APSI 2018 AP Physics 1: Chapter 3.2

3.2 Gravity and Kepler’s Solar System Warm Up If you drop a piece of paper and a book from the same height, which object will hit the ground first? If you put the piece of paper on top of the book and drop them together, why does the paper fall at the same rate as the book? What can you conclude about the rate at which objects fall? ____________________________________________________________________________________________ ____________________________________________________________________________________________ ____________________________________________________________________________________________

Falling Objects

If gravity is the only force acting on a body, the body is in free fall. If there is no friction, each of the bodies in Figure 3.2.1 is in free fall. Each is accelerating downward at rate  g , which is independent of the mass of the body and has a magnitude of 9.80 m/s2 near Earth’s surface.

a=g

a=g

a=g (a)

(b)

(c)

Figure 3.2.1 In all three situations, the only force causing the acceleration of the free-falling body is gravity.

“What goes up must come down.” This simple truth has been known for centuries.

Any unsupported object will fall to the ground. According to legend, Isaac Newton (1642–1727) was sitting under an apple tree when he saw an apple fall to the ground. He looked up at the Moon and wondered, “Why should the Moon not fall down, as well?”

Might the Moon be in “free fall”? If it is, then why does it not fall to Earth like other The Moon in Free Fall unsupported bodies? Newton created a diagram like the one in Figure 3.2.2 to explain why the Moon circles Earth without “falling down” in the usual sense.

Imagine you are at the top of a mountain that is high enough so that there is essentially no air to offer resistance to the motion of a body projected horizontally from the top of the mountain. If a cannon is loaded with a small amount of gunpowder, a cannonball will be projected horizontally at low speed and follow a curved path until it strikes the ground at A. If more gunpowder is used, a greater initial speed will produce a curved path ending at B.

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A B

Planet Earth

C D

Figure 3.2.2 Newton’s diagram explaining

why the Moon doesn’t fall to Earth.

With increasing initial speeds, paths ending at C and D will be achieved. If the initial speed is just high enough, the path will have a curvature parallel with Earth’s curvature, and the cannonball will circle Earth for an indefinite period. The cannonball will orbit Earth, just like the Moon.

Isaac Newton actually anticipated artificial Earth satellites. Of course, the technology for sending a satellite into orbit was not available 300 years ago. To place an artificial satellite in orbit 500 km above Earth’s surface requires a horizontal speed of approximately 7.6 km/s. It will take such a satellite approximately 90 minutes to complete one orbit.

The Moon was our first satellite. No one knows how or why it attained its orbit, but it is in a nearly circular path around this planet. The Moon is, indeed, in free fall. It does “fall toward Earth.” Its horizontal speed is great enough that it completes its orbit with no danger of colliding with Earth’s surface  just like the cannonball in Figure 3.2.2 but at a much greater altitude. The Moon’s mean distance from Earth is 3.84 × 105 km. The orbit is actually slightly elliptical; hence mean (average) distance.

Could the centripetal force needed to keep the Moon in a near-circular orbit around Earth Gravitational Force, be the same force that makes an apple fall from a tree? What would exert this force of Earth, and the Moon gravity on the Moon?

Since every known thing on Earth experiences the pull of gravity, Newton was certain that Earth itself exerted a gravitational force on objects near its surface. At Earth’s surface, gravity makes objects accelerate at a rate of 9.8 m/s2. What would the acceleration due to gravity be at a distance as far away as the Moon?

Treating the Moon as a body in circular orbit, the acceleration due to gravity is just the centripetal acceleration of the Moon. Therefore, ac =

4 2 R 4 2 (3.84 10 8 m) = = 2.7 10 –3 m s 2 T2 [(27.3 d)(24 h d)(3600 s h)]2

If this value of ac was really the magnitude of g at the Moon’s distance from Earth, it is much smaller than the value of g at Earth’s surface. Newton was convinced that, since F = ma, the force of gravity causing this acceleration must decrease rapidly with distance from Earth.

In Investigation 3.2, you will be provided with data about most of the planets in the solar system. Given their orbital radii and their periods of revolution, you will calculate their centripetal accelerations toward the Sun. You will then use graphical analysis to determine the nature of the relationship between ac, caused by the force of gravity of the Sun on the planet, and distance R from the Sun.

Johannes Kepler

Johannes Kepler (1571–1630) was a German mathematician who had worked in the astronomical laboratory of Danish astronomer Tycho Brahe (1546–1601). Brahe is famous

for his precise observations of the positions of almost 800 stars and his accurate records of the positions of planets over a period of two decades. Brahe did all his work without a telescope! The telescope had not yet been invented.

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Kepler was fascinated with planetary motions, and devoted his life to a search for mathematical patterns in their motions. To find these patterns he relied completely on the observations of Tycho Brahe. Kepler formulated three laws describing the orbits of the planets around the Sun. 1. Each planet orbits the Sun in an elliptical path, with the Sun at one of the two foci of the ellipse.

Kepler’s Three Laws of Planetary Motion 2. A line joining the centre of the Sun and the centre of any planet will trace out equal areas in equal intervals of time (Figure 3.2.3). If time intervals T2–T1 and T4–T3 are equal, the areas traced out by the orbital radius of a planet during these intervals will be equal. (The diagram is not to scale.)

T3

Sun

F1

T2

F2

T1

T4 Figure 3.2.3 Kepler’s second law

3. For any planet in the solar system, the cube of its mean orbital radius divided by the square of its period of revolution is a constant.

R3 =K T2

Kepler’s three laws can be neatly summarized in half a page of this text, but remember that to arrive at these laws required many years of work by this dedicated mathematician, not to mention the 20 years spent by Tycho Brahe making his painstaking observations of the heavenly wanderers, the planets.

In Investigation 3.3, you will calculate Kepler’s constant K for the solar system.

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Sample Problem 3.2.1 — Kepler’s Laws Planet Xerox (a close copy of planet Earth) was discovered by Superman on one of his excursions to the far extremes of the solar system. Its distance from the Sun is 1.50 × 1013 m. How long will this planet take to orbit the Sun?

What to Think About

How to Do It

1. Collect data needed to answer the question.

Earth’s orbital radius, Re = 1.5 x 1011 m Earth’s period of revolution, Te = 1.0 a Xerox’s orbital radius, Rx = 1.50 x 1013 m

2. You could solve the problem using Kepler’s constant derived in Investigation 4.2.2. Using Kepler’s third law, you would simply solve for Tx. Another way to solve the problem is to use data for Earth (above) and the fact that RX3 Re3 = = K Sun TX2 Te2

T x2 =

3. The planet Xerox has a period of revolution around the Sun of 1.0 × 103 a.

R x3 2 • Te R e3

[ 1.5 10 13 m ]3 T = •[ 1.0a ]2 11 3 [ 1.5 10 m ] 2 T x = 1.0 106 a 2 2 x

T x = 1.0 10 3 a

Practice Problems 3.2.1 — Kepler’s Laws 1. A certain asteroid has a mean orbital radius of 5.0 × 1011 m. What is its period of revolution around the Sun?

2. A satellite is placed in orbit around Earth with an orbital radius of 2.0 × 107 m. What is its period of revolution?

Use the facts that the Moon’s period of revolution is 2.36 × 106 s and its orbital radius is 3.84 × 108 m.

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