CHAPTER 7 Gravitation

CHAPTER

7

Gravitation

Practice Problems

7.1 Planetary Motion and Gravitation pages 171?178

page 174 1. If Ganymede, one of Jupiter's moons, has a period of 32 days, how many units are there in its orbital radius? Use the information given in Example Problem 1.

TTGI 2 rrGI 3

rG 3 (4.2 units)3 13.28 dd aayyss 2

3 2 3.41 03 un its3

29 units

2. An asteroid revolves around the Sun with a mean orbital radius twice that of Earth's. Predict the period of the asteroid in Earth years.

TTEa 2 rrEa 3 with ra 2rE

Ta

rrEa 3TE2

2rrEE

3

(1.0

y)2

2.8 y

3. From Table 7-1, on page 173, you can find that, on average, Mars is 1.52 times as far from the Sun as Earth is. Predict the time required for Mars to orbit the Sun in Earth days.

TTME 2 rrME 3 with rM 1.52rE

Thus, TM

rrME 3TE2

1.5rE2 rE

3

(365

days)2

4 .68 105 days2

684 days

4. The Moon has a period of 27.3 days and a mean distance of 3.90105 km from the center of Earth. a. Use Kepler's laws to find the period of a satellite in orbit 6.70103 km from the center of Earth.

TTMs 2 rrMs 3

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Chapter 7 continued

Ts

rrMs 3TM2

63..790011 0035 kkmm

3

(27.3

days)2

3 .781 03 days2

6.15102 days 88.6 min b. How far above Earth's surface is this satellite?

h rs rE 6.70106 m 6.38106 m 3.2105 m 3.2102 km

5. Using the data in the previous problem for the period and radius of revolution of the Moon, predict what the mean distance from Earth's center would be for an artificial satellite that has a period of exactly 1.00 day.

TTMs 2 rrMs 3

rs

3 rM3

TTMs

2

3 (3.90105 km)3

12.70.03 ddaayyss

2

3 7 .961 013 k m3

4.30104 km

Copyright ? Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

Section Review

7.1 Planetary Motion and Gravitation pages 171?178

page 178 6. Neptune's Orbital Period Neptune orbits the Sun with an orbital radius of 4.4951012 m, which allows gases, such as methane, to condense and form an atmosphere, as shown in Figure 7-8. If the mass of the Sun is 1.991030 kg, calculate the period of Neptune's orbit.

T 2 Grm3S

(4.4951012 m)3

2 (6.671 011 N m2/kg2) (1.991 030 kg)

5.20109 s 6.02105 days

Figure 7-8

7. Gravity If Earth began to shrink, but its mass remained the same, what would happen to the value of g on Earth's surface?

The value of g would increase.

142 Solutions Manual

Physics: Principles and Problems

Chapter 7 continued 8. Gravitational Force What is the gravitational force between two 15-kg packages that are 35 cm apart? What fraction is this of the weight of one package? Fg G mrE 2m (6.671 01(10.N35m m2)k2g2)( 15 kg)2

1.2107 N Because the weight is mg 147 N, the gravitational force is 8.21010 or 0.82 parts per billion of the weight.

9. Universal Gravitational Constant Cavendish did his experiment using lead spheres. Suppose he had replaced the lead spheres with copper spheres of equal mass. Would his value of G be the same or different? Explain.

It would be the same, because the same value of G has been used successfully to describe the attraction of bodies having diverse chemical compositions: the Sun (a star), the planets, and satellites.

10. Laws or Theories? Kepler's three statements and Newton's equation for gravitational attraction are called "laws." Were they ever theories? Will they ever become theories?

No. A scientific law is a statement of what has been observed to happen many times. A theory explains scientific results. None of these statements offers explanations for why the motion of planets are as they are or for why gravitational attraction acts as it does.

11. Critical Thinking Picking up a rock requires less effort on the Moon than on Earth. a. How will the weaker gravitational force on the Moon's surface affect the path of the rock if it is thrown horizontally?

