster . 2 Lstar d1 d2 160 , 000 5 92 , 000 pc. 1 (b) The visible disk is only about 30,000 pc in diameter, so the halo is at least three times larger than the visible disk. Review: Looking at Figure 20.8, we find that the halo extends to about this distance. 37. Setup: See Figure 20.18. Solve: The stellar orbits around the galactic center show that there is an object of around 4 million solar masses, but also that this object must be smaller than 6,000 AU. If one imagines 4 million stars in a dense cluster, those stars would collide in tens of thousands of years, meaning the object could not be a star cluster. The only other reasonable way to compact that much mass into that small of a space is in a black hole. Review: Note that in some respects this is a process of deduction and exclusion rather than one of direct proof. Science often works that way. 38. Setup: Terrestrial planets form out of heavy material (elements other than H and He), so they will be located only around stars that contain an appreciable amount of that material. Solve: (a) dcluster 5 230 pc

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Solve: Terrestrial planet formation requires star formation from enriched gas, which is mostly in the inner disk of the galaxy. Planet formation is probably less likely in the outer disk and certainly unlikely in the halo stars, which are very metal poor. Review: From this argument, it is unlikely to find planets in the halo or halo objects, such as old globular clusters. However, we expect that nearby stars, which formed around the same time as the Sun, could host many planets. 39. Setup: For this problem, we will use E 5 mc2 and solve E it for mass m 5 2 . c Solve: (a) Using E 5 mc2 to solve for the rest energy of a proton, we find E 5 1.7 3 10227 kg  (3 3 108 m/s)2 5 1.53 3 10210 J. (b) The relativistic mass of the fast-moving proton with energy of 100 J is 100 J m5 5 1. 11 3 10215 kg. (c) The rel8 2 (3 3 10 m/s) 1. 11 3 10215 kg 5 6. 54 3 1011 or 1. 7 3 10227 kg 654 billion times more massive than its rest mass. Review: Fast-moving particles gain so much energy that when they collide with another particle the energy goes into creating an entire zoo of new particles. This is the principle behind particle accelerators. 40. Setup: This problem essentially requires us to compute distance d 5 vt, where v is the speed and t is the time. Solve: 100 million years is 3.15 3 1015 s, so in this time, the cosmic ray has traveled m d 5 vt 5 (1 2 10 2 24)  3 3 108  3. 15 3 1015 s 5 s (1 2 10224)  9.45 3 1023 m, whereas the photon has traveled the full 9.45 3 1023. The difference in distance is 10224 · 9.45 3 1023 m 5 0.945 m. Review: For all practical purposes, ultrahigh-energy cosmic rays and photons travel the same speed and arrive at the same time. However, unlike the photon, this particle has enough energy to create billions of new particles when it collides with, for example, an atom in our upper atmosphere. A3 , where 41. Setup: Kepler’s third law says P 2 5 M m m is orbiting M, mass is in solar masses, distance is in AU, and period is in years. Because the black hole is far more massive than a star, we can ignore m and solve ativistic proton is

for P 5 A 3 / M . Solve: 0.02 ly is 1,265 AU, so Kepler’s law gives P 5 1, 2653 / 5 3 106 5 20 yrs .

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154 ◆ Chapter 20 The Milky Way: A Normal Spiral Galaxy Review: This star orbits so quickly that we can see it complete more than one orbit during a human lifetime. 42. Setup: Use Kepler’s third law in general form a3 P2 5 , where m is orbiting M, masses are in M m solar masses, period is in years, and distances are in AU. Solve: Convert units to AU and solar masses: the black hole is 4 3 106 M and the orbital distance is 828 AU. Kepler’s third law gives a period of 11.9 years. The circumference of the orbit is 2r 5 5,205 AU, so the star 5, 205 AU 5 437. 4 AU/yr . These aren’t travels v 5 11. 9 yr very conventional units. It turns out that 1 kilometer per second (km/s) is about 1 parsec per million or 0.2 AU/yr. So using this last relationship, v 5 2,187 km/s. Review: 2,000 km/s is a very high speed, a little less than 1 percent the speed of light. 43. Setup: A black hole has a Schwarzschild radius 2GM R s 5 2 5 2. 96 M km , where M is in solar masses. c Solve: Our supermassive black hole “weighs in” at 4 million solar masses, so its event horizon is this times 2.96 kilometers, or 1.18 3 107 AU 5 0.073 AU. Review: Note that this supermassive black hole has a size much smaller than even the orbit of Mercury. That is tiny. 44. Setup: Use the equation for circular velocity: v circ 5

GM . r 2

v 2 r 2000  1 . 5 3 10 5 , where the Solve: M 5 G 6. 67 3 10220 number on the bottom is Newton’s gravitational constant in unconventional units of km3 kg–1 s–2. This yields 9 3 1036 kg 5 4.5 million solar masses. Review: We see that stars orbiting at a few thousand km/s are probably orbiting compact objects with millions of solar masses, that is, supermassive black holes. 45. Setup: The volume of a solid disk is πR2h where R is the radius and h is the height. The volume of a rectangular doughnut (or torus) is 2πRwh, where R is the inner radius, w is the width, and h is the height. The

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habitable zone extends from 7 to 9 kpc, so has a width of 2 kpc. Solve: If we assume that the disk extends to a radius of 15 kpc (diameter of 30 kpc) and has a thickness of 1 kpc, then the total volume of the disk is π · 152 3 1 5 707 kpc3. The habitable “doughnut” will have a volume 2   7 3 2 3 1 5 88 kpc3. The habitable zone contains 88 5 0. 12 or 12 percent of the total disk. 707 Review: If we believe that stars that host habitable planets can only form in these regions, there is still an extremely large percentage of a galaxy in which life could form. Using the Web 46. A natural lightscape refers to “resources and values that exist in the absence of human-caused light at night.” It is increasingly rare for people to see the Milky Way because of a combination of pollutants in the air and urbanization that produces light at night even in areas that are remote, such as many national parks. The National Parks Service considers the natural landscape at night (including the sky) to be just as important of a resource to protect and treasure as the landforms of the parks themselves. 47. (a) There is a change in wavelength of the selected pictures of the APOD movie because we can only see into the center of the galaxy in the infrared. Answers will vary on recent images of the Milky Way and whether you can see the Milky Way from your location. (b) The galactic center can only be seen in x-rays and infrared because these wavelengths are needed to peer through all the dust. Answers will vary on reports of images from the Chandra X-ray Observatory. 48. Rapidly moving stars in a small area have such short periods because of a large, massive central object. A combination of the mass of the central object and the small size of the orbit constrains the object to be a black hole. Answers will vary on new results from the two research groups. 49. You will perform the Milky Way Project exercise at the Zooniverse to find galactic bubbles. Report results. 50. Answers will vary. Report on the impact of the story in 2012, whether astronomers disputed the study, and whether it is still receiving attention.

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Chapter 20

Exploration Following is a plot of all the clusters listed in Table 20.1, similar to Figure 20.25. 202.5

180

157.5

225

135

247.5

112.5

270

292.5

67.5

315

1. The geometric center of these clusters appears to be about 10 kiloparsecs from Earth. 2. The geometric center of these clusters appears to be about 22.5 degrees. 3. If the Sun were at the center of the galaxy, then the clusters would be distributed randomly around the center of the graph. These clusters are not centered on the Sun but, rather, are offset, showing that the Sun is not at the center of the globular cluster distribution.

45 337.5

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22.5

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CHAPTER 21

The Expanding Universe INSTRUCTOR’S NOTES Chapter 21 covers the Big Bang theory. If you used the fourth edition of this textbook, the material in this chapter was in Chapter 19 of the former edition. Major topics include

▶ the Cosmological Principle (homogeneity and isotropy) ▶ the universe has no center ▶ how Hubble’s Law provided the foundations of the Big Bang theory ▶ Hubble flow, redshift, and the expansion of spacetime ▶ the Cosmic Microwave Background (CMB)

My favorite quote about cosmology is from Dr. Mike Seeds at Franklin & Marshall College: “The Universe is very big, but it is described by a small set of rules and . . . we have found a way to figure out the rules—a method called science.” By the time we reach cosmology in an introductory course, we have learned all the rules and the methods that we need to understand the history of the universe back to an infinitesimal fraction of a second after it began. Unfortunately, this is also the time in the semester when the frenzy of final papers and term projects and impending exams can lead students to lose sight of the proverbial big picture. Cosmology—the history of our universe—really is the big picture. I like to pause for a few minutes of reflection at the beginning of this material to help students recognize that they are now in that enviable position of not just seeing, but also understanding, that big picture. A common frustration in astronomy is the difficulty in doing classroom demonstrations of physical situations, and cosmology, although dealing with the very largest scales, actually lends itself well to demos. One of my favorite visual explanations of the need for both homogeneity and isotropy can be made with a dozen 1- or 2-foot lengths of PVC pipe, all taped together. The pipes are homogeneous, but their appearance is extremely different when viewed side versus edge on. Large rubber sheets can be purchased from Grainger or McMaster-Carr, allowing you to do a live demonstration of Figure 21.5a with a permanent marker, some pennies, glue, and a handful of volunteers.

With all due respect to the authors of the textbook, my students and I have always found Figure 21.2, which asks the reader to visualize ants living on paper clips stuck to a rubber band being stretched, sufficiently complicated that it obfuscates the analogy of galaxies in an expanding universe. I much prefer to use the two and three-dimensional illustrations in Figure 21.5, to elucidate how cosmological redshift is due to expansion of the universe, not movement of galaxies through it. A real-life example is to have the class think about baking a cake with raisins or chocolate chips inside. The chips are embedded within the batter and the batter expands; from the point of view of any raisin/chip, every other raisin/chip would be moving away from it. No chips, however, are flying through the batter. Of course, you should preface this with the limitations that the cake stops rising and the pan has edges, but the analogy still holds. The cosmic microwave background (CMB) isotropy maps that are shown in Figure 21.13 are both pretty and dramatic, but explaining how they are interpreted to yield clues about cosmic parameters can be quite a challenge. You might use the Planck CMB Simulator (http://www.strudel.org.uk/planck) to provide students a hands-on sense of how the amounts of normal matter, dark matter, and dark energy change the appearance of the CMB. However, the real utility of the CMB comes from examining its power spectrum, which I fear may be a little too challenging to explain in a short one-page exercise. I encourage you to show the simulation with the power spectrum and explain how clustering of the anisotropies yields very precise measurements of the cosmological parameters. You and your students may also find it useful to peruse Dr. Wayne Hu’s website on the CMB at http://background.uchicago.edu.

DISCUSSION POINTS

▶ Ask students to come up with their own experiences of

environments that are homogenous and/or isotropic. Do we expect to find these in nature? ▶ Is an expanding and evolving spacetime consistent with relativity from Chapter 18? From that chapter, what can we expect to have the greatest effects on how the universe evolves? 157

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158 ◆ Chapter 21 The Expanding Universe

▶ The Hubble Constant has been a value of considerable

and consistent debate, with values ranging from 30 to 500 km/s/Mpc. Why has a consistent measurement of this value been elusive? Can personal bias effect its measurement? Observational biases? ▶ It is often difficult for students to comprehend the meaning of the Big Bang. The concept that space is stretching/ expanding (Figure 21.5) and the effects that has on light waves (Figure 21.7) and on the background blackbody energy of the CMB (Figure 21.12) have been proved by Hubble’s law and the observations of the cosmic microwave background (Figure 21.13). Connect all these figures with Wien’s law to provoke discussion among students about the properties of the Big Bang and its observable consequences. ▶ Arno Penzias and Robert Wilson were awarded the 1979 Nobel Prize in Physics for the discovery of the CMB, and John Mather and George Smoot were awarded the 2006 Nobel Prize in Physics for the Cosmic Background Explorer (COBE) measurements of the CMB. These awards recognize the importance of providing real answers to the fundamental questions addressed by cosmology. Discuss some more recent examples of the importance of investigating cosmology, such as the discovery of the accelerated expansion of the universe using Type Ia supernovae by Saul Perlmutter, Brian Schmidt, and Adam Riess, for which they were awarded the 2011 Nobel Prize in Physics.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 21 and are available from the free Student Site (digital.wwnorton.com/Astro5). These animations are also integrated into assignable Smartwork5 online homework exercises. Hubble’s Law This animation uses Hubble’s observed expansion of galaxies as a starting point for an exploration of galaxy expansion, expansion of the universe, the cosmic microwave background radiation as a resulting redshift, and the Big Bang. Text reference: Section 21.2 Big Bang Nucleosynthesis This is a discussion of how many of the elements of the periodic table were made during the Big Bang, by referencing nuclear reaction rates and elemental abundances. Text reference: Chapter 21 “Origins”

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ASTRONOMY IN ACTION VIDEOS These videos are a mixture of live demos and mini lectures, enabling students to prepare for class or review what they have learned. All videos are available on the free Student Site (digital.wwnorton.com/Astro5) and offline versions can be found on the USB drive. Assignable assessment questions can be found in Smartwork5 and the Coursepack. Expanding Baloon Universe Dr. Palen uses a baloon and stickers to show how expanding space leads to Hubble’s Law. Text reference: Section 21.2 Observable vs. Actual Universe A discussion of how more of the universe becomes observable to use over time. Text reference: Section 21.2

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. b, a. Isotropy is uniformity in direction; homogeneity is uniform in location. 2. (d) The Big Bang happened everywhere because it was the beginning of the entire universe. 3. (c) The scale factor shows how distances change in time. 1 . 4. (b) Scale factor and redshift are related by R u 5 1 z 5. (a) Temperature in the universe cools over time so a cool blackbody today would have been very hot at early times. Reading Astronomy News 1. Before the discovery of the CMB, evidence of the Big Bang came in many forms, including Hubble’s redshiftdistance relationship and stellar chemical abundances. 2. Theoretical investigations of a Big Bang scenario predicted the existence of a background radiation between 2 and 5 K; the discovery of the CMB confirmed this prediction. 3. Answers will vary. “Thermal echo of the universe’s explosive birth” is perhaps a wordy and overly-technical description because it implies that the reader understands both thermal and echo and is perhaps misleading as well, as it implies the event was an explosion. One generally views an explosion as traveling into the space around it, whereas the Big Bang was an explosion of space itself. 4. Bell Labs was actually supporting research to investigate background microwave signals for a practical purpose

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Chapter 21 The Expanding Universe ◆ 159

of using microwave signals to determine the distances to satellites. The discovery of the CMB was therefore accidental. However, Bell Labs has a long history of supporting scientific exploration and advancement because as the cliché goes, “a rising tide raises all ships.” Scientific advances almost always benefit the public, even if that benefit is not clearly foreseeable at the time. Test Your Understanding 1. (b) Homogeneity means the distribution of matter is uniform on the largest scales. 2. (d) Isotropy means the distribution of matter looks the same from all viewing directions. 3. (c) Galaxies are so far away that we cannot resolve individual objects. 4. (d) Although it is true that the Milky Way appears to be at the center of the universe, any other observer would see the same thing, proving that we cannot be at its center. 5. (d) Galaxies are not literally moving at speeds greater than light; it just appears so because of the expansion of space. 6. b, d. Although a, c, and e are explained by the Big Bang theory, they were not predictions that it made; rather they were the basis from which it was developed. 7. (a) The “Hubble time” is the slope of the Hubble law and is an estimate of the age of the universe. 8. All answers apply. 9. (a) If repeated measurements contradict a prediction from a theory, the theory is wrong. 10. (c) Although (a) and (b) are true, they are not part of the cosmological principle. 11. (d) Our galaxy is embedded within space; that is, it is expanding away from other galaxies. However, our galaxy itself is not expanding because it is bound together by gravity. 12. (c) The scale factor is a measure of how much a given distance has changed over time due to Hubble expansion. 13. (d) Although we call it a “bang,” the Big Bang was less of an explosion and more the first moment of the universe. 14. (b) Isotropy is when something looks the same in all directions. 15. (e) Carbon was too massive to have been created during Big Bang Nucleosynthesis; the repulsion of the protons was greater than the thermal energy available at the time. Thinking about the Concepts 16. (a) Yes, because everything looks the same in all directions. (b) Yes, because everything would look the same no matter where you were standing.

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17. Galaxies are fixed in space, but space itself is expanding. Imagine space filled with a mesh or grid. The galaxies are not moving within the grid, but the space between gridlines gets larger. See Figure 21.5 18. According to the Cosmological Principle, an astronomer in another galaxy located anywhere in the Universe would see all other galaxies moving away with the same redshift-distance law as we observe. 19. Quite the opposite, Hubble’s law implies that there is no center of the universe because all galaxies are moving away from each other. As a result, each observer thinks they are at the center of the universe, meaning there is no center at all. 20. The scale factor tells us how the distance between galaxies has changed over time. If we imagine two galaxies as being embedded in space and space has expanded over time, then the scale factor tells us the expansion rate of the universe. 21. The expansion rate of the universe does not make the Sun or the Milky Way bigger, because these are objects that are held together by their own gravitational attraction. The “Hubble flow” of expansion only happens on the largest scales in which gravitational attraction is very weak. 22. Answers will vary. 23. Answers will vary. Verified predictions of the Big Bang include Hubble’s redshift-distance relationship, the chemical abundances of stars and the interstellar medium, the presence of the cosmic-microwave background, inhomogeneities, and fluctuations within the CMB. 24. Regardless of your position in the universe, all distant galaxies recede from you as the universe expands. If you claim that a distant galaxy is moving rapidly away from you, an observer in that distant galaxy would say that you are actually rapidly moving away from him or her. In other words, every observer in the universe perceives himself or herself to be at the center of the universe’s expansion. Another answer is that the Big Bang took place everywhere; everyone thinks they are at the center. 25. As long as recessional velocities remain well below the speed of light, the more complex relativistic equation reduces to the simple form of vr 5 cz. As recessional velocities approach the speed of light, the simple approximation breaks down, and we must use the more complete relativistic version of the relationship. According to Hubble’s law, the most distant galaxies with the highest redshifts have recessional velocities approaching the speed of light. 26. The fact that the CMB displays a blackbody spectrum today indicates that the universe was a blackbody at early times when the spectrum was formed. This is what the Big Bang theory predicted, and this observation helped to rule out the Steady State theory.

