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CHAPTER 6: Gravitation and Newton’s Synthesis
Solutions to Assigned Problems
7.
The distance from the Earth’s center is r  REarth  300 km  6.38  106 m  3  105 m 
6.68  106 m  2 sig fig  . Calculate the acceleration due to gravity at that location.
g G
M Earth
r
2
G
M Earth
r
2

 6.67  1011 N m 2 kg 2

5.97  1024 kg
 6.68  10 m 
6
2
 8.924 m s 2
 1" g " 
 8.924 m s 2 
 0.91g's
2 
 9.80 m s 
This is only about a 9% reduction from the value of g at the surface of the Earth.
8.
We are to calculate the force on Earth, so we need the distance of each planet from Earth.
rEarth  150  108   106 km  4.2  1010 m
rEarth   778  150   106 km  6.28  1011 m
Venus
Jupiter
rEarth  1430  150   10 km  1.28  10 m
6
12
Saturn
Jupiter and Saturn will exert a rightward force, while Venus will exert a leftward force. Take the
right direction as positive.
M Earth M Jupiter
M M
M M
FEarth-  G
 G Earth2 Saturn  G Earth2 Venus
2
rEarth
rEarth
rEarth
planets
Jupiter

2

 GM Earth
Saturn
318
  6.28  1011 m 2


 6.67  1011 N m 2 kg 2
Venus
95.1

1.28 10 m 
  5.97  10 kg   4.02 10
Sun
M Earth M Sun
2
Earth
Sun
r
And so the ratio is FEarthplanets
2
12
2
24
The force of the Sun on the Earth is as follows.
FEarth-  G


2
10
 4.2 10 m  
0.815

22

m 2  9.56  1017 N  9.6  1017 N
 5.97 10 kg 1.99 10

1.50 10 m 
24

 6.67  1011 N m 2 kg 2
11
30
kg
2
  3.52 10
22
N
FEarth-  9.56  1017 N 3.52  1022 N  2.7  105 , which is 27 millionths.
Sun
19. The expression for g at the surface of the Earth is g  G
mE
rE2
. Let g  g be the value at a distance
of rE  r from the center of Earth, which is r above the surface.
(a) g  G
mE
2
E
r
 g  g  G
mE
 rE  r 
2
mE
G

rE2  1 

r 
rE 
2
G
mE 
r 
2

r 
1
 g 1  2  


r 
rE 
rE 

2
E
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g  2 g
r
rE
(b) The minus sign indicated that the change in g is in the opposite direction as the change in r. So,
if r increases, g decreases, and vice-versa.
(c) Using this result:
r
1.25  105 m
g  2 g
 2  9.80 m s 2 
 0.384 m s 2  g  9.42 m s 2
rE
6.38  106 m
Direct calculation:
5.98  1024 kg 

mE
11
2
2
g  G 2   6.67  10 N m kg 
 9.43m s 2
2
6
5
r
 6.38  10 m  1.25  10 m 
The difference is only about 0.1%.
23. The shuttle must be moving at “orbit speed” in order for the satellite to remain in the orbit when
released. The speed of a satellite in circular orbit around the Earth is shown in Example 6-6 to be
vorbit  G
M Earth
r
v G
.
M Earth
 G
r
M Earth
 REarth  680 km 

 6.67 10
11
2
N m kg
2

 5.98 10
24
kg

 6.38 10 m  6.8 10 m 
6
5
 7.52  103 m s
27. The speed of an object in a circular orbit of radius r around mass M is given in Example 6-6 by
v  G M r , and is also given by v  2 r T , where T is the period of the orbiting object. Equate
the two expressions for the speed and solve for T.
G
M
2 r

r
T  2

T
r3
GM
1.86  10 m 

N m kg  7.35  10 m 
3
6
 2
 6.67  10
11
2
2
22
7.20  103 s  120 min
30. The speed of an object in an orbit of radius r around the Earth is given in Example 6-6 by
v  G M Earth r , and is also given by v  2 r T , where T is the period of the object in orbit.
Equate the two expressions for the speed and solve for T. Also, for a “near-Earth” orbit, r  REarth .
G
M Earth
r
T  2

2 r
T
3
REarth
GM Earth
 T  2
r3
GM Earth
 6.38 10 m 
 5070 s  84.5 min
N m kg  5.98  10 m 
6
 2
 6.67 10
11
2
3
2
24
No , the result does not depend on the mass of the satellite.
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165
Gravitation and Newton’s Synthesis
Chapter 6
35. Consider the lower left mass in the diagram. The center of the orbits is
the intersection of the three dashed lines in the diagram. The net force
on the lower left mass is the vector sum of the forces from the other
two masses, and points to the center of the orbits. To find that net
force, project each force to find the component that lies along the line
towards the center. The angle is   30 .
F G
M2
l
 Fcomponent  F cos   G
2
Fnet  2G
M2
l
towards
center
M2
3

