Instructor Solution Manual

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13
NEWTON’S THEORY OF GRAVITY
Conceptual Questions
13.1. Newton’s third law tells us that the forces are equal. They are also clearly equal when Newton’s law of gravity
is examined: F12 = Gm1m2 /r 2 has the same value whether m1 = Earth and m2 = sun or vice versa.
13.2. The force of the star on planet 1=
is F1 G
M s m1
r12
. The force of the star on planet 2=
is F2 G
M s m2
r2 2
. Since
m2 = 2m1, and r2 = 2r1,
F2
=
F1
M s (2m1 )
2 1
(2r1 ) 2
= =
M s m1
4 2
G 2
r1
G
13.3. (a) The force of the Earth on the first satellite is F1 = G
M e m1
r12
=
F2 G
, while
M e m2
r2 2
. Since r1 = r2 ,
F1 m1 1000 kg 1
=
=
= .
F2 m2 2000 kg 2
(b) With=
Fnet ma
=
, F1 m1a1 and F2 = m2 a2 , so
a1  F1  m2   1   2000 kg 
=
=

   =
 1
a2  F2  m1   2   1000 kg 
The free-fall acceleration is the same for objects of different mass.
13.4. The astronauts can be “weightless” at any distance because an object is said to be weightless if it is in free fall
(as in orbit). For the gravitational force to become zero, the spacecraft would have to be an infinite distance away.
13.5. The hammer will not fall to the Earth. The astronaut, shuttle, and hammer are all in free fall around the Earth
(in an orbit), so the hammer has the same acceleration as the astronaut and does not move away from him.
13.6. The acceleration due to gravity is=
g G
M
R2
g1 20 m/s 2 , M
=
. With =
2 2 M1, and R2 = 2 R1,
M2
(2M1 )
M 1 1
=
g2 G =
G
=
G 21 =
g1 10 m/s 2
 =
2
2
R2
(2 R1 )
R1  2  2
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13-1
13-2
Chapter 13
13.7. The gravitational potential energy is negative because we choose to place the zero point of potential energy at
infinity (U at ∞= 0). With this choice, the gravitational potential energy is negative because the conservative force of
gravity is attractive. The two masses will gain kinetic energy as they approach each other.
13.8. The escape speed from Planet X is
=
vescape X
2GM X
=
10,000 m/s.
RX
Planet Y is twice as dense as Planet X but the same size so has twice the mass. Therefore
2G (2M X )
=
RX
=
vescape Y
=
2vescape X 14,142 m/s
13.9. The Earth’s new orbital period would be about 11.9 years. For circular orbits, Kepler’s third law relates orbital
 4π 2  3
period to radius and can be written
=
T2 
r . Here M is the mass of the body being orbited (the Sun in this
 GM 


problem.) The orbital period is independent of the mass of the orbiting body.
13.10. For a circular orbit, v = GM/r (Equation 13.22). As r decreases, v increases, so the satellite speeds up as
the satellite spirals inward.
Exercises and Problems
Section 13.3 Newton’s Law of Gravity
13.1. Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size
and mass and that of either the sun or the earth, a human body can be treated as a particle.
GM s M y
GM e M y
Solve: Fs on you =
=
and Fe on you
2
rs-e
re2
Dividing these two equations gives
2
2
Fs on y
Fe on y
 M  r   1.99 × 1030 kg  6.37 × 106 m 
= s  e  =
=6.00 × 10-4

24
11



 M e  rs-e   5.98 × 10 kg  1.50 × 10 m 
13.2. Model: Assume the two lead balls are spherical masses.
Gm1m2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(10 kg)(0.100 kg)
= 6.7 × 10−9 N
r2
(0.10 m) 2
(b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is
Solve: (a) F1 on 2 = F2 on 1 =
=
6.67 × 10−9 N
(0.100 kg)(9.8 m/s 2 )
= 6.8 × 10−9
Assess: The answer in part (b) shows how small is the gravitational force between two lead balls separated by 10 cm
compared to the gravitational force on the 100 g ball.
13.3. Model: Model the sun (s), the moon (m), and the earth (e) as spherical masses.
Solve: Fs on m =
GM s M m
2
rs-m
and Fe on m =
GM e M m
2
re-m
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Newton’s Theory of Gravity
13-3
Dividing the two equations and using the astronomical data from Table 13.2,
2
2
Fs on m  M s  re-m   1.99 × 1030 kg  3.84 × 108 m 
= 

 = 2.18

 = 
11
Fe on m  M e  rs-m   5.98 × 1024 kg 
 1.50 × 10 m 
Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.
GM M
GM e M p
rs-p
re2
s p
Solve: Fsphere on particle =
13.4.
=
and Fearth on particle
2
Dividing the two equations,
Fsphere on particle
Fearth on particle
2
2
 M   r   5900 kg   6.37 × 106 m 
-7
=  s   e  = 
 = 1.6 × 10
 