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160 ◆ Chapter 21 The Expanding Universe 27. The tiny variations in the CMB are the microscopic fluctuations of structure that grew over time into the large-scale structure we observe and live in today. Studying the properties of those fluctuations gives us very strong tests of the Big Bang theory. 28. The observed abundances of isotopes confirm the predictions of Big Bang nucleosynthesis, which are an important test of the Big Bang theory. 29. Nuclear fusion requires high temperatures to overcome the repulsion of protons from each other, and high densities so that many collisions happen in a given amount of time. As nuclei grow larger, that repulsion increases; thus, higher temperatures and densities are needed to build nuclei into heavier elements. However the universe expanded and cooled as it built He from H; thus, conditions were never appropriate for the He to fuse into anything heavier in any meaningful amounts. 30. A star only converts a percentage of its mass of hydrogen into helium over its lifetime, and much of that helium is never released back into space as it is locked up in neutron stars, black holes, and white dwarfs. Stellar evolution therefore tells us that the amount of helium today is larger than could have been produced inside stars and must therefore have been “primordial” in nature (produced before stars). Applying the Concepts 31. Setup: Area of a square of size s is s2 and volume is s3. We can use these scaling relationships to answer the questions. Solve: If a square doubles in size, its area increases by 22 5 4 . The volume of a cube whose side has doubled in size increases by 23 5 8. Review: One can also solve this by “brute force” by considering a square or cube of length 1 growing to length 2. One finds the same answer. 32. Setup: Error bars are shown for individual data points, but are they used when one shows a range of values? If a range of values is shown, does the thickness or the lines or range indicate the uncertainty? Solve: In Figure 21.12, individual data are not shown, but rather the ranges of acceptable values. As such, error bars are not shown per se, but lines and regions have actual thicknesses that show the accuracy of the measurements. The range of overlap of these various regions of various sizes give the overall uncertainty in the conclusions. Although Figure 21.12 does not show tiny errors, a variety of observations are all consistent, which in itself is a strong statement of confidence.

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Review: Error bars are not the only way to show uncertainty or to gauge whether results are significant. 33. Setup: Observations are marked “observed” in Figure 21.14, whereas curves and regions that have comments like “would have” or “if the” tells us that these are predictions. Solve: (a) The vertical yellow bar and slanted lines/ curves show predictions of theory. (b) The horizontal pastel bars and vertical black line show observation. (c) The observations and predictions match up exquisitely. For example, the predicted helium abundance and density cross the observational region where the actual density and abundances are shown. Review: A strong test of a theory is when the observations and predictions from a number of different tests all match up. 34. Setup: This is an exercise in unit conversion and all the necessary conversion factors have been given. Solve: 1 1 5 H 0 70 km/s/Mpc 5

13 yr 1 Mpcs  106 pc 3. 09 3 10 km pc 3. 16 3 10 7 s 70 km Mpc

51. 4 3 1010 yr.

If the Hubble constant increases to 75 km/s/Mpc, the age will decrease to 13 billion years. Review: The universe is approximately 14 billion years or 14 Gyr. 35. Setup: This is an exercise in unit conversion, and all the necessary conversion factors have been given. Solve: For a Hubble constant of 55, the Hubble time is 1 1 5 H 0 50 km/s/Mpc 5

13 yr 1 Mpcs  106 pc 3. 09 3 10 km pc 3. 16 3 10 7 s 50 km Mpc

51 . 96 3 1010 yr, whereas f o r a value of 100, 1 1 = H 0 100 km/s/Mpc 5

13 yr 1 Mpc  s 106 pc 3. 09 3 10 km Mpc pc 3. 16 3 107 s 100 km

5 9. 7 9 3 109 yr.

Review: Note that this is a very large age range (almost a factor of 2), which explains why there was great uncertainty about many of the specific details of the universe’s evolution. This also shows how hard it was to make these measurements.

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1 and relates the 1 z expansion of the universe (size) to the observable redshift. Solve: A redshift of two corresponds to a scale factor of one-third; thus, the universe has expanded by a factor of three since the light was emitted. Review: Note this implies that the universe was infinitely small (RU 0) when the universe first began (z ). 1 37. Setup: The scale factor R u 5 and relates the 1 z expansion of the universe (size) to the observable 36. Setup: The scale factor R u 5

redshift. Solve: A redshift of eight corresponds to a scale factor of one-ninth; thus, the universe has expanded by a factor of nine since the light was emitted. Review: Note this implies that the universe was infinitely small (RU 0) when the universe first began (z ).  38. Setup: Redshift is defined as z 5 . To find the size  of the universe at the time this light was emitted, use 1 the scale factor R u 5 . 1 z 547. 2  121. 6 Solve: The redshift is z 5 5 3.5. 121. 6 At the time this light was emitted, the universe was 1 RU 5 5 0 . 22 or 22% of its present size. 1  3.5 Review: Recall that the universe was smaller at higher redshift. 39. Setup: Hubble’s law states that v 5 H0 d, which we can v 1 . The scale factor R u 5 rearrange to give d 5 . 1 z H0 287 , 000 km/s 5 4 ,100 Mpc. km/s 70 Mpc (b) 4,100 Mpc is 4,100 3 3.26 5 13,366 Mly, so the look-back time is 13.37 billion years. If we assume a Hubble time look-back time of 13.7 billion years, then the quasar formed 13.7–13.37 billion years 5 0.33 billion years after the Big Bang. (c) The scale factor 1 5 0 . 147. of the universe at that time was R u 5 15. 82 Review: Note that this galaxy was not actually around only 330 million years after the Big Bang because at large distances, the simplistic treatment we have used here is not appropriate. The actual age is about Solve: (a) d 5

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990 million years, as computed by a cosmological calculator such as the one found at http://astro.ucla. edu/~wright/CosmoCalc.html. 2. 9 3 106 nm K . 40. Setup: Wien’s law is T 5  peak Solve: Reading off of Figure 21.12, the peak of the spectrum occurs just past 1 mm, around 1.1 mm. Plugging this into Wien’s law, we find 2 . 9 3 106 nmK 5 2 . 6 K . The actual temperature T5 1 . 1 3 106 nm of the CMB is 2.73 K, so this was very close. Review: We could have done much better if we had a more accurate spectrum. Nonetheless, we see that a blackbody this cool has a very long peak, almost into the radio region of the electromagnetic spectrum. 41. Setup: For speeds that are not “too fast” (that is, nonrelativistic, or less than about 10 percent the speed of light), the redshift formula can also be written as v z 5 r . If it turns out that z is less than zero, then z is a c blueshift rather than a redshift. 368 km/s Solve: z 5 5 0. 00123 . 300 , 000 km/s Review: This is a very small blueshift because we are dealing with a speed that is very small compared with that of light. 42. Setup: Dividing the mass density (kg/m3) by the mass of a hydrogen atom (kg) yields the number of atoms per cubic meter. 4 3 1028 kg/m 3 5 0 . 24 atoms/m 3 . Solve: 1. 66 3 1027 kg Review: This is about one atom each per 4 cubic meters, which is about the volume of a small shower stall. The average density of the universe is very low! 43. Setup: To compute the ratio of the density of the universe to that of air on Earth, simply divide the two. 1. 2 kg/m 3 5 3 3 1027 . Solve: 4 3 1028 kg/m 3 Review: Earth’s atmosphere is 3 billion billion billion times denser than the average density of the universe. 1 44. Setup: The scale factor RU 5 and volume scales 1 z with size cubed. Solve: At a redshift of z 5 10, the scale factor will be 1/11, so the universe would be 113 5 1,331 times denser than today, for a total of 5.3 3 10–25 kg/m3.

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162 ◆ Chapter 21 The Expanding Universe Review: Redshift of 10 is very high and very far back in time, but even then, densities were very low compared to ordinary matter. We see that the universe was very dense for only a short period of time. 1 gives the relative 45. Setup: The scale factor RU 5 1 z size of the universe at any redshift z when light was emitted. Solve: In the case of when the CMB was emitted, the 1 scale factor RU 5 5 9. 9 3 104 or about 11000 0.001 the size of the universe today. Review: The universe was certainly much smaller at the time the CMB was emitted than today! Using the Web 46. Einstein (and most of his contemporaries) believed the universe was in “steady state.” Lemaître was one of the first to propose a definite beginning to the universe. The steady-state theory proposes that the universe has always looked as it looks now and that there is a steady creation of new matter to keep the overall density constant. 47. The age of the Universe, around 13.7 to 13.8 Gyr, is much older than the age of our solar system (about 4.5 to 5 Gyr) and also older than the oldest globular clusters (10 to 12 Gyr). The current value will vary with time. Based on the 2012 data release by the Planck satellite, the age is 13.772 Gyr. Based on the 2013 data release, it is 13.82 Gyr.

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48. In his talk, Dr. Wilson judged the discovery of the CMB as his most important discovery, but he never explicitly states why. However, as he notes, this discovery brought cosmology to the forefront of the world’s scientific attention and was quite possibly the greatest step science has made toward answering the question “how did we get here.” He and his colleague, Dr. Penzias, identified the source of this radiation somewhat by happenstance when a mutual friend connected their discovery with researchers at Princeton (led by Dr. Dicke), who had predicted the existence of this signal. 49. Answers will vary. Report on the impact of the story in 2012, whether astronomers disputed the study, and whether it is still receiving attention. 50. You will complete the Hubble law exercise, in which you will measure the angular size and recessional velocity of many galaxies and plot these to measure the Hubble constant. Be sure to answer the 10 questions at the bottom of the lab. Exploration 1. My graph trends upward at a roughly constant slope. 2. My reference galaxy does not move. 3. The trend line would not change if I picked a different galaxy. 4. “More stuff ” means it was harder for the elastic to expand. This mimics the extra mass in some regions that created clusters of galaxies, which opposed the expansion.

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CHAPTER 22

Cosmology INSTRUCTOR’S NOTES Chapter 22 covers the physics of the Big Bang and evolution of the universe. Major topics include

▶ the critical density ▶ the cosmological constant and dark energy ▶ the geometry of the universe ▶ the fate of the universe ▶ inflation ▶ the physics of the first few seconds of the Big Bang ▶ string theory and multiverses Most of the material in this book lends itself to a fairly linear and predictable path of science. Option one: “We needed to understand something, so we used some physics, made a few tests, and here is how it works.” Option two: “People had one world view, someone poked a few holes in it, someone else developed a new theory that has stood the test of time, and here is how it works.” The story of cosmology is full of stutters and missteps and continues to unfold. “We had one theory, and then someone poked a number of holes in it, so we formed a new theory, and here is how it works. By the way, we threw some of that theory out until someone poked a few more holes, so we brought that back and here is how it works now. Then someone thought of this problem and came up with this solution, and now here is how it works.” And on it goes, until we deliver the ultimate insult to the student who seeks certainty: “Ninety-five percent of the universe is made up of stuff we do not understand, and we may never have any idea of its future.” Because one of my primary learning goals throughout this material is the process of science, I relish the opportunity to walk students through this messy history because real science is messy! And some of this chapter is recent enough that you can recount the history firsthand. I remember looking down my nose at the first ΛCDM cosmological simulations, wondering “why-oh-why would anyone waste their time with that old nonsense?” and sitting at a conference room table watching the closed-circuit first report from WMAP that pretty well proved me wrong. I actually find it comforting to know that we can know so much and so little about the universe at the same time. It imposes on us a duty

to continue and a sense of humility in our successes. Plus, what can more inspiring at the end of a course to be told by your professor “and yes, maybe one of you will be the one in 30 years explaining to the world why this all happens.” In the introductory astronomy curriculum, cosmology can be one of the most opaque subjects for nonscience students, which is problematic because it is also a subject to which they look forward the most. Part of this is due to the scope: we cover all of space and all of time, from most unimaginably small zillionths of a second to hundreds of billions parsecs. Talk about information overload. So, my advice to the novice instructor is simple: go slow and tie it all together. Whether as the authors of this textbook or of your lectures, we all have to make artificial dividing lines in the content of what is really a seamless story. This book splits the subject into three highly interrelated chapters, of which this is the middle. But your students may still be used to a chapter as a self-contained, single-serving learning unit, and they may therefore miss the big picture of this big picture. One could just as easily have put this chapter’s discussion of expansion of the universe (Sections 22.1 and 22.2) at the end of the last chapter and moved its last section on the CMB (Section 21.4) to the next chapter (Large Scale Structure), and then made an entirely new chapter on the first 3 minutes. The point is, we divide and categorize out of necessity to teach the classes, but we must make the connections explicit for our students because this is too new for them to do it themselves. My personal preference is to preface the whole discussion with the disclaimer that this material will be like a Quentin Tarantino film—multiple overlapping timelines that will all come together at the end.

DISCUSSION POINTS

▶ Do we have a firm prediction for the future of the uni-

verse, or is it uncertain? Given the mysterious nature of dark energy, does it seem likely that we ever could know the ultimate fate of the universe with any certainty? ▶ The concept of inflation has been put forward as a way to solve the flatness and horizon problems in cosmology. This process calls for two points in space to travel apart 163

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164 ◆ Chapter 22 Cosmology at velocities greatly exceeding the speed of light. Discuss how inflation can happen without violating the theory of relativity. ▶ This chapter opens with a discussion of how gravity affects the expansion of the universe; then we learn that dark energy is working against gravity. Similarly, Einstein first proposed, then retracted the cosmological constant, and it is now favored again. Do these evolving views represent a weakness or strength in the process of science? ▶ Discussing the universe at the very earliest times requires imagining states that are unknown to current physics. Should these be considered scientific theories, or scientific ideas? ▶ In Chapters 16 and 17, students learned that stars derive their luminosity mainly from nuclear fusion of hydrogen into helium. Use Big Bang nucleosynthesis to explain why hydrogen is so abundant in the universe, and ask students whether all the stars that exist now in the universe or that have ever existed since the Big Bang could have been able to change the fact that hydrogen is still the most abundant element.

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (c) If there is enough mass to overcome gravity, the universe will collapse. The critical density is the amount of mass needed for that to happen. 2. (c) At the largest distances, objects are moving away from us faster than Hubble’s law predicts. 3. Inflation solves the horizon and flatness problems. 4. (d) According to our theories, the antimatter annihilated with matter, but a small asymmetry of matter existed, leaving the atoms that fill our universe today. 5. (b) At present, there is no way to test the idea of parallel universes, so it remains a hypothesis. Reading Astronomy News 1. Milky Way dust obscures, reddens, and reflects light including that from the CMB. When light is reflected, it can become polarized. 2. Astronomers run experiments at the South Pole because there is a long period of uninterrupted nighttime observation, and the humidity is very low. 3. Inflation is the exponential growth of the universe at very early times (around 10233 sec) needed to solve the flatness and horizon problems. 4. As a scientific theory, inflation has made predictions of observations that would confirm its validity. It is therefore important to seek out those observations.

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5. Scientific discoveries must be repeatable and stand up to the circumspection of others. These are two important hallmarks of the process of science. Test Your Understanding 1. (d) The density of the universe () dictates whether it will be open, closed, or flat. 2. (c) “Dark energy” is the name that has been given to the “stuff ” that causes acceleration. The cosmological constant is the mathematical treatment of that effect. 3. (a) The CMB is uniform to about 1 part in 100,000. 4. (b) String theory has not yet made testable predictions, which leaves it as an “idea.” 5. a-f-b-e-j-g-d-i-h-c. This is shown in Figure 22.13, which gives a graphical chronology of the universe. 6. (a) Prior to the epoch at which the CMB formed (“decoupling”), matter and radiation were in equilibrium, which means that light could not travel very far before hitting a particle. This made the universe opaque. 7. (c) The critical density is the density at which the universe changes from open to closed. 8. (b) The electromagnetic force depends on charges and their motions. 9. (b) See Figure 22.7, which shows triangles and circumferences for three different geometries. 10. (b) Protons were the first to form; these were built up during Big Bang nucleosynthesis into nuclei. Much later, the universe cooled enough for electrons to recombine with the nuclei to form neutral atoms. 11. (d) Quarks are the particles out of which baryons (protons, neutrons) and mesons are made. 12. (b) The density of the universe is consistent with  5 1, which means it is flat, even though dark energy seems to dominate, which means the expansion of the universe is accelerating (but not inflating). 13. (a) This is the process of pair production in reverse. 14. (b) High-redshift supernovae show us that very distant objects are moving away from us faster than predicted by the Hubble law predictions. 15. (b) There are many statements of the anthropic principle but the essence is that everything happened such that we are here to see it today. Thinking about the Concepts 16. To make the universe collapse, the average density of matter would have to be higher than the critical density, and the forces of gravity from this matter would

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have to overpower that of any dark energy, which would cause the universe to be pushed apart. 17. When we look at distant galaxies and supernovae, we see that their recessional velocity increases with distance faster than the linear relationship of Hubble’s law, suggesting that there is a nonzero cosmological constant. 18. Dark energy is vacuum energy, that is, a repulsive force that results from the energy present in totally empty space. 19. Just as our planet, galaxy, and so on are not being torn apart by the expansion of space, so too are they stable against the repulsive force of dark energy: gravity holds them together. 20. Hubble time does not take into account the changing dynamics that result from having different cosmological densities, a cosmological constant, or dark energy. 21. For the universe to be flat today, it needed to be flat for all time, meaning even at the earliest times in its evolution. This is such a specialized situation that it seems unlikely, which leads one to ask, “How was it that the universe was so flat despite all the possible outcomes that it could have had?” 22. Matter and information did not travel faster than light; space itself expanded so fast that the distances between points in space expanded faster than at the speed of light. There was no causal connection between these points, however. 23. The early universe was hot and dense, which are situations that are replicated (or that we try to replicate) within the realm of high-energy physics. At these high temperatures and densities, all sorts of exotic forms of matter can exist, which may have had substantial influences on how the early universe evolved. 24. (a) Gravity fights against expansion, so if gravity changes with time, the universe might close on itself or accelerate faster than predicted. (b) Electric force holds or repels charged particles. If this changed with time, the structure of atoms and processes of fusion and fission would be very different from today. 25. A theory may be incomplete, but that does not mean it is inaccurate. Instead, there is just more to be learned. However, in learning more, one might find a better theory that replaces the current, incomplete one. At present, the standard model is the most successful theory of particle physics, so the pieces that are missing are considered to be small compared to the total body of predictions that the theory makes. 26. If a photon contains more energy than the combined mass energy of a particle and its antiparticle, then the photon can become a particle-antiparticle pair that fly off in different directions, conserving mass energy and momentum.