3G
2
3
2
r


l 2
M2
l 2 2
l 2
The net force is causing centripetal motion, and so is of the form Mv 2 r . Note that r cos   l 2 .
Fnet  2G
v
M2
l
2
3
2
 3G
M2
l
2
Mv 2

r

Mv 2
l
 2 cos  

Mv 2
l

3
3G
M2
l
2

Mv 2
l

3
GM
l
44. (a) Use Kepler’s third law to relate the Earth and the hypothetical planet in their orbits around the
Sun.
T
planet
TEarth    rplanet rEarth 
2
Tplanet  TEarth  rplanet rEarth 
3/ 2
3

 1 y  3 1
3/ 2
 5.20 y  5 y
(b) No mass data can be calculated from this relationship, because the relationship is massindependent. Any object at the orbit radius of 3 times the Earth’s orbit radius would have a
period of 5.2 years, regardless of its mass.
46. (a) In a short time  t , the planet will travel a
distance vt along its orbit. That distance is
essentially a straight line segment for a short
dF
dN
time duration. The time (and distance moved)
during  t have been greatly exaggerated on the v N t
Sun
diagram. Kepler’s second law states that the
area swept out by a line from the Sun to the
planet during the planet’s motion for the  t is
the same anywhere on the orbit. Take the areas
swept out at the near and far points, as shown on the diagram, and approximate them as
triangles (which will be reasonable for short  t ).
 Area  N   Area  F

1
2
 vN t  d N  12  vF t  d F

vF t
vN vF  d F d N
(b) Since the orbit is almost circular, an average velocity can be found by assuming a circular orbit
with a radius equal to the average distance.
11
11
1
2 r 2 12  d N  d F  2 2 1.47  10 m  1.52  10 m 
vavg 


 2.973  104 m s
7
T
T
3.16  10 s
From part (a) we find the ratio of near and far velocities.
vN vF  d F d N  1.52 1.47  1.034
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
166
Physics for Scientists & Engineers, 4th Edition
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For this small change in velocities (3.4% increase from smallest to largest), we assume that the
minimum velocity is 1.7% lower than the average velocity and the maximum velocity is 1.7%
higher than the average velocity.
vN  vavg 1  0.017   2.973  104 m s 1.017   3.02  10 4 m s
vF  vavg 1  0.017   2.973  104 m s  0.983  2.92  10 4 m s
53. (a) The acceleration due to gravity at any location at or above the surface of a star is given by
gstar  G M star r 2 , where r is the distance from the center of the star to the location in question.
gstar  G
M sun
2
Moon
R
1.99 10 kg   4.38 10

1.74 10 m 
30

 6.67  1011 N m 2 kg 2
6

7
2
m s2

(b) W  mgstar   65 kg  4.38  107 m s 2  2.8 109 N
(c) Use Eq. 2-12c, with an initial velocity of 0.
v 2  v02  2a  x  x0  
v

2a  x  x0  
2 4.38  107 m s 2
 1.0 m   9.4  10
3
m s
56. The speed of an object in an orbit of radius r around the Moon is given by v  G M Moon r , and is
also given by v  2 r T , where T is the period of the object in orbit. Equate the two expressions
for the speed and solve for T.
G M Moon r  2 r T
T  2
r3
GM Moon
 7.1  103 s

 2

 RMoon  100 km 
GM Moon
1.74 10 m  110 m 
 6.67 10 N m kg  7.35 10
6
3
 2
11
2
5
2
3
22
kg

2.0 h 
65. Find the “new” Earth radius by setting the acceleration due to gravity at the Sun’s surface equal to
the acceleration due to gravity at the “new” Earth’s surface.
g Earth  gSun 
new
GM Earth
2
rEarth

GM Sun
2
rSun
 rEarth  rSun
new
M Earth
M Sun

 6.96  108 m

5.98  1024 kg
1.99  1030 kg
new
 1.21  106 m , about
69. If the ring is to produce an apparent gravity equivalent to that of
Earth, then the normal force of the ring on objects must be given by
FN  mg . The Sun will also exert a force on objects on the ring.
See the free-body diagram. Write Newton’s second law for the
object, with the fact that the acceleration is centripetal.
 F  FR  FSun  FN  mv2 r
1
5
the actual Earth radius.
FSun
Sun
FN
Substitute in the relationships that v  2 r T , FN  mg , and
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167
Gravitation and Newton’s Synthesis
Chapter 6
FSun  G
M Sun m
r2
, and solve for the period of the rotation.
FSun  FN  m v r  G
2
T
4 2 r

M Sun
G 2 g
r
M Sun m
r2
 mg 
4 2 mr
T2

 G
4 2 1.50  1011 m
 6.67 10
11
N m 2 kg 2

M Sun
r2
g 
4 2 r
T2

1.99 10 kg   9.80 m s
1.50 10 m 
30
11
2
2
 7.77  105 s  8.99 d
The force of the Sun is only about 1/1600 the size of the normal force. The force of the Sun could
have been ignored in the calculation with no significant change in the result given above.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. 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.
168