Download 1 - Ruthsc

Chapter 12: Gravity
Answers to Even-Numbered Conceptual Questions
–7
2.
A person passing you on the street exerts a gravitational force on you, but it is so weak (about 10 N or less)
that it is imperceptible.
4.
As the tips of the fingers approach one another, we can think of them as two small spheres (or we can
replace the finger tips with two small marbles if we like). As we know, the net gravitational attraction
outside a sphere of mass is the same as that of an equivalent point mass at its center. Therefore, the two
fingers simply experience the finite force of two point masses separated by a finite distance.
6.
No. A satellite must be moving relative to the center of the Earth to maintain its orbit, but the North Pole is
at rest relative to the center of the Earth. Therefore, a satellite cannot remain fixed above the North Pole.
8.
Yes. The rotational motion of the Earth is to the east, and therefore if you launch in that direction you are
adding the speed of the Earth’s rotation to the speed of your rocket.
10.
As the astronauts approach a mascon, its increased gravitational attraction would increase the speed of the
spacecraft. Similarly, as they pass the mascon, its gravitational attraction would now be in the backward
direction, which would decrease their speed.
12.
It makes more sense to think of the Moon as orbiting the Sun, with the Earth providing a smaller force that
makes the Moon “wobble” back and forth in its solar orbit.
Solutions to Problems and Conceptual Exercises
1.
Picture the Problem: Three systems contain different masses that are separated by different distances.
Strategy: Use the Universal Law of Gravity (equation 12-1) to determine the ranking of the force magnitudes.
Solution: 1. Apply equation 12-1:
FA  G
2. Repeat for system B:
FB  G
m1m2
m2

G
rA2
r2
m  2m 
G
m2
2r 2
 2r 
 2m  3m 
3m2
FC  G
G 2
2
2r
 2r 
 4m  5m 
20m2
FD  G
G
2
9r 2
 3r 
3. Repeat for system C:
4. Repeat for system D:
2
5. By comparing the forces we arrive at the ranking B < A < C < D.
Insight: Changing the magnitude of the separation distance has the greatest effect upon the force.
2.
Picture the Problem: The two apples attract each other gravitationally.
Strategy: Use the Universal Law of Gravity (equation 12-1) to find the force between the apples.
Solution: 1. (a) Apply equation 12-1:
F G
 0.16 kg  0.16 kg 
m1m2
  6.67  1011 N  m2 /kg 2 
 2.7  1011 N
2
2
r
0.25
m


2. (b) Repeat for the new distance:
F G
 0.16 kg  0.16 kg 
m1m2
  6.67  1011 N  m2 /kg 2 
 6.8  1012 N
2
2
r
0.50
m


Insight: Doubling the distance between the apples cut the gravitational force by a factor of 4.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 1
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
3.
Picture the Problem: The two bowling balls attract each other gravitationally.
Strategy: Use the Universal Law of Gravity (equation 12-1) to find the force between the bowling balls, then solve the
same equation for distance to answer part (b).
 6.1 kg  7.2 kg 
m1m2
  6.67 1011 N  m2 /kg 2 
 5.2 109 N
2
2
r
 0.75 m 
Solution: 1. (a) Apply equation 12-1:
F G
2. (b) Solve equation 12-1 for r:
Gm1m2
r

F
 6.67 10
11
N  m 2 /kg 2   6.1 kg  7.2 kg 
2.0  109 N
 1.2 m
Insight: Increasing the distance between the balls from 0.75 m to 1.2 m decreased the force from 5.2 nN to 2.0 nN.
4.
Picture the Problem: The Earth and the satellite attract each other gravitationally.
Strategy: Use equation 5-5 to find the weight of the satellite on the Earth’s surface, and then equation 12-1 to find the
gravitational force on the satellite while it is in orbit.
Solution: 1. (a) Apply equation 5-5:
Ws  mg   480 kg   9.81 m/s 2   4700 N  4.7 kN
2. (b) Apply equation 12-1 directly:
F G
 480 kg   5.97 10
mM E
  6.67  1011 N  m2 /kg 2 
2
2
r
35 106 m 
24
kg 
 0.16 kN
Insight: The distance between the satellite and the center of the Earth has increased by a factor of 5.5 so that the
gravitational force on the satellite has decreased by a factor of 5.5 2 = 30.
5.
Picture the Problem: You and the asteroid attract each other gravitationally.
Strategy: Estimate that your mass  70 kg. Apply equation 12-1 to find the gravitational force between you and Ceres.
Solution: Apply equation 12-1:
F G
8.7 1020 kg   70 kg   0.021 N
m1m2
11
2
2

6.67

10
N

m
/kg


2
r2
14 106 m 
Insight: If you stood on the surface of Ceres (radius 500 km) the force would be 16 N (3.6 lb).
6.
Picture the Problem: The apple and the orange attract each other gravitationally.
Strategy: Use the Universal Law of Gravity (equation 12-1) to find the force between the two fruits.
Solution: 1. (a) Apply equation 12-1:
F G
 0.11 kg  0.24 kg 
m1m2
  6.67  1011 N  m2 /kg 2 
 2.4  1012 N
2
2
r
 0.85 m 
2. (b) The force the apple exerts on the orange is equal and opposite to the force the orange exerts on the apple, so its
magnitude must be 2.4 1012 N .
Insight: Halving the distance between the two fruits would quadruple the force between them, but it would still be tiny.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 2
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
7.
Picture the Problem: The spaceship is attracted gravitationally to both the Earth and the Moon.
Strategy: Use the Universal Law of Gravity (equation 12-1) to relate the attractive forces from the Earth and the Moon.
Set the force due to the Earth equal to twice the force due to the Moon when the spaceship is at a distance r from the
center of the Earth. Let R  3.84 108 m, the distance between the centers of the Earth and Moon. Then solve the
expression for the distance r.
Solution: 1. (a) Set FE  2 FM using
equation 12-1 and solve for r:
G
ms mE
mm
 2G s M 2
2
r
R  r
mE  R  r   2mM r 2
2
R  r  2mM mE r
r
R
1  2mM mE

3.84  108 m
1
 2  7.35 10
22
kg   5.97 10 24 kg 
 3.32  108 m
2. (b) The answer to part (a) is independent of the mass of the spaceship because the spaceship’s mass is included in the
force between it and both the Moon and the Earth, and so its value cancels out of the expression.
Insight: The distance in part (a) is the same for any mass, and corresponds to about 52 Earth radii or about 86% of the
distance R between the Earth and the Moon. The two forces are equal at 3.46×10 8 m or about 90% of R.
8.
Picture the Problem: All three masses attract each other gravitationally.
Strategy: Add the gravitational force on each mass due to the other two masses using equation 12-1.
Solution: 1. (a) Add
the forces:
M
M EMS
M M
M 
 G E2 M  GM E  2 S  2 M 
2
rE-S
rE-M
 rE-S rE-M 


2.00  1030 kg
7.35 10 22 kg 
  6.67  10 11 N  m 2 /kg 2  5.97  10 24 kg  

 1.50  1011 m 2  3.84  108 m 2 


FE  G
FE  3.56  10 22 N toward the Sun
2. (b) Add the forces:
M
MSM M
M M
M 
 G E2 M  GM M  2 S  2 E 
2
rS-M
rE-M
r
r
 S-M E-M 


2.00  1030 kg
5.97 10 24 kg 
  6.67  10 11 N  m 2 /kg 2  7.35  10 22 kg  

 1.50  1011  3.84  108 m 2  3.84 108 m 2 


FM  G
FM  2.40  10 20 N toward the Sun
3. (c) Add the forces:
M
MSM M
M M
M 
 G S2 E  GM S  2 M  2 E 
2
rS-M
rS-E
 rS-M rS-E 


7.35  1022 kg
5.97  1024 kg 
  6.67  1011 N  m 2 /kg 2  2.00 1030 kg  

 1.50  1011  3.84  108 m 2 1.50  1011 m 2 


FS  G
FS  3.58  1022 N toward the Earth-Moon system
Insight: Note that if you compare the two terms that contribute to FM , you will see that the Sun exerts a force on the
Moon (4.38×1020 N) that is 2.2 times larger than the force the Earth exerts on the Moon (1.98×1020 N).
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 3
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
9.
Picture the Problem: Both the Earth and the Sun exert attractive
gravitational forces on the Moon. The three objects are arranged as shown.
Strategy: Use the Universal Law of Gravity (equation 12-1) to find the
components of the force acting on the Moon. The Earth exerts a downward
force on the Moon and the Sun exerts a force toward the left. The net force
is therefore at an angle θ below the line that connects the Moon to the Sun.
 7.35 1022 kg  2.00 1030 kg   4.36 1020 N
M M MS
11
2
2

6.67

10
N

m
/kg


2
2
rM-S
1.50 1011 m 
Solution: 1. Use equation
12-1 to find FS :
FS  G
2. Now find FE :
7.35 1022 kg  5.97 1024 kg 

MMME
11
2
2
FE  G 2
  6.67 10 N  m /kg 
 1.98 1020 N
2
11
rM-E
3.84

10
m


3. Add the components:
F  Fx 2  Fy 2 
4. Find the direction θ :
  tan 1
Fy
Fx
 4.36 10
 tan 1
20
N   1.98 1020 N   4.79 1022 N
2
2
1.98 1020 N
4.36  1020 N
 24.4 toward the Earth off the line from the Moon to the Sun
Insight: Note that the Sun exerts a force on the Moon (4.36×1020 N) that is 2.2 times larger than the force the Earth
exerts on the Moon (1.98×1020 N).
10. Picture the Problem: Both the Earth and the Moon exert attractive gravitational
forces on the Sun.
Strategy: Use the Universal Law of Gravity (equation 12-1) to find the components
of the force acting on the Sun. The Earth exerts a force downward and to the right
force on the Sun and the Moon exerts a force toward the right. The net force is
therefore at an angle  below the line that connects the Moon to the Sun. Because
the distance between the Moon and Earth (3.84×10 8 m) is small compared with the
distance between the Sun and Earth (1.50×1011 m), we’ll approximate the distance
to be 1.50×1011 m for both the Moon and the Earth.
Solution: 1. Use equation
12-1 to find FM :
FM  G
 7.35 1022 kg  2.00 1030 kg   4.36 1020 N
M M MS
11
2
2

6.67

10
N

m
/kg


2
2
rM-S
1.50 1011 m 
2. Now find FE :
FE  G
 2.00 1030 kg 5.97 1024 kg   3.54 1022 N
MSM E
11
2
2

6.67

10
N

m
/kg


2
2
rS-E
1.50 1011 m 
3. Use trigonometry to
find  :
8
 rE-M 
1  3.84  10 m 

sin


  0.147 toward the Earth off the line from the
11
 1.50 10 m 
 rS-M 
Sun to the Moon
  sin 1 
4. Find FE,x :
FE,x  FE cos    3.54  1022 N  cos  0.147   3.54  1022 N
5. Find FE,y :
FE,y  FE sin    3.54  1022 N  sin  0.147   9.08  1019 N
6. Add the components:
F

F
M
 FE,x   FE,2y
 4.36 10
2
20
 3.54 1022 N    9.08 1019 N   3.58 1022 N
2
2
Insight: Note that the Sun exerts a force on the Earth that is 81 times larger than the force the Sun exerts on the Moon,
because the Earth is 81 times more massive than the Moon and the distances to the Sun are approximately the same.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 4
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
11. Picture the Problem: The three masses are arranged at the vertices of an equilateral triangle
as shown in the figure at right.
Strategy: Each mass will be gravitationally attracted to the other two masses. The vector sum
of the forces will be along a line toward the midpoint of the other two masses, due to
symmetry. An examination of the geometry reveals that this force equals 2F cos30 , where F
is the gravitational force between any two masses. Use this relation together with equation
12-1 to find the force exerted on one of the masses.
Solution: 1. (a) Find 2F cos30 , where
F  Gm 2 r 2 from equation 12-1:
m
r
F
60°
m
m

m2
N  m 2   6.75 kg 
cos 30  2  6.67 1011
cos 30

2
r
kg 2  1.25 m 2

2
F  2G
 3.37 109 N
2. (b) If the side lengths are doubled the gravitational forces will be reduced by a factor of four.
Insight: Doubling the masses will quadruple the gravitational force, because in this case the gravitational force depends
upon the square of the mass of each object and the square of the separation distance.
12. Picture the Problem: The 2.0-kg mass is gravitationally attracted to the
other three masses.
Strategy: Add the gravitational forces using the component method of
vector addition. Use equation 12-1 and the geometry of the problem to
determine the magnitudes of the forces. Let m1  1.0 kg, m2  2.0 kg,
m3  3.0 kg, and m4  4.0 kg.
Solution: 1. (a) Find the x component of the force on m2 :
m1m2
mm
 G 22 4
2
r12
r24
m1m2
m2 m4
 G 2 G 2
r12
r24
Fx  G
  6.67  1011
Fx  1.3  108 N
2. Find the y component
of the force on m2 :
m2 m3
mm
 G 22 4
2
r23
r24
m2 m3
m2 m4
 G 2 G 2
r23
r24
Fy  4.5  10 8 N
3. Use the components of F to
find its magnitude and direction.
 m1 m4 r12 