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27. The Planck era was the earliest time of the early universe, when all the laws of physics that we know were combined in a single way. Others define the Planck era as the early time of the universe during which we cannot explain what was going on because energies were too high and timescales were too small. 28. Grand unified theories (GUTs) attempt to unite the electroweak and strong nuclear forces. Theories of everything (TOEs) attempt to unify suitable GUTs with gravity. One basic distinction is that GUT forces are inherently quantum mechanical in nature, whereas gravity is described with general relativity, which is not quantum mechanical but a theory of geometry. 29. Superstring theory is an attempt to explain physical phenomena; however, it is very difficult to test its proposals, and generally, a theory must be testable and disprovable. 30. Answers will vary. As one example, if the nuclear force were stronger in a parallel universe, then heavier nuclei than Helium might have been created during the Big Bang. Applying the Concepts 31. Setup: Logarithmic axes increase by factors of 10 (number spacing is not uniform), whereas linear ones increase by single values. Solve: (a) The vertical axis is logarithmic. (b) The top label is linear. (c) The bottom axis is logarithmic. (d) The relationship between relative brightness and time is not linear, but the fainter something is, the older it is. Review: Logarithmic axes are a very useful way to pack a large range of numerical information into a single graph. 32. Setup: Acceleration means that objects become larger faster, that is, the curve will become more vertical with time. Solve: (a) The red curve becomes horizontal, whereas the blue curve trends toward vertical. (b) The red curve shows a universe that expands more slowly with time; this is deceleration. The blue one shows the universe expanding more quickly with time; this is acceleration. (c) A straight line would indicate that the universe will expand forever. Review: Note that the future of the universe can be fit by models that study the past. 33. Setup: For data to be consistent with a model, the prediction (that is, the curves in this case) has to pass through most of the points, or there should be very little scatter between the points and the line. Solve: (a) The right-hand side of the graph is the future. Because it has not happened yet, there can be

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166 ◆ Chapter 22 Cosmology no data. (b) The red, orange, and green models are excluded. (c) Data go back roughly 9 billion years. (d) 9/13.7 5 0.66 or two-thirds of the age of the universe. Review: Note that there is one datum that is consistent with a universe that collapses. It is important to understand whether this datum is reliable or flawed, and if reliable, we must explain its discrepancy with the other points. 34. Setup: Logarithmic axes increase by factors of 10 (number spacing is not uniform), whereas linear ones increase by single values. Inflation occurred during the time period indicated by the orange bar. Solve: Figure 22.8 has logarithmic axes. During inflation, the universe increased about 1037 in size. At the beginning of the inflationary period, the universe was about 1037 times smaller than in the noninflationary model. Review: We see here that the universe truly did inflate exponentially in size. 35. Setup: Review Chapter 1, which discusses the process of science. Solve: Given the excellent agreement of data from a large array of different experiments, astronomers would probably expect that new observations that propose radically different values must be carefully studied to ensure that there are no mistakes or errors in the measurements. If these data prove to be robust, then all the data taken until now must be studied, and if all are robust, the theory of the Big Bang must be considered flawed. Review: The strength of science is that it accommodates new, reliable findings and does not reject them, no matter how controversial the findings are. 36. Setup: Logarithmic axes increase by factors of 10 (number spacing is not uniform), whereas linear ones increase by single values. Solve: (a) The time axis is logarithmic. (b) Density drops from 1099 to 10226 or 125 orders of magnitude. (c) Temperature drops from 1032 to 2.73, or about 31 orders of magnitude. Review: Over 14 billion years (give or take), the universe has dropped dramatically in temperature and density. Note that most of that happened after the first few hundred thousand years. 37. Setup: Uncertainty in one variable yields a direct uncertainty in the next if there is a simple linear or inverse relationship. Solve: Because the Hubble constant is good to 4 percent, so is the age of the universe (roughly). With H0 5 70 km/s/Mpc, the Universe has an age of 13.6 Gyr; 4 percent error yields minimum and maximum ages of 13.1 and 14.2 Gyr, respectively. Review: It is astounding to the author of these solutions that we can know the history of the universe to

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this precision. I hope the reader appreciates this amazing result as well. 38. Setup: The critical density is 8 3 10227 kg/m3, and an H atom has a mass of 1.7 3 10227 kg. Number density (the number of atoms per cubic meter is simply the mass density divided by the mass of a single atom). 8 3 10227 kg / m 3 5 atoms per cubic meter. Solve: 1. 7 3 10227 kg Review: The typical densities in the galactic interstellar medium are a few atoms per cubic centimeter, so we see that this density is a million times smaller. Remember, space is very big and the space between galaxies is very large, so we expect the answer to be very small. 39. Setup: As given in the problem, the scale factor can be  solved for RU 5 3 0 .  Solve: Using values in the text, 3 3 10228 5 6. 2 3 10210 . 1.23 Review: Note that this means the universe was about a billionth of its current size. 40. Setup: Because there are two particles and two gamma rays, each gamma ray is formed from the complete conversion of one particle’s mass into energy, so we use Einstein’s mass-energy formula, E 5 mc2. Solve: E 5 mc2 5 (1.67 3 10227 kg)(3 3 108 m/s)2 5 1.5 3 10210 J. Review: Although this might not seem like a lot of energy, consider that one gram of hydrogen contains 6 3 1023 protons, so its annihilation would result in about 1014 J being given off. 41. Setup: At 2.7 K, each photon has a wavelength around 2 , 900 m  K 1 mm using Wien’s law. Because T hc E 5 , the CMB photon has an energy of about  2 3 10222 J. Knowing how many photons there are and E 5 mc2, we can compute the total energy and then equivalent mass present in the CMB. Solve: If there are 500 million CMB photons for each H atom, then the total energy in these 500 million photons is 500 3 106  2 3 10222 J 5 10213 J. This corresponds to an equivalent mass of 10213 J E m5 2 5 10230 kg. Because the 16 2 2 c 9 3 10 m /s mass of a proton is about 10227 kilograms, we see that the CMB contains about 1,000 times less equivalent mass than all the hydrogen atoms in the universe. We can ignore it when considering the density of the universe. RU 5 3

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Review: It is important to note that even though there are almost a billion times more CMB photons than atoms, their total energy is insignificant compared to that mass, because each photon’s energy is so low. 42. Setup: Rather than worry about how much energy is given off per atom, simply use E 5 mc2 where the mass is 2 grams. Solve: (a) E 5 0.002 kg  9 3 1016 m2/s2 5 1.8 3 1014 J. (b) This is approximately 10% of energy released in the detonation of a 1-megaton hydrogen bomb. Review: Remember that although the energy per atom might be small, there are about 6 3 1023 atoms in a gram of hydrogen, so 1 gram’s annihilation results in a huge amount of energy. 43. Setup: We need to know how many protons are in the detector. Water has a molar mass of 18 grams per mole. Because there are 6.02 3 1023 molecules per mole, there are roughly 3.3 3 1022 molecules in 1 gram of water. With a total capacity of 20 million kilograms, or 20 billion grams, there are about 6.7 3 1032 water molecules in the detector. Water is made of two hydrogen atoms and one oxygen atom, which have a total of 1 1 1 1 6 5 8 protons, so the total number of protons in Super-Kamiokande is around 5.4 3 1033 protons. Solve: No decays in 5 years limits the lifetime to around 5. 5.4 3 1033 5 2.7 3 1034 years. Review: This problem is oversimplified and the real limit is around 1033 years, but it still shows this particular GUT theory is ruled out. 44. Setup: The circumference of a circle equals 2πr in flat space, is more than this value in open space and is less in closed space. Solve: A gravitational well is a closed space, so the circumference is less than 2πr. Review: This is another way to think about gravitational time dilation because the planet close to the star travels a shorter distance than expected for a flat universe. 45. Setup: This is just an exercise in comparing the order of magnitude of two numbers. 1.2 5 1. 2 3 1026 Solve: Our atmosphere is 227 9. 9 3 10 times denser than the average density of the universe. This is a factor of 120,000,000,000,000,000,000,000,000. Review: How big is this? According to science writer David Blatner, the Earth’s beaches contain about grains of sand. This number is still 10 million times smaller than the difference we found above. Using the Web 46. Major stages after the Big Bang include inflation, symmetry breaking, electron-positron annihilation,

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and Big Bang Nucleosynthesis. Only hydrogen, helium, and deuterium were made, as well as a tiny bit of lithium. Heavier elements were not made because there was not enough thermal energy to overcome the Coulomb repulsion from these heavier nuclei. 47. The dark-energy survey will use optical imaging to study four probes of dark energy: Type Ia supernovae, baryon acoustic oscillations, galaxy clustering, and weak gravitational lensing. These will study acceleration of the universe, clustering statistics to constrain the history of the cosmic expansion rate, the growth rate of structure in the universe, and the clumping/ distribution of dark matter. Answers will vary on news, the project status, and any results. 48. (a) Particles may have had no mass when they formed in the Big Bang. The Higgs boson was responsible for the transfer of mass to particles as they interacted with the Higgs field. Answers will vary on what is new in the search for this boson. (b) Exotic decays are better found with human eyes because the Large-Hadron Collider’s computer algorithms were not written to find offcenter vertices. 49. Answers will vary. Report on arguments for and against multiverses. 50. Answers will vary. Discuss whether the arguments made in the series are compelling and whether the science is explained in an understandable way to you. Exploration 1. There are probably 60-100 different particle tracks. 2. The magnetic field pointed perpendicular to the page. 3. Answers will vary. 4. A charged particle with a nearly straight track was probably both high energy (high speed) and massive. 5. (a) The looser spiral indicates a faster electron. (b) The looser spiral indicated a higher energy. 6. (a) A bright track indicates a slower particle. (b) A tight spiral indicates a slower speed and thus should be brighter. (c) If a looser track is also brighter, the particle was traveling slowly; therefore, it must have had a higher mass than the particle in the tight track. 7. If both a proton and electron’s tracks curve by a very small amount, then the lower-mass particle (electron) must be traveling much faster than the higher-mass one (proton). 8. A neutral particle could decay into a charged particle, which then decayed back into a neutral particle, to produce a track with a definite beginning and end. 9. (a) If there is no trace away from the track at the location of a kink then the emitted particle is neutral. (b) Because a particle was emitted, mass and energy were taken away from the original particle. (c) Answers will vary.

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CHAPTER 23

Large-Scale Structure in the Universe INSTRUCTOR’S NOTES Chapter 23 covers how the universe went from a big bang to today. Major topics include

▶ types of large-scale structures ▶ the role of gravity and dark matter ▶ hot versus cold dark matter ▶ the first generation of stars and galaxies ▶ cosmological evolution of galaxies ▶ comparison of observations with simulations I like to use a two-pronged approach to teaching structure formation. On the one hand, we have the macroscopic view. We know what our region of the universe currently looks like (Figure 23.21) and how it appeared about 380,000 years after the Big Bang (see Figure 21.13). On the other hand, we have the microscopic view. We know what happens to particles at most of the high energies and densities predicted by the Big Bang (Section 22.4). Cosmology, in general, and the formation of structure in particular stitch these all together into a coherent story. If you recall from my Instructor’s Notes in the last chapter, I think it important to make these connections clear for your students; otherwise, they might not relate the formation of helium nuclei and quantum fluctuations to the filaments and voids we observe today across decades of Mpc. To me, the most dramatic evidence that we have a fairly solid understanding of the history of the universe is the comparison between the large-scale structure we observe and that which we model through computer simulations. Figures 23.19 through 23.21 show two-dimensional view of this, but it is not hard to find three-dimensional views as well, such as http://vimeo.com/29769051. There are many interesting “fly-throughs” of the Millennium Simulation (http://www.mpa-garching.mpg.de/galform/virgo/ millennium/) that show how well our cosmological models have reproduced structure on the largest and smallest scales. The OpenCL NBody Cosmic Web is a wonderful movie of structure formation from a seemingly homogenous background of matter (https://www.youtube.com/ watch?v5tpgrNctOkws). Perhaps my favorite is the movie at https://www.youtube.com/watch?v574IsySs3RGU, which puts together a number of simulations and media,

including the links above, along with some useful annotation. It may be important to remind students that these simulations do not prove our theories are true, but they do suggest that our physical understanding is fairly accurate. During the discussion of dark matter, your students may find it interesting and amusing to know a little more about the two hypothetical categories of dark matter: MACHOs (Massive Compact Halo Objects) and WIMPs (Weakly Interacting Massive Particles). The former have been long sought using microlensing surveys, whereas the latter are one of the prized targets of particle accelerators and high-energy particle detectors. But after a long course full of boring and mundane names of objects, it helps to add a little levity to the class with these two humorous names. An interesting discussion of them can be found at http://www.astro.caltech.edu/~george/ay21/ eaa/eaa-wimps-machos.pdf. As a final wrap up to this section, the textbook makes a careful note in Section 23.3 to compare galaxy formation to star formation. This is a perfect example of how the same laws and principles of physics can be used to understand things on vastly different scales of size and time. Were it not for this universality of physics, a class such as ours would not even be possible, so please be sure to stress that to your class.

DISCUSSION POINTS

▶ Compare images of the CMB fluctuations (see

Figure 21.13) with those of large-scale structure observed today (Figure 23.21). Ask students to enumerate the steps through which the former evolved into the later. ▶ This chapter presents the results of a number of simulations (Figures 23.9, 23.19 through 23.21). Discuss why simulations are useful, and explore their limitations in testing scientific theories. ▶ Connect the previous epochs (for example. Figure 23.22) with what may happen in the distant future to discuss what is special about our current era in the evolution of the universe. ▶ The model of galaxy formation from cold dark matter requires the presence of dark matter clumps as seeds 169

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170 ◆ Chapter 23 Large-Scale Structure in the Universe for the collapse of galaxies. Taking this model to the extreme, discuss what would happen in a universe with no cold dark matter at all. ▶ When comparing galaxy and star formation there are many parallels, but what are some distinct differences? As examples: Can star and planet formation be considered analogous? Is a halo like an Oort cloud? Does the center of a galaxy exert as much influence as the central star of a solar system?

END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. c-b-d-a. 2. (b) The distribution of dark matter in the early universe caused the large-scale structure we see today. 3. (c) The earliest generation of stars had no heavy elements at all. 4. (b) At high redshift, there were more irregular galaxies, indicating the spirals took time to grow. Reading Astronomy News 1. Laniakea is about 500 million light-years across, or about 150 Mpc. 2. It was hard to discover our supercluster because we are inside the object looking outside of it, which makes it hard to determine the detailed structure of the surroundings, much like the difficulty of discovering the properties of our own galaxy. 3. No, the Milky Way and Local Group sit near the far edge of the supercluster. 4. To map out more distance superclusters, astronomers need to know the component galaxies’ positions and velocities, which is challenging for distant objects. 5. At this scale the universe is not yet homogenous and isotropic. It is, however, on the scale of looking at many superclusters. Test Your Understanding 1. g-b-a-c-e-f-d. See Section 23.1. 2. (a) Gravity is the dominant force in all processes of formation of macroscopic astronomical objects (planets, stars, galaxies, etc.). 3. (b) Sometimes called the “bottom up” building scenario, small things combine to form bigger ones, whether planets from dust or galaxies from star clusters. 4. (d) Hot and cold dark matter particles move at different speeds (that is, have different energies or

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temperatures) and are believed to be dominantly made of different species; also, the fast-moving particles are not easily influenced by gravitational fields. 5. (c) This is done through weak lensing of distant galaxies by foreground clusters. 6. (a) We expect the universe to expand forever and all matter to eventually decay. 7. (c) Clusters have more galaxies (and therefore mass) than groups. 8. (d) As long as the galaxy is moving with the Hubble flow, redshift gives recessional velocity by the Doppler equation and distance by the Hubble law. 9. (d) Galaxies and stars both form “pancakes” during collapse. 10. (d) Dark matter particles do not interact electromagnetically so they cannot easily get rid of their energy by, for example, collisions or radiation; therefore, they will not slow down to collapse the way ordinary matter can, say, in accretion disks. 11. (d) We believe that ellipticals result from the merging of two or more spiral galaxies. 12. (d) During the Dark Ages, photons could not travel freely, but instead traveled a very short distance before hitting a free electron. 13. (a) For there to be no stars with zero heavy elements observed, such first generation stars must be mostly all dead or exploded today, meaning star formation had to begin very early on. 14. (d) With the first generation of stars came intense radiation from the high-mass stars and subsequent supernovae, which re-ionized much of the gas in the universe. 15. a-c-b. We know from cosmological studies that some 75 percent of the universe is dark energy; the other 25 percent or so is matter, and of that, only 4 to 5 percent is ordinary; the rest is dark matter. Thinking about the Concepts 16. If we could see the early universe when galaxies were first forming, it would be smaller and more active. There was a large amount of hot gas but little dust. Protogalaxies were forming the first generations of stars, and supernovae were frequent as the earliest massive stars ended their brief lives. Gas was still collapsing into galactic disks, so there were few (if any) large galaxies with regular structure. Instead smaller, irregularly shaped protogalaxies dominated the visible universe. Many young galaxies were merging with each other, making irregular galaxies a common sight. Supermassive black holes were forming, and galactic centers were much more luminous across the entire electromagnetic spectrum.