24


0
50
m
M
r
.
 e   s-p   5.98 × 10 kg  

13.5. Model: Model the woman (w) and the man (m) as spherical masses or particles.
Solve: Fw on m = Fm on w =
GM w M m
2
rm-w
=
(6.67 × 10-11 N ⋅ m 2 /kg 2 )(50 kg)(70 kg)
(1.0 m) 2
= 2.3 × 10-7 N
13.6. Model: Model the earth (e) as a sphere.
Visualize:
The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re + rs =6.37 × 106 m + 0.30 × 106 m =6.67 × 106 m away
from the center of the earth.
Solve: (a) Fe on s =
GM e M s
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.0 kg)
( Re + rs )
(6.67 × 106 m) 2
=
2
= 9.0 N
Section 13.4 Little g and Big G
13.7. Model: Model the sun (s) as a spherical mass.
Solve: (a) gsun surface
=
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 274 m/s 2
Rs2
(6.96 × 108 m) 2
GM
(6.67 × 10-11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
(b) gsun at earth = 2 s =
=5.90 × 10-3 m/s 2
rs-e
(1.50 × 1011 m) 2
13.8. Model: Model the moon (m) and Jupiter (J) as spherical masses.
Solve: (a) g moon surface =
GM m
Rm2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)
(1.74 × 10 m)
6
2
= 1.62 m/s 2
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13-4
Chapter 13
(b) g Jupiter surface=
GM J (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
=
= 25.9 m/s 2
RJ2
(6.99 × 107 m) 2
13.9. Model: Model the earth (e) as a spherical mass.
Visualize: The acceleration due to gravity at sea level is 9.83 m/s 2 (see Table 13.1) and Re =6.37 × 106 m (see
Table 13.2).
Solve: g observatory =
GM e
( Re + h)
2
=
GM e
2
Re (1 + h/Re ) 2
=
g earth
(1 + h/Re ) 2
= (9.83 − 0.0075) m/s 2
Here gearth = GM e /Re2 is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the value
9.83 m/s 2 . Solving for h,
 9.83

2.4 km
h=
− 1 Re =

9
8225
.


13.10. Model: Model the earth (e) as a spherical mass.
Solve: Let the acceleration due to gravity be 3.0gsurface when the earth is shrunk to a radius of x. Then,
GM e
GM e
=
and 3.0 gsurface
2
Re
x2
=
gsurface
3
GM e
Re2
=
GM e
x
⇒
2
x=
Re
=0.58 Re
3.0
The earth’s radius would need to be 0.58 times its present value.
13.11. Model: Model Planet Z as a spherical mass.
Solve: (a) g Z surface =
GM Z
⇒ 8.0 m/s 2 =
(6.67 × 10−11 N ⋅ m 2 /kg 2 ) M Z
RZ2
(b) Let h be the height above the north pole. Thus,
gabove N pole =
GM Z
( RZ + h)
2
=
(5.0 × 106 m) 2
GM Z
=
RZ2 (1 + h/RZ ) 2
g Z surface
(1 + h/RZ )
2
=
⇒ M Z = 3.0 × 1024 kg
8.0 m/s 2
 10.0 × 10 m 
1 +