  Gm2  2  3 
r24 

 r12


 4.0 kg  0.20 m 
 1.0 kg

2
2
N  m /kg   2.0 kg  


2
2
2 3/ 2
 0.20 m    0.10 m   
  0.20 m 

 

Fy  G
  6.67  10 11
F  Fx 2  Fy 2 
cos 
 r12

 r24
sin 
 m3 m4 r14 

  Gm2  2  3 
r24 

 r23


 4.0 kg  0.10 m 
 3.0 kg

2
2
N  m /kg   2.0 kg  


2
2
2 3/ 2
 0.20 m    0.10 m   
  0.10 m 

 

 r14

 r24
1.3 10
8
N    4.5  108 N   4.7  108 N
2
2
8
 Fy 
N
1  4.5  10
  tan 
  74 below horizontal and to the left
8
 1.3  10 N 
 Fx 
  tan 1 
4. (b) If the sides of the rectangle are all doubled, all forces are reduced by a factor of 22  4; the directions of the
forces are unchanged.
Insight: Doubling all the masses will quadruple the gravitational force, because the gravitational force depends upon
the product of the masses of each object. Therefore, the force would stay exactly the same if we doubled all the masses
and doubled the length of the sides of the rectangle.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 5
Chapter 12: Gravity
James S. Walker, Physics, 4th Edition
13. Picture the Problem: The three objects lie along a line as depicted in the
diagram at right. The net force on object 2 is zero.
1
2
F12
3
F23
x
Strategy: Let m1 be an arbitrary unit of mass, then let m2  17 m1 and
0
m3  14 m1 . Use the Universal Law of Gravity (equation 12-1) with
x
D
r12  x and r23  D  x to determine the position x that results in zero net
force on object 2.
Solution: 1. Let  F12  F23  0 and
substitute for r12 , r23 and m3 :
G
mm
m1m2
 G 22 3
2
r12
r23
1
m
m3
m1

 4 1 2
2
2
x
 D  x  D  x
m1  D 2  2 Dx  x 2   14 m1 x 2
2. Rearrange the expression to obtain a quadratic formula:
4 x 2  8Dx  4 D 2  x 2
3x 2  8Dx  4 D 2  0
 8D   4  3  4 D 2 
2  3
b  b 2  4ac  8D  
x

2a
3. Apply the quadratic formula to find x:

2
8D  16 D 2 8D  4 D 8D  4 D

,
 2 D,
6
6
6
4. Because x  2 D is not between masses 1 and 3, we reject that solution and say that x 
2
3
2
3
D
D is the solution.
Insight: Object 3 has less mass than object 1, so in order for it to attract object 2 with as much force as does object 1,
object 2 must be closer to object 3 than it is to object 1.
14. Picture the Problem: The acceleration of gravity at a planet’s surface is determined by its mass and radius.
Strategy: Apply a formula similar to equation 12-4 to find the acceleration of gravity on the surfaces of Mercury and
Venus. Use the data included in Appendix C to find the masses and radii of the planets.
 0.0553  5.97 10 kg   3.70 m/s
 2.440 10 m 
 0.816  5.97 10 kg   8.87 m/s

 6.052 10 m 
Solution: 1. (a) Solve g P  G M P RP2
for Mercury:
g M   6.67 1011 N  m2 /kg 2 
2. (b) Solve g P  G M P R for Venus:
g V   6.67 10
2
P
24
2
2
6
24
11
N  m /kg
2
2
2
6
2
Insight: Although the mass of Mercury is smaller than Mars’, its smaller radius and higher density results in a nearly
identical acceleration of gravity at the surface. The acceleration of gravity on Venus is similar to that on the Earth.
15. Picture the Problem: The acceleration of gravity at an altitude h above the Earth’s surface is reduced due to the
increased distance from the center of the Earth.
Strategy: Set the acceleration of gravity at an altitude h equal to one-half the acceleration at h = 0 using equation 12-4,
and solve for h.
Solution: Set g h  g0 and solve for h:
GM E

1  GM E 


2  RE2 
 RE  h 
2
 RE  h   2 RE2
2
h


2  1 RE 


2  1  6.37 106 m   2.64  106 m
Insight: This altitude is over ten times higher than the 200-km altitude of the International Space Station.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 6
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
16. Picture the Problem: The bowling balls are in contact with each other so that their centers are separated by one
diameter.
Strategy: A sphere acts like a point mass at the sphere’s center, so apply Newton’s Law of Universal Gravitation
(equation 12-1) to determine the force between the two bowling balls when their centers are 2×11 cm = 22 cm apart.
 6.7 kg 
m2
  6.67 1011 N  m2 /kg 2 
 6.2 108 N  62 nN
2
2
r
0.22
m


2
Solution: Apply equation 12-1 directly:
F G
Insight: If the bowling balls had the same mass (these are approximately 15-lb balls) but half the radius, the attractive
force would be four times larger, or about 250 nN.
17. Picture the Problem: The acceleration of gravity due to the Earth, at a distance equal to the Moon’s orbit radius, is
diminished because of the increased distance to the center of the earth.
Strategy: Use equation 12-4 to find the acceleration of gravity, except replace the radius of the Earth with the distance
to the center of the Earth.
Solution: Replace RE
with r in equation 12-4:
g G
ME
5.97 1024 kg
  6.67 1011 N  m2 /kg 2 
 0.00270 m/s 2
2
2
8
r
3.84 10 m 
Insight: Another way to solve this question is to realize the Moon orbits the Earth at a distance of about 60RE , so the
acceleration of gravity due to the Earth at that location is
1
602
g
1
3600
 9.81 m/s   0.00273 m/s , almost correct!
2
2
18. Picture the Problem: The acceleration of gravity at a planet’s surface is determined by its mass and radius.
Strategy: Apply a formula similar to equation 12-4 to find the acceleration of gravity on the surface of Titan. Use the
mass and radius data given in the problem.
Solution: Solve g T  G M T RT2 for Titan:
gT   6.67 1011 N  m2 /kg 2 
1.35 10 kg   1.36 m/s
 2.57 10 m 
23
6
2
2
Insight: This acceleration of gravity isn’t much different from the 1.67 m/s2 found on the surface of Earth’s Moon.
Titan has more mass but also a larger radius than our Moon, and the two characteristics yield a similar surface gravity.
19. Picture the Problem: The mass experiences a gravitational attraction to the Earth that depends upon its distance from
the center of the Earth.
Strategy: Use Newton’s Universal Law of Gravitation (equation 12-1) to find the distance from Earth’s center that
would produce the given force (weight) for a 4.0-kg mass. Then use Newton’s Second Law to find the acceleration.
Solution: 1. (a) Solve equation 12-1 for r:
r

2. (b) Solve Newton’s Second Law for a:
a
GM E m
F
 6.67 10
11
N  m2 /kg 2  5.97 1024 kg   4.6 kg 
2.2 N
 2.9 107 m
F 2.2 N

 0.48 m/s 2
m 4.6 kg
3. (c) Because the gravitational force is inversely proportional to r 2 , doubling r reduces F by a factor of 4.
4. (d) Doubling r also reduces the acceleration by a factor of 4 because the force has decreased by that factor and the
mass has not changed.
Insight: The distance 2.9×107 m is about 4.6RE . If the mass were on the Earth’s surface, it would weigh 45 N (10 lb).
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 7
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
20. Picture the Problem: Both the Earth and the Moon exert a force on other masses in their vicinity according to
Newton’s Universal Law of Gravitation.
Strategy: Use a ratio of gravitational forces to find the mass of the Moon in terms of the mass of the Earth. Note that if
you solve Newton’s Universal Law of Gravitation (equation 12-1) for m2 you find m2  F r 2 Gm1 .
Solution: Determine the ratio mM mE
by solving equation 12-1 for m2 :
mM FM rM2 Gm1 FM rM2



mE
FE rE2 Gm1
FE rE2
 16 FE   14 rE 
2

FE rE2
1
1
 mM 
mE
96
96
Insight: Using a ratio can be a powerful tool for solving a question like this one, where few hard data are given, only
relationships between various quantities. The exact ratio is 7.35 1022 kg 5.97 10 24 kg  1 81.2.
21. Picture the Problem: The volcano on Io spews material at high speed straight upward. The mass slows down, rising to
a height of 5.00 km before coming to rest momentarily under the influence of Io’s gravitation.
Strategy: Use conservation of energy to relate the initial kinetic energy of the ejected material to its potential energy at
the maximum altitude. This relation will allow the calculation of the surface gravity of Io. Then use equation 12-4 and
the given radius of Io to find the mass of Io.
Solution: 1. (a) Use 12 mvi2  mghf to find g , and use g  GM R2 to find M .
2. (b) Set Ei  Ef and solve for g:
1
2
mvi2  mghf
134 m/s 
v2
g i 
 1.80 m/s 2
2hf 2  5.00 103 m 
2
2
6
gR 2 1.80 m/s 1.82 10 m 
M

 8.94 1022 kg
G
6.67 1011 N  m 2 /kg 2
2
3. Solve equation 12-4 for M:
Insight: The use of conservation of energy to find g is equivalent to solve equations 4-6, v 2  v02  2 g y , for g. This
approach assumes that g is constant over the 5.00 km that the ejected material rises. This is a pretty good assumption,
because you will get the same answer to 3 significant figures if you apply the more exact method of section 12-5. The
mass of Io according to NASA’s Solar System Exploration web site is 8.9316×10 22 kg.
22. Picture the Problem: Both the Earth and the Moon exert a gravitational force on the spaceship, but in opposite
directions.
Strategy: Use the Universal Law of Gravity (equation 12-1) to relate the attractive forces from the Earth and the Moon.
Set the force due to the Earth equal to the force due to the Moon when the spaceship is at a distance r from the center of
the Earth. Let R  3.84 108 m, the distance between the centers of the Earth and Moon. Then solve the expression for
the distance r.
Solution: 1. (a) Set FE  FM using
equation 12-1 and solve for r:
G
ms mE
mm
 G s M2
r2
R  r
mE  R  r   mM r 2
2
R  r  mM mE r
r
R
1  mM mE

3.84  108 m
1
 7.35 10
22
kg   5.97 10 24 kg 
 3.46  108 m
2. (b) The net gravitational force on the spaceship and the astronauts will steadily decrease, reaching zero at the location
found in part (a), and then gradually increase in the opposite direction (toward the Moon). However, the astronauts are
accelerated at the same rate as the spaceship and so they will appear to float, not walk.
Insight: The distance in part (a) is about 90% of the distance between the Earth and the Moon! The Earth’s gravity is
the dominant force through most of the journey to the Moon.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 8
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
23. Picture the Problem: The spherical asteroid exerts a gravitational force on nearby objects.
Strategy: Use equation 12-4 to find the acceleration of gravity at the surface of the asteroid using the given
information. Then set the gravitational force equal to the centripetal force required to keep an object in orbit just above
the surface of the asteroid. Let the centripetal force be written F  macp  m 2 R according to equation 10-13. Then
use equation 10-5, T  2  , to determine the rotational period of the satellite that would produce orbital motion for
loose rocks at the surface.
Solution: 1. (a) Apply eq. 12-4 to the asteroid:
2. (b) Set Fgravity  Fcentripetal , substitute   2 T
(equation 10-5) and solve for T:
g A   6.67 1011 N  m 2 /kg 2 
G
3.35 10 kg   6.2 10
1.9 10 m 
15
4
4
2
m/s 2
 4 2 
mA m
2