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Chapter 23

17. If voids were filled with considerable dark matter, then we would expect to see gravitational lensing of galaxies behind those voids. We would also have expected galaxies to have formed inside the dark matter voids. So it is much more likely that they are fairly empty of mass. 18. In the absence of luminous matter, we could detect a dark matter galaxy only from the gravitational influence it exerts on its surroundings. If there were a luminous companion galaxy that either orbited or merged with a dark matter galaxy, we could detect it. If a dark matter galaxy passed in front of a luminous background object, we could infer its presence from gravitational lensing. 19. A gas cloud collapses to form a star, or many stars, and each protostar spins itself up such that an accretion disk feeds the central object. Similarly, a huge ensemble of gas can collapse and spin up to form disk-like protogalaxies. However, once stars begin forming in a galaxy, that formation process provides feedback that can change how the galaxy continues to form. Also, galaxy formation is sensitive to the distribution of masses around it. 20. Quantum-sized (ultramicroscopic) fluctuations in density after the Big Bang are believed to be the origin of large-scale structure. 21. The density of normal matter in the universe is too low to trigger its collapse into galaxies. Without dark matter to boost the local mass density in cluster-size regions, all of the normal matter in the universe would still reside in the form of hydrogen and helium gas clouds. 22. The correct sequence is (a). We see evidence for merging events, in our galaxy (with small dwarfs) and with larger galaxies, which supports the notion that small things formed first and agglomerated to make larger ones. We also find evidence in very distant (that is, old) galaxies being much smaller and less massive than those observed today. 23. Because inhomogeneities in the density of normal matter were never strong enough to cause the structure seen today, there had to be another source of gravity to cause those seeds of structure seen in the CMB to grow into the actual structure we now observe. 24. A supercluster is composed of many clusters, spanning sizes up to many hundreds of Mpc and with much more mass. Also, clusters are small enough that the galaxies are orbiting each other uniformly, whereas a supercluster has not had time to “relax” into this state. We are part of the local “group” which is a little too poor in large galaxies to be called a cluster. We are part of the “local supercluster,” also containing the Virgo Cluster, as our motion through space is also being affected by the Virgo Cluster.

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25. Just as a cloud of gas spins up and flattens as it collapses, so does a protogalactic cloud. In particular, uneven gravity from all the masses around the protogalaxy can help to give it a little spin at the beginning, which is amplified as the cloud collapses into a galaxy. 26. Observational signs of dark matter include the rotation curves of galaxies; the velocities of galaxies in groups that are higher than can be caused by visible mass; the high temperature of intragalaxy gas in clusters; and gravitational lensing. 27. As we look back in time, we should see more spirals than ellipticals; as we look back even further, we should see galaxies that have not yet formed spiral arms, and galaxies should have a lot of massive stars in them because they have just formed and are not yet old enough to have exploded. This is indeed what is observed. 28. Hot dark matter could not be gravitationally bound enough to form the halos needed to contain galaxies and to explain galactic rotation curves. 29. Fluctuations in the mass distribution caused mass to flow toward them, and as more mass attracts even more mass, these small overdensities soon grew into large halos. Because mass was being pulled away from the general distribution into these overdensities, the result was a foamy structure of voids surrounded by filaments and walls. Inside these halos, galaxies collapsed and formed and were then bound together into groups and clusters seen today. 30. The nuclear forces are too short ranged to work over cosmic distances, and electromagnetic forces only work for charged particles; most matter in the universe is neutral. This leaves only gravity. Applying the Concepts 31. Setup: Figure 23.3a gives us redshift on the top side of the “pie” wedge. We can compute recessional velocity from this using v 5 cz, where z is the redshift. The wall galaxies have a redshift around 0.035. Solve: v 5 0.035  3 3 108 m/s 5 1.05 3 107 m/s. Review: The highest redshift objects observed tend to lie around z 5 6 and above. This figure shows a slice of the universe only out to redshift 0.05, so we see that structure is still small on cosmic scales. 32. Setup: The universe is homogeneous and isotropic on the largest scales, whereas Figure 23.3 shows structures that are clearly not either. However, remember that if the whole universe looks like this single slice, then any direction and position will appear the same to any observer. Solve: The Great Attractor may be massive, but it is still local. On the largest scales, we do not see preferential or inhomogeneous structure.

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172 ◆ Chapter 23 Large-Scale Structure in the Universe Review: Structures such as clusters and walls are still small on the cosmic scale and do not violate the expectation that the universe is homogeneous and isotropic. 33. Setup: For particles to give up their energy, they must either radiate or collide with other particles. Solve: Dark matter particles do not interact electromagnetically, so they cannot emit radiation, and collisions are very rare (believe it or not, most “collisions” are not literal impacts but the interactions of the electromagnetic fields of the two particles); therefore, there is no straightforward mechanism for dark matter to cool. On the other hand, ordinary matter readily loses energy through emission processes (lines and blackbody) as well as via collisions; thus, ordinary matter can collapse to the tiny sizes needed to form stars. Review: Note that when dark matter clouds “heat up” during collapse into galaxies and clusters, these particles still are gravitationally bound and are not considered “hot dark matter.” 34. Setup: Stellar evolution tells us how a star evolves once it is born. The star-formation history is the process and feedback from continued star formation over a long period. Solve: Stellar evolution is quite different from starformation history. A single star is fairly easy to follow along its life because there are just a few simple principles that dictate how the star evolves. Star-formation histories, on the other hand, depend on how much gas and dust there is, where it is, how many heavy elements are in it, how many massive versus low-mass stars are made, how the massive-star supernovae effect the surrounding space, and so on. This is not a failure of a theory but because the process of star formation depends on so many different variables. Review: This explains why galaxies are a different subject of study than stars, even though galaxies are mostly made of stars. 35. Setup: For the universe to be closed, we need a mass density around 10226 kg/m3. Knowing that 5 percent of the universe is neutrinos and that their number density is 300 million per cubic meter, we can directly estimate their mass by dividing the appropriate mass density by the number density. Solve: The neutrino mass density is around 0.05 3 10226 kg/m3 5 5 3 10228 kg/m3, so the mass of one neutrino is roughly 5 3 10228 kg/m 3 2 3 10236 kg. 300 3 106 v / m 3 Review: This is about 1 million times less than the mass of an electron, truly a small (and almost negligible) mass.

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36. Setup: This question can be answered using orders of magnitude, knowing that our galaxy is about 1012 solar masses and if a group has a handful of large galaxies, some small ones, and dark matter; a cluster is a handful of groups plus dark matter; and a supercluster is a handful of clusters plus dark matter. Solve: (a) 1013 solar masses; (b) 101421015 solar masses; (c) 101621018 solar masses. Review: Note that the total masses are about 10 times larger than the mass of a single galaxy because dark matter can be up to 90 percent of the mass of a given structure. 37. Setup: The problem states that black hole decay rate scales t a M3, so to compare the lifetimes for two black holes, compute a ratio. 3 t1 M 1 Solve: 5 3 , or the supermassive black hole evapt 2 M2 3

 3 3 106  5 1 018 times more time. orates in   3  Review: In other words, it takes an awfully long time for a supermassive black hole to evaporate. 38. Setup: Ellipticals appear yellow or red, and we know the stars are old, so how old are typical red mainsequence stars? Solve: Based on their colors, ellipticals must be composed of low mass like our Sun. This gives them an age of 5 to 10 billion years. If galaxies first formed about 1 billion years after the Big Bang, then realistically there were 3 to 8 billion years for merging of spirals. Review: This is a significant fraction of the lifetime of the universe, which suggests that all the galaxies that were going to form already have, and now we are simply watching them evolve. 39. Setup: For part (a), remember that redshift makes galaxies look redder. For part (b), count the number of galaxies and assume it is typical of any point in space. Solve: (a) The highest-redshift galaxies will look extremely red and dim. (b) Using the cosmological principle, we will assume that this patch of space is typical of any other patch, so that the average number of galaxies here is the same as any patch of the same size. Estimating that there are about 10,000 galaxies in the UDF, there are ten million times 10,000 or 100 billion galaxies in the observable universe. Review: Section 23.1 says there are several hundred billion galaxies in the observable universe. This is consistent with our estimate. 40. Setup: For this exercise, let’s just count galaxies in the red and blue areas. Because it is hard to know what is

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Chapter 23

a galaxy and what is a star, I am going call a galaxy anything that is not a small dot or smudge. Solve: In the top cluster, I count about 40 galaxies. In the bottom cluster, I count about 120 galaxies. Review: The Bullet Cluster is especially important to studies of dark matter, as the collision of the two clusters created a separation between the dark matter and ordinary matter. Studies suggest that the ratio of dark to normal matter is at least 10 to 1. 41. Setup: If fluctuations were 10 times larger, gravity in those regions would have been significantly stronger than it actually was, and vice versa for the fluctuations being weaker. Solve: Larger fluctuations leading to larger concentrations of mass and stronger gravity would have probably accelerated the formation process of structure. If all galaxies formed earlier, there might be far fewer spirals than there are today because many more would have collided to become ellipticals. On the other hand, if structure formation were slowed down, there might be far fewer ellipticals because the spirals have not yet had time to collide. Review: One could also pose this question as: What would happen if the strength of gravity changed? 42. Setup: Figure 23.16 shows the star-formation rate as a function of redshift. Solve: The early universe is high redshift, meaning far to the right. Star formation now is about 0.01, whereas at the peak it is close to 0.2, so it was about 20 times higher at its peak. Review: It stands to reason that star formation was higher in the past, because many stars in our galaxy had to be formed to generate the heavy elements needed to make Earth. 43. Setup: As astrophysicist Rocky Kolb is keen to say, the definition of simulation is to deceive, so one must always remember that a simulation is not a test of reality but a test of whether our physical theory produces something consistent with reality. Solve: Although the panels differ in their actual structure, both show filaments and voids with both having roughly the same size and frequency. As such, this simulation appears to show that the physical model is consistent with reality. Review: You should be suspect of a simulation that exactly reproduces reality, because then you have to ask whether the person who wrote and ran it cheated. 44. Setup: To answer this question, use Figure 23.16, by comparing redshift on the bottom with age on the top. The dropout of a spectrum occurs at 121.6 nm, and the redshifted dropout occurs at (see Working It

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 Out 23.2) 1 z 5 obs , or solving for observed waveem length, obs 5 121. 6 nm(1  z) . Solve: At redshift 0.5, the universe was about 9 Gyr old. At z 5 4, the universe was about 1.5 years old. At z 5 10, the universe was about 0.4 Gyr old. The “dropout” of a spectrum at redshift (z 5 0.5) would be seen at 121. 6 nm(1. 5) 5182. 4 nm. Similarly, at redshift 4, the dropout is at 608 nm, and at redshift 10, it is at 1,338 nm. The first can only be seen from space, the latter two can be observed from the ground. Review: We can confirm the problem with Table 23.1, which lists age versus redshift. At redshift 0.5, the universe was 8.6 Gyr old; at redshift yr 4, it was 1.6 Gyr, and at redshift 10, it was 480 million. Pretty close to what we measured off the graph. 45. Setup: To estimate the mass of a cluster, we use the distance and velocity of an orbiting object as shown in Working It Out 23.1. Note that we must convert Mpc 3. 1 3 1019 km to km, so 4 Mpc 3 5 1. 2 3 1020 km . Mpc 2 20 v 2 r (800 km/s) (1. 2 3 10 kg ) 5 5 Solve: M 5 G 6. 67 3 10220 km 3 /(kg s2 ) 1.2 3 1045 kg, or converting to solar masses M 1. 2 3 1045 kg 5 5 . 8 3 1014 M . 30 2 310 kg Review: Compare this to Working It Out 23.1, in which a slightly closer galaxy orbiting at a slightly higher speed yielded roughly the same amount, typical of cluster masses. Using the Web 46. (a) Scientists are using these maps not only for public outreach but also because these data help retrace the history of the universe. Projects include how much of the universe is made of dark matter, what are the compositions of the nearest 500,000 stars, and what were the scale of baryonic acoustic oscillations from early times. The fly-through of the survey provides a three-dimensional visualization of the actual structure of the matter distribution. (b) The DES survey will image 5000 deg2 of the southern sky to probe the origin of the accelerating universe. Observations will study Ia supernovae, baryon accoustic oscillations, galaxy xlusters, and weak lensing. Current results will vary. 47. The Bolshoi simulation is the most accurate cosmological simulation of large-scale structure and the evolution of the universe ever made. These simulations are

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174 ◆ Chapter 23 Large-Scale Structure in the Universe critical for testing the various models of cosmology so that those whose physics disagrees with observations can be ruled out. Answers will vary whether big Bolshoi has produced results. 48. Answers will vary; do a lab from Galaxy Crash and provide the answers to all questions. 49. Cosmological and galaxy-evolution models make predictions about the universe at early times, but we have to make observations of objects at those times (that is, high redshift) to confirm or deny those predictions. This includes both understanding the properties of those objects and how common they are compared with objects that are closer and older. 50. The ALMA Observatory is an array of radio telescopes to observe the universe in millimeter and below millimeter wavelengths. It is going to be the largest telescope at these wavelengths, and located 5,000 meters in elevation, it will suffer very little from terrestrial sources of noise. The array will study the “Dark Ages” by observing extremely high-redshift galaxies, whose optical emissions have been redshifted into the radio regime. The Deep Field will therefore discover many times more high-redshift galaxies (z is greater than 1.5) than were discovered with the Hubble Deep Field because the Deep Field is looking for the extremely highredshifted emission that the Hubble Deep Field could not see. Answers will vary about news and new reports.

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Exploration 1. The proton was formed about 100 seconds after the Big Bang, when three quarks came together and were not subsequently split apart by the high-energy photons that were also present. 2. That proton was part of a large cloud of other H and He atoms, which gravitationally collapsed due to instability to form the first clusters and stars. 3. As a C-atom, that proton would be expelled in stellar winds from a low-mass star to become a planetary nebula. 4. That C-atom, incorporated into a molecular cloud, would collapse under the self-gravity of the cloud to form a protostar and accretion disk. Angular momentum of the atom kept it spinning around the central protostar, whereas gravity caused the cloud to collapse. Then the atom stuck to another atom through electromagnetic forces to form the beginning of a dust grain that would grow into a planet. 5. Answers will vary considerably. In broad brushstrokes, the proton is formed about 3 minutes after the Big Bang, becomes part of a protogalactic clump after a few million years, becomes a star after maybe a billion years, is ejected in a supernova as a carbon atom about 8 billion years ago, and becomes part of our Solar System about 5 billion years ago.

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CHAPTER 24

Life INSTRUCTOR’S NOTES Chapter 24 summarizes our search for life beyond Earth. Major topics include

▶ origins and evolution of life on Earth ▶ evolution and change ▶ biological and physical expectations for the presence of life elsewhere in the universe

▶ current and previous searches for signs of intelligent life ▶ expectations for the fate of life on Earth

A nontrivial part of our jobs is to engender positive attitudes toward science in our courses, as today’s students are tomorrow’s taxpayers and voters who will help dictate the extent to which public money funds the sciences. NASA has been extremely successful in securing funding through its Cosmic Origins program, and these public investments have paid off well. The search for life beyond Earth makes astronomical research personal for each of us; thus, covering the material in this chapter is in no small way an important part of public relations between the astronomical community and general public. The last chapter of any text is an easy one to skip, given the time constraints of our courses, but I highly encourage you to devote at least one class lesson to how we are looking for life beyond Earth, both in our own Solar System and out in the far reaches of our galaxy. Your students have grown up in a time of remarkable activity on the subject of life beyond Earth. First, NASA announced on December 2, 2010, that bacteria in California’s Mono Lake were using arsenic in place of phosphorus in their DNA. In other words, it seemed that life could exist with an entirely new chemistry. Following in the best practices of science, a number of studies tried to replicate these findings, and by summer 2012, two other studies claimed to have debunked this remarkable finding. On December 12, 2013, researchers using the Hubble Space Telescope announced the first discovery of a water geyser from Jupiter’s moon Europa, providing fairly conclusive evidence that liquid water exists beneath

the surface (rather than slushy ice). On April 3, 2014, NASA announced that the Cassini spacecraft and its Deep Space Network found evidence strongly suggesting that there is a huge ocean of liquid water underneath the surface of Saturn’s moon Enceladus. On July 23, 2015, researchers using the Kepler space telescope confirmed the discovery of the very first Earth-sized exoplanet (Kepler-186f) orbiting in its parent star’s habitable zone. There are growing observational signs of water vapor in exoplanets as well, and it has now been reported in both Jupiter and Neptune-sized exoplanets. The chances are very good that by the time you read this, the field will have made many more exciting discoveries. As a final note, the textbook authors have ended this chapter with a number of Fermi problems. The end of the course may be too late to ask students to attack them, and it may not be that instructive to just have a given student do one problem. I have turned these into a fun extra-credit project in which I assign one problem each week or two and give a small amount of extra credit to anyone who turns in a genuine effort at answering the question. After experimenting with different amounts of credit to offer, I have found that the right amount of incentive comes in at about one lab grade (if your course has a lab section) or one-half of a homework assignment for all of the problems put together.