Horizontal throwing requires the same effort because the inertial character, F ma, of the rock is involved. The mass of the rock depends only on the amount of matter in the rock, not on its location in the universe. The path would still be a parabola, but it could be much wider because the rock would go farther before it hits the ground, given the smaller acceleration rate and longer time of flight. b. If the thrower accidentally drops the rock on her toe, will it hurt more or less than it would on Earth? Explain.

Assume the rocks would be dropped from the same height on Earth and on the Moon. It will hurt less because the smaller value of g on the Moon means that the rock strikes the toe with a smaller velocity than on Earth.

Copyright ? Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

Physics: Principles and Problems

Solutions Manual 143

Chapter 7 continued

Practice Problems

7.2 Using the Law of Universal of Gravitation pages 179?185

page 181 For the following problems, assume a circular orbit for all calculations.

12. Suppose that the satellite in Example Problem 2 is moved to an orbit that is 24 km larger in radius than its previous orbit. What would its speed be? Is this faster or slower than its previous speed?

r (h 2.40104 m) rE (2.25105 m 2.40104 m) 6.38106 m 6.63106 m

v GmrE

(6.671011 Nm2/kg2)(5.971024 kg) 6.6 3106 m

7.75103 m/s, slower

13. Use Newton's thought experiment on the motion of satellites to solve the following. a. Calculate the speed that a satellite shot from a cannon must have to orbit Earth 150 km above its surface.

v GmrE (6.67(61. 038111N06 mm2/kg12.)5 (5.91705 1m 0)24 kg)

7.8103 m/s b. How long, in seconds and minutes, would it take for the satellite to complete

one orbit and return to the cannon?

T 2 Grm3E 2 (6.67(61. 3081110N6 mm2/kg12.5) (5.19075m1 0)324 kg)

5.3103 s 88 min

Copyright ? Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

14. Use the data for Mercury in Table 7-1 on page 173 to find the following. a. the speed of a satellite that is in orbit 260 km above Mercury's surface

v GmrM

r rM 260 km 2.44106 m 0.26106 m 2.70106 m

v (6.671 011 2N.7 m02/k1g026) m(3.301 023 kg)

2.86103 m/s b. the period of the satellite

144 Solutions Manual

T 2 Grm3M 2 (6.671 011(2N.7 m02/1k0g62)m (3).3301 023 kg)

5.94103 s 1.65 h Physics: Principles and Problems

Chapter 7 continued

Section Review

7.2 Using the Law of Universal of Gravitation pages 179?185

page 185 15. Gravitational Fields The Moon is 3.9105 km from Earth's center and

1.5108 km from the Sun's center. The masses of Earth and the Sun are 6.01024 kg and 2.01030 kg, respectively. a. Find the ratio of the gravitational fields due to Earth and the Sun at the

center of the Moon. Gravitational field due to the Sun: gS G rmSS2 Gravitational field due to Earth: gE G rmEE2

ggSE mmSE rrES22

((26..001100 3204 kkgg))(( 31..951100 58 kkmm))22

2.3 b. When the Moon is in its third quarter phase, as shown in Figure 7-17, its

direction from Earth is at right angles to the Sun's direction. What is the net gravitational field due to the Sun and Earth at the center of the Moon?

Moon

Copyright ? Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

Sun Earth

Figure 7-17

Because the directions are at right angles, the net field is the square root of the sum of the squares of the two fields. gS Grm 2S

(6.6710 1(11N.5m120 k1g1 2m)()22.0 1030 kg) 5.9103 N/kg Similarly, gE 2.6103 N/kg

gnet ( 5.91 03 N/kg)2 (2.6 10 3 N/kg)2

6.4103 N/kg

16. Gravitational Field The mass of the Moon is 7.31022 kg and its radius is 1785 km. What is the strength of the gravitational field on the surface of the Moon? g GrM2 (6.671 01(11.N78 m521k0g32 )m(7).231 022 kg)

1.5 N/kg, about one-sixth that on Earth

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