m

R

m


 2 R
R2
 T

T
4 2 R 3

GmA
 6.67 10
4 2 1.9 104 m 
11
3
N  m 2 /kg 2  3.35 1015 kg 
 3.5  104 s  9.7 h
Insight: This rotational period corresponds to a surface speed of 3.4 m/s at the equator. In other words, if the asteroid
were not rotating at all, a person at the equator could launch themselves into orbit by running at 3.4 m/s (7.7 mi/h)!
24. Picture the Problem: The orbital speed of the Earth is greatest around January 4 and least around July 4.
Strategy: Consider the characteristics of elliptical orbits when answering the conceptual question.
Solution: 1. (a) The speed of a satellite in an elliptical orbit is greatest when the orbit distance is least. We conclude
that the distance from the Earth to the Sun on January 4 is less than its distance from the Sun on July 4.
2. (b) The best explanation is II. The Earth sweeps out equal area in equal time, thus it must be closer to the Sun when
it is moving faster. Statements I and III are each false.
Insight: It may seem surprising that the Earth is closest to the Sun during the wintertime in the northern hemisphere.
However, the seasons on the Earth are due to the tilt of the Earth’s axis, not its distance from the Sun.
25. Picture the Problem: A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine is fired
in such a way that its speed increases rapidly by a small amount.
Strategy: Consider the principles explained in Figure 12-13 when answering the conceptual question.
Solution: 1. (a) Firing the rocket engine will increase the kinetic energy of the satellite while keeping its gravitational
potential energy constant. As shown in Figure 12-13, this additional kinetic energy will be converted into gravitational
potential energy at a farther distance from the Earth, a point that will then be the apogee of the new orbit. We conclude
that the apogee distance will increase (it will be larger than r).
2. (b) As shown in Figure 12-13 the elliptical orbit produced by firing the rocket engine has a perigee distance equal to
the original circular orbit radius r. We conclude that the perigee distance will stay the same.
Insight: The rocket motor produces a torque on the satellite with respect to the center of the Earth, increasing its
angular momentum.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 9
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
26. Picture the Problem: A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine is fired
in such a way that its speed decreases rapidly by a small amount.
Strategy: Consider the principles explained in Figure 12-13 when answering the conceptual question.
Solution: 1. (a) Firing the rocket engine will decrease the kinetic energy of the satellite while keeping its gravitational
potential energy constant. As shown in Figure 12-13, this means that the satellite no longer has sufficient kinetic energy
to maintain a circular orbit and it will fall to a closer distance from the Earth, a point that will then be the perigee of the
new orbit. The elliptical orbit produced by firing the rocket engine has an apogee distance equal to the original circular
orbit radius r. We conclude that the apogee distance will stay the same (it will be equal to r).
2. (b) As discussed above the perigee distance will decrease.
Insight: The rocket motor produces a torque on the satellite with respect to the center of the Earth, decreasing its
angular momentum.
27. Picture the Problem: The average distance from the Earth to the Moon is increasing at the rate of 3.8 cm per year.
Strategy: Consider Kepler’s third law of orbits when answering the conceptual question.
Solution: 1. (a) Kepler’s third law states that T   constant  r 3 2 . Increasing the orbit radius r will therefore increase the
orbit period T. We conclude that the length of the month will increase.
2. (b) The best explanation is I. The greater the radius of an orbit, the greater the period, which implies a longer month.
Statements II and III are each false.
Insight: The angular momentum of the Earth-Moon system is not conserved due to tidal friction, which converts some
of the mechanical energy of the system into thermal energy through the motion of ocean waters on Earth and the
deformation of the rocks of both the Earth and the Moon.
28. Picture the Problem: The Apollo capsule orbited the Moon at an altitude of 110 km above the Moon’s surface.
Strategy: Use equation 12-7 to determine the period of orbit using the mass and radius of the Moon from the inside
back cover of the text.
Solution: Apply equation 12-7:
 2  3 / 2
T 
r
 GM 
M 



2

 1.74  106  110  103 m 3 / 2

 6.67  1011 N  m 2 /kg 2 7.35  10 22 kg  

 
 
T  7140 s  1.98 h
Insight: This period turns out to be a bit larger than the 1.44-h orbit period of a satellite that is 110 km above the
Earth’s surface.
29. Picture the Problem: The satellite orbits the Earth with a period of 24.0 h and an altitude of 35,800 km.
Strategy: Set the gravitational force between the satellite and the Earth equal to the centripetal force required to keep
the satellite moving in a circular path, and solve for the speed. Use equation 6-15, fcp  mv2 r , for the centripetal
force. Use the mass and radius of the Earth given in the inside back cover of the text.
Solution: Set Fgravity  Fcentripetal and solve for v:
G
MEm
 RE  h 
2
m
v
v2
 RE  h 
G ME

RE  h
 6.67 10
11
N  m 2 /kg 2  5.97 10 24 kg 
6.37 106  3.58 107 m
 3070 m/s  3.07 km/s
Insight: The speed of an object in low Earth orbit is about 7.91 km/s. But this isn’t a low orbit, because the altitude of
3.58×107 m is about 5.62 Earth radii above the surface, or a total orbit radius of 6.62 RE .
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 10
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
30. Picture the Problem: The planet orbits the star Iota Horologii according to Kepler’s Third Law.
Strategy: Use a ratio to find the orbit radius of the extrasolar planet by comparing its orbit with the Earth’s orbit around
the Sun.
Solution: Make a ratio using Kepler’s Third
Law (equation 12-7) and solve for r:
r 
 
 rE 
32

T GM s 2
TE GM s 2
T 
r  rE  
 TE 
23

T
TE
 320 d 
 1.50  1011 m  

 365 d 
23
 1.4  1011 m
Insight: If we knew the period of the planet were exactly 320 days, we could keep an extra significant figure and report
1.37×1011 m as the orbit radius. Note that if we use Astronomical Units (1 AU = 1.50×10 11 m) to describe the orbit
radius and years to describe the orbit period, Kepler’s Third Law can be written T 2  r 3 (see problem 87).
31. Picture the Problem: The moon Phobos travels in a circular obit around Mars.
Strategy: Use equation 12-7 and the mass and radius of Mars given in Appendix C to determine the orbit period.
 2  3 / 2
T 
r
 GM 
M 



2

 9378  103 m 3 / 2

 6.67  10 11 N  m 2 /kg 2 0.108  5.97  10 24 kg  

 
 
T  27,500 s  7.64 h
Solution: Apply equation 12-7 directly:
Insight: Mars rotates on its axis about every 24.62 hours, so the moon Phobos would cross the Martian sky about 3
times a day!
32. Picture the Problem: The moon Ganymede travels in a circular obit around Jupiter.
Strategy: Solve Kepler’s Third Law (equation 12-7) for the mass of Jupiter.
Solution: Solve
equation 12-7 for M J :
2 3
2
1.07 109 m 

2
 2  r


MJ  

 1.90 1027 kg


5
11
2
2
 T  G  6.18 10 s  6.67 10 N  m /kg
3
Insight: Jupiter is 318 times more massive than the Earth!
33. Picture the Problem: The tiny moon Dactyl travels around 243 Ida in an approximately circular orbit.
Strategy: Solve Kepler’s third law (Equation 12-7) for the mass of 243 Ida, using the orbit distance and period given in
the problem.
Solution: 1. (a) Solve Kepler’s third law (Equation 12-7) for the mass of 243 Ida, using the orbit distance and period
given in the problem.
2 3
2
89 103 m 

2
 2  r




 8.9 1016 kg


11
2
2
T
G
19
h

3600
s/h
6.67

10
N

m
/kg




3
2. (b) Solve equation 12-7
for M 243 Ida :
M 243 Ida
Insight: This asteroid is a little bit bigger than Manhattan. The moon Dactyl is only about 1.4 km in diameter.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 11
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
34. Picture the Problem: The GPS satellite travels around the Earth in a circular orbit.
Strategy: Use equation 12-7 to find the orbit period of the GPS satellite, then set the speed of the satellite equal to the
circumference of its orbit divided by its orbit period.
Solution: 1. (a) Solve equation 12-7:
2. (b) The speed is the circumference
divided by the period:
 2  3 / 2
T 
r
 GM 
E 



2

 2.0  107  6.37  106 m 3 / 2

 6.67  1011 N  m 2 /kg 2 5.97  1024 kg  

 
 
T  43000 s  12 h
v
7
2 r 2  2.637 10 m 

 3900 m/s  3.9 km/s
T
4.3 104 s
Insight: The orbits of these GPS satellites are designed so that at least 3 satellites are overhead every spot on Earth at
any one time.
35. Picture the Problem: The two satellites travel in circular orbits about the Earth.
Strategy: Set the gravitational force between the satellite and the Earth equal to the centripetal force required to keep
the satellite moving in a circular path, and solve for the speed. Use equation 6-15, fcp  mv2 r , for the centripetal
force. Use the mass and radius of the Earth given in the inside back cover of the text.
Solution: 1. (a) Kepler’s Third Law (equation 12-6) indicates that the orbit period grows faster than the orbit radius.
The orbit speed is proportional to the orbit radius divided by the period, and so the orbit speed is proportional to
v  r T  r r 3 2  1 r . That means the orbit speed decreases as the radius increases, so satellite 2 will have the
greater orbital speed.
2. (b) Set Fgravity  Fcentripetal and
solve for v:
G
MEm
 RE  h 
v
2
m
G ME

RE  h
v2
 RE  h 
G ME

RE  RE
 6.67 10
11
N  m 2 /kg 2  5.97  1024 kg 
2  6.37  106 m
 5590 m/s  5.59 km/s
3. (c) Repeat step 2 for the new h:
v
G ME

RE  h
G ME

RE  2 RE
 6.67 10
11
N  m 2 /kg 2  5.97 1024 kg 
3  6.37 106 m
 4560 m/s  4.56 km/s
Insight: The satellite with the larger orbit radius in part (c) had the smaller orbit speed, as predicted.
36. Picture the Problem: The satellites travel in circular orbits around the Earth.
Strategy: Use equation 12-7 to find the periods of the satellites. Use the mass and radius of the Earth given in the
inside back cover of the book. When the altitude h  RE , the orbit radius r  RE  h  2RE .
 2 
3/ 2
Solution: 1. (a) Apply equation 12-7: T  
  2 RE 
 GM 
E 



2
 2  6.37  106 m 3 / 2

 6.67  1011 N  m 2 /kg 2 5.97  10 24 kg  

 
 
T  14,300 s  3.98 h
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 12
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
2. (b) Repeat for r  3RE
 2 
3/ 2
T 
  3RE 
 GM 
E 



2
 3  6.37  106 m 3 / 2

 6.67  1011 N  m 2 /kg 2 5.97 10 24 kg  

 
 
T  26,300 s  7.31 h
3. (c) The periods do not depend on the mass of the satellite because the satellite mass cancels out of the equation as
shown on page 390.
4. (d) The periods depend inversely on the square root of the mass of the Earth.
Insight: The fact that the orbit period of an object is independent of its mass, as discussed in part (c), is ultimately due
to the equivalence of gravitational and inertial mass. This equivalence lays the foundation for Einstein’s General
Theory of Relativity (see section 29-8).
37. Picture the Problem: Deimos and Phobos travel in circular orbits about Mars.
Strategy: Solve Kepler’s Third Law (equation 12-7) for the orbit radius of Deimos. Using the data given in
Appendix C, M M  0.108  5.97 1024 kg  6.45 10 23 kg.
Solution: 1. (a) Because T  r 3/ 2 and Deimos has the greater period, Deimos is farther from Mars than is Phobos.
2/3
2. (b) Solve equation 12-7 for r:
 T GM M 
r 

 2



 1.10  105 s 



 6.67 10
N  m 2 /kg 2  6.45 10 23 kg  


2

11
2/3
r  2.36  107 m
Insight: The orbit radius of Deimos, 23,600 km, is larger than the 9378 km orbit of Phobos (as given in problem 31), as
predicted in part (a).
38. Picture the Problem: The two companion stars Centauri A and Centauri B orbit
one another in circular paths.
Strategy: Set the centripetal force required to keep the star traveling in a circular
path equal to the gravitational force of attraction between the two stars. Set the
orbit speed equal to the circumference of the path divided by the orbit period.
Then solve the resulting expression for the mass of the stars. Let mA  mB  m,
A
cm
B
F
r
and let the stars be separated by distance d  2r, where r is the radius of the
orbital path.
Solution: Set Fgravity  Fcentripetal , let
Fgravity  G
v   d T , and solve for m:
mA mB
m2
v2
 G 2  Fcentripetal  m
2
r
d
d
d 2 v 2 2d   d 
2 2 d 3




rG
G  T 
GT 2
2
m

2 2  3.45  1012 m 
3
 6.67 1011 N  m 2 /kg 2  2.52 109 s 
2
 1.91 1030 kg
Insight: These two stars have a very similar mass to that of our Sun (2.00×1030 kg). Their orbit radii are about 23 times
larger than the Earth’s and their orbit periods are about 80 years.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 13
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
39. Picture the Problem: The two companion stars Centauri A and Centauri B orbit one another in circular paths.
Strategy: The speed of the stars in their orbits is the circumference C of the orbit divided by the period T.
12
C  d   3.45 10 m 
v 

 4300 m/s  4.30 km/s
T
T
2.52 109 s
Solution: Solve for v   d T :
Insight: This speed is significantly less than the Earth’s orbital speed of 29.9 km/s.
40. Picture the Problem: Sputnik I orbited the Earth along an elliptical path.
Strategy: Use equation 12-8 and the distance data given in the problem to find the gravitational potential energy at
perigee and at apogee and take the difference. The potential energy will be largest in magnitude (the most negative) at
the closest distance (perigee), so that the satellite loses potential energy (and gains kinetic energy) as it goes from
apogee to perigee.
Solution: Use equation
12-8 to find U :
U  U perigee  U apogee
 G
M Em 
M m
  G E 
rp
ra 