DISCUSSION POINTS

▶ Using the list of requirements for life as we know it, and

bearing in mind what we know about the history of the Solar System, discuss the possible places in our Solar System where life (as we envision it) could have originated around the Sun. ▶ Self-replicating chemical reactions become more complex through a very inefficient process of mutation and natural selection. Connect this with the fact that the rise of complex life-forms on Earth did not start until about 540 million years ago. Discuss the implications of this

175

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176 ◆ Chapter 24 Life inefficient biological evolution for the search of life in the Milky Way. ▶ The speed of evolution depends on the rate of reproduction. Consider bread yeast, which can reproduce every 100 minutes given the right nutrients. Some bakers claim to have been using the same yeast for over a century. Given the timescales and probability for mutations, assess the likelihood that the yeast they grow today are are actually identical to those of years-gone-by. ▶ Figure 24.16 illustrates the distances and extent of habitable zones around different types of stars. Discuss the advantages and disadvantages of living around stars of different spectral types. Consider spectral types O, A, G, and M. ▶ We live in the disk of a giant barred spiral galaxy. Discuss the pros and cons of living in different types of galaxies. Consider a giant elliptical and an irregular galaxy. ▶ Discuss each of the seven factors than enter into the Drake equation. Is there any factor missing? What can be done to decrease the uncertainty in estimating any of the factors?

NEBRASKA SIMULATIONS Developed at the University of Nebraska–Lincoln, these Interactive Simulations enable students to manipulate variables and work toward understanding physical concepts presented in Chapter 24. All simulations are available on the free Student Site (digital.wwnorton.com/Astro5), and offline versions can be found on the USB drive. Cirmcumstellar Habitable Zone This simulation shows the expected habitable zone around a variety of stars. It also shows how this zone changes over the star’s lifetime. Text reference: Section 24.3 Milky Way Habitable Zone Explorer This module shows the likelihood of surviving a supernova, and the abundance of heavy elements, as a function of position in the Milky Way, to help estimate the regions of our galaxy that would be most likely to support a long-term stable stellar system. Text reference: Section 24.3

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END-OF-CHAPTER SOLUTIONS Check Your Understanding 1. (d) Extremo- means “extremes” and -phile means “loving.” 2. (a) Carbon is highly reactive, which makes it favorable for life. 3. (d) Uranus has no surface on which to land, no liquid water, no source of warm internal heat, and an extremely low temperature. 4. (a) As we learned in stellar evolution, the mass and age of a star determine its luminosity and temperature, which in turn determine the distances about the star in which liquid water could exist. 5. (b) The Drake equation is unlike others found in this text, because it is not used to calculate an exact quantity, but rather to provide guidelines about probability. Reading Astronomy News 1. Current censuses of nearby stars suggest 10 to 20 percent of stars host Earth-like planets in the habitable zones. Given the number of stars present, the odds are in our favor that some of these host some form of life. 2. Signatures of life include looking for atmospheric water vapor, carbon dioxide, and other molecules that suggest the presence of life. 3. Imaging from space is required because there is so little background light and no sources that blur that light, providing high sensitivity to low levels of light and very high spatial resolution. 4. Answers will vary. As of autumn 2015, TESS is scheduled to launch in 2017 and JSWT in late 2018. 5. Answers will vary. I believe many people will be skeptical and claim it is a conspiracy, whereas many will be ecstatic. However, I doubt it will lead to world peace. Test Your Understanding 1. All are correct. 2. (b) Our easiest way to look for life beyond Earth is to look in locations that have properties that support life as we know it; water is one of the most universal needs of all life on Earth.

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Chapter 24

3. (c) The Urey-Miller experiment showed that energy input can create the building blocks of proteins out of commonly found materials. 4. (c) On its own, a mutation is just a change in DNA. Whether it is beneficial, detrimental, or inconsequential to the life form depends on how that mutation plays out in the organism’s life. 5. All of the items listed are needed for the origination of life. 6. (a) The conditions given mean that a system will change over time. 7. (d) Another way to put this is that absence of evidence is not evidence of absence. 8. (c) The “explosion” was of biodiversity during the Cambrian geological period. 9. (d) A habitable zone is the region around a star where water (the solvent we believe is most useful for life) can be in a liquid state on the planet. 10. (c) Prokaryotes do not have cell nuclei. 11. (b) Thermo- means “heat” and -phile means “loving.” 12. (b) Currently, we are only able to send robots to other regions of the Solar System to probe those places for signs of current or former life. 13. (c) For a planet or moon to host life, we believe it must experience a relatively steady environment for a long time (billions of years). O and B stars expire too soon to meet this criterion. 14. (a) The first simple forms of life, converting carbon dioxide to oxygen, appeared billions of years ago. 15. (d) This is the scientific definition of theory. Thinking about the Concepts 16. As carbon-based life forms, all forms of life on Earth that we know of have and use DNA (and RNA). Furthermore, the evolution of life depends on changes in DNA and RNA being propagated from parent to child. 17. Amino acids, the building blocks of DNA, were probably built from simpler elements and molecules, using energy from the Sun or lightning as a catalyst. 18. Many early life forms were probably extremophiles, living in very hot, acidic, or alkaline conditions. 19. Answers will vary. 20. Cyanobacteria slowly broke CO2 molecules apart to form O2. This required about 4 billion years to reach today’s oxygen levels.

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21. The Cambrian explosion was a burst of biodiversity that happened around 500 million years ago. At pre sent, it is unclear what caused it. Some ideas include a decrease in predators, a warming of Earth, or the oxygen levels reaching high enough levels for modern life to evolve. 22. Plants and forests use photosynthesis, which relies on CO2 and which was abundant far earlier than oxygen, thus explaining why they appeared first on Earth. Once they converted a reasonable amount of the atmosphere to O2, large animal life could evolve and flourish. 23. The region of space around a star where we think life could form in Earth-like conditions is called the “habitable zone.” In broad terms, the boundaries are defined by the distances at which water would be a solid (too cold) or gas (too hot), but the details also have to do with the energy output from the star, as well as the properties of the planet, such as albedo and the greenhouse strength. 24. For life to arise, we believe there needs to be a reasonably stable environment, abundant building blocks, a “solvent” in which the chemistry could work, and a lot of time. 25. Evolution is certainly under way today. If evolution follows natural selection, then our evolution should allow us to live better in our current or future circumstances. The possibilities are endless here, but it is a good exercise for the student to think about it. 26. H and He are mostly from the Big Bang. All the other atoms in your body are from giant-branch winds and supernovae. 27. Absence of evidence is not evidence of absence. Just because two probes did not find the signs of life they were looking for in two isolated regions does not prove or disprove the presence of life on Mars. 28. If we find that Earth is the only planet in the Solar System capable of supporting life, then f1 (the fraction of worlds on which life arises), fc (the fraction of living worlds on which intelligence arises), and L (the likelihood that an advanced civilization exists today) are all likely to be small. If we do find nonterrestrial life elsewhere in the Solar System, then f1 must be large, forcing both fc and L to be extremely small if intelligence is rare.

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178 ◆ Chapter 24 Life 29. The second law of thermodynamics applies only to a closed system. Living organisms do not violate the second law because the Sun is constantly pouring energy onto Earth’s surface. Therefore, the order that is generated in living creatures comes at the expense of even greater disorder in the Sun. 30. Main-sequence stars slowly become more luminous over time. In a few billion years, the Sun’s luminosity will scorch Earth. In the short term, human impact on our climate may do us in.

Applying the Concepts 31. Setup: The Drake equation reads N 5 R* 3 fp 3 ne 3 fl 3 fi 3 fc 3 L, where R* is the star formation rate, fp is the fraction that forms planetary systems, ne is the fraction of places per system that could host life, fl is the fraction of places per system that do support life, fi is the fraction of those places where life becomes intelligent, fc is the fraction of that life that forms civilization, and L is the number of years that advanced civilization exists. The Kepler mission is looking for planets around stars and is measuring their distances from their central stars. Solve: Given Kepler’s mission, it will place limits on the number of stars with planets, the number of planets around those stars, and their distances. Thus, this mission could help to constrain fp and perhaps ne, if one makes assumptions about the habitable zone around each star. If most stars have planets with habitable zones, then fp will be close to one, which will increase N. Review: Note that Kepler may not discover all the planets around a given star (or any planets for some stars); thus, its sample is incomplete and one must be careful to draw too many strong conclusions from its findings. 32. Setup: Figure 24.18 is the famous radio signal sent from Arecibo in 1974. The top row, we are told, represents the numbers 1 through 10, where the first three rows are the numbers and the bottom row is just a placeholder to show us that a number is being given there. Solve: We are asked how three digits either being on or off can count to 10. This means the signal is using binary notation. In this system, the number

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N 5 1 3 n1 1 2 3 n2 1 4 3 n3 where n1 is the bottom row, n2 is the middle row, and n3 is the top row; each number equals one if there is a dot present and zero if not. The right-most number is a single dot on the bottom, which means 1 3 1 1 2 3 0 1 4 3 0 5 1. Moving left, the next number has no dot in the bottom column but one dot in the middle, which means 1 3 0 1 2 3 1 1 4 3 0 5 2, and so on. Review: Binary is the language of computers, which use “bits” of either zero or one (on or off, respectively) to designate numbers. 33. Setup: Replication means doubling. If we have one organism that replicates in the first second, we will have two; the next second we have four, then eight, and so forth, such that the total population is 2t, where t is the number of seconds. Solve: Using the formula above, there will be 210 5 1,024 organisms. Review: Replication may start off slowly, but this growth accelerates quite rapidly. This is why we call it exponential growth. 34. Setup: Replication means doubling. If we have one organism that replicates in the first second, we will have two; the next second we have four, then eight, and so forth, such that the total population is 2t, where t is the number of seconds. Solve: We wish to know how many seconds must elapse to reach a population of 1,024. Thus, our formula is 1,024 5 2t, which we can solve either with logarithms or by inspection. By inspection, the answer is 10 seconds. With logs, the answer is log(1, 024) 5 t log(2) or t 5

log(1, 024) 5 10 . log(2)

Review: Replication may start off slowly, but this growth accelerates quite rapidly. This is why we call it exponential growth. 35. Setup: According to the problem, the doubling time is roughly 70/p, where p is the percentage rate of increase per year. Solve: If the human population increases 1 percent per year, then the doubling time is about 70/1 5 70 years. If the population doubles in 70 years, then it will quadruple in 140 years. Review: The population today is about 6 billion. Thus, by 2085 we expect about 12 billion people on Earth and 24 billion people by 2155.

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Chapter 24

36. Setup: At one second, there is one particle. At two seconds, two particles. At three seconds, four particles, and at n seconds there are 2n–1 particles. Solve: Here is a linear graph of the growth, followed by a logarithmic plot. Linear plot 3E17

Number of molecules

2.5E17 2E17 1.5E17 1E17 5E16 0 0

Time (sec)

Logarithmic plot 1E+18

Number of molecules

1E+15 1E+12 1E+09 1E+06 1E+03 1E+00 0

30 Time (sec)

Review: Exponential growth proceeds slowly at first but then hits a population explosion where the population grows into millions of billions after a minute. 37. Setup: For a self-replicating object, at first there is one particle. After one doubling time, two particles. At three doubling times, four particles, and so on, such that at n doubling times there are 2n particles. If you start with N0 particles, then you end up with N0 3 2n particles.

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12 3 60 min 5 36 doubling times. 20 min So, there will be 10 3 236 5 6.87 3 1011 or 687 billion bacteria in your body. Review: The body does not respond instantly to an infection because the bacteria (or virus) has to replicate itself to the point that there are large enough numbers for the body to react. We call this the incubation time. 38. Setup: One generation doubles the number of molecules, so n generations makes 2n molecules. Solve: If the probability of an error is 1 in 100,000, then we need 100,000 molecules to be made to get one error, on average. Because 217 5 131,072, it would take about 17 generations. Review: For some organisms, this can be less than a day (consider yeast, for example) which shows just how quickly mutations can occur! 39. Setup: Answers will vary considerably. The Drake equation reads N 5 R* 3 fp 3 ne 3 fl 3 fi 3 fc 3 L. Let’s take optimistic values from the text: R* 5 7, fp 5 1, ne 5 2, fl 5 1, fi 5 1, fc 5 1, L 5 106. Pessimistic values could be R* 5 7, fp 5 0.5, ne 5 1, fl 5 0.1, fi 5 0.1, fc 5 0.1, L 5 103. Solve: The product of the optimistic factors is 14 million, that is, there should be 14 million advanced civilizations. The product of the pessimistic factors is about three, that is, there should be about three advanced civilizations in our galaxy. Review: With optimistic values, our galaxy will be teeming with life, although because there are billions of stars, this means that civilization is still fairly rare, with only 1 in 20 to 100 stars having the possibility of hosting life. The pessimistic values would suggest we are the only advanced civilization in our galaxy. 40. Setup: The Drake equation reads N 5 R* 3 fp 3 ne 3 fl 3 fi 3 fc 3 L, where L is the lifetime of a civilization. Solve: Because N depends linearly on L, N changes with L. Using optimistic and pessimistic values of L from problem 39, L ranges from a million to a thousand; thus, we see that N can easily vary by three orders of magnitude based just on the uncertainty in this one parameter alone. Short lifetimes make the chances of contacting extraterrestrials quite small because we both need to be alive at right times so one can transmit and the other can receive. Review: Remember that we have only been transmitting and listening for about 100 years, so we must bear in mind that communicating with extraterrestrials requires an enormous congruence in timing. Solve: 12 hours is

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180 ◆ Chapter 24 Life 41. Setup: In Kelvin, water freezes at 273 K and boils at 373 K. Solve: The annotated figure looks like this:

Temprature (K)

800 600 400

BOIL FREEZE

200 0 0.1

1.0

Distance from the Sun (AU)

From this figure, no planet is predicted to lie inside the habitable zone, although measurable temperatures show that Earth is solidly inside the freeze-boil region, whereas all other terrestrial planets are either fully or partly outside of this—that is, Mars is far too cold, Venus is far too hot, and Mercury has too much of a temperature swing in both directions. Review: The notion of a habitable zone is very fluid, because it does not account for rotation of a planet, albedo, or the green house effect. 42. Setup: The habitable zone is where the average temperature of a planet from its star is between the freezing and boiling points of water, 273 K and 373 K, respectively. Solve: Mercury’s average temperature is hotter than boiling, and its temperature range is far above and below this range. Thus, only a small part of the planet may be inside this temperature range at a given time. With such wild temperature swings such as these, one would not consider Mercury to be inside the habitable zone. Review: Being in the habitable zone means not only having an average planetary temperature that is between boiling and freezing but also that the excursions away from this average are not too large. 43. Setup: Given the period of a planet, we can determine the orbit from Kepler’s second law, P2 5 a3, where period is in years and distance in AU. Based on Figure 24.16, the habitable zone around the Sun is roughly between Venus’s and Mars’s orbits, if we include nonrunaway greenhouse effects. 87 5 0. 24 years. This Solve: Eighty-seven days is 365 yields a semimajor axis of a 5 P2⁄3 5 (0.24)2⁄3 5 0.38 AU. This is basically at the location of Mercury, which is outside the habitable zone.

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10.0

100.0

Review: Bear in mind that there could be extenuating circumstances that allowed this planet to harbor life; thus, although we say it is not in the expected habitable zone, it does not mean this hypothetical planet necessarily cannot host life. 44. Setup: When considering where to send a signal to search for life, does it make sense to target a single star, or a huge number at once? Solve: Being a globular cluster with only low-mass stars, M13 provided a single “target” with many solar-type stars, which we hoped would maximize our chances of finding life when we beamed out our radio signal in 1974. Review: Unfortunately, M13 is metal poor and about 11 billion years old, so it is not certain whether planets with sufficient heavy elements to form life would exist in large numbers, and if life did evolve, whether it would still exist today. 45. Setup: Recall that the Solar System formed about 5 billion years ago, so the life-forms that did the seeding must be older than this at a minimum. But to get into the pre-solar nebula, they had to have formed elsewhere and then traveled through space, which probably required the explosion of a nearby star. Solve: Simple life on Earth required a billion or more years to form, so these seeds probably took a similar length of time. Knowing that more massive stars than our Sun live shorter life spans but also are much hotter and luminous (too much UV radiation could be good or bad for life), we might propose that the star needs to have a mass similar to the Sun, that is, 2 solar masses or so. This means the star formed about 8 to 10 billion years ago. Review: One has to ask, were there enough metals in the prestellar nebula to provide life-building chemicals in its planets to seed Earth with life?