5.0 × 106 m 

6
2
= 0.89 m/s 2
Section 13.5 Gravitational Potential Energy
13.12. Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as an
isolated system, so mechanical energy is conserved.
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Newton’s Theory of Gravity
13-5
Visualize:
Solve: A height of 15 m is very small in comparison with the radius of earth or Mars. We can use flat-earth gravitational potential energy to find the speed with which the astronaut can throw the ball. On earth, with yi = 0 m and
vf = 0 m/s, energy conservation gives
1 mv 2
f
2
+ mgyf =12 mvi2 + mgyi
⇒ vi = 2 ge yf = 2(9.81 m/s 2 )(15 m) =
17.2 m/s
Energy is also conserved on Mars, but the acceleration due to gravity is different.
g Mar’s surface =
GM M
2
RM
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.42 × 1023 kg)
(3.37 × 106 m) 2
= 3.77 m/s 2
On Mars, with yi = 0 m and vf = 0 m/s, energy conservation is
1 mv 2
f
2
+ mgyf =
1 mv 2
i
2
+ mgyi
⇒
yf =
vi2
(17.2 m/s) 2
=
= 39 m
2 g m 2(3.77 m/s 2 )
13.13. Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter form an
isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, allows an object to escape to an infinite distance from Jupiter (or, in general, from its partner object).
Visualize:
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13-6
Chapter 13
Solve: The energy conservation equation K 2 + U 2 = K1 + U1 is
1 m v2
2 o 2
−
GM J mo 1
GM J mo
=2 mov12 −
r2
RJ
where RJ and M J are the radius and mass of Jupiter. Using the asymptotic condition v2 = 0 m/s as r2 → ∞,
0 J =12 mov12 −
GM J mo
RJ
⇒ v1 =
2GM J
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
=
=6.02 × 104 m/s
7
RJ
6.99 × 10 m
Thus, the escape velocity from Jupiter is 60.2 km/s.
13.14. Model: Model the earth (e) as a spherical mass. Compared to the earth’s size and mass, the rocket (r) is
modeled as a particle. This is an isolated system, so mechanical energy is conserved.
Visualize:
Solve: The energy conservation equation K 2 + U 2 = K1 + U1 is
1 m v2
2 r 2
−
GM e mr 1
GM e mr
=2 mr v12 −
r2
Re
In the present case, r2 → ∞, so
1 m v2
2 r 2
v2=
=12 mr v12 −
(1.5 × 104 m/s) 2 −
GM e mr
Re
⇒ v22 =
v12 −
2GM e
Re
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
= 10 km/s
6.37 × 106 m
13.15. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The minimum
launch speed for escape, which is called the escape speed, allows an object to escape to an infinite distance from the
sun, where the object will have slowed to zero speed with respect to the sun.
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Newton’s Theory of Gravity
13-7
Visualize:
We denote by mp the mass of the probe. M s is the sun’s mass, and Rs-p is the separation between the centers of the
sun and the probe.
Solve: The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM s mp 1
GM s mp
1
mpv22 = mpv12 r2
Rs-p
2
2
Using the condition v2 0 m/s asymptotically as r2 → ∞,
=
GM s mp
1
2
mpvescape
=
Rs-p
2
⇒ vescape =
2GM s
2(6.67 × 10-11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
=4.21 × 104 m/s
Rs-p
1.50 × 1011 m
13.16. Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated, the
mechanical energy of the object on both the planet and the earth is conserved.
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13-8
Chapter 13
Visualize:
Let us denote the mass of the planet by M p and that of the earth by M e . Your mass is m0 . Also, acceleration due to
gravity on the surface of the planet is g p and on the surface of the earth is ge . Rp and Re are the radii of the planet
and the earth, respectively.
Solve: (a) We are given that M=
gp
P 2 M e and =
Since g p =
GM p
Rp2
and g e =
GM p
Rp2
GM e
Re2
1
4
ge .
, we have
1 GM e 1 G ( M p / 2)
=
=
4 Re2
4
Re2
⇒
Rp = 8 Re = 8(6.37 × 106 m) =1.80 × 107 m
(b) The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM p m0 1
GM p m0
1
m0v22 −
= m0v12 −
r2
Rp
2
2
Using v2 = 0 m/s as r2 → ∞, we have
GM p m0
1
2
=
m0vescape
Rp
2
vescape =
2GM p
Rp
=
2G (2 M e )
=
Rp
4(6.67 × 102 11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
1.80 × 107 m
= 9.41 km/s
Section 13.6 Satellite Orbits and Energies
13.17. Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle.
Visualize: The asteroid, having mass ma and velocity va , orbits the sun in a circle of radius ra . The asteroid’s time
period is Ta = 5.0 earth years =1.58 × 108 s.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Newton’s Theory of Gravity
13-9
Solve: The gravitational force between the sun (mass = M s ) and the asteroid provides the centripetal acceleration
required for circular motion.
2
GM s ma ma va2
=
ra
ra2
⇒
GM s  2π ra 
ra
= 
 ⇒=
ra
 Ta 
1/3
 GM sTa2 

2 

 4π

Substituting G =6.67 × 10−11 N ⋅ m 2 /kg 2 , M s =1.99 × 1030 kg, and the time period of the asteroid, we obtain
=
ra 4.4 × 1011 m. The velocity of the asteroid in its orbit will therefore be
v=
a
2π ra 2π (4.4 × 1011 m)
=
= 1.7 × 104 m/s
Ta
1.58 × 108 s
Solve: We give the answer to two significant figures because the asteroid period is given to two significant figures.
13.18. Model: Model the sun (s) and the earth (e) as spherical masses.
Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs-e . The sun’s and
the earth’s masses are denoted by M s and me , respectively.
Solve: The gravitational force provides the centripetal acceleration required for circular motion.
GM s me meve2 me (2π rs-e ) 2
=
=
2
rs-e
rs-e
rs-eTe2
3
4π 2rs-e
4π 2 (1.50 × 1011 m)3
=
=2.01 × 1030 kg
Ms =
GTe2
(6.67 × 10-11 N ⋅ m 2 /kg 2 )(365 × 24 × 3600 s) 2
Assess: The tabulated value is 1.99 × 1030 kg. The slight difference can be ascribed to the fact that the earth’s orbit
isn’t exactly circular.
13.19. From Kepler’s third law, the orbital period squared is proportional to the orbital radius cubed: T 2 ∝ r 3 .
Thus, at ry = 4rx , Ty2 ∝ (4rx )3 =64rx3 ∝ (8Tx ) 2 . Therefore T=
1600 earth days
y 8Tx . A year on Planet Y is 8 × 200 =
long.
13.20. Model: Model the star (s) and the planet (p) as spherical masses.
Solve: From the planet’s acceleration due to gravity, we find its mass to be
gp =
Mp =
g p Rp2
G
GM p
Rp2
(12.2 m/s 2 )(9.0 × 106 m) 2
=
6.67 × 10−11 N ⋅ m 2 /kg 2
= 1.5 × 1025 kg
(b) From the planet’s orbital period, we find the mass of the star Omega to be
 4π 2  3
T2 =
r
 GM 
s