1 1
 GM E m   
r r 
p 
 a
1
1


  6.67  1011 N  m 2 /kg 2  5.97  10 24 kg  83.5 kg  


3
3
 7330  10 m 6610  10 m 
U   4.94  108 J
Insight: The satellite loses 494 MJ of gravitational potential energy and gains 494 MJ of kinetic energy. This occurs as
the spacecraft changes its speed from about 7750 m/s at apogee to about 8450 m/s at perigee.
41. Picture the Problem: A spacecraft travels from the Earth to the Moon.
Strategy: Consider the values of the gravitational potential energy of a spacecraft at the surface of the Earth and at the
surface of the Moon when answering the conceptual question.
Solution: 1. (a) The gravitational potential energy of a spacecraft is at a lower value (more negative) on the surface of
the Earth than it is on the surface of the Moon. In order to travel from the Earth to the Moon, the energy difference must
be supplied by the rocket engine, but the spacecraft gains kinetic energy on its way from the Moon to the Earth. We
conclude that the amount of energy required to get a spacecraft from the Earth to the Moon is greater than the energy
required to get the same spacecraft from the Moon to the Earth.
2. (b) The best explanation is I. The escape speed of the Moon is less than that of the Earth; therefore, less energy is
required to leave the Moon. Statements II and III are each false.
Insight: Note that you must essentially “escape” from the Earth to get to the Moon, and this takes much more energy
than is required to “escape” from the Moon, with its much weaker gravity. This is why an enormous Saturn V rocket
was required to get to the Moon, but only a small rocket on the lunar lander was required to lift off the lunar surface.
42. Picture the Problem: The four masses are positioned at the corners of a
rectangle, as indicated in Figure 12-24.
Strategy: As in Example 12-5, the total gravitational potential energy is
determined by summing the potential energy of each pair of masses. In this
case the pairs are 1−2, 1−3, 1−4, 2−3, 2−4, and 3−4. Use equation 12-9 to
sum the contributions of these six terms to the total gravitational potential
energy. Let x = 0.20 m represent the length of the rectangle, y = 0.10 m the
height of the rectangle, and d 
 0.20 m 
2
  0.10 m   0.22 m be the
2
diagonal of the rectangle.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 14
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
Solution: 1. (a) Add the six terms
that contribute to U total :
m m m m m m m m m m m m 
U total  G  1 2  1 3  1 4  2 3  2 4  3 4 
d
y
y
d
x 
 x
   6.67  1011
 1.0 kg  2.0 kg  1.0 kg  3.0 kg 




0.20 m
0.22 m




1.0
kg
4.0
kg
2.0
kg
3.0
kg







N  m 2 /kg 2   

0.10 m
0.10 m




 2.0 kg  4.0 kg   3.0 kg  4.0 kg  





0.22 m
0.20 m


U total  1.5  108 J
2. (b) The answer to part (a) would quadruple if all the masses were doubled because each term involves the product of
two masses.
3. (c) The answer to part (a) would double if each side of the rectangle were halved in length because each term is
inversely proportional to the distance between masses.
Insight: The potential energy is a tiny −15 nJ because the masses are relatively small and so are the attractive forces
between them.
43. Picture the Problem: An object is located at the surface of the Earth and later at an altitude of 350 km.
Strategy: Use equation 12-8 to find the gravitational potential energy of the object as a function of its distance
r  RE  h from the center of the Earth. Then take the difference between the values at h = 0 and h = 350 km and
compare it with the approximate change in potential, U  mgh.
Solution: 1. (a) Calculate
M m
at h = 0:
U  G E
RE  h
U 0    6.67 1011 N  m2 /kg 2 
5.97 10
24
kg   8.8 kg 
6.37 106 m
5.97 10
24
kg  8.8 kg 
  5.5 108 J
2. (b) Calculate U at h = 350 km:
U h    6.67 1011 N  m2 /kg 2 
3. (c) Take the difference U :
U  U h  U 0    5.50 108 J     5.21108 J   2.9 107 J
4. Compare with mgh:
U  mgh  8.8 kg   9.81 m/s 2  350 103 m   3.0 107 J
6.37 10  350 103 m
6
  5.2 108 J
Insight: The two calculations of U (without any rounding) differ by about 5%. The mgh calculation is an
approximation because it assumes the value of g is constant over the 350 km, when in fact it gets smaller as the distance
from the Earth’s center increases.
44. Picture the Problem: The two basketballs are initially touching and are then separated from each other.
Strategy: When the basketballs are touching, their centers are one diameter or 0.24 m apart. Find the difference in
gravitational potential energy between when they are touching and when they are separated by using equation 12-9.
m2 
m2 
1
2 1
  G
  Gm   
r2 
r1 
r
r
 1 2
Solution: 1. (a) Determine
an expression for U :
U  U 2  U1   G
2. Calculate U for r2  1.0 m:
1
1 
2
11
U   6.67 1011 N  m2 /kg 2   0.59 kg  

  7.4 10 J
 0.24 m 1.0 m 
3. (b) Calculate U for r2  10.0 m:
1
1 
2
11
U   6.67 1011 N  m2 /kg 2   0.59 kg  

  9.4 10 J
0.24
m
10.0
m


Insight: The work required to separate these basketballs is tiny, only 94 pJ to separate them by 10.0 m. This is
equivalent to raising the center of mass of one of the basketballs by 16 pm, less than the 100-pm diameter of a typical
atom!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 15
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
45. Picture the Problem: The rocket is given enough kinetic energy to completely escape the Earth or the Moon.
Strategy: The rocket completely escapes the Earth when it is infinitely far away, which is when its gravitational
potential energy is zero. Set the kinetic energy equal to the magnitude of the initial (negative) gravitational potential
energy in order to find the energy needed to escape. Use the radius and mass data for the Earth and Moon given in the
inside back cover of the book.
Solution: 1. (a) Calculate
K  U   U i  U i for the Moon:
K  Ui  G
MMm
RM
  6.67  10
2. (b) Calculate K  Ui for the Earth:
K  U i  G
 7.35 10 kg   39, 000 kg   1.110

1.74 10 m 
22
11
N  m /kg
2
2
MEm
RE
  6.67  1011 N  m 2 /kg 2 
11
6
J
 5.97 10 kg  39, 000 kg   2.4 10
 6.37 10 m 
24
12
6
J
Insight: It takes 22 times more energy to escape the Earth because of its larger mass. The larger radius of the Earth
actually makes it a bit easier to escape, but in the end the fact that the Earth has 81 times more mass makes it harder to
escape the Earth than the Moon.
46. Picture the Problem: The Earth suddenly shrinks to half its current diameter but its mass remains constant.
Strategy: Consider the expression for escape speed when answering the conceptual question.
Solution: 1. (a) The escape speed is determined by ve  2GM E RE . From this expression we can see that reducing
the radius while keeping the mass of the Earth the same will increase the escape speed.
2. (b) The best explanation is III. The force of gravity would be much stronger on the surface of the compressed Earth,
leading to a greater escape speed. Statements I and II are each false (they are inconsistent with the expression for ve).
Insight: Another approach is to note that in this case the rocket starts closer to the center of the Earth, and therefore
experiences a greater attractive force. It follows that a greater speed is required to overcome the increased force.
47. Picture the Problem: A rocket is launched vertically to a height h above the Earth’s surface.
Strategy: Consider the expressions for the mechanical energy of an orbiting rocket when answering the question.
Solution: 1. A rocket must supply energy to change its gravitational potential energy when it moves from the Earth’s
surface to an altitutde h. If it is to attain orbit the rocket needs an additional amount of kinetic energy in order to have
sufficient horizontal speed. We conclude that the energy required to launch a rocket vertically to a height h is less than
the energy required to put the same rocket into orbit at the height h.
Insight: It can be shown that a rocket in circular orbit has half as much kinetic energy as the magnitude of its
gravitational potential energy. That is, for circular orbits U   GMm r and K  GMm 2r .
48. Picture the Problem: The satellite loses gravitational potential energy and gains kinetic energy as it moves from
apogee (largest distance) to perigee (smallest distance to the center of the Earth).
Strategy: Find the change in kinetic energy as the satellite moves from perigee to apogee, and set it equal to minus the
change in the gravitational potential energy. Then use the change in gravitational potential energy, together with the
perigee distance, to find the apogee distance.
Solution: 1. Set Ea  Ep and solve for U a :
Ka  U a  Kp  U p
U a  Kp  Ka  U p

GMm 1 2 1 2  GMm 
 2 mvp  2 mva   


ra
rp 

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 16
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
2
2
1 1 va  vp
 
ra rp 2GM
2. Divide both sides by  GMm and solve for 1 ra :
 3640 m/s    4460 m/s 
1


7
2.00  10 m 2  6.67  1011 N  m 2 /kg 2  5.97  1024 kg 
2
2
 4.17  108 m 1
ra  1  4.17 108 m1   2.40 107 m  2.40 104 km
3. Now invert to find ra :
Insight: This problem is very similar to Active Example 12-2. As expected, the apogee distance is greater than the
perigee distance. If the satellite had a mass of 1000 kg, the kinetic energy difference would be K p  K a  3.32 GJ.
49. Picture the Problem: The impact on Mars gives the meteorite enough kinetic energy to escape the planet.
Strategy: Use an analog to equation 12-13 to find the escape speed from the surface of Mars. Use the mass and radius
information given in Appendix C.
Solution: Apply equation 12-13 for Mars:
2  6.67 1011 N  m 2 /kg 2  0.108  5.97 1024 kg 
2GM M

RM
3.394 106 m
ve 
 5030 m/s  5.03 km/s
Insight: The smaller radius of Mars makes it a bit more difficult to escape that planet when compared with the Earth,
but its smaller mass more than compensates and its 5.03 km/s escape speed is much smaller than the 11.2 km/s for the
Earth.
50. Picture the Problem: The Millennium Eagle starts from rest at
point A and then is accelerated by the gravitational force of
attraction from the two 3.50×1011 kg asteroids.
Strategy: Its speed at B is determined by its kinetic energy at B,
which equals the potential energy lost. The potential energy at
points A and B is twice the potential energy due to a single
asteroid. Find twice the change in potential energy due to the
approach of the Millennium Eagle to a single asteroid and set it
equal to the change in kinetic energy. Then use K  12 mv2
(equation 7-6) to find the speed of the spacecraft.
Solution: 1. Find the distance
to an asteroid from point A:
2. Set EA  EB and solve
for K B :
rA 
 3000 m   1500 m 
2
 3350 m
KA  U A  KB  U B
0  2G
1
2
3. Multiply by 2 m and
solve for vB :
2
mM 1 2
mM
 2 mvB  2G
rA
rB
1 1
mvB2  2GmM   
 rB rA 
1 1
vB2  4GM   
 rB rA 
1 
 1
vB  4  6.67  1011 N  m 2 /kg 2  3.50 1011 kg  


 1500 m 3350 m 
 0.185 m/s
Insight: The problem assumes the asteroids stay in place, although in reality they would attract each other and collide.
The occupants would have to be very patient; with the engine turned off it would take about 28 hours for the spacecraft
to travel the 3.00 km from point A to point B!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 17
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
51. Picture the Problem: The projectile rises from one Earth radius at launch to two Earth radii at the highest point of its
travel.
Strategy: Use conservation of energy to determine the initial kinetic energy required to change the distance of the
projectile from RE to 2RE from the center of the Earth. Then use equation 7-6 to find the initial speed of the projectile.
Ki  U i  K f  U f
Solution: 1. Set Ei  Ef and solve for Ki  12 mvi2 :
1
2
mvi2  G
mM E
mM E
 0G
RE
2 RE
1
2
vi 
2. Now multiply by 2 m and solve for vi :
mvi2 
GM E

RE
GmM E
RE
 6.67 10
 1
1  
 2
11
N  m 2 /kg 2  5.97 1024 kg 
6.37 106 m
 7910 m/s  7.91 km/s
Insight: It is not practical to achieve a launch speed this large (Mach 23!). Instead, spacecraft are accelerated gradually
and over a large distance by the impulse from a rocket motor.
52. Picture the Problem: The projectile rises from one Moon radius at launch to 365 km above the surface at the highest
point of its travel.
Strategy: Use conservation of energy to determine the initial kinetic energy required to change the distance of the
projectile from RM to RM  h from the center of the Moon. Then use equation 7-6 to find the initial speed of the
projectile.
Ki  U i  K f  U f
Solution: 1. Set Ei  Ef
and solve for Ki  mv :
1
2
2
i
2. Now multiply by 2 m
and solve for vi :
1
2
mvi2  G
mM M
mM M
 0G

RM
RM  h
1
2
 1
1 
mvi2  GmM M 


 RM RM  h 
 1
1 
vi  2GM M 


 RM RM  h 


1
1
 2  6.67  1011 N  m 2 /kg 2  7.35  10 22 kg  


6
6
3
 1.74  10 m 1.74  10  365  10 m 
vi  988 m/s
Insight: If instead you assume the acceleration of gravity is constant over the 365 km of the projectile’s flight (see
problem 43) you obtain a minimum launch speed of 1090 m/s. The actual minimum launch speed is less because the
acceleration of gravity decreases with increasing h.
53. Picture the Problem: The projectiles are given enough kinetic energy to escape the planet.
Strategy: Use an analog to equation 12-13 to find the escape speed from the surface of Mercury and Venus. Use the
mass and radius information given in Appendix C.
Solution: 1. (a) Apply equation 12-13
for Mercury:
2. (b) Apply equation 12-13 for Venus:
ve 
2  6.67 1011 N  m 2 /kg 2  0.0553  5.97 1024 kg 
2GM M