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Chapter 24

Using the Web 46. (a) Answers will vary. Report on new findings on extremophiles. Astronomers care about extremophiles because these are forms of life that evolved in hostile circumstances, such as those expected around other planets and moons. (b) Answers will vary. Discuss recent topics of interest from the “Life, Unbounded” blog. Include whether this is based on new data or a new theory. 47. (a) Answers will vary. Discuss whether Cassini is still operational and what of interest has been found about moons. (b) Curiosity includes a host of research apparatus to search for fossilized evidence for life on Mars. Astrobiologists are very interested in whether life evolved, to what stage, and whether it could still be present in some form. Answers will vary on recent results. (c) MAVEN will explore Mars’s upper atmosphere, including how molecules like water and carbon dioxide have been lost in space. This provides important clues about when and how Mars became geologically dead. Answers will vary on recent news. 48. Answers will vary with time. At the end of 2012, the exoplanet catalog listed zero exoplanets that were confirmed to be habitable, and seven that had the potential, with 30 additional moons that had the potential to harbor life. This yielded a lower limit of about 1.6 percent of exoplanets being habitable, which would be the fraction fl in the Drake equation. 49. (a) Answers will vary. Provide recent news on a discovery of a planet in the habitable zone, and a planet with an interesting atmosphere. (b) Gaia’s scientific mission is to produce a three-dimensional chart of our galaxy by making radial velocity measurements of about 1 billion stars. This is expected to reveal tens of thousands of exoplanets because a star’s radial velocity changes reveal the gravitational tugs of its planets. Answers will vary on what is new with the satellite and whether it has been launched. 50. The addition of more planets almost always destabilizes the first planet, whether they are placed inside or outside the habitable zone. Lower-mass planets destabilize less than higher masses. The best solution for stability is for the distances between the planets to be as large as possible. Exploration 1. There are about 6 billion people on Earth today, and there were about 5 billion on Earth around 1990. If we assume the population grew by 20 percent in 20 years,

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Life ◆ 181

then it grows about 1 percent per year. So let’s assume if it grew by 60 million people last year and if each person weighs about 100 pounds, then the mass went up by about 6 billion pounds. 2. A horse is much larger and more active than a person. I need about 2,000 calories per day, and because a horse does a lot more work, I am going to guess that a horse needs 20,000 calories per day. If a horse lives 20 years, or 7,000 days, then it needs roughly 20,000 3 7,000 5 140 million calories over its life. 3. Let’s assume there are 300 million Americans, and each person eats two potatoes per week. At the grocery store, I think I buy about three potatoes per pound, so let’s round our number to say that each person eats 1 pound per week, or 52 pounds per year. Then, the total consumption is 300 3 106 3 52 5 15.6 billion pounds of potatoes per year. Note that http://www.potato2008.org lists the United States as consuming about 17 million tons of potatoes, or about 34 billion pounds, so my solution is close! 4. Let’s assume the average human weighs 100 pounds, which is made up entirely of cells. If we assume a cell is pure water (the body is mostly water) and is on average a sphere of radius 10 mm, then a cell has a volume of 4.2 3 10215 m3. Water has a density of 1,000 kg/m3, so a cell weighs about 4.2 3 10212 kg. There are about 2 pounds per kilogram, so a cell weighs about 9 3 10212 pounds. Let’s round that up to 10211 pounds per cell. Then, dividing the average weight of a person by the weight of a cell yields 100/10211 5 1013 cells. Wikipedia says humans have about 1014 cells, so this was not far off! 5. Let’s forget inflation for the moment and assume that I will work for 50 years, earning an average of $75,000 per year (I am just a college professor!). Then I will make about $3.75 million over my lifetime. For this problem, I am going to assume a 40-hour workweek at 50 weeks per year, meaning I work 2,000 hours per year, or 100,000 hours in my life. Then my salary would be about $37.50 per hour. 6. I have a wife and two kids and take out about 10 bags of trash per week. I am estimating that they weigh 10 pounds each, meaning my family goes through 100 pounds of garbage per week. If there are 300 million Americans, and each family has four people in it, then there are about 75 million families, meaning we all produce about 7.5 billion pounds of trash per week. With 52 weeks per year, this comes out to 390 billion pounds of trash per week. Note that http://www. cleanair.org lists that each person in the United States

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182 ◆ Chapter 24 Life produced, on average, 4.5 pounds of trash per day (but this includes industry and business) for a total of 4.5 3 365 3 300 3 106 5 500 billion pounds per year. Not far off from my estimate! 7. I cut my hair once per month and take off about an inch. So I need to convert 1 inch per month into feet per hour: in ft month day 1 5 1024 ft per hr month 12 in 30 days 24 hr 8. Let’s assume again that there are 300 million Americans and that each person takes up a space of 1 square foot (my shoulders are much longer than 1 foot but my width is less, so I am averaging down). Then we would all take up about 300 million square feet. There are about 28 million square feet in a square mile, so we could all pack into a space of 11 square miles, or about the size of Manhattan. 9. Earth is 70 percent water, so let’s assume only 30 percent of the land mass is usable. Earth has a radius of about 6,400 km, or a surface area of 4  3 (6 , 400 , 000 m)2 5 5 3 1014 m 2 , of which 30 percent is usable, or about 1014 m2 of land (I rounded down). Then, if each person used 1 square meter, 1014 people, or 100 trillion, could live on Earth. 10. To estimate how much carbon dioxide is given off by a car per year, we should find out how many gallons of gas are used per year and how much carbon dioxide is produced per gallon. A typical lease for a car is 3 years for 36,000 miles, so let’s assume a person drives 12,000 miles per year, and the car makes 20 miles per gallon (for each efficient car there is a gas-guzzling SUV). Thus, a person uses about 600 gallons of gas per year. A gallon of water weighs about 8 pounds

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(a pint is a pound the whole world) and gas is a hydrocarbon, but gas burns oxygen. Carbon and oxygen have about the same atomic weight, so burning gas will use 1 pound of carbon for each 2 pounds of oxygen. I will guess that of 1 gallon of gas, 5 pounds are carbon, so a total of 15 pounds of CO2 are released per gallon of gas. (The EPA lists 19.4 pounds of CO2 per gallon, by the way). This means the car releases about 600 3 15 5 9,000 pounds of CO2 per car in a year. The EPA estimates that an average passenger car releases 5.48 metric tons of CO2 per year, or 12,000 pounds per year. 11. Earth may be 70 percent water on the surface, but it is mostly rock and iron. Water has a density of 1,000 kg/m3, and iron is about 5,500 kg/m3, so I will assume an average density of 5,000 kg/m3 . Earth’s radius is 6,400 km, 4 so its volume is  3 (6 , 400 , 000 m)3 5 1021 m 3 3 for a total mass of about. 5,000 kg/m3 3 1021 m3 5 5 3 1024 kg. The actual mass is 5.9 3 1024 kg. 12. There are a lot of costs associated with having a car. Here are all of the ones that I can think of, estimated for a year. I assume an average driver puts on 12,000 miles per year, which yields 600 gallons of gas if the car makes 20 mpg. At $4 per gallon, this is $2,400. Insurance is about $1,000. Routine maintenance can vary between a cheap oil change ($50) and five new tires ($1,000 or more), so I assume about $1,200 in maintenance. You might also spend $100 in parking meters and lots, $50, for an embarrassing speeding or parking ticket, $30 for registration with the DMV, $40 in car washes, $40 in tolls, and I am going to toss in $250 extra for things I cannot think of. This comes to a grand total of $4,110 per year.

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PA RT I I :

Answers to Starry Night Workbook Exercises

183

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The Norton Starry Night Workbook: Exercise Summaries and Activity Answers EXERCISE 1: THE CELESTIAL SPHERE This exercise illustrates the daily motion of the celestial sphere. It animates the apparent motion of the sky as seen from Earth’s North and South poles. Students will learn the most basic operations of Starry Night: selecting objects to view, changing the orientation of their gaze, and altering the flow of time.

ANSWER KEY Activity 1: Directions on the Sky 1. In this view, west is toward the right, and east is toward the left. 2. From the North Pole, the apparent motion is parallel to the horizon. Activity 2: Direction of Rotation 3. From the North Pole, the apparent motion of the sky is counterclockwise. 4. Polaris is located in the constellation of Ursa Minor. Activity 3: View from the South Pole 5. From the South Pole, the apparent motion of the sky is clockwise. 6. The south celestial pole is in the constellation Octans. 7. Stars at the North and South poles appear to move in opposite directions. This is because the point of view of the observer is inverted, so that a counterclockwise motion as seen at the North Pole appears clockwise as seen at the South Pole.

EXERCISE 2: EARTH’S ROTATION PERIOD Here, the students will become familiar with the appearance of the celestial sphere from an intermediate latitude. The instructor may wish to direct the students to pick a particular viewing location for this exercise. The students will determine Earth’s rotation period by finding the average time between meridian crossings for a star. The exact value of Earth’s rotation period (23 hours 56 minutes 4 seconds), is called the sidereal day. Many

students are surprised that the rotation period is not a full 24 hours. A number of the exercises for Starry Night require that the students observe in intervals that are multiples of 1 sidereal day, so it is important that the students learn this concept.

ANSWER KEY Activity 1: Circumpolar Constellations 1. From typical latitudes in the continental United States, the circumpolar stars are in the constellations of Camelopardalis, Cassiopeia, Cepheus, Draco, Ursa Major, and Ursa Minor. You might want to point out to students that all the constellations in the Northern Hemisphere are circumpolar as seen from the North Pole. This fact was demonstrated in the exercise on the celestial sphere. Activity 2: Rising and Setting Constellations 2. The answer will depend on the constellation selected. Small constellations take about an hour to rise, whereas larger ones (Orion, Pegasus) may take 3 hours or more. Activity 3: Earth’s Rotation Period 3. The students are asked to determine the interval between meridian crossings a few times and average the answer to smooth out any inaccuracies in their measurements. 4. The length of the sidereal day is 23 hours 56 minutes 4 seconds. 5. The average amount by which stars cross the meridian earlier each day is 24 hours minus the length of the sidereal day, or 3 minutes 56 seconds. 6. A star that crosses the meridian at midnight tonight would cross 4 minutes earlier tomorrow night, at 11:56 p.m. Thirty days later, the star would cross the meridian 120 minutes earlier (30 3 4), at 10 p.m. This implies that any particular star will continue to transit earlier and earlier until it is eventually only above the horizon during the day. But also, the star would again 185

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186 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers continue to transit earlier until a year later, when it would again transit at midnight.

EXERCISE 3: MOTION OF THE SUN ALONG THE ECLIPTIC This exercise demonstrates that the Sun moves eastward along the ecliptic each day, and asks the students to note which constellation the Sun appears in at different times of the year. The exercise reinforces the concept of Earth’s changing position in its orbit around the Sun, causing constant changes in the apparent position of the Sun in the celestial sphere as viewed from Earth. Most students know something of popular astrology, if only their birth sign. The students will compare the actual direction of the Sun with the locations listed on many calendars and see that the dates of the astrological signs have their origins more than 2,000 years ago.

ANSWER KEY Activity 1: The Sun and the Zodiac 1. Dates on which the Sun entered the modern constellation boundaries: Constellation

Date Sun Enters

Days Spent

Gemini

6/22/2007

Cancer

7/21/2007

Leo

8/11/2007

Virgo

9/17/2007

Libra

11/01/2007

Scorpius

11/23/2007

Ophiuchus

11/30/2007

Sagittarius

12/19/2007

Capricornus

1/20/2008

Aquarius

2/17/2008

Pisces

3/12/2008

Aires

4/19/2008

Taurus

5/14/2008

Gemini

6/21/2008

—

Activity 3: Dates in the Distant Past 3. The approximate year that the Sun entered Gemini on May 21 is 900 BC (allow ±1 century). 4. The Sun only appears to pass through the zodiac constellations because the ecliptic plane of Earth’s orbit around the Sun lines up with these constellations. The Sun never passes through Ursa Major because that constellation is not in the ecliptic plane.

EXERCISE 4: MOTION OF THE MOON This exercise asks the students to measure the synodic and sidereal periods of the Moon and to compare their measurements to the values quoted in the reading. The instructor may wish to stress that any set of measurements will be accurate only to a particular level. The students should be encouraged to be honest about their measurements and not to try to land on the stated periods. By comparing their values to the actual ones, the students will get a feel for the accuracy of their measurements. There are more accurate ways of measuring the synodic and sidereal periods using Starry Night, but this exercise is designed to produce answers accurate only to about a quarter of a day.

ANSWER KEY Activity 1: Time of Moonrise on Successive Nights 1. The exact times will depend on the viewing location. The instructor may wish to specify a particular viewing location so that all students come up with the same times. 2. The average time interval will be about 50 minutes. Activity 2: The Moon’s Sidereal Period 3. The students should derive numbers between 27 and 28 days. In the exercise, they work in steps of 1 day, so the fraction of a day over 27 that they derive will depend on how they interpolate between positions of the Moon on successive nights. 4. The length of the sidereal month is 27.3 days.

Allow 6 1 day for the student estimates.

Activity 2: Astrological Dates 2. Students will notice that (a) the current constellation boundaries mean that the Sun spends different intervals in the various constellations; (b) there is a difference of about 1 month between the traditional dates and the current dates; and (c) the Sun spends almost 3 weeks in Ophiuchus, which is not one of the traditional zodiacal constellations.

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Activity 3: The Moon’s Synodic Period 5. As in answer 3, the exact value will depend on how the students interpolate. 6. The length of the synodic month is 29.5 days. 7. The difference between the Moon’s sidereal and synodic periods is about a little more than 2 days. The synodic period is longer because as the Moon orbits Earth, the Earth-Moon system is moving around the Sun. It takes 27.32 days for the Moon to complete

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one orbit around Earth (sidereal period), but in that time, Earth has moved to a different position in its orbit around the Sun. The Moon must therefore move a little farther in its orbit to line up with the Sun again, making this synodic period 29.5 days.

EXERCISE 5: EARTH AND MOON PHASES This exercise explores the relationship between the phases of the Moon as seen from Earth and the phases (shading)

Date and Time

Moon Phase (shadow = dark)

Phase Description

Activity 2: Earth Phases 1-3.

Earth Shading (shadow = dark)

2B Waning Gibbous Earth

3A Jan. 16, 2016 21:54:18 UT

Last/3rd Quarter Earth

1st Quarter Moon

4A Waxing Gibbous Moon

Jan. 19, 2016 21:42:30 UT

Waning Crescent Earth

5A Jan. 23, 2016 21:26:47 UT

Full Moon

New Earth

6A Waning Gibbous Moon

Jan. 27, 2016 21:11:03 UT

Waxing Crescent Earth

7A Last/3rd Quarter Moon

Jan. 31, 2016 20:55:19 UT

Shading Description

Full Earth

Waxing Crescent Moon

Jan. 13, 2016 22:06:06 UT

1st Quarter Earth

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Activity 1: Moon Phases

New Moon

Feb. 4, 2016 20:39:36 UT

ANSWER KEY

1A Jan. 9, 2016 22:21:49 UT

of Earth as seen from the Moon. Students will predict the best conditions to observe phases of the Earth from the Moon.

Waning Crescent Moon

Waxing Gibbous Earth

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188 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 4. The shading (phases) of Earth appears to be opposite or the reverse of the Moon phase shading. A New Moon is a full Earth for the same date and time. Similarly, a waxing crescent Moon phase is accompanied by a complementary waning gibbous Earth phase. The first quarter Moon features a last/third quarter Earth. A waning crescent Moon is opposed by a waxing gibbous Earth, and so on. 5. To see Earth completely, it should be in its full phase, which corresponds to the new Moon phase. 6. It depends on how well the observers can be protected from the extreme cold on the surface of the Moon, in the total darkness of a new Moon. Observing might not be practical during the extreme cold of the new Moon Date and Time

Moon Phase (shadow = dark)

Phase Description

New Moon

Jan. 13, 2016 22:06:06 UT

Waning Gibbous Earth

Last/3rd Quarter Earth

1st Quarter Moon

4A Jan. 19, 2016 21:42:30 UT

Waxing Gibbous Moon

Jan. 23, 2016 21:26:47 UT

Waning Crescent Earth

5A Full Moon

Jan. 27, 2016 21:11:03 UT

New Earth

6A Waning Gibbous Moon

Jan. 31, 2016 20:55:19 UT

Waxing Crescent Earth

7A Last/3rd Quarter Moon

1st Quarter Earth

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Full Earth

Waxing Crescent Moon

Jan. 16, 2016 21:54:18 UT

Shading Description

Feb. 4, 2016 20:39:36 UT

Earth Shading (shadow = dark) 1B

1A Jan. 9, 2016 22:21:49 UT

phase. Since there is no atmosphere on the Moon, heat is conducted only by radiation and conduction. Furthermore, Bruce Crater is near the lunar equator, so the surface experiences the most extreme high temperatures during the lunar day time such as when there are waxing and waning as phases. The temperature can rise to almost 390K (water boils at 373K) during the daytime, and sink to as low as 95K (water freezes at 273K) in the nighttime. Selecting a phase when the most moderate temperatures occur, around dawn as the Sun first rises for the lunar day, which is a little over 27 Earth days long, might be the best time to observe. This procedure was used during the Apollo lunar landings from 1969 to 1972.

Waning Crescent Moon

Waxing Gibbous Earth

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EXERCISE 6: SUNRISE ON MARS

ANSWER KEY

In this exercise, students will observe and document sunrises and sunsets along with related objects from the surface of mars. Students will also predict the seasons based on the shifting location of the sun along the Martian horizon.

Activity 1: Sunrises from Mars

1/ 9/13/2015 9/14/2015

Sunrise time Stopped time 22:0:47/ 0:05:03 UT

List other objects observed near the Sun along the sunrise path to the horizon

Sketch of Sun and any objects at sunrise (show and label horizon with directions) 30˚

Earth Venus Mercury

V Sun

75˚ 80˚

60˚ 2/ 3/13/2016

18:06:30 UT/ 20:06:57 UT

E Me

Observation step/ Date

Sunrise Table:

10˚

0˚

90˚

100˚ 30˚

Mercury Venus

10˚ E 60˚ 3/ 9/13/2016

16:59:37/ 19:04:38 UT

80˚

30˚ 10˚

80˚

14:23:27/ 16:25:18 UT

10:48:36/ 12:52:17 UT

90˚ 100˚

0˚ 108˚ 120˚ 30˚

Jupiter Earth Venus Saturn

10˚ SE 100˚

5/ 9/13/2017

103˚ 120˚

0˚ 140˚ 30˚

Jupiter Saturn Mercury Venus

10˚ E 60˚

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100˚

Earth Mercury Venus E

4/ 3/13/2017

0˚

90˚

65˚

90˚ 80˚

0˚ 100˚

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190 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers Sunset Table:

Observation step/ Date 1/ 9/13/2015 9/14/2015

Sunset time Stopped time 10:31:17/ 12:31:29

List other objects observed near the Sun along the sunset path to the horizon

Sketch of Sun and any objects at sunrise (show and label horizon with directions)

Venus Earth Mercury

V E

30˚ 10˚ NW 280˚

2/ 3/13/2016

7:28:26/ 9:29:02 UT

Me 288˚ 300˚

0˚ 320˚ 30˚

Earth Mercury Venus

10˚ NW 0˚ 280˚

3/ 9/13/2016

04:39:16/ 6:34:30 UT

292˚ 300˚

320˚ 30˚

Earth Jupiter Mercury

10˚ W 240˚

4/ 3/14/17

2:13:29/ 4:12:50 UT

0˚

270˚ 252˚ 280˚ 260˚ 30˚

Venus Earth

10˚ W 240˚ 5/ 9/14/17

0:12:27/ 2:28:05 UT

252˚ 260˚

270˚

280˚ 30˚

Venus Mercury Earth

0˚

10˚

E NW 280˚

293˚ 300˚

0˚ 320˚

1. They are the planets rising or setting. The names of these may be seen by rolling the cursor over the objects or by turning on the planet labels.