Ms =
4π 2r 3
GT
2
=
4π 2 (2.2 × 1011 m)3
(6.67 × 10
2 11
N ⋅ m 2 /kg 2 )(402 × 24 × 3600 s) 2
= 5.2 × 1030 kg
13.21. Model: Model the planet and satellites as spherical masses.
Visualize: Please refer to Figure EX13.21.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13-10
Chapter 13
Solve: (a) The period of a satellite in a circular orbit is T = [4π 2r 3/(GM )]1/2 This is independent of the satellite’s
mass, so we can find the ratio of the periods of two satellites a and b:
r 
Ta
=  a
Tb
 rb 
3
3/2
Satellite 2 has r2 = r1, so T2= T=
so T3 (3/2)
=
T1 459 min.
1 250 min. Satellite 3 has r3 = (3/2) r1,=
(b) The force on a satellite
is F GMm/r 2 . Thus the ratio of the forces on the two satellites a and b is
=
2
Fa  rb   ma 
=  

Fb  ra   mb 
2
Satellite 2 has r2 = r1 and m2 = 2m1, =
so F2 (1)
=
(2) F1 20,000 N. Similarly, satellite 3 has r3 = (3/2)r1 and
2
so F3 (2/3)=
(1) F1 4440 N.
m3 = m1, =
(c) The speed of a satellite in a circular orbit is v = (GM/r ) 2 , so its kinetic energy =
is K
1=
mv 2
2
GMm/2r . Thus the
ratio of the kinetic energy of two satellites a and b is
K a  rb  ma 
=  

K b  ra  mb 
Satellite 3 has r3 = (3/2)r1 and m3 = m1, so K1/K3 =(3/2)(1/1) =3/2 =1.50.
13.22. Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle.
Visualize: The satellite, having mass ms and velocity vs , orbits the sun with a mass M S in a circle of radius rs .
Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration for
circular motion. Newton’s second law gives
GM Sms
rs2
=
msvs2
rs
Because vs = 2π rs /Ts where Ts is the period of the satellite, this equation simplifies to
GM S
rs2
=
(2π rs ) 2
⇒ rs3 =
Ts2rs
GM STs2
4π 2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(24 × 3600 s) 2
4π 2
⇒ rs = 2.9 × 109 m
13.23. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle.
Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs . The radius of the circular orbit is
denoted by rs and the mass of the earth by M e .
Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular
motion:
GM e ms
rs2
m v2
= s s
rs
GM
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
⇒ rs = 2 e =
=1.32 × 107 m
vs
(5500 m/s) 2
2π Rs 2π (1.32 × 107 m)
Ts =
=
=1.51 × 104 s = 4.2 h
vs
5500 m/s
13.24. Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle.
Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs around
Mars. Let us denote mass of Mars by M m .
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Newton’s Theory of Gravity
13-11
Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion:
GM m ms
rs2
=
msvs2 ms (2π rs ) 2
=
rs
rsTs2
1/3
 GM mTs2 
⇒ rs = 
 4π 2 


Using G =6.67 × 10−11 N ⋅ m 2 /kg 2 , M m =6.42 × 1023 kg, and Ts =
(24.8 hrs) =
(24.8)(3600) s =
89,280 s, we obtain
rs =2.05 × 107 m. Thus, altitude =rs − Rm =1.72 × 107 m. The speed is the orbital circumference divided by the
period, or
vs =
2π rs 2π (2.05 × 107 m/s)
=
= 1.44 km/s
89,280 s
Ts
13.25. Solve: We are given M1 + M=
2 150 kg which means M=
1 150 kg − M 2 . We also have
(8.00 × 10−6 N)(0.20 m) 2
−6
=
8
.
00
×
10
N
⇒
M
M
=
=4798 kg 2
1
2
(0.20 m) 2
6.67 × 10−11 N ⋅ m 2 /kg 2
GM1M 2
Thus, (150 kg − M 2 ) M 2 =
4798 kg 2 or M 22 − (150 kg) M 2 + (4798 kg 2 ) =
0. Solving this equation gives M=
2 103.75 kg
and 46.25 kg. The two masses are 104 and 46 kg.
13.26. Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 5.0 kg mass
(m3 ) is on the y-axis and the 10.0 kg mass (m2 ) is on the x-axis.
Solve: (a) The forces acting on the 20.0 kg mass (m1 ) are

gm1m2 ˆ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(10.0 kg) ˆ
Fm2 on=
=
i
=
i 1.33 × 10−6 iˆ N
m1
r122
(0.10 m) 2

gm1m3 ˆ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg) ˆ
Fm3 on=
=
j
=
j 1.67 × 10−7 ˆj N
m1
r132
(0.20 m)2



−6 ˆ
−−6
7ˆ
N
=
Fon m1 Fm2 on m1 + Fm3 on=
m1 1.33 × 10 i N +1.67 ×10 j N ⇒ Fon=
m1 1.34 × 10