RM
2.44 106 m
 4250 m/s  4.25 km/s
ve 
2GM V

RV
2  6.67 1011 N  m 2 /kg 2  0.816  5.97  1024 kg 
6.052  106 m
 10400 m/s  10.4 km/s
Insight: The radius of Mercury equals 0.383RE . If instead its radius were 0.0553RE and it had the same mass, the
escape velocity from Mercury would be 11.2 km/s, the same as it is from Earth.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 18
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
54. Picture the Problem: The comet gains gravitational potential energy and loses kinetic energy as it moves from perigee
(smallest distance) to apogee (largest distance to the center of the Earth).
Strategy: Find the change in gravitational potential energy as the comet moves from perigee to apogee, and set it equal
to minus the change in the kinetic energy. Then use the change in kinetic energy, together with the speed at perigee, to
find the speed at apogee.
Solution: 1. (a) As the comet moves from perigee to apogee, it gains gravitational potential energy and loses kinetic
energy. Therefore we expect its speed at apogee to be less than its 54.6 km/s speed at perigee.
2. (b) Set Ea  Ep
Ka  U a  Kp  U p
Ka  Kp  U p  U a
and solve for K a :
1
2
3. Multiply both sides
by 2 m and solve for va :
 GMm   GMm 
mva2  12 mvp2   



rp  
ra 

1 1
va2  vp2  2GM   
r r 
p 
 a
va 
 54.6 10
3
m/s 
2
1
1


 2  6.67  1011 N  m 2 /kg 2  2.00 1030 kg  


12
10
 6.152  10 m 8.823 10 m 
 783 m/s
Insight: This problem is very similar to Active Example 12-2. As expected the apogee speed is less than the perigee
speed. If the comet has a mass of 2.6×1014 kg, the kinetic energy difference would be Kp  Ka  3.9 1023 J.
55. Picture the Problem: The Lunar Module loses gravitational potential energy and gains kinetic energy as it crashes to
the surface of the Moon.
Strategy: Set the mechanical energy at the jettison point equal to the mechanical energy at the crash point. The Lunar
Module will be traveling faster than 1630 m/s because it will gain kinetic energy as it loses gravitational potential
energy.
Solution: 1. Set Ei  Ef and
simplify the expression:
2. Now solve for vf :
Ki  U i  Kf  U f
M m
M m
1
mvi2  G M  12 mvf2  G M
2
RM  h
RM
2
GM
2
GM
M
M
vi2 
 vf2 
RM  h
RM
 1
1 
vf  vi2  2GM M 


 RM RM  h 
2GM M h
 vi2 
RM  RM  h 

1630 m/s 
2

2  6.67  10 11 N  m 2 /kg 2  7.35 10 22 kg 110 103 m 
1.74 10 m 1.85 10 m 
6
6
vf  1730 m/s  1.73 km/s
Insight: After falling 110 km toward the Moon, the Lunar Module is only traveling 6.1% faster than at jettison. If it fell
from that height from rest, it would be traveling 579 m/s when it impacted the lunar surface.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 19
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
56. Picture the Problem: The projectile is launched at 11.2 km/s from the surface of the Earth, but slows down as it
converts its initial kinetic energy into gravitational potential energy.
Strategy: Set the mechanical energy at the launch equal to the mechanical energy when the speed is equal to the onehalf the escape speed. Then determine the gravitational potential energy and therefore the distance such a projectile
would be from the center of the Earth.
Solution: 1. Set Ei  Ef and simplify the
expression by multiplying both sides by 2 m :
Ki  U i  Kf  U f
M m
M m
1
mvi2  G E  12 mvf2  G E
2
RE
rf
2GM E 2GM E
2

 vf  vi2
rf
RE
2. Substitute vi  ve  2GM E RE and
2GM E 2GM E  1 2GM E   2GM E 

 4


rf
RE
RE   RE 

1
1

 rf  4 RE  4  6.37  106 m   2.55 107 m
rf 4 RE
vf  12 ve 
1
2
2GM E RE and solve for rf :
Insight: It may be surprising that the projectile travels a distance of 4 Earth radii before it has slowed down from the
escape speed (11.2 km/s) to half that value (5.6 km/s). It will be at 9RE before its speed decreases to 13 ve .
57. Picture the Problem: The planet is ten times more massive and has one-tenth the radius of Earth. A projectile at its
surface is given sufficient kinetic energy to escape the planet.
Strategy: Use a ratio to compare the escape speed on the new planet with the escape speed on Earth. Use equation
12-13 to form the ratio.
Solution: Write a ratio of the escape speeds:
ve,new
ve,Earth

2GM new Rnew
2GM E RE

M new RE

M E Rnew
10M E  RE
M E  101 RE 
 100  10
ve,new  10 ve,Earth
Insight: Although the speed only needs to be increased by a factor of 10, the initial kinetic energy of a projectile needs
to be 100 times larger on this new planet in order for the projectile to escape that planet’s gravity.
58. Picture the Problem: The projectile is launched at 1050 m/s from the surface of the Moon, but slows down as it
converts its initial kinetic energy into gravitational potential energy.
Strategy: Set the mechanical energy at the launch equal to the mechanical energy when the speed is equal to the onehalf the launch speed. Then determine the gravitational potential energy and therefore the distance such a projectile
would be from the center of the Moon.
Solution: 1. Set Ei  Ef and
simplify the expression by
multiplying both sides by 2 m :
2. Substitute vf  12 vi and solve for rf :
1
2
Ki  U i  Kf  U f
M m
M m
mv  G M  12 mvf2  G M
RM
rf
2GM M 2GM M

 vf2  vi2
rf
RM
2
i
2GM M 2GM M 1 2 2

 4 vi  vi
rf
RM
3 2
v
1
1

 4 i
rf RM 2GM M
3
1
4 1050 m/s 

6
11
1.74  10 m 2  6.67  10 N  m 2 /kg 2  7.35 10 22 kg 
2

1
1

 rf  2.04  106 m
rf 2.04  106 m
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 20
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
h  rf  RM  2.04  106  1.74  106 m  3.0  105 m  300 km
3. Subtract RM from rf to find h:
Insight: The speed is cut in half after traveling a distance equivalent to 17% of the Moon’s radius (that is, rf  1.17 RM ).
Note that this distance depends upon the initial speed, so that a large initial speed corresponds to a large rf .
59. Picture the Problem: The radius of the Sun shrinks, but its mass remains the same, until the escape speed from its
surface equals the speed of light.
Strategy: Set the escape speed (equation 12-13) equal to the speed of light, and solve for the radius R.
Solution: Set ve  c and solve for R:
vesc  c 
R
2GM S
R
11
2
2
30
2GM S 2  6.67  10 N  m /kg  2.00  10 kg 

2
c2
 3.00 108 m/s 
 2.96 km
Insight: This radius corresponds roughly to the size of a single mountain on Earth’s surface! When a black hole forms,
the theory indicates that there is nothing to prevent a further collapse under its own weight, and the entire black hole
shrinks into an infinitesimally small volume.
60. Picture the Problem: The two baseballs attract each other gravitationally, losing potential energy and gaining kinetic
energy as they approach each other.
Strategy: Find the potential energy when the balls are released from rest and the potential energy when the balls are
separated by 145 m. The difference in the potential energies is the gain of kinetic energy of both balls. Because the
balls have the same mass, they each gain half the kinetic energy.
Solution: 1. (a) Set Ei  Ef
and solve for v:
Ki  U i  Kf  U f
0G
m2
m2
 2  12 mv 2   G
ri
rf
1 1
mv 2  Gm 2   
 rf ri 
1 1
v  Gm    
 rf ri 
 6.67 10
11
1 
 1
N  m 2 /kg 2   0.148 kg  


 145 m 395 m 
v  2.07  107 m/s
2. (b) If the mass of each ball is doubled, the speed found in part (a) would increase by a factor of
2 as can be seen by
the formula in step 1 that shows v  m .
Insight: Doubling the initial separation distance to 790 m but keeping everything else the same results in a speed of
2.35×10−7 m/s, about 14% faster than the original problem. To increase the speed you are better off doubling the mass!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 21
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
61. Picture the Problem: The person jumps straight upward from the surface of an asteroid with the same initial speed that
she has when she jumps from the surface of the Earth and rises to a height h.
Strategy: Because the height that people can jump (elevating their center of mass about half a meter) is tiny compared
with the radius of the Earth, we can safely claim the acceleration of gravity is constant over the height of the jump and
determine the initial speed of the jump by using either the energy methods of Chapter 7 or the projectile methods of
Chapter 4. Then set this initial speed equal to the escape speed of an asteroid by using equation 12-13, relate the mass
to the volume and density of the asteroid using equation 15-1, and solve for the asteroid radius RA .
Solution: 1. Use equations 4-6 to find
the jumping speed v0 :
2. Set the jumping speed equal to the
escape speed:
3. Now substitute M A  VA    34  RA3 
v 2  v02  2 g y
0  v02  2 gh  v0  2 gh
2GM A
RA
2G
2 gh 
MA
RA
2 gh 
gh 
(equation 15-1) and solve for RA :
G
  34  RA3   34  G  RA2
RA
RA 

3  9.81 m/s 2   h
3gh

4 G 
4  6.67 1011 N  m 2 /kg 2  3500 kg/m 3 
1.00 10 m  h
7
Insight: If a person can jump 0.50 m vertically on the Earth, they could escape from an asteroid of radius 2.2 km by
simply leaping vertically with the same speed!
62. Picture the Problem: You are 1 million miles away from a black hole whose mass is a million times that of the Sun.
Strategy: The gravitational force from the black hole pulls harder on the portion of your body that is closer to it than it
does on the portion of your body that is furthest from it. This produces a tidal force on your body that tends to stretch it
out along a radial line from the black hole. Estimate your mass to be 70 kg and your length to be 1.8 m (5 ft 11 in).
Solution: 1. (a) Apply the given equation
to find the tidal force on your body:
F  4GmMa r 3
 4  6.67  10
11
N  m /kg
2
2
 70 kg  106  2.00 1030

10
6
kg  1.8 m 
mi  1609 m/mi 
3
F  16 N  3.6 lb
1/ 3
2. (b) Solve the given expression for the
distance r from the black hole if F  10mg :
 4GmMa 
r

 F 
1/ 3
 4GmMa 


 10mg 
1/ 3
 4GMa 


 10 g 
 4  6.67  1011 N  m 2 /kg 2  2.00  1036 kg  1.8 m  


10  9.81 m/s 2 


r  2.1 108 m  1.3  105 mi
1/ 3
Insight: At a million miles (about 4 times the distance from the Earth to the Moon) the tidal force is just a few pounds,
but at 130,000 miles (about half the distance from the Earth to the Moon) the stretching force is 10 times your weight!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 22
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
63. Picture the Problem: The dumbbell is aligned radially with the center of the Earth and its center is a distance r from
the center of the Earth.
Strategy: The gravitational force of attraction is a little bit stronger of the end of the dumbbell that is closest to the
Earth and a little weaker on the farthest end. This difference in force tends to stretch the dumbbell along its length. Use
Newton’s Universal Law of Gravitation (equation 12-1) to find the difference in force between the two ends of the
dumbbell.
Solution: 1. Use equation 12-1 to find F :
2. Now if r  a,
1
r  a
2

1
r  a
2

4a
:
r3
F  G
mM E
r  a
2
G
 1
1 
 GmM E 


2
2
  r  a   r  a  
mM E
r  a
2
 4a  4GmM E a
F  Ftidal  GmM E  3  
r3
r 
Insight: The tidal force decreases with 1 r 3 , faster than the magnitude of the gravitational force decreases ( 1 r 2 ). It is
significant only when the masses are large (such as with the Moon and the Earth) or the separation distance is small.
64. Picture the Problem: The two spherical masses are touching each other and are aligned radially with the center of the
Earth. Their center of mass is a distance r from the center of the Earth.
Strategy: Use Newton’s Universal Law of Gravitation to find the force between the two masses, substituting the
expression m  34  a3  given in the problem that relates the mass of the spheres to their radii a and density  . Then
set that force equal to the tidal force expression, F  4GmM E a r 3 found in problem 63 and solve for r to find the
Roche limit.
Solution: 1. (a) Use equation 12-1 for the two masses
separated by distance 2a and let m  34  a3  :
3a  
m2
F G
G
2
(2a)
4a 2
2. (b) Set the force from step 1 equal to F
and solve for r:
G
3
4
2
4
9
G 2 a 4  2
m 2 4GmM E a

