Explorations Introduction to Astronomy 8th Edition Arny Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/explorations-introduction-to-astronomy-8th-edition-arny-solutions-manual/
Chapter 7
The Moon
CHAPTER 7 THE MOON Answers to Thought Questions 1. A newly paved road has no potholes. A very old road that hasn’t been paved for a long time has many. Similarly, the very old lunar surface is heavily cratered, but regions that have been covered with lava flows more recently (a sort of cosmic ―re-paving‖) have many fewer craters. 2. Bergmann’s rule suggests that larger animals are better adapted to existing in the cold than smaller ones—we might expect this means they are less likely to freeze to death—so therefore the larger animals must stay warm more easily. Larger animals have a larger volume, and a larger ratio of volume to surface area, so they lose heat less quickly. This is the same idea as is used in the chapter to explain the difference in temperature between the Earth’s interior and the Moon’s interior. 3. Since the Earth and the Moon were both hot and melted at the time the material that makes up the Moon was splashed into orbit, if comets brought water to the Earth it makes sense it would have to happen after the formation of the Moon. Therefore the Moon should also have been bombarded with comets and have had significant water deposited on it, which does not seem to be the case. A counter-argument: the lack of water on the Moon could also be explained by the fact that it would quickly evaporate in the low or zero pressure atmosphere, and then the Moon’s low gravity would be unable to hold onto the water molecules, which would be lost to space fairly quickly. Evidence of water in some deep, dark craters adds weight to this possibility. 4. On Earth, wind, rain, etc. quickly erode footprints. On the Moon, no such processes occur. Thus, footprints last until obliterated by meteor impacts - a very, very slow process. 5. If the Moon were not in synchronous rotation, we would still observe the same phases, but the particular parts of the lunar surface lit up or in shadow would vary (we’d see what is now the ―far side‖ sometimes). If the Earth and Moon were both in synchronous rotation, the phases would look much the same as they do today, but would only be visible from the half of the Earth that could see the Moon. In synchronous rotation, the same half of the Earth would always face the Moon. (The period of the phases would also be different because for synchronous rotation the distance to the Moon would not be the same as what it is now). 6. Students should make a reasoned argument for future missions. Among other results, missions to the moon have allowed us to (1) place (retro) reflectors on the surface, to make exact measurements of the distance; (2) place seismic detectors on the surface, to learn about Moonquakes and the Moon’s internal structure to a greater detail than remote observing permits; (3) orbiting craft provide detailed gravitational data to map the Moon’s internal density; (4) the return of lunar rock samples provided critical and specific evidence about how the Moon’s crust is similar and different to the Earth’s, which strongly affected our theories of the origin of the Moon (sec. 7.4); (5) missions allowed us to photograph the far side of the Moon, revealing fewer maria and in combination with other results allowing us to determine the offset of the Moon’s center from the crust (fig. 7.8); (6) probes in 2009
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Chapter 7
The Moon
proved the presence of ice in shadowed craters near the Moon’s poles. This data has significantly constrained models of the Moon’s origin, composition, and structure. 7. If the day were 12 hours long, there would still be 2 tidal bulges and so the time between the high and low tides would be halved. Currently, as shown in Figure 7.19, if there is a low tide at 6 a.m., there is a high tide at noon. With the rotation period of the Earth halved, low and high tides would be approximately 3 hours apart instead of approximately 6 hours apart. 8. As the Moon recedes from the Earth, its gravitational impact is lessened and the tides will be shorter (think about the solar tides—the Sun is much more massive but so far away the tides are smaller than the lunar ones). If the Moon were twice as far from the Earth, there would still be two tides each day (assuming the Earth’s rotational period is about the same). 9. The tides occur about an hour later each day for the same reason that the Moon rises about an hour later each day: the Moon is moving in its orbit. It takes the Earth about an hour to catch up to this motion. Since the what tide it is depends on the orientation of the Earth and Moon (and where you are), it’s actually a little more than 6 hours between each tide, closer to 6 hours and 13 minutes, and closer to an hour more for the same tide the next day. Answers to Problems 1. Moon’s mass to the Earth’s mass: just more than 1% mMoon 734.9 10 20 kg 73.49 10 21 73 0 . 012 10 3 12 10 3 0.012 m Earth 6 5.97 10 24 kg 5.97 10 24 Moon’s radius to the Earth’s radius: a bit more than a 25%: rMoon 1.738 103 km 174 101 174 0.27 10 1 2.7 10 1 0.27 3 2 rEarth 6.378 10 km 64 64 10 2. If the Mare Serentitatis has an angular diameter of 5 arc minutes, its linear diameter can be calculated according to the formula D/2d = A/360°, which relates angular diameter (A), true diameter (D), and distance (d) . Thus, D = 2dA/360°. To use the formula, we must express A in degrees, not arc minutes. We do this by recalling that 1 degree contains 60 arc minutes. Inserting this value and the Moon’s distance in the formula, we get: D = 2dA/360° = 2(384,000 km)(5' )(1°/60')/360° = 558 km. Note: the angular diameter has been rounded slightly, and the mare is not exactly circular, so the value may differ slightly from literature values. 3. The crater Tycho is 88 km wide. D = 2dA/360°, so