(
) 
 1.33 × 10−6 N 
1 Fon m1 x
 
 tan 1 
θ= tan −−
=
= 83°
 1.67 × 10−7 N 
 ( Fon m ) y 


1



−6
Thus the force is Fon m1 = 1.3 × 10 N, 83° cw from the +y-axis
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13-12
Chapter 13
(b) The forces acting on the 5.0 kg mass (m3 ) are


Fm1 on m3 =
− Fm3 on m1 =
−1.67 × 10−7 ˆj N
Fm2 on=
m3

Fm2
on m3
gm2 m3 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(10.0 kg)(5.0 kg)
=
= 6.67 × 10−8 N
2
(0.20 m) 2 + (0.10 m) 2
r23
8
N)sinφ iˆ − (6.67 × 10 8 N)cosφ ˆj
= (6.67 × 10−−
 10 cm  ˆ
 20 cm  ˆ
−8
= (6.67 × 10−8 N) 
 i − (6.67 × 10 N) 
j
 22.36 cm 
 22.36 cm 
8
N)iˆ − (5.966 × 10 8 N) ˆj
= (2.983 × 10−−



8ˆ
Fon m3 = Fm1 on m3 + Fm2 on m3 = 2.983 × 10−−
i N − 2.264 × 10 7 ˆj N
8
2
=
(2.983 × 10−−−
N) 2 + (−2.264 × 10 7 N)
2.284 × 10 7 N

 ( Fon m ) x 
 2.983 × 10−8 N 
−1
 tan −1 
θ ′ = tan   3 =
=
 7.51°
−7



 2.264 × 10 N 
 ( Fon m3 ) y 
Fon=
m3

Thus Fon m3 = 2.3 × 102 7 N, 7.5° ccw from the -y-axis.
13.27. Visualize:
 5 
Solve: The angle θ= tan −1  = 14.04°. The distance r1 =r2 = (0.050 m) 2 + (0.200 m) 2 =0.206 m. The forces
 20 
on the 20.0 kg mass are

Mm
F1= G 2 1 (− sinθ iˆ + cosθ ˆj )
r1

Mm2
F2 G 2 (sinθ iˆ + cosθ ˆj )
=
r2
Note m1 = m2 and r1= r2 . Thus, the net force on the 20.0 kg mass is

 
Mm
Fnet = F1 + F2 = 2G 2 1 cosθ ˆj
r1
=
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg)cos(14.04°) ˆ
j
(0.206 m) 2
= 3.0 × 10−7 ˆj N
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Newton’s Theory of Gravity
13-13
13.28. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs
of masses.
U = U12 + U13 + U 23
mm
mm
mm
=
−−−
G 1 2 G 1 3 G 2 3
r12
r13
r23
With m1 =
20.0 kg, m2 =
10.0 kg, m3 =
5.0 kg, r12 =
0.20 m, r13 =
0.10 m, and r23 = 0.2236 m,
U =−1.48 × 10−7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there
are no vector calculations required.
13.29. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs
of masses.
U = U12 + U13 + U 23
mm
mm
mm
=
−−−
G 1 2 G 1 3 G 2 3
r12
r13
r23
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13-14
Chapter 13
With m1 =20.0 kg, m2 =m3 =10.0 kg, r12 =r13 = (0.050 m) 2 + (0.200 m) 2 =0.206 m, and r23 = 0.100 m,
U =−1.96 × 10−7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there
are no vector calculations required.
13.30. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will make an
angle of θ with the vertical. The distance between the sphere centers is therefore going to be less than 1 m. The freebody diagram shows the forces acting on the lead sphere.
Solve: We can see from the diagram that the distance between the centers is d =
1.000 m − 2 L sin θ . Each sphere is
in static equilibrium, so Newton’s second law is
0 ⇒ T sin θ =
Fgrav − T sin θ =
Fgrav
∑ Fx =
∑ Fy = T cosθ − mg = 0 ⇒ T cosθ = mg
Dividing these two equations to eliminate the tension T yields
Fgrav Gmm/d 2 Gm
sin θ
θ
= tan
=
=
= 2
cosθ
mg
mg
d g
We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to be
enough to affect the value of Fgrav , the gravitational attraction between the two masses, so we’ll evaluate the right
side of this equation by using 1.00 m for d. This gives
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(100 kg)
10
=6.80 × 10−−
⇒ θ =(3.90 × 10 8 )°
tan θ =
(1.00 m) 2 (9.81 m/s 2 )
This small angle causes the two spheres to move closer by 2 L sin θ = 1.4 × 10−7 m = 0.00000014 m. Consequently,
the distance between their centers is d =
0.99999986 m.
13.31. Visualize:
We placed the origin of the coordinate system on the 20 kg sphere (m1 ). The sphere (m2 ) with a mass of 10 kg is
20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses m1
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Newton’s Theory of Gravity
13-15
and m2 . This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As the
gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of m1 or to
the right of m2 , will experience gravitational forces from m1 and m2 pointing in the same direction, thus always
leading to a nonzero force.
Solve:
Fm1=
on m Fm2
on m
m1m
m2m
20
10
⇒ G=
G
⇒
=
2
2
2
x
x
(0.20 − x)
(0.20 − x) 2
10 x 2 − 8 x + 0.8 = 0 ⇒
x = 0.683 m and 0.117 m
The value x= 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2 . Thus, the
mass should be placed 11.7 cm to the right of the larger mass. To two significant figures, this is 12 cm.
13.32. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle.
Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity ( galtitude ) is 10%
of the surface value ( gsurface ).
Solve: (a) Since galtitude= (0.10) gsurface , we have
GM e
( Re + h) 2
h = 2.162 Re
=(0.10)
GM e
Re2
⇒ ( Re + h) 2 =
10 Re2
⇒ h = (2.162)(6.37 × 106 m) = 1.377 × 107 m = 1.4 × 107 m
(b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the
earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with
velocity vs ,
msvs2
=
( Re + h) 2 ( Re + h)
GM e ms
GM e
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
4.5 km/s
⇒ vs =
=
=
Re + h
(6.37 × 106 m + 1.377 × 107 m)
13.33. Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air resistance. This
is an isolated system, so mechanical energy is conserved.
Visualize:
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13-16
Chapter 13
Solve: (a) The conservation of energy equation K 2 + U g2 =K1 + U g1 is
1
GM e mo 1
GM e mo
mov22 −
= mov12 −
Re
Re + y1
2
2
 1
1 
2GM e  −