4a 2
r3
1/ 3
 16a 3 M E 
r 

 m 
1/ 3
 16a 3 M 
 4 3 E 
 3a  
1/ 3
 12M E 
 

  
24
 12M S  12  95.1 5.97 10 kg  


rS  


  3330 kg/m3 

   
1/ 3
3. (c) Find the Roche limit for Saturn using the
mass given in Appendix C:
 8.67 107 m
Insight: Inside of the Roche limit the tidal force that tends to pull masses apart is larger than the gravitational force that
tends to hold them together. That is why the small rocks in Saturn’s rings don’t coalesce into large moons.
65. Picture the Problem: You weigh yourself on a scale inside an airplane flying due east above the equator. The airplane
then turns around and heads due west with the same speed.
Strategy: Consider the speed of the aircraft relative to the center of the Earth when answering the conceptual question.
Solution: If you fly to the east, which is the direction of the Earth’s rotation, you have a greater speed relative to the
center of the Earth than if you fly to the west. As a result, the centripetal force required to maintain your circular motion
is greater, and your apparent weight is less. We conclude that the reading on the scale will increase slightly when the
airplane turns around and heads due west with the same speed relative to the Earth’s surface.
Insight: There would be no difference at all in the scale reading if the plane were flying due north and then turned
around and flew due south. In both such cases the airplane has the same speed relative to the center of the Earth.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 23
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
66. Picture the Problem: Three objects are arranged in the manner shown at right.
Strategy: Use the Universal Law of Gravitation to determine the relative
magnitudes of the forces experienced by each object.
Solution: 1. Find the force on object A:
FA 
2. Find the force on object B:
FB 
3. Find the force on object C:
FC 
GM  2M 
L2

GM  3M 
G  2M  3M 
L2
G  3M  2M 
L2
 2L 


2

11 GM 2
4 L2
GM  2M 
L2
GM  3M 
 2L 
2
4

GM 2 16 GM 2

4 L2
L2
27 GM 2
4 L2
4. By comparing the magnitudes of the forces we arrive at the ranking, A < B < C.
Insight: If the object C were instead to switch places with object B it would experience a force of
either object A

7
2

GM 2 L2 or object B

13
2

6
2
GM 2 L2 , less than
GM 2 L2 , so the new ranking would be C < A < B.
67. Picture the Problem: Three objects are arranged in the manner shown at right.
Strategy: Use the Universal Law of Gravitation to determine the relative
magnitudes of the accelerations experienced by each object. Refer to the previous
problem for the force magnitudes.
Solution: 1. Find the acceleration on object A:
aA 
FA 11 GM 2 L2 11 GM 33 GM



mA
4
M
4 L2
12 L2
2. Find the acceleration of object B:
aB 
FB 4GM 2 L2
GM 24 GM

2 2 
mB
2M
12 L2
L
3. Find the force on object C:
aC 
FC 27 GM 2 L2 27 GM


mC
4
3M
12 L2
4. By comparing the magnitudes of the accelerations we arrive at the ranking, B < C < A.
Insight: If the object C were instead to switch places with object B it would experience an acceleration of GM L2 , less
than either object A

14
4

GM L2 or object B

13
4

GM L2 , so the new ranking would be C < B < A.
68. Picture the Problem: The Moon’s path is affected by the gravitational attraction of
both the Earth and the Sun, but the force exerted by the Sun on the Moon is more
than twice the force exerted by the Earth on the Moon.
Strategy: Consider the relative magnitudes of the forces exerted on the Moon by
the Sun and by the Earth to answer the conceptual question.
Solution: 1. The net force acting on the Moon must point toward the Sun. This is due to the fact that the Sun exerts
more force on the Moon than does the Earth. This fact is mentioned in Conceptual Question 12 and can be verified by
the following calculations:
2. Find the force on the
Moon by the Sun:
FM-S

M M
 G M2 S  6.67 1011 N  m 2 /kg 2
rM-S
 7.35 10 kg  2.00 10

1.50 10 m 
22
11
30
kg

2
 4.35 1020 N
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 24
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
3. Find the force on the
Moon by the Earth:
FM-E

M M
 G M2 E  6.67 1011 N  m 2 /kg 2
rM-E
 7.35 10 kg 5.97 10

3.84 10 m 
22
8
24
kg

2
 1.98 1020 N
Insight: The path of the Moon must always be curved toward the Sun because the net gravitational force on the Moon is
always pointed toward the Sun. As shown in the diagram above, the Moon’s path curves more sharply toward the Sun
when both the Sun and the Earth pull inward on the Moon, and curves only slightly toward the Sun when the Moon is
pulled in opposite directions by the Sun and the Earth, as is the case during the new-moon position.
69. Picture the Problem: A satellite goes through one complete orbit of the Earth.
Strategy: Consider the mechanical energy of the orbiting satellite when answering the conceptual question.
Solution: 1. (a) In one complete orbit, a satellite returns to its starting point, resulting in zero net displacement. Since
gravity is a conservative force, the net work done on the satellite by the Earth’s gravitational force is zero.
2. (b) No, the answer to part (a) is independent of the shape of the orbit—all that mattered was that the orbit was closed.
Insight: For an elliptical orbit gravitational potential energy is converted to kinetic energy and back again, but the net
work done by gravity is zero. For a circular orbit the force of gravity does no work whatsoever because the force is
always perpendicular to the direction of motion.
70. Picture the Problem: The Skylab spacecraft fell back to Earth out of orbit due to the friction it experienced in the upper
reaches of the Earth’s atmosphere.
Strategy: Consider the mechanical energy of the orbiting satellite when answering the conceptual question.
Solution: As the radius of Skylab’s orbit decreased its speed did increase because the gain in its kinetic energy due to
the loss of gravitational potential energy was much larger than the loss of its kinetic energy due to friction. Another
approach is to recall that T   constant  r 3/2 (Kepler’s Third Law) and that v  2 r T (circular motion). It follows
that v   constant  r 1 2 , and therefore the speed increases with decreasing radius.
Insight: You might think that friction would slow Skylab—just like other objects are slowed by friction—but by
dropping Skylab to a lower orbit, and freeing up gravitational potential energy, friction is ultimately responsible for an
increase in speed.
71. Picture the Problem: The three masses are positioned as indicated in the
figure at right.
Strategy: As in Example 12-5, the total gravitational potential energy is
determined by summing the potential energy of each pair of masses. In this
case the pairs are 1−2, 1−3, and 2−3. Use equation 12-9 to sum the
contributions of these three terms to the total gravitational potential energy.
Solution: Add the six terms
that contribute to U total :
1
1.00
2
2.00
3
x
3.00 m
 mm
mm
mm 
U total   G  1 2  1 3  2 3 
 x2  x1 x3  x1 x3  x2 
   6.67  10 11
 1.00 kg  2.00 kg  1.00 kg  3.00 kg  



1.00 m
2.00 m

N  m 2 /kg 2  


2.00 kg  3.00 kg 




1.00 m


U total   6.34  1010 J
Insight: The potential energy is a tiny − 0.634 nJ because the masses are relatively small and so are the attractive forces
between them.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 25
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
72. Picture the Problem: The astronaut jumps straight upward, momentarily comes to rest, and falls back to the surface of
the planet.
Strategy: The height of the astronaut’s jump is very small compared with the radius of the planet, so we assume that the
acceleration of gravity is constant during the jump. Use conservation of mechanical energy (equation 8-8) to find an
expression for the acceleration of gravity as a function of the initial speed and the maximum jump height, and set it
equal to GM R 2 according to equation 12-4. Then solve for the mass of the planet.
Solution: 1. Set Ei  Ef and solve for g:
Ki  U i  K f  U f
1
2
mvi2  0  0  mgh
g  vi2 2h
vi2 GM
 2
2h
R
2. Now let g  GM R 2 and solve for M:
 3860 103 m   3.10 m/s 
R 2 vi2
M 

 1.85  1024 kg
2Gh 2  6.67 1011 N  m 2 /kg 2   0.580 m 
2
2
Insight: As the height h increases for the same initial speed vi , the mass of the planet decreases. A less massive planet
will not accelerate the astronaut as much and she will rise to a greater height.
73. Picture the Problem: The Sun, Moon, and Earth are arranged as indicated
in the diagram at right.
Strategy: Use Newton’s Law of Universal Gravitation (equation 12-1)
together with the data given in the inside back cover of the book to find the
forces exerted by the Earth and by the Sun on the Moon.
Solution: 1. (a) Apply equation 12-1:
FE  G
MMME
rE-M 2
  6.67  1011 N  m 2 /kg 2 
 7.35 10 kg  5.97 10
 3.84 10 m 
22
8
24
kg 
2
FE  1.98  1020 N
2. (b) Apply equation 12-1 again:
FS  G
M M MS
rS-M 2
  6.67  1011 N  m 2 /kg 2 
 7.35 10 kg  2.00 10 kg 
1.50 10 m    3.84 10 m 
22
11
30
2
8
2
FS  4.36  1020 N
3. (c) As the force exerted by the Sun is more than twice the force exerted by the Earth on the Moon, it makes more
sense to think of the Moon as orbiting the Sun, with a small effect due to the Earth.
Insight: It takes twice as much force to keep the Moon orbiting the Sun once per year as it does to keep the Moon
orbiting the Earth once a month.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 26
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
74. Picture the Problem: The three masses attract each other gravitationally
and are positioned as shown in the figure at right.
Strategy: Determine the magnitudes of the forces from the 2.00 kg ( F2 )
and 3.00 kg ( F3 ) masses using Newton’s Law of Universal Gravitation.
Use the magnitudes and directions of the two forces to add them using the
component method of vector addition. Combine the components of the
vector sum to find the magnitude and direction of the net force on the
1.00-kg mass. Let downward and to the left be the positive directions.
1.00 kg  2.00 kg 
m1m2
  6.67 1011 N  m2 /kg 2 
 1.33 1012 N
2
r2
10.0 m 
Solution: 1. Find the magnitude of F2 :
F2  G
2. Find the magnitude of F3 :
F3   6.67 1011 N  m2 /kg 2 
3. The horizontal components
subtract and the vertical
components add:
1.00 kg  3.00 kg 
 2.00 1012
2
10.0 m 
N
F1x  F3 x  F2 x   2.00 1012 N  cos 60  1.33 10 12 N  cos 60
 3.35 1013 N (to the left)
F1 y  F3 y  F2 y   2.00 1012 N  sin 60  1.33 1012 N  sin 60
 2.88 1012 N (downward)
4. Combine the components
to find the magnitude and
the direction:
F1  F1x 2  F1 y 2 
  tan 1
 3.35 10
13
N    2.88  1012 N   2.90  1012 N
2
2
 2.88  1012 N 
 tan 1 
  83.4 below horizontal, to the left
13
F1x
 3.35  10 N 
F1 y
Insight: The net force points more toward the 3.00-kg mass because it exerts the larger force on the 1.00 kg mass.
75. Picture the Problem: The masses are initially at the vertices of an equilateral triangle 10.0 m on a side. They are then
released from rest and collide together at the center.
Strategy: Find the potential energy of the three masses upon release and at the instant they collide. There are three
identical terms from each of the 3 pairs of masses. When they collide, the center-to-center distance between pairs of
spheres is 2×0.0714 m = 0.143 m. The loss of potential energy equals the gain in kinetic energy, which is shared
equally among the three identical masses. Determine the speed of each mass from its kinetic energy when it collides
with the other two.

11
2
2  5.95 kg 
  7.08 10 10 J
  3   6.67 10 N  m /kg 
10.0
m

Solution: 1. Find U i
using equation 12-9:
 Gm 2
Ui  3 
r

2. Find U f in a similar fashion:
U f  3   6.67 1011 N  m 2 /kg 2 
3. Set Ei  Ef to find K f :
Ki  U i  K f  U f
4. Set K f  3  mv
2 Kf
v 
3m
1
2
2
 and solve for v:
2
 5.95 kg 
0.143 m
2
  4.95 10 8 J
Kf  U i  U f  0    7.08 1010 J     4.95 108 J   4.88 108 J
2
2  4.88 108 J 
2 Kf
 v

 7.39 105 m/s  73.9  m/s
3m
3  5.95 kg 
Insight: You’d have to be patient to watch this experiment unfold; each mass travels about 5.7 m before colliding at the
center, and at an average speed of 12  73.9  m/s  , the trip will take almost 43 hours!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 27
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
76. Picture the Problem: The asteroid was at rest infinitely far away from the Earth, and then picked up speed as it
approached, losing gravitational potential energy and gaining kinetic energy.
Strategy: Find the gravitational potential energy at an infinite distance (zero) and at the distance of closest approach
(using equation 12-8). The loss of potential energy equals the gain in kinetic energy, from which the speed of the
asteroid at the distance of closest approach can be determined. The kinetic energy can be determined using equation 7-6
when the mass of the asteroid is estimated. Density   m V is defined in equation 15-1.
Solution: 1. (a) Set Ei  Ef and solve for v:
Ki  U i  K f  U f
K f  Ki  U i  U f  0  0  U f
  GmA M E 
1
mA v 2   

2
r


v
2  6.67 1011 N  m 2 /kg 2  5.97 1024 kg 
2GM E

r
 73,600 mi 1609 m/mi 
 2590 m/s  2.59 km/s
2. (b) Let m   V    34  R 3  and find K f :
K f  12 mv 2  12  V v 2  12   34  R 3  v 2

1
2
 3330 kg/m     
3
4
3
1
2
 2000 m   2590 m/s   4.7 1019 J
3
2
Insight: The kinetic energy of the asteroid at closest approach is equivalent to 8,400 one-megaton nuclear weapons!
The energy would be even larger (8.7×1020 J or 156,000 bombs) if the asteroid actually struck the Earth because it
would have lost even more gravitational potential energy as it approached the Earth’s center.
77. Picture the Problem: The planet has the same density (mass/volume) as the Earth but half the radius.
Strategy: The acceleration of gravity for any planet is given by equation 12-4, g  GM R 2 . Density   m V is
defined in equation 15-1. Combine these concepts to find a relation between the acceleration of gravity and the radius
of a planet for a given density  .
Solution: 1. (a) Substitute m   V    34  R 3  into equation 12-4:
GM G V G  34  R