2 © 2017 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 7
The Moon
A = (360°)(D)/(2d) = (360°) (88 km)/( 2×384,000 km) = 0.01313°. and 0.01313° × 60'/1° = 0.78’, or 0.78 arc minutes, or 47 arc seconds. Theoretically, the human eye can manage a resolution of about 20 arc seconds, or 1/3 of an arc minute (see Chapter 5 problems). At least some people should be able to see this crater with the naked eye, under ideal conditions, and possibly with a filter if the Moon is very bright. 4. To calculate the Moon’s density (), divide its mass, M = 7.349 × 1020 kg = 7.349 × 1023 g, by its volume. Assuming it is a sphere, its volume is 4r3/3. Thus, 4r3/3). Now insert the values of M and r = 1738 km = 1738 × 105 cm = 1.7 × 108 cm to find, = 7.3 × 1025 g/[4(1.7 × 108 cm)3/3] = {7.3/[(4 × (1.7)3]} × 1025 g / (108 cm)3 = 0.35 × 1025-(8×3) g/cm3 = 0.35 × 10 g/cm3 = 3.5 g/cm3. The density of iron is 7.9 g/cm3. The very low value of the Moon’s density suggests there is a much smaller fraction of iron in the Moon than in the Earth (average density of 5.5 g/cm3). 5. If the Moon were made of incompressible Swiss, it’s average density would equal the density of Swiss cheese, or 1.1 g/cm3. Since /V, then M = V. So, M = V = 4r3/3 = (1.1 g/cm3) × 4(1.7 × 108 cm)3/3 = 2.4 × 1025 g = 2.4 × 1022 kg This is of course about one-third of the Moon’s actual mass (7.3 × 1025 g). An equally valid and simpler solution to this problem is to notice that the ratio of the density of Swiss cheese to the actual density must be the same as the ratio of the Moon’s mass if made of cheese to the Moon’s actual mass and calculate accordingly: Mcheese = (1.1 g/cm3/ 3.5 g/cm3) Mreal = (1.1/3.5) 7.3 × 1025 g = 2.3 × 1025 g. (The slight difference here is a result of rounding). 6. The Lunar Reconnaissance Orbiter orbits the Moon 50 km above the surface with an orbital period of 113 minutes. Assuming a circular orbit, the modified form of Kepler’s 2 3 2 Third Law (see Chapter 3) gives us m + M = 4 d /GP , with M = mass of Moon, m = mass -11 3 2 of spacecraft, d = radius of orbit, and G = 6.67 × 10 m /(kg sec ). Expressing d in meters and P in seconds so that units will cancel, P = 113 × 60 = 6780 sec d = 1738 + 50 km = 1788 km Inserting these values in the law gives 3 © 2017 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 7
The Moon
2
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m + M = 4 × (1788 × 10 m) /( (6780s) × 6.67 × 10 m /(kg s ) ) 22 = 7.36 × 10 kg. However, the mass of the spacecraft, m, is so tiny compared to the Moon’s mass that we can ignore it. Thus, the measurement of the Moon’s 22
mass is 7.36 × 10 kg (very close to the value in the Appendix). 7. Round trip light-travel time to the Moon and back is 2.56 seconds. The speed of light, c, 5 is 3 × 10 km/s. Using d = vt, with t = 2.56 sec / 2 for the time to the Moon, 5 D = ct = 3 × 10 km/s × 2.56/2 s = 384,000 km. Conversations will be clumsy because of the extra 2.56s between when you finish asking a question and when you hear the reply. 8. The surface area of a sphere is 4r2, and the volume is 4/3r3. The ratio of the surface area to the volume is therefore 4r 2 3 . 3 4 r 3 r So for the Moon, 3/r = 3/1738 km = 0.0017 km-1. For the Earth, 3/r = 3/6378 km = 0.00047 km-1. The Moon’s SA to V ratio is 3.6 times larger than the Earth’s – naively we might estimate the Moon would cool to the same temperature 3.6 times faster than the Earth. 9. The rate at which the length of the day increases is 0.002 s/century. To find out when the Earth’s day was 5 hours long, we would need to consider how long it would take for the day to lengthen by 19 hours, or 19 × 60 min/hr × 60 s/min = 68,400s. 68,400s = 0.002s/century × time passed, so time passed = 68,400s / (0.002 s/century) = 3.42 × 107 centuries = 3.42 × 109 years = 3.4 billion years Answers to Self-Test 1. (d) The maria were formed after the highlands (and after the bombardment of the highlands). 2. (b) The Moon’s weak gravity makes it hard to hold onto an atmosphere (atoms and molecules are sufficiently heated to move fast enough to escape). 3. (a) We would observe both sides. 4. (c) Its mantle is cold and rigid. 5. (d) High tide to low tide is about 6 hours (high to high is 12 hours, half a day). 4 © 2017 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 7
The Moon
6. (a) When it is high tide locally, the Moon is pulling you ―up‖ the most.
5 © 2017 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.