 Re Re + y1 
v2
=
1
1


= 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) 
−
 =3.02 km/s
 6.37 × 106 m 6.87 × 106 m 
(b) In the flat-earth approximation, U
=
g mgy. The energy conservation equation thus becomes
1
1
mov22 + mo gy2 = mov12 + mo gy1
2
2
v2 = v12 + 2 g ( y1 − y2 ) = 2(9.81 m/s 2 )(5.00 × 105 m − 0 m) =3.13 km/s
(c) The percent error in the flat-earth calculation is
3130 m/s − 3020 m/s
≈ 3.6%
3020 m/s
13.34. Model: Consider the object as a particle and take the planet to be a spherical mass.
Solve: Conservation of mechanical energy of the object gives
−G
Mm 1 2
Mm
= mv − G
R+h 2
R
The object’s mass drops out. Solving for the speed as it hits the ground,
=
v
Assess: Compare this=
to v
1 
1
=
2GM  −

R R+h
=
2 gh
2GMh
R2
R+h−R
=
2GM 

 R ( R + h) 
2GMh
R ( R + h)
, which is the result if the potential U = mgh is used.
13.35. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an isolated system, so mechanical energy is conserved.
Visualize:
A pictorial representation of the before-and-after events is shown.
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Newton’s Theory of Gravity
13-17
Solve: After using v2 = 0 m/s, the energy conservation equation K 2 + U 2 = K1 + U1 is
0 J−
GM e mp
GM e mp
1
= mpv12 −
Re + h
Re
2
The projectile mass cancels. Solving for h, we find
−1
1
v12 
h =  −−
Re = 4.2 × 105 m

 Re 2GM e 
13.36. Model: Model the earth as a spherical mass and the meteoroids as point masses.
Visualize:
Solve: (a) The energy conservation equation K 2 + U 2 = K1 + U1 gives
1 mv 2
2
2
−
GM e m 1 2 GM e m
=2 mv1 −
Re
rm
1/2

 1
1 
⇒ v2 =v12 + 2GM e  −  

 Re rm  
1/2

1
1


2
2
2
24
−11
−

(2000 m/s) + 2(6.67 × 10 N ⋅ m /kg )(5.97 × 10 kg) 
6
8
 6.37 × 10 m 3.85 × 10 m  

=
1.1 × 104 m/s =
11 km/s
The speed does not depend on the meteoroid’s mass.
(b) This part differs in that r2 =
Re + 5000 km =
1.137 × 107 m. The shape of the meteoroid’s trajectory is not important for using energy conservation. Thus
1 mv 2
2
2
−
GM e m
GM e m
=1 mv12 −
Re + 5000 km 2
rm
1/2