 34 G  R
R2
R2
R2
3
g
2. As can be seen from the expression in step 1, the acceleration of gravity is linearly proportional to the radius R of a
planet for a given density  . If the radius of the planet is half the radius of the Earth but the density is the same, the
acceleration of the gravity at the surface of the planet will be less than it is on Earth.
g
3. (c) Because the radius is cut in half, g is cut in half too:
1
2
9.81 m/s   4.91 m/s
2
2
Insight: The overall mass of the planet is reduced when the volume is reduced but the density remains the same.
78. Picture the Problem: The planet has the same mass as the Earth but half the radius.
Strategy: The acceleration of gravity for any planet is given by equation 12-4, g  GM R 2 . Use this relation to
determine the change in g when the radius is cut in half while the mass remains the same.
Solution: 1. (a) Use a ratio to find g new :
2
2
g new GM Rnew
Rold


 4  g new  4 g old
2
2
g old
GM Rold
 12 Rold 
2. The acceleration of gravity on this planet is more than the acceleration due to gravity on the Earth.
3. (b) As shown in step 1, g is quadrupled:
g  4  9.81 m/s2   39.2 m/s2
Insight: The mass is the same for this new planet but every object on its surface is closer to the center of the planet, and
the gravitational force is thus increased fourfold.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 28
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
79. Picture the Problem: The satellite is in a circular orbit at a height h above the surface of the Earth.
Strategy: This satellite is a distance RE  h from the center of the Earth. Set the gravitational force of attraction
between the satellite and the Earth (equation 12-1) equal to the centripetal force required to keep it in circular orbit
(equation 6-16). Solve the resulting expression for the orbit speed v.
 v2 
MEm
m
G
2
 RE  h 
 RE  h 
Solution: Set Fcentripetal  Fgravity and solve for v:
v2 
GM E
 v
RE  h
GM E
RE  h
Insight: The higher the altitude h, the slower the orbital speed. The Moon orbits the Earth with a speed of about
1.0 km/s, much slower than the 7.9 km/s orbit speeds of near-Earth satellites.
80. Picture the Problem: The stars orbit their common center of mass as
depicted in the figure at right.
Strategy: The stars exert forces on each other according to Newton’s
Universal Law of Gravitation. The forces they exert on each other must be
equal and opposite by Newton’s Third Law. Set the forces equal to each
other and to the centripetal force required to keep them moving in circular
paths.
Solution: 1. Set the centripetal forces equal to each other:
m1a1  m2 a2
m1  v r1   m2  v22 r2 
2
1
m1
2. Now let v  2 r T and solve for m2 m1 :
 
2 r1 2
T
r1

m2
 
2 r2
T
2
r2
m1r1  m2 r2
m2 r1
r
1
  1 
m1 r2 2r1
2
Insight: The more massive star m1 moves in a smaller orbit than the less massive star m2 .
81. Picture the Problem: The stars orbit their common center of mass as
depicted in the figure at right.
Strategy: The stars exert forces on each other according to Newton’s
Universal Law of Gravitation (equation 12-1). The forces they exert on
each other must be equal and opposite by Newton’s Third Law. Set the
gravitational force between the two stars equal to the centripetal force
required to keep star 1 moving in a circular path (equation 6-16).
 v 2  Gm1m2
m1  1  
2
 r1   r1  r2 
Solution: 1. Set Fcentripetal  Fgravity :
2. Divide both sides by m1 and substitute v  2 r T and r2  2r1 :
 2 r1 T 
r1
2

Gm2
 r1  2r1 
2
4 2 r1 G  12 m1 

 T
2
T2
 3r1 
3. Let m2  12 m1 (from problem 80) and solve for T:
72 2 r13
Gm1
Insight: The period increases with increasing orbit distance r1 and decreases with increasing mass m1 .
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 29
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
82. Picture the Problem: The comet gains gravitational potential energy and loses kinetic energy as it moves from perigee
(smallest distance) to apogee (largest distance to the center of the Earth).
Strategy: Use the mass (9.8×1014 kg), perigee and apogee distances, and the speeds at perigee and apogee to determine
the angular momentum of the comet using equation 11-12. From problem 54 we find that ra  6.152  1012 m,
va  783 m/s, rp  8.823 1010 m, and vp  54, 600 m/s.
Lp  mvp rp   9.8  1014 kg   54, 600 m/s  8.823 1010 m 
Solution: 1. (a) Apply equation 11-12:
 4.7  1030 kg  m 2 s
La  mva ra   9.8  1014 kg   783 m/s   6.152 1012 m 
2. (b) Repeat for aphelion:
 4.7  1030 kg  m 2 s
Insight: The angular momentum of an orbiting object is conserved. This is the foundational principle behind Kepler’s
Second Law, the law of equal areas.
83. Picture the Problem: The orbit period of a geosynchronous satellite matches the rotational period of the planet.
Strategy: This problem is similar to Active Example 12-1. Set the orbit period equal to the rotational period of the
planet and solve for distance r and then the altitude h. The rotational period of Mars is
24.6229 h  3600 s/h  8.86424 104 s. Use the mass and radius data given in Appendix C.
4 2 r 3
GM
1/ 3
 GMT 2 
r 
2 
 4 
1/ 3
  6.67  10 11 N  m 2 /kg 2  0.108  5.97  10 24 kg 8.86424 10 4 s 2 




4 2


r  2.05  107 m
Solution: 1. Square both sides of
equation 12-7 and solve for r:
T2 
2. Solve r  R  h for h:
h  r  R  2.05  107 m  0.3394  107 m  1.71 107 m
Insight: This distance is less than half of the geosynchronous orbit radius of 4.22×107 m for the Earth because although
the rotation periods of the two planets are nearly equal, Mars has much less mass than does the Earth.
84. Picture the Problem: The satellite travels in a circular orbit with a radius 1000 miles greater than that of a
geosynchronous satellite.
Strategy: Use Kepler’s Third Law in the form of equation 12-7 to relate the orbit period to the orbit radius. In this case
the orbit radius is RE  h  6.37 106 m   23,300 mi 1609 m/mi   4.39 107 m.
Solution: 1. (a) Because its altitude is higher than geosynchronous, it has a smaller orbit speed (see problem 79), and its
period is greater than 24 hours. See also Kepler’s Third Law (equation 12-7).
2. (b) The satellite lags behind the Earth’s eastward rotation and thus moves westward.
3. (c) Apply equation 12-7 directly:
 2  3 / 2
T 
r
 GM 
E 



2

 4.39  106 m 3 / 2

 6.67  1011 N  m 2 /kg 2 5.97 10 24 kg  

 
 
T  91, 600 s  25.4 h
Insight: A careful analysis of the uncertainties shows that in order to have an orbit period that is within 1.0 minute of
24.00 hours, the altitude of the satellite must be within 12 miles of the 22,300 mi target altitude!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 30
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
85. Picture the Problem: The Millennium Eagle starts at point A and then
is accelerated by the gravitational force of attraction from the two
3.50×1011 kg asteroids, arriving at point B with a speed of 0.985 m/s.
Strategy: Its speed at B is determined by its kinetic energy at B, which
equals the potential energy lost. The potential energy at points A and B is
twice the potential energy due to a single asteroid. Find twice the change
in potential energy due to the approach of the Millennium Eagle to a single
asteroid and set it equal to the change in kinetic energy. Then use
K  12 mv2 (equation 7-6) to find the speed of the spacecraft at point A.
Solution: 1. Find the distance
to an asteroid from point A:
2. Set EA  EB and solve
for K B :
rA 
2
 1500 m   3350 m
2
KA  U A  KB  U B
1
2
mvA2  2G
1
2
3. Multiply by 2 m and
solve for vA :
 3000 m 
mM 1 2
mM
 2 mvB  2G
rA
rB
1 1
mvB2  12 mvA2  2GmM   
 rB rA 
1 1
vA2  vB2  4GM   
 rB rA 
vA 
 0.905 m/s 
2
1 
 1
 4  6.67  1011 N  m 2 /kg 2  3.50 1011 kg  


1500
m
3350
m

 0.886 m/s
Insight: The problem assumes the asteroids stay in place, although in reality they would attract each other and collide.
In this problem the occupants would have to be very patient; at an average speed of 0.895 m/s it would take about 56
minutes for the spacecraft to travel the 3.00 km from point A to point B!
86. Picture the Problem: The Moon is attracted gravitationally to both the Earth and the Sun.
Strategy: Use Newton’s Law of Universal Gravitation to find the force of attraction between the Moon and the Sun
when the Moon is at its farthest distance from the Sun (at the full moon phase) and compare it to the force of attraction
between the Moon and the Earth.
Solution: 1. Find FE
using equation 12-1:
FE  G
MEMM
RE-M 2
  6.67  1011 N  m 2 /kg 2 
2. Find the minimum force
between the Moon and the Sun:
FS,min  G
 5.97 10 kg  7.35 10
 3.84 10 m 
MSM M
RS M 2
  6.67  1011 N  m 2 /kg 2 
24
8
22
kg 
2
 1.98  1020 N
 2.00 10 kg  7.35 10 kg   4.34 10
1.50 10  3.84 10 m 
30
22
11
8
20
2
N
3. By comparing the forces in steps 1 and 2 we can see that even at its weakest, the Sun’s pull on the Moon is greater
than the Earth’s pull on the Moon.
Insight: The force required to keep the Moon in orbit around the Sun once a year is greater than the force required to
keep the Moon in orbit around the Earth once a month!
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 31
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
87. Picture the Problem: Kepler’s Third Law relates the orbit period to the orbit distance or radius.
Strategy: Start with equation 12-7 and then convert the units for each of the variables in order to determine the
constant C.
Solution: 1. Begin with equation
12-7, which applies to any mass
that is orbiting the Sun:
 2  3 / 2
T 
r
 GM 
S 



2


GM S   6.67  1011 N  m 2 /kg 2  2.00 1030 kg  


 5.44 1010 s  m 3 2
2
2. Determine the proportionality
constant in SI units:
3. Convert units if T is in years and
r is in A.U.:

 3.16  107 s 
 1.50  1011 m  
10
3 2
T  yr  

5.44

10
s

m
r
AU








yr
AU





 3.16  107 s 
3/ 2
7
3 / 2

 T  yr    3.16  10 s  AU  r  AU 
yr


yr 
3/ 2

T  yr   1.00
r AU 
3/2  
AU 

3/ 2
4. We conclude the constant C  1.00 and Kepler’s Third Law can be compactly written, T  r 3 2 if T is measured in
years and r is measured in astronomical units.
Insight: This version of Kepler’s Third Law is based upon the fact that the Earth orbits the Sun at a distance of one
A.U. and has a period of one year.
88. Picture the Problem: The satellite travels in a circular orbit about the Earth.
Strategy: Set the centripetal force required to keep the satellite in orbit equal to the gravitational force of attraction
between the satellite and the Earth. Solve for v 2 and multiply by 12 m to determine the kinetic energy of the satellite.
Finally, find the difference in mechanical energy between the 12,600-mile orbit and the 25,200-mile orbit to find the
energy required to boost the orbit.
Solution: 1. (a) Set Fcentripetal  Fgravity and solve for v 2 :
2. Multiply by
1
2
m to find K:
 v2 
M m
M
m    G E2
 v2  G E
r
r
 r 
GM E m
2r
11
6.67

10
N  m 2 /kg 2  1720 kg   5.97 10 24 kg 

K  12 mv 2 

2 12, 600 mi  1609 m/mi 
K  1.69  1010 J  16.9 GJ
3. (b) Find an expression for the total mechanical
energy of a satellite in circular orbit:
4. Take the difference in mechanical energy
between the two orbits:
E  K U 
GMm
Mm
GMm
G

.
2r
r
2r
1 1
1
E   GMm   
2
 rf ri 
1
   6.67  1011 N  m 2 /kg 2  5.97  1024 kg  1720 kg 
2

1
1
1


 



 25, 200 mi 12, 600 mi   1609 m/mi  
E  8.45  109 J  8.45 GJ
Insight: The change in mechanical energy is a combination of the additional potential energy required to boost the
satellite into a higher orbit and the reduced kinetic energy required because the orbit speed is less.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 32
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
89. Picture the Problem: The space shuttle travels in a circular orbit about the Earth at an altitude of 250 km.
Strategy: Use the formula derived in problem 79 to determine the orbital speed of the shuttle. Then solve the relation,
v  2 r T for T in order to find the orbit period of the space shuttle.
Solution: 1. (a) No, the orbit speed is independent of mass, as seen in problem 79 and elsewhere.
2. (b) Use the equation derived in
problem 79 to find the orbit speed:
v
GM E