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Activity 2: Predicting Seasons from Sunrise/-Set Data 2. Difference a. Summer Solstice

1/3/2016

-------------------

b. Autumnal Equinox

7/4/2016

6 months, 1 day

c. Winter Solstice

11/28/2016

4 months, 24 days

d. Spring Equinox

5/5/2017

5 months, 7 days

e. Summer Solstice

11/20/2017

6 months, 15 days

Total

22.6 months

3. Northern Hemisphere. 4. The greater eccentricity of the orbit of Mars causes the seasons to be unequal.

EXERCISE 7: PRECESSION Here we focus on the precession of Earth’s rotation axis. The students will measure the angular separation between the north celestial pole (NCP) and Polaris at different epochs and will see that the current separation is close to the smallest possible value. The text emphasizes the effect of precession on the times of the spring and fall equinoxes; we touch on that concept in Exercise 3, on the motion of the Sun along the ecliptic.

ANSWER KEY Activity 1: The Distance Between Polaris and the North Celestial Pole 1. The separation in 2007 is about 0 degrees 42 minutes (or 0° 429). Allow an error of a few minutes, as the students may land on slightly different positions near the north celestial pole. Activity 2: When Polaris is Closest to the Ncp 2. The minimum separation will occur about the year 2100 and will be approximately 0° 289. 3. This is about 67 percent of the current value. Activity 3: The Separation in the Past 4. In the year AD 1, the separation was almost 12° (about 11° 459)! 5. The following table shows a few bright stars that were closer than Polaris to the north celestial pole in the year AD 1. The last column displays the separation in degrees and minutes. If they ignore the instructions to examine the stars in the stick figures for Ursa Minor

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and Draco, some students may zoom in and identify faint stars that are much closer to the NCP. Constellation

Star

Separation

Ursa Minor

Kochab

8° 209

Zeta Ursae Minoris

8° 329

Kappa Draconis

9° 109

Draco

6. Though Polaris was about the same distance from the NCP as Kochab at this time, Polaris is the brighter star. 7. The change in the position of the north celestial pole affects all stars, not just Polaris. The stars stay in the same positions relative to each other, so the constellation patterns themselves are not affected, but if the NCP moves away from Polaris, it must move closer to other stars.

EXERCISE 8: KEPLER’S LAWS This exercise reinforces a student’s knowledge of Kepler’s laws by creating an asteroid with a semimajor axis of 1 AU, but with a very elongated orbit. In the first part, the students will calculate the orbital period and then verify it by inspection. They will see that the period is independent of the eccentricity. The elongated orbit for the asteroid makes Kepler’s second law more apparent than it would be for the orbits of the major planets, which are nearly circular. This exercise views the orbits of Earth and the asteroid from a position above Earth’s ecliptic pole. This helps illustrate that Earth’s distance from the Sun does not vary much during the year. Many students get the opposite impression from figures that show Earth’s orbit from the side, which make it appear much more eccentric, thereby causing confusion when they learn the causes of the seasons.

ANSWER KEY Activity 1: Kepler’s Third Law 1. The length of the semimajor axis is a = 1 AU. From Kepler’s harmonic law, P2 = a3, so P = 1 year. 2. The orbital period is independent of the eccentricity e. This sometimes surprises students because a circular orbit looks so different from a highly elongated orbit. 3. With a = 4 AU, a3 = 64, so P = 8 years. Activity 2: The Period of X’s Orbit 4. The period determined by measurement should be very close to 1 year, in agreement with the value from Kepler’s harmonic law.

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192 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 5. The asteroid X orbits fastest at perihelion. 6. Kepler’s law of equal areas describes the speed of motion. It states that a line connecting the Sun to an object sweeps out equal areas in equal times. Activity 3: Inside and Outside 7. The asteroid X spends about 3 months inside Earth’s orbit. If the students are clever with the zoom tool and the time flow rate, they might show that the time is about 3 months 2 days. The asteroid spends about 9 months outside Earth’s orbit. Activity 4: Kepler’s Second Law 8. The distance at perihelion is 0.25 AU. The distance at aphelion is 1.75 AU. 9. The semimajor axis for a = 2 AU is twice the value of a, or 2 AU. This agrees with the measured values in questions 9 and 10, because these values add to the longest dimension across the orbit (that is, the length of the major axis). 10. Asteroids that may cross Earth’s orbit typically have very eccentric orbits. Kepler’s second law says that eccentric orbits have greater variation in orbital speeds. The asteroids would move more slowly when they were farther away from the Sun, spending more of their time outside of Earth’s orbit. Searches for these types of asteroids should be focused away from the Sun, outside of Earth’s orbit.

EXERCISE 9: FLYING TO MARS In this exercise, the students create an asteroid with an orbit that carries it from the vicinity of Earth to near Mars. The asteroid imitates a spacecraft, particularly the two Mars exploration rovers Spirit and Opportunity, which were launched on June 10 and July 7, 2003, respectively, arriving on Mars in January 2004. They followed an orbit that was similar to the ones created in this exercise. Some students may be familiar with the phrase “launch window,” which is an interval of time during which a spacecraft must be launched in order to reach its destination. Here, the students will learn that it is rather complicated to put a spacecraft into the right orbit to arrive at another planet on time. They will need to use their judgment about whether the orbit they create gets the spacecraft close enough to Mars. The instructions guide the students in creating a Hohmann transfer orbit, named after its discoverer (1925). Such an orbit usually minimizes the amount of fuel needed to reach the other planet.

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The complications are greater, naturally, if the spacecraft has a human crew. The last part of the exercise asks students to think of trade-offs involved in crewed spaceflight. They simply cannot fly past Mars and expect Earth to be there to receive them!

ANSWER KEY Activity 1: The Transfer Orbit 1. e = 0.21. 2. a =1.28. 3. w = 269°. 4. L = 217°. These values are approximate. Students should get values that are within ±0.02 for e and a, and within ±5° for w and L. Activity 2: Time to Fly to Mars 5. The spacecraft arrives near Mars around January 2004. Activity 3: Coming Home 6. The spacecraft returns to Earth in late November 2004. Allow about 1 month leeway, as students may pick different orbits. Earth is not there at the time; it is on the opposite side of the Sun. 7. The orbital period is approximately 1.4 years. The exact value depends on the value of the semimajor axis of the orbit that students pick. 8. There are two choices: (a) fly back on a different orbit, which requires more fuel; or (b) wait on Mars for several months until Earth would be in the right place for a return.

EXERCISE 10: THE MOONS OF JUPITER January 2010 was the 400th anniversary of Galileo’s first observations of Jupiter through his simple telescope. This exercise illustrates the observations that Galileo made when seeing Jupiter for the first time to show some of the challenges he faced in identifying the four large moons and their patterns. The students will use their observations of the moons to deduce the patterns of their motions with regard to distance from Jupiter, orbital speeds, and orbital periods.

ANSWER KEY Activity 1: Galileo’s Observations 1. Students will be observing the orientation of the Galilean moons and recording them in the table.

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determine the date of both maximum tilt and the next ring plane crossing.

Date

Time

Sketch

1/07/1610

8:00 p.m.

* (2 moons) *

1/08/1610

8:00 p.m.

***

1/10/1610

8:00 p.m.

1/11/1610

8:00 p.m.

* *

1/12/1610

8:00 p.m.

* * *

1/13/1610

8:00 p.m.

***

1/14/1610

8:00 p.m.

(2 moons) *

* *

O * (2 moons)

2. Galileo may have only noticed three moons at first for two possible reasons: (a) in three of the observations ( January 7, 11, and 14), two moons were so close together that they appeared as one; and (b) in the earlier observations, one moon was quite far from Jupiter and may have been mistaken for a background star. Activity 2: The Orbits of Jupiter’s Moons 3. When the moons are shown in motion, you can see that they pass alternately in front of and behind Jupiter. Also, the moons seem to slow down when approaching their apparent farthest point from Jupiter, meaning that they are moving either more directly toward or away from us, rather than directly across our field of view, as when they are apparently closer to Jupiter. 4. Io stays the closest to Jupiter. Callisto moves the farthest from Jupiter. 5. Io orbits with the fastest speed. Callisto orbits with the slowest speed. 6. The elongation dates and times will vary widely among students, but the orbital periods should all be consistent. The period of Io is about 1 day 18 hours; Europa is about 3 days 12 hours; Ganymede is about 7 days 4 hours; and Callisto is about 16 days 18 hours. 7. The farther a moon is from Jupiter, the slower it moves in its orbit and the longer it takes to complete one orbital period.

EXERCISE 11: THE RINGS OF SATURN Starting in 2009 and for the following few years, Saturn’s rings were inclined at a very small angle. If students observed Saturn through a telescope during these years, they may have wondered why the rings were not as spectacular as they typically look in photographs. This exercise has students investigate the changing orientation of Saturn’s rings caused by the tilt of the planet and its orientation relative to Earth’s orbit. Specifically, they will be guided through the ring plane crossing of 2009 and then continue into the future to watch the changing orientation of the rings and

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ANSWER KEY Activity 1: Finding Saturn’s Ring Plane Crossing 1. On January 1, 2009, the rings were very close to edge-on, so students would not have had the very obvious view of the rings they might have expected. When the rings are seen in this orientation, they are much less bright than they might be otherwise. This can make it easier to see dimmer objects such as small moons nearby. 2. On September 4, 2009 (plus or minus a day), the rings appeared edge-on. Because they were not visible, the effect from question 1 was enhanced, allowing even dimmer and smaller objects to be most easily seen at that time. 3. The rings must be extremely thin. Even when zoomed in to maximum, the ring still barely appears as a faint line across the planet. Activity 2: Saturn’s Rings at Maximum Inclination 4. Saturn appears to grow larger and smaller as Earth’s orbit around the Sun brings us closer to and farther from Saturn. The size can be seen to fluctuate over a period of 1 year. When Earth is between Saturn and the Sun, we are at our closest, and Saturn appears largest. When Earth is on the opposite side of the Sun from Saturn, we are at our farthest, and Saturn appears smallest. 5. This answer will be much harder for the students to gauge accurately. Sometime in 2016 to early 2017 would be acceptable. 6. The diameter of the (visible) rings is more than twice the diameter of the planet. 7. The year Saturn’s rings were seen edge-on was 2009 and the year of maximum tilt was found to be 2016 or 2017. Assuming that the timing of the next edge-on view of the rings follows a pattern, then a student might reasonably predict that the next occurrence would be 7 or 8 years later, sometime between 2023 and 2025. Activity 3: Finding the Next Ring Plane Crossing 8. The rings will next appear edge-on on March 23, 2025. 9. The interval between ring plane crossings was about 15 years 6 months. Using the interval between the first crossing and maximum tilt to be halfway (about

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194 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 7 years), the student might have predicted the next crossing as sometime in 2023. The book does state that the ring plane crossings take place every 15 years, but there is not one precise interval because of other effects in the planets’ orbits.

EXERCISE 12: PLUTO AND KUIPER BELT OBJECTS In 2006, the International Astronomical Union adopted new guidelines for use of the term planet, and Pluto was no longer on the list. This change was widely reported in the popular press; less well covered was the long-known connection between Pluto and other distant, icy objects in the outer Solar System. These Kuiper Belt objects (KBOs) are being discovered regularly, and each new one shows that Pluto is related to the KBOs and not the giant gas planets. In this exercise, students will show that Pluto’s orbit is similar to those of the largest-known KBOs by comparing the orbital eccentricity and inclination angles.

ANSWER KEY Activity 1: Perihelion and Aphelion of Neptune and Pluto 1. Pluto was at perihelion around August 1989. 2. Its distance at perihelion was 29.656 AU. 3. The distance of Neptune at the same time was 30.2 AU. Activity 2: Orbits in the Outer Solar System 4. Pluto’s orbit is more like that of the KBOs. It has an eccentricity that is very different from zero and a large inclination. The following table includes data for Neptune and Pluto: Object

Eccentricity

Inclination

Neptune

0.01

1.7°

Pluto

0.25

17°

1998 WA31

0.43

9.5°

2003 EL61

0.19

28°

2004 XR190

0.08

46°

2005 FY9

0.16

29°

5. The classical planets are all confined to be very close to the ecliptic plane of the Solar System. However, Pluto and other similar objects have eccentric orbits that are inclined widely from this plane, and so could be found in almost any part of the sky.

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EXERCISE 13: ASTEROIDS This exercise explains some of the characteristics of asteroid orbits that are illustrated in Figure 9.14 of the text. The students first examine the orbits of four large asteroids in the main belt between Mars and Jupiter and then look at the orbit of the near-Earth asteroid Bacchus.

ANSWER KEY Activity 1: Kepler’s Laws Revisited 1. Mercury and Venus orbit inside Earth’s orbit. 2. Kepler’s third law: P2 = a3, where P is the period and a is the length of the semimajor axis. Activity 2: Orientation of the Orbital Planes of the Asteroids 3. The orbits range from nearly circular to rather elliptical. Ceres and Vesta have eccentricities that are about the same as Mars (0.09); Pallas’s orbit is quite elliptical. 4. The asteroid orbits have a range of inclinations from the plane of the Solar System. Activity 3: Asteroids that Pass Near Earth 5–6. Bacchus did not strike Earth because its orbit is inclined to Earth’s orbital plane. 7. The asteroids are much smaller than planets, so their orbits are more easily influenced through gravitational interactions with the planets, causing the smaller asteroids to be flung into more eccentric and inclined orbits.

EXERCISE 14: THE MAGNITUDE SCALE AND DISTANCES The magnitude scale describing the brightness for stars typically needs careful explanation because of its nonintuitive, reversed scale. Students need to be reminded that smaller positive and larger negative numbers are progressively brighter stars. The method of estimating distances using absolute and apparent magnitude comparison is a very useful tool. It can be done mathematically using the distance modulus formula, but to get just a rough estimate, this comparison method works very well and stresses the student’s reasoning skills rather than math. If a star’s absolute and apparent magnitudes are equal, then the star must be at 10 pc. If the apparent magnitude is smaller (brighter) than the absolute magnitude, then the

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star must be closer than 10 pc. If the apparent magnitude is larger (dimmer) than the absolute magnitude, then the star must be farther than 10 pc, and the greater the difference, the greater the distance. Star

The stars of Hercules were chosen to complement the H-R diagram exercise in which they were used previously and because they represent a good range of brightness, luminosities, and distances.

Apparent Magnitude

Absolute Magnitude

Brightness Ranking

Luminosity Ranking

Epsilon Herculis

3.90

0.41

11 (tie)

Eta Herculis

3.46

0.78

Gamma Herculis

3.71

–0.18

Iota Herculis

3.81

–2.11

9 (tie)

Kornephoros

2.75

–0.53

1 (tie)

Lambda Herculis

4.40

–0.87

Mu Herculis

3.40

3.78

Omicron Herculis

3.81

–1.34

9 (tie)

5 (tie)

Phi Herculis

4.21

–0.02

Pi Herculis

3.15

–2.12

Rasalgethi

2.75

–2.60

1 (tie)

Rho Herculis

4.12

–1.34

5 (tie)

Sarin

3.09

1.18

Sigma Herculis

4.16

–0.66

Tau Herculis

3.90

–1.03

11 (tie)

Theta Herculis

3.84

–2.73

Xi Herculis

3.68

0.59

Zeta Herculis

2.87

unknown

ANSWER KEY Activity 1: Apparent and Absolute Magnitudes of Stars in Hercules 1. Apparent and absolute magnitudes of the stars. Activity 2: Using the Magnitude Scale 2–3. Brightness and luminosity rankings are given in the previous table. 4. The brightest stars are not necessarily the most luminous. The stars may be bright because they are closer. Activity 3: Comparing Distances Using Magnitudes 5. There are three pairs of stars with the same apparent magnitude: Rasalgethi and Kornephoros; Iota and Omicron; and Tau and Epsilon. For each pair, the star with the smaller absolute magnitude is the one that is farther away (Rasalgethi, Iota, and Tau). A star with a smaller absolute magnitude is a more luminous star. For a more luminous star to look the same brightness

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as a less luminous star, the more luminous star must be farther away. 6. Omicron and Rho have the same absolute magnitude, which means they have the same luminosity. If they were side by side at the same distance, they would appear the same brightness. Because Rho has a larger apparent magnitude (it looks dimmer), it must be farther away. 7. The brightest star is not necessarily the closest. Brightness depends on both distance and luminosity. A very bright star could be far away but very luminous. Remember that two stars are tied for the greatest brightness. They cannot be the same distance away because they have different absolute magnitudes. 8. Absolute magnitude is a measure of how bright a star would look if it were 10 pc (32 light-years) away. If a star’s apparent magnitude is less than (or brighter than) its absolute magnitude, this means it is closer than 10 pc. Mu Herculis is the only star in the list that has a smaller apparent magnitude than absolute magnitude, so it must be the closest. The rest of the stars all have apparent magnitudes greater than their absolute magnitudes, which means they are all farther away than 10 pc. The greater the difference in these numbers, the greater the distance

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196 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers of the star. Theta Herculis has the greatest difference and is therefore the farthest away.

EXERCISE 15: STARS AND THE H-R DIAGRAM In this exercise, students gather data on luminosity and temperature for two groups of stars and transfer those data to an H-R diagram. The first group consists of bright stars selected from the constellation Hercules. They will learn that most of the bright stars are very luminous and are typically red giants and supergiants. The second group consists of stars that are located near the Sun. Most of these stars are main-sequence stars that are cooler and less luminous than the Sun. This exercise reinforces the concepts presented in Figure 10.19 of the text. Selection is an important idea here. The result of an experiment can depend on how the data are gathered. Picking the bright stars in the sky is not the best way to learn about typical stars: the stars are bright either because they are luminous or because they are close to the Sun. To determine which stars are the most common, we have to sample all stars in the local volume of space and assume that the results from that sample are typical.