1
1 
⇒ v2 =v12 + 2GM e 
− 

 Re + 5000 km rm  
= 8.9 × 103 m/s = 8.9 km/s
13.37. Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated system,
so mechanical energy is conserved.
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13-18
Chapter 13
Visualize:
In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential energy.
In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both kinetic energy
and potential energy.
Solve: The conservation of energy equation K f + U f = Ki + U i gives
1 2 GMm GMm
mvf −−
= 0 J+0 J
2
rf1
rf 2
=
vf
4GM
=
rf
4(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
= 32,600
=
m/s 33 km/s
0.50 × 1012 m
Assess: Note that the final velocity of 33 km/s does not depend on the mass of the comet.
13.38. Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated system, so
mechanical energy is conserved.
Visualize: The radius of the asteroid is M a and its mass is Ra .
Solve: The conservation of energy equation K f + U f = Ki + U i for the asteroid gives
1 2 GM a m 1 2 GM a m
mvf −
= mvi −
Ra + r 2
Ra
2
The minimum speed for escape is the one that will cause you to stop only when the separation between you and the
asteroid becomes very large. Noting that vf → 0 m/s as r → ∞, we have
2GM
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1014 kg)
vi2 = a =
Ra
2.0 × 103 m
2.58 m/s
⇒ vi =
That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on the
earth. The conservation of energy equation gives
1
0 J − mvi2 =
−mg (0.50 m) ⇒ vi =
2(9.8 m/s 2 )(0.50 m) =
3.13 m/s
2
This means you can escape from the asteroid.
13.39. Model: The projectile is a particle. The earth and moon are spherical masses.
Solve: The projectile is attracted to both the moon and earth. Its final velocity and potential energy are zero. Since
the projectile is fired from the far side of the moon, its initial distance from the center of the earth is the earth-moon
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Newton’s Theory of Gravity
13-19
distance Re-m plus the radius of the moon Rm . Let the projectile have mass m. The conservation of mechanical energy equation is
M em
M m
1 2
-G m =
mvescape - G
0 J+0 J
Rm
2
( Re-m + Rm )

Me
M 
2
=
+ m
vescape
2G 
 Re-m + Rm Rm 
From Table 13.2, Re-m =3.84 × 108 m, Rm =1.74 × 106 m, M e =5.98 × 1024 kg, and M m =7.36 × 1022 kg, which
gives vescape =
2.78 km/s.
Assess: The escape velocity does not depend on the mass of the object which is trying to escape.
13.40. Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also use the
law of conservation of momentum for our system.
Visualize:
Solve: The conservation of momentum equation pfx = pix is
M (vfx )1 + 2 M (vfx ) 2 =
0 kg m/s ⇒ (vfx )1 =
−2(vfx ) 2
The equation for mechanical energy conservation K f + U f = Ki + U i is
1
1
1
0.8GM
G ( M )(2M )
G ( M )(2M )
[2(vfx ) 2 ]2 + (vfx ) 22 =
M (vfx )12 + (2M )(vfx ) 22 −
=
−
⇒
2
2
2R
10 R
2
R
GM
GM
(vfx ) 2 =−0.516
⇒ (vfx )1 =−2(vfx ) 2 =1.032
R
R
The heavier asteroid has a speed of 0.516(GM/R )1/2 and the lighter one a speed of 1.032(GM/R )1/2 .
13.41. Model: Gravity is a conservative force, so we can use conservation of energy.
Visualize:
The planets will be pulled together by gravity and each will have speed v2 as they crash and the separation between
their centers will be 2R.
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13-20
Chapter 13
Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential and
kinetic energy. Thus,
K2 + U 2 =
GMM
GMM
1
1
= K1 + U1 = 0 J −
Mv22 + Mv22 −
r2
r1
2
2
=
v2
1 1
GM  − 
 r2 r1 
Because the planet is “Jupiter-size,” we’ll use M = M Jupiter = 1.9 × 1027 kg and r2 = 2 RJupiter = 1.4 × 108 m. Inserting
these values into the expression above gives the crash speed of each planet as v2 = 3.0 × 104 m/s.
Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant acceleration
motion. The formulas from constant-acceleration kinematics do not work for problems such as this.
13.42. Model: Model the distant planet (P) as a spherical mass.
Solve: The acceleration at the surface of the planet and at the altitude h are
GM
gsurface = 2 P
R
1
MP
1 GM
and galtitude = gsurface ⇒ G
= 2P
2
2
2 R
( R + h)
( R + h ) 2 =2 R 2
⇒
R + h = 2 R ⇒ h =( 2 − 1) R =0.414 R
That is, the starship is orbiting at an altitude of 0.414R.
13.43. Model: The stars are spherical masses.
Visualize:
Solve: The starts are identical, so their final speeds vf are the same. They collide when their centers are 2R apart.
From energy conservation,
2

 1
M2
  M 
3  −G
3  Mvf2  − 3  G
 =

9


(5.0 × 10 m)   2
  2 R 

1
 1

vf2 2GM 
=
−

9
 2 R 5.0 × 10 m 
vf = 3.7 × 105 m/s
13.44. Model: The stars are spherical masses. They each rotate about the system’s center of mass.
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Newton’s Theory of Gravity
13-21
Visualize:
Solve: (a) The stars rotate about the system’s center of mass with the same period: T1= T2 = T . We can locate the
center of mass by letting the origin be at the smaller-mass star. Then
r1 = rcm =
(2.0 × 1030 kg)(0 m) + (6.0 × 1030 kg)(2.0 × 1012 m)
2.0 × 1030 kg + 6.0 × 1030 kg
= 1.5 × 1012 m
Mass m2 undergoes uniform circular motion with radius r2 = 0.5 × 1012 m due to the gravitational force of mass m1
at distance R = 2.0 × 1012 m. The gravitational force is responsible for the centripetal acceleration, so
2
=
F
grav
Gm1m2
m2v22 m2  2pp
r2 
4 2 m2r2
=
= =
m2acentripetal
 =