RE  h
 6.67 10
11
N  m 2 /kg 2  5.97  1024 kg 
6.37 106  250 103 m
 7760 m/s  7.76 km/s
3. (c) Solve v  2 r T for T:
T
6
3
2 r 2  6.37 10  250 10 m 

 5360 s  1.49 h
v
7.76 103 m/s
Insight: If the shuttle were to be boosted to a higher orbit, its speed would be smaller and its orbit period greater.
90. Picture the Problem: The object travels in a circular orbit of radius r around the Earth.
Strategy: Set the gravitational force of attraction between the satellite and the Earth (equation 12-1) equal to the
centripetal force required to keep it in circular orbit (equation 6-16). Solve the resulting expression for the orbit speed
v. Find the kinetic energy by using equation 7-6 and the gravitational potential energy using equation 12-8 in order to
show that the total mechanical energy equals (−1) times the kinetic energy.
Solution: 1. (a) Set Fcentripetal  Fgravity and solve for v:
mv 2 GmM E

r
r2
GM E
v
r
2. (b) Calculate E, substituting the expression
v 2  GM E r from step 1:
E  K U
GM E m
r
E  12 mv 2   v 2  m   12 mv 2   K
 12 mv 2 
3. (c) Yes, this result applies to an object orbiting the Sun. It applies in general to any object that is in a circular orbit
around a much larger mass, and that is not significantly influenced by other objects or forces.
Insight: Verify for yourself that if there were two identical objects orbiting each other, as discussed in problem 38, the
velocity would be v  GM 4r for each mass and the energy would be E   32 mv2  3K .
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 33
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
91. Picture the Problem: The stars orbit their common center of mass as depicted in
the figure at right.
Strategy: The stars exert forces on each other according to Newton’s Universal
Law of Gravitation (equation 12-1). The forces they exert on each other must be
equal and opposite by Newton’s Third Law. Set the gravitational force between the
two stars equal to the centripetal force required to keep the star with mass 2m
moving in a circular path (equation 6-16).
2m  0   m  d 
Solution: 1. Find the distance the mass 2m is
from the center of mass:
r2 m 
2. Set Fcentripetal  Fgravity and substitute r2m  13 d :
 2 m   12 m  
2m  m
 13 d
 v2 
G  2m  m
 3d 
d2
 2  d 3 
Gm
v 
 
3d
 T

2
3. Divide both sides by 6m d and
2
substitute v2m  2  13 d  T :
4 2 d 2 Gm

 T
3d
9T 2
4. Now solve for T:
4 2 d 3
3Gm
Insight: This situation is identical to the one described in problems 80 and 81, only the results are cast in terms of d
instead of r1 .
92. Picture the Problem: The stars orbit their common center of mass as
depicted in the figure at right.
Strategy: The stars exert forces on each other according to Newton’s
Universal Law of Gravitation (equation 12-1). Set the gravitational force
between the two stars equal to the centripetal force required to keep the star
with mass 2m moving in a circular path (equation 6-16).
Solution: 1. Find the length of a
side of the triangle:
2. Find the force exerted on one
star by the other two:
L  2  R cos30


GM 2
F  2
 cos 30
2
  2 R cos 30  

GM 2
2 R 2 cos 30
3. Set Fcentripetal  Fgravity :
 v2 
GM 2
M 
2
 R  2 R cos 30
4. Multiply both sides by R M
GM
 2 R 
v2  
 
2 R cos 30
 T 
2
and substitute v  2 R T :
5. Now solve for T:
4 2 R 2
GM

T2
2R 3 2


 T
4 3 2 R3
 2
GM
3 R3
GM
Insight: The larger the masses of the stars, the larger the force and acceleration and the smaller the period.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 34
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
93. Picture the Problem: The satellite of mass m travels in a circular orbit of radius r around the planet of mass M.
Strategy: Set the gravitational force of attraction between the satellite and the planet (equation 12-1) equal to the
centripetal force required to keep it in circular orbit (equation 6-16). Solve the resulting expression for the orbit speed v
and find the kinetic energy by using equation 7-6.
mv 2 GmM

r
r2
GM
v2 
r
Solution: 1. Set Fcentripetal  Fgravity and solve for v2:
2. Calculate K  12 mv2 :
1
2
mv 2  K 
GMm
2r
Insight: Verify for yourself that if there were two identical planets orbiting each other, as discussed in problem 38, the
velocity v  GM 4r for each mass and E   32 mv2  3K .
94. Picture the Problem: The Millennium Eagle travels along the x axis
and in the positive x direction, passing between two asteroids as
indicated in the figure at right.
Strategy: Find the distance between the Millennium Eagle and one of
the asteroids when the spacecraft is at any position x. Then use
Newton’s Universal Law of Gravitation to determine the x component
of the force on the spacecraft. The y components of the forces will
cancel as long as the Millennium Eagle remains on the x axis. Let
ya represent the 1500-m distance of each asteroid from the x axis.
Solution: 1. Find the distance
r to one of the asteroids:
r 2  x 2  ya2
2. Find the cosine of the
angle θ in the figure:
cos  
3. Now write an expression
for the total force Fx on the
Millennium Eagle in the
x-direction:
4. Insert the numerical values:
x

r
x
x  ya2
2
Fx  2 Fgravity cos   2  cos  
  x   GMm 

 2

 x 2  y 2   x 2  ya2 
a


2GMmx
Fx  
3/ 2
 x 2  ya2 
Fx  

GMm
r2
2  6.67 1011 N  m 2 /kg 2  3.50 1011 kg  2.50 107 kg  x
 1.17 10
9
 x 2  1500 m 2 


x
N  m2 
3/ 2
 x 2  1500 m 2 


3/ 2
5. A plot of the force as a function of x is drawn above.
Insight: The gravitational force of the asteroids always tends to push the spacecraft toward the origin. If the
Millennium Eagle has a sufficiently small initial speed, it will oscillate back and forth about the origin.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 35
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
95. Picture the Problem: The satellite travels in an elliptical path with the center of the Earth at one focus.
Strategy: Set the mechanical energy at apogee equal to the mechanical energy at perigee, and use the resulting
expression together with the given data to find the mass of the Earth.
Solution: Set Ea  Ep and solve for M:
Ka  U a  Kp  U p
mM 1 2
mM
1
mva2  G
 2 mvp  G
2
ra
rp
M
M
va2  2G
 vp2  2G
ra
rp
M 
va2  vp2
2G

1
ra
 r1p

 3990 m/s    4280 m/s 
2

2  6.67  1011 N  m 2 /kg 2 
M    10
24

1
2.41107 m
2
 2.251107 m

kg
Insight: The result from the given data is 2.0% higher than the known value of 5.97×10 24 kg. The difference stems
from rounding errors that are magnified whenever you subtract two large numbers to obtain a smaller number, as is the
case in both the numerator and the denominator of the above expression.
96. Picture the Problem: The roughly spherical comet Wild 2 has
a radius of 2.7 km, and the acceleration due to gravity on its
surface is 0.00010g. The two curves in the figure at right show
the surface acceleration as a function of radius for a spherical
comet with two different masses, one of which corresponds to
comet Wild 2.
Strategy: Use the given radius and surface acceleration of the
comet to select the appropriate curve.
Solution: The point (2.7 km, 0.00010g) is a point on curve II,
not curve I.
Insight: The indicated point is also very close to the density of ice, so we can conclude the comet Wild 2 is composed
of primarily ice.
97. Picture the Problem: The comet Wild 2 is a roughly spherical body of radius 2.7 km and a surface gravitational
acceleration of 1.0×10−4g.
Strategy: Use equation 12-4 to determine the mass of the comet and equation 12-13 to find the escape speed.
4
2
g RC2  9.8 10 m/s   2700 m 
MC 

 1.11014 kg
G
6.67 1011 N  m2 /kg 2
2
Solution: Solve equation 12-4 for M C :
Insight: Although this seems like a very large number (the comet would weigh 110 billion tons on Earth!) it is a very
small mass compared to the Earth. The comet’s mass is about 2.2% of the mass of all the water in Lake Michigan
(4.9×1015 kg).
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 36
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
98. Picture the Problem: The comet Wild 2 is a roughly spherical body of radius 2.7 km and a surface gravitational
acceleration of 1.0×10−4g.
Strategy: Use the mass of the comet from the previous problem and equation 12-13 to find the escape speed.
Solution: Apply equation 12-13 to the comet:
ve 
2GM C

RC
2  6.67  1011 N  m 2 /kg 2 1.1 1014 kg 
2700 m
 2.3 m/s
Insight: A person could easily leap from the comet nucleus and leave it forever! Even more fun, he could launch
himself into orbit around the comet with a speed of 1.6 m/s! (Verify for yourself using the results of problem 79.)
99. Picture the Problem: The comet Wild 2 is a roughly spherical body of radius 2.7 km and a surface gravitational
acceleration of 1.0×10−4g. We imagine that it has a small satellite that orbits at 5.4 km from the center of the comet.
Strategy: Use the mass of the comet and Kepler’s Third Law (equation 12-7) to find the orbit period.
Solution: Apply equation 12-7:
T
2
G MC
r3 2 
2
 6.67 10
11
 5400 m 
32
N  m2 /kg 2 1.11014 kg 
 2.9 104 s  8.1 h
Insight: This orbit period is much longer than the 4.0 h orbit period of a satellite that orbits at twice the radius of the
Earth. The period of the International Space Station (in low Earth orbit) is 1.5 h.
100. Picture the Problem: The planets orbit the Sun according to Kepler’s Third Law.
Strategy: Solve Kepler’s Third Law (equation 12-7) for r to determine the orbital radius that corresponds to an orbit
period of 150 days × 86,400 seconds/day = 1.30×10 7 s.
Solution: Solve equation 12-7 for r:
T2 
4 2 3
r
GM S
GM ST 2
r3

4 2
3
 6.67 10
11
N  m 2 /kg 2  2.00  1030 kg 1.3 107 s 
2
4 2
 8.3  1010 m  0.55 A.U.
Insight: The planet has a shorter orbital period than does the Earth, so by Kepler’s Third Law it must have a smaller
orbit radius.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 37
James S. Walker, Physics, 4th Edition
Chapter 12: Gravity
101. Picture the Problem: The Earth orbits the Sun according to Kepler’s Third Law.
Strategy: Use Kepler’s Third Law (equation 12-7) to determine the orbit period (length of a year) of the Earth if the
Sun’s mass were to double.
Solution: 1. (a) Equation 12-7 predicts that if the mass of the Sun were to increase, the orbit period (length of a year) of
the Earth would decrease.
2. (b) Use a ratio to find Tnew :
32
Tnew 2 r

Told
2 r 3 2
Tnew 
Told
2

GM S,new
GM S,old
365 d
2

M S,old
M S,new

M S,old
2M S,old
1

2
 258 d
3. (c) If the mass of the Earth were to double, but the orbit radius were to remain the same, the length of a year would
stay the same because Kepler’s Third Law indicates the orbit period is independent of the mass of the orbiting object.
Insight: If the Earth were to remain at the same orbit radius when the Sun’s mass is doubled, the orbital speed of the
Earth would have to increase. See problem 90 for a derivation of the orbit speed v  GM S r .
102. Picture the Problem: The spacecraft is launched straight upward by Jules Verne’s cannon at a high initial speed.
Strategy: Use the expression for the escape speed of an object (equation 12-13) to determine the new escape speed that
would be necessary if the Earth’s mass were to double while its radius remains the same.
Solution: 1. (a) An examination of equation 12-13 indicates the escape speed is proportional to the square root of the
Earth’s mass. The escape speed of a rocket would therefore increase if the mass of the Earth were to double.
2. (b) Use a ratio to find the new ve :
ve,new
ve,old

2GM E,new RE
2GM E,old RE

M E,new
M E,old

2M E,old
M E,old
 2
ve,new  2ve,old  2 11.2 km/s   15.8 km/s
3. (c) If the mass of the rocket were to double, but the mass and radius of the Earth were to remain the same, the escape
speed of the rocket would stay the same because it is independent of the rocket’s mass as indicated by equation 12-13.
Insight: Doubling the rocket’s mass does mean that it would require twice as much kinetic energy to escape the Earth,
however, even though the escape speed would stay the same.
103. Picture the Problem: The spacecraft is launched straight upward by Jules Verne’s cannon at a high initial speed.
Strategy: Use the expression for the escape speed of an object (equation 12-13) to determine the new escape speed that
would be necessary if the Earth’s radius were to be cut in half while its mass remains the same.
Solution: 1. (a) An examination of equation 12-13 indicates the escape speed is inversely proportional to the square
root of the Earth’s radius. The escape speed of a rocket would therefore increase if the mass of the Earth were to
decrease.
2. (b) Use a ratio to find the new ve :
ve,new
ve,old

2GM E RE,new
2GM E RE,old

RE,old
RE,new

RE,old
1
2
RE,old
 2
ve,new  2ve,old  2 11.2 km/s   15.8 km/s
Insight: Cutting the radius of the Earth in half while keeping its mass the same would increase the density of the Earth
(see equation 15-1) by a factor of 23  8! The objects on the Earth’s surface would be half the distance to the Earth’s
center and therefore weigh 4 times as much as they do now.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12 – 38