Activity 2: Luminosities and Temperatures of Nearby Stars 2. Data for nearby stars: Star

Luminosity

Temperature (K)

Sun

5,770

HIP71683

1.8

5,715

HIP71681

0.61

5,075

HIP54035

0.03

3,483

HIP32349

8,858

HIP16537

0.38

5,136

HIP37279

7.5

6,800

HIP108870

0.26

4,604

HIP439

0.03

3,572

Activity 3: The H-R Diagram 3. Bright stars in Hercules are shown as filled points and nearby stars are shown as open points. The solid line is the main sequence. 10,000

ANSWER KEY Activity 1: Luminosities and Temperatures of Bright Stars in Hercules

1,000

1. Data for Hercules: Luminosity

Temperature (K)

Epsilon Herculis

9,008

Eta Herculis

5,025

Gamma Herculis

103

7,389

2,015

15,551

Kornephoros

200

4,928

Lambda Herculis

883

3,635

5,574

Omicron Herculis

352

9,008

Phi Herculis

109

9,262

Pi Herculis

2,823

3,628

Rasalgethi

2,022

4,303

Rho Herculis

350

8,969

Sarin

8,473

Sigma Herculis

188

8,980

Tau Herculis

566

13,789

Theta Herculis

3,741

3,831

4,965

unknown

5,701

Iota Herculis

Mu Herculis

Xi Herculis Zeta Herculis

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100 Luminosity (Suns)

Star

.01 15,000

12,000

9,000

6,000

3,000

Temperature (Kelvin)

4. Most bright stars are located away from the main sequence and are giants or supergiants. The nearby stars are mostly located near the main sequence.

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5. The nearby stars are more representative of “average” stars. The brighter stars are mostly very luminous stars that are rarer, as can be seen by the fact that there are none near the Sun. Cool, low-luminosity stars are very common, as can be seen by the fact that the Sun is surrounded by them.

EXERCISE 16: NEBULAE: THE BIRTH AND DEATH OF STARS The birth and death of stars both take place in nebulae. In this exercise, students are investigating the differences in the appearance, shape, and size of two types of nebulae. These nebulae often provide some of the most beautiful images in astronomy. By investigating these images more closely, students will be looking for patterns that will help them tell these types of nebulae apart in the future.

ANSWER KEY Activity 1: Emission Nebulae 1. Emission nebulae generally look pink because they are all composed of mainly the same gas, hydrogen. 2. Emission nebulae generally encompass many stars. 3. Emission nebulae are mostly found along the plane of the Milky Way. Activity 2: Planetary Nebulae 4. Planetary nebulae are roughly round or double-lobed in shape. 5. Each planetary nebula contains one star at the center. 6. Planetary nebulae are smaller than emission nebulae. You must zoom in further to see the shape, and they only encompass one star, whereas emission nebulae encompass many. 7. Planetary nebulae have a much wider variety of colors because they are composed of a larger fraction of elements other than hydrogen. 8. At the low magnification of early telescopes, planetary nebulae typically looked like small fuzzy circles, which is similar to how planets appeared. They were named because of this similarity, even though they were later found to have nothing to do with planets.

EXERCISE 17: PULSARS AND SUPERNOVA REMNANTS Massive stars die violent deaths. This exercise investigates the remnants these supernova explosions leave behind. The supernova identification portion may be challenging

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because of the nature of the images used. All Type II (core collapse) supernovae should have a neutron star or black hole remaining in the center, but sometimes these are hard to detect. If a pulsar is created, then it may not even be detected if the poles do not point near the direction of Earth. If the star is hard to detect, it may be mistaken for a Type Ia (white dwarf explosion) supernova, in which the central star is completely destroyed. The determination is done more accurately by looking at the spectrum of the supernova, but this activity gives a basic idea of what astronomers are looking for.

ANSWER KEY Activity 1: Pulsars 1. Most pulsars are near the cloud of the Milky Way and in the general direction of the center of the galaxy. This cloud represents where most of the stars of the galaxy are distributed. As pulsars are the remnants of dead stars, they should be found where the concentration of stars is located. 2. The constellation Aquila appears to have the greatest amount of pulsars. This constellation is just to one side of the center of the galaxy. When we look in this direction, we are looking into the largest concentration of stars in our galaxy. 3. The period of the Crab Nebula pulsar is 0.0334 seconds. This works out to about 30 times per second. Activity 2: Supernova Remnants 4. Table of supernova remnants and justifications: Supernova Remnant

Type Ia or II

Crab Nebula

Type II

There is a pulsar remaining in the center

Tycho

Type Ia

No central star

Cassiopeia A

Type II

Star faintly visible in the center

Vela Pulsar

Type II

Pulsar in the center

B1509-58

Type II

The brightness at the center is a pulsar

Cygnus Loop

Type Ia

No central star

Justification

Cassiopeia A is a Type II, but students will likely classify it as a Type Ia, as the neutron star remaining in the center is very dim and not actually shown in the program. B1509-58 is a Type II as well. The bright center of the image is caused by a young, energetic pulsar, but this pulsar is not marked in Starry Night.

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198 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 5. The Cygnus Loop is much larger in diameter than the other remnants. This means it has been expanding for much longer and is older than the other remnants. 6. Not all supernova remnants have a star remaining in their centers. Type Ia supernovae are the result of a white dwarf destroying itself, leaving only the debris cloud behind. 7. Not all pulsars have supernova remnants surrounding them. All pulsars originally must have been surrounded by a debris cloud, but this cloud expands over time and will eventually dissipate into space as it ages, leaving the pulsar behind.

EXERCISE 18: GALAXY CLASSIFICATION Classifying galaxies within the elliptical or spiral sequence is somewhat subjective depending on how a student interprets the shape. There is no set division between a spiral Sb and Sc for example, so there could be a wide variety of answers to some of the galaxy classifications. Students should be encouraged to make use of the Hubble tuning fork diagram in Figure 14.2 to aid in their classifications. The answer table provided for Activity 4 gives the most commonly accepted classifications, which in some cases like M64 (Sa/b) indicate that there is not necessarily agreement even among professional astronomers.

ANSWER KEY Activity 1: Spiral Galaxies 1. M101 looks spiral in shape, with dusty arms spiraling into a more compact, brighter, yellowish-orange center. 2. The stars in the two different areas are of different temperatures. The light of hotter, bluer stars dominates the arms, whereas cooler, orange to red stars dominate the central area. 3. M31 appears to be the same basic type of galaxy, but viewed at more of an angle. 4. The view of NGC 4565 shows a spiral galaxy completely on edge. In this view, the arms are shown to be very thin compared to their diameter. 5. This question is a bit subjective, but anything less than one-tenth should be acceptable. Activity 2: Elliptical Galaxies 6. Elliptical galaxies are generally round in shape, without any other distinguishing features. 7. This galaxy appears more yellowish-orange, which means that cooler stars are dominating the light.

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Activity 3: Irregular Galaxies 8. The LMC has no similarity in shape but does appear mostly bluish, similar to the arms of a spiral galaxy. Activity 4: Galaxy Classification 9. Table of galaxies in the Messier catalog: Galaxy

Galaxy Type

LMC

Irr

M87

M101

M32

M51

Sb/c

M59

M60

M63

SBb/c

M64

Sa/b

M65

Sa/b

M74

M82

Irr or Sa

Galaxy

Galaxy Type

M85

M90

M104

M105

SMC

Irr

10. We see our galaxy as a cloud surrounding us on all sides, with a slightly greater density in one particular direction. This is what would be expected if you were inside of a disk of stars that had a more concentrated center, like a spiral galaxy.

EXERCISE 19: QUASARS AND ACTIVE GALAXIES In this exercise, students will use skills similar to the magnitudes exercise to estimate relative distance using apparent and absolute magnitudes. Quasars are some of the most luminous objects in the universe, yet none of them are apparently very bright. This tells us that the distances to all quasars must be huge. As all quasars are very far away, this also means that they existed earlier in the history of the universe. No quasars and few active galactic nuclei are seen to exist near the Milky Way. At one time, the universe must have been a very active place, perhaps due to more galaxy interactions.

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ANSWER KEY Activity 1: Quasars 1. Quasars appear to lie mainly above and below the dusty plane of the Milky Way. This implies that the quasars are outside of our galaxy. 2. As there are several quasars for the students to choose from, there will be a variety of different answers. Apparent magnitudes range from 14.1 to 15.11 and absolute magnitudes range from –23.5 to –29.6. The averages for these magnitudes will fall somewhere inside these ranges. Redshifts are mostly around 0.1, with one at 1.178. 3. If these quasars were located at 10 pc from our Solar System, they would look roughly as bright as the Sun does in our daytime sky (some would be a little dimmer; two would be even brighter). 4. These quasars have a very large difference between their apparent and absolute magnitudes. Their absolute magnitudes are all large negative numbers, which means they are highly luminous. The apparent magnitudes are all larger positive numbers, which means they are very dim. For a highly luminous object to appear very dim, it must be very far away. The difference between the absolute and apparent magnitudes is much larger than the example given for M13, so the quasars must be much farther away than 25,000 light-years. Activity 2: Active Galaxies 5. As in question 2, the answers for the active galaxy information may vary among students. Apparent magnitudes will fall somewhere between 11 and 15, and absolute magnitudes will range from –17 to –21, with averages for both falling somewhere inside these ranges. Redshifts will be between 0.001 and 0.01. 6. Active galaxies are less intrinsically bright than quasars. 7. The difference between the apparent and absolute magnitudes for active galaxies is large, but smaller than it is for quasars; therefore, active galaxies must be closer than quasars. 8. Quasars generally have greater redshifts. According to Hubble’s law, the greater the redshift, the greater the distance. This agrees with the above estimate of quasars being more distant than active galaxies, which have smaller redshifts.

EXERCISE 20: VIEWS OF THE MILKY WAY By viewing the Milky Way in all possible types of radiation, we learn much more about its true structure. Visible light is an especially poor choice for viewing the galaxy because it is easily blocked by the large amounts of dust that exist within the galactic disk. Other wavelengths of light, such as infrared, easily penetrate this dust to reveal much more of

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the galaxy. This is especially helpful when probing the nature of the very center of the galaxy, which is hidden behind copious amounts of dust, gas, and other stars.

ANSWER KEY Activity 1: The Milky Way in Visible Light 1. The Milky Way appears to be a diffuse cloudy band that encircles the Solar System. It appears brighter and denser in one particular direction. 2. The fact that the cloud appears to be confined to one common plane means that the overall shape must be flat in one dimension and is likely circular like our Solar System. That this surrounds us on all sides means that we are in this plane as well. The direction that looks brighter is likely the center of this shape, meaning we are not at the center. 3. The center of the galaxy appears to be in the direction of Sagittarius/Scorpius, as this direction is the brightest and densest. 4. The plane of the Milky Way is at a large angle to the plane of the Solar System. Activity 2: The Center of the Milky Way in Other Wavelengths 5. IR radiation is brightest toward Sagittarius/Scorpius. This matches the earlier observation in visible light. 6. Gamma, X-rays (diffuse), near-IR, molecular hydrogen, radio (408 MHz). 7. Visible light (too many clouds), atomic hydrogen. 8. The conditions at the center of our galaxy are likely very intense because of the high-energy radiation we see from that direction. But it is also cloudier in that direction, which blocks some types of radiation (such as visible light). Activity 3: Other Parts of the Milky Way 9. Some possible X-ray sources are the Crab Nebula (Taurus), Cassiopeia A (Cassiopeia), Tycho (Cassiopeia), Vela Pulsar (Vela), and Eta Carina (Carina). 10. Most of these objects also emit considerable amounts of gamma rays and radio (408 MHz). This tells us that these objects must be intensely energetic and hot. 11. Supernova remnants are concentrated along the plane of the Milky Way, mostly in the direction of the center of the galaxy, in the region of most intense gamma-ray emission. Because the density of stars is greater in the direction of the galactic center, supernovae are more likely to happen in this direction, so we should see both the high-energy radiation from them as well as the remnants they leave behind.

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EXERCISE 21: GLOBULAR CLUSTERS

Activity 3: The Distance to Galactic Center

This exercise re-creates how Harlow Shapley determined the location of the center of our galaxy in 1915. The globular clusters used to make the plot were chosen to give an equal number above and below the galactic plane. Also, the longitudes of the chosen clusters are within roughly 10° of the longitude of the accepted galactic center. While this is not a truly random assortment, the point is still made. A brief explanation of how to plot points on the polar graph will likely be needed, although two clusters are plotted for the students on their answer sheet. There will be some variation in location of points and where students choose the center to be, but the results should be fairly consistent.

5. Plot of globular cluster locations relative to the Sun. 6. The apparent center should be roughly between 25,000 and 30,000 light-years from the Sun. 7. The distance given in the textbook is 27,000 light-years. 75° 60°

45°

30°

ANSWER KEY Activity 1: The Distribution of Globular Clusters 1. Globular clusters are mostly located near the plane of the Milky Way, with almost all concentrated in the direction of the apparent center of the galaxy. 2. Globular clusters are most numerous around the constellations Sagittarius and Scorpius. 3. The Milky Way appears larger and brighter in this direction.

15°

0°

Sun

−15°

Activity 2: The Distances to Globular Clusters 4. Table of globular cluster data:

−30°

Cluster

Distance (ly)

Latitude (°)

NGC 7099

26,000

–46

NGC 5897

41,000

NGC 5904

24,000

NGC 6093

32,000

NGC 6121

7,000

NGC 6144

33,000

NGC 6171

20,000

NGC 6235

32,000

NGC 6284

47,000

NGC 6316

35,000

NGC 6333

26,000

NGC 6453

36,000

–3

NGC 6522

25,000

–3

NGC 6544

8,000

–2

NGC 6652

31,000

–11

NGC 6681

29,000

–12

NGC 6723

28,000

–17

NGC 6809

17,000

–23

NGC 6864

61,000

–25

26_IM_93899_AK_183–202.indd 200

−45°

−60° −75°

EXERCISE 22: THE NEIGHBORHOOD OF THE SUN One of the most fruitful ways of finding stars near the Sun is to look for stars with a large proper motion. The proper (or tangential) motion is the rate of change of the position of a star. Stars can have a large proper motion if they are moving rapidly through space or if they are located close to the Sun. The text emphasizes the importance of high-velocity stars as tracers of the Milky Way’s halo. On average, halo stars have high velocities relative to the Sun. Samples of high-velocity stars pick out many halo stars, but they also pick out a few nearby stars in the Milky Way’s disk. The examples in this exercise are all disk stars.

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The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 201

ANSWER KEY

Activity 1: Finding Nearby Stars

Activity 1: Stepping to the Nearest Stars

1. Rigil Kentaurus is a nearby star because it has a large motion. The rate of motion is given by v/D, where D is the distance and v the tangential velocity. 2. Rigil Kentaurus is about 4 light-years from our Sun. 3. The answer will depend on what stars were picked. The students will generally find that apparently stationary stars will have distances that are much larger than 4 light-years, the distance to Rigil Kentaurus.

1. The current location is 500 AU from Earth. Even the nearest stars are (300,000 AU)/(500 AU) = 600 times farther away. 2. At this distance, Neptune’s orbit is visible, but small. Earth’s orbit is too close to the Sun to be visible unless we zoom in to high magnification. 3. If the students notice any stars moving, they will pick ones with distances less than about 8 light-years.

Activity 2: The Fastest-Moving Star

Activity 2: Leaving the Milky Way

4. Barnard’s star is currently in the constellation Ophiuchus. 5. Barnard’s star will be in Hercules about 3,400 years from now. 6–7. Barnard’s star is 5.9 light-years away from the Sun. Students might reasonably conclude that Barnard’s star is closer than Rigil Kentaurus because it has a higher tangential motion. 8. Tangential velocity is the other factor besides distance that determines a star’s apparent motion. Barnard’s star has a higher tangential velocity than Rigil Kentaurus.

4. The Milky Way is a spiral galaxy. It has a central bulge surrounded by a flat disk containing spiral arms. 5. There are several possibilities other than the example Fornax. Others that might be sighted include the Ursa Minor dwarf galaxy, the Sagittarius dwarf elliptical galaxy, Sextans, UGC, and possibly the Large Magellanic Cloud and the Small Magellanic Cloud.

EXERCISE 23: BEYOND THE MILKY WAY Here, students leave the Solar System and head out into deep space. In imitation of the movie Powers of Ten, the distance will jump by factors of 100 at each step. On occasion, students will pause to look around. It is worth stressing just how fast distances increase with geometric jumps. Only a few jumps take us from the small to the large. Intervals of a factor of 100 correspond to half the width of a fingertip, one ordinary step, the length of a football field, a 10-km footrace, a three-hour car trip, and a third of the distance to the Moon. At the end of the exercise, the viewpoint is located 200 Mly from Earth. As large as this distance seems, it is still only about 1.5 percent of the way to the most distant objects in the universe.

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Activity 3: Distant Galaxies 6. The galaxies are not uniformly distributed. There are many dense clusters and several areas with fewer galaxies. Students may notice the zone of avoidance along the plane of the Milky Way. There are undoubtedly many galaxies there, but the catalogs do not include them because their light is blocked by dust in the plane of the Milky Way. 7. We assume that the galaxies are distributed homogeneously and isotropically; that is, they are uniform in all places and in all directions. If this assumption is true, it applies only on much larger-distance scales than we can see in Starry Night. 8. The recession velocity is given by v = Hd, where d is in units of Mly. For d = 200 Mly, the velocity is 4,400 km/s.

09/02/16 10:56 am