2
r2
r2  T 
R
T2
1/2
1/2
2
 4pp

 4 2 (0.5 × 1012 m)(0.5 × 1012 m) 2

T=
r2 R 2  = 

11
2
2
30
−
 Gm1

 (6.67 × 10 N ⋅ m /kg )(2.0 × 10 kg) 
= 7.693 × 108 s = 24 years
(b) The speed of each star
is v (2π r )/T . Thus
=
v1=
2π r1 2π (1.5 × 1012 m)
=
= 12.3 km/s
T
7.693 × 108 s
v2 =
2π r2 2π (0.5 × 1012 m)
=
= 4.1 km/s
T
7.693 × 108 s
13.45. Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated system,
so mechanical energy is conserved.
Visualize: The initial position of the lunar lander (mass = m1 ) is at a distance =
r1 Rm + 50 km from the center of
the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is
=
r2 Rm + 300 km.
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13-22
Chapter 13
Solve: The external work done by the thrusters is
1 ∆U
Wext =
∆Emech =
g
2
where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
=
ri Rm + 50 km to the final orbit rf =
Rm + 300 km. Thus
Wext =
1  −−
GM m m
GM m m  GM m m  1 1 
−

=
 − 
rf
ri
2
2  ri rf 

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(4000 kg) 
1
1

−


2
 1.79 × 106 m 2.04 × 106 m 
= 6.7 × 108 J
13.46. Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an isolated system, so the mechanical energy is conserved.
Visualize: The space shuttle (mass = ms ) is at a distance of r1 =
Re + 250 km.
Solve: The external work done by the thrusters is
1 ∆U
Wext =
∆Emech =
g
2
where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
=
r1 Re + 250 km to the final orbit rf =
Re + 610 km. Thus
Wext =
1  −−
GM e m
GM e m  GM e m  1 1 
−

=
 − 
ri
2  rf
2  ri rf 

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(75,000 kg) 
1
1

−


6
6
2
 6.62 × 10 m 6.98 × 10 m 
= 1.2 × 1011 J
This much energy must be supplied by burning the on-board fuel.
13.47. Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated system, so
mechanical energy is conserved.
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Newton’s Theory of Gravity
13-23
Visualize:
The orbital radius of the satellite is
r = Ra + h = 8,800 m + 5,000 m = 13,800 m
Solve: (a) The speed of a satellite in a circular orbit is
1/2
v=
GM  (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) 
= 

r
13,800 m


= 7.0 m/s
(b) The minimum launch speed for escape (vi ) will cause the satellite to stop asymptotically (vf = 0 m/s) as
rf → ∞. Using the energy conservation equation K 2 + U 2 = K1 + U1, we get
GM a ms 1
GM a ms
1
msvf2 −
msvi2 −
=
rf
Ra
2
2
vescape
=
2GM a
=
Ra
J
⇒ 0 J − 0=
GM a
1 2
vescape −
Ra
2
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg)
= 12 m/s
8800 m
13.48. Model: Model the moon as a spherical mass and the satellite as a point mass.
Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own axis.
This time happens to be 27.3 days.
Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration necessary for
circular motion around the moon. Therefore,
GM m m
 2π 
mr
=
=
ω 2 mr  
2
r
 T 
=
r3
2
GM mT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(27.3 × 24 × 3600 s) 2
=
4π 2
4π 2
r =8.84 × 107 m
Since=
r Rm + h, then h =r − Rm =8.84 × 107 m − 1.74 × 106 m =8.67 × 107 m.
13.49. Model: Model the earth as a spherical mass and the satellite as a point mass.
Visualize: The satellite is directly over a point on the equator once every two days. Thus, T =2Te =2 × 24 ×
3600 s =
1.728 × 105 s.
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13-24
Chapter 13
Solve: From Kepler’s laws applied to a circular orbit:
 4π 2  3
T2 =
r
 GM 
e

r3
=
GM eT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.728 × 105 ) 2
=
4π 2
4π 2
r =6.71 × 107 m
Assess: The radius of the orbit is larger than the geosynchronous orbit.
13.50. Visualize: Please refer to Figure P13.50.
Solve: The gravitational force on one of the masses is due to the star and the other planet. Thus
G
Mm
r2
+
Gmm
(2r ) 2
=
mv 2 m  2π r 
= 

r
r T 
G
m  4π 2 r 2
M + =

r
4
T2
2
⇒
GM Gm 4π 2r 2
+
=
r
4r
T2
1/2
 4π 2 r 3

1
⇒=
T 

 G ( M + m/4) 
13.51. Solve: (a) Taking the logarithm of both sides of v p = Cu q gives
log C
q
[log(v p ) =p log v] =[log(Cu q ) =log C + q log u ] ⇒ log v = log u +
p
p
But x = log u and y = log v, so x and y are related by
q
log C
=
y  x +
p
 p
(b) The previous result shows there is a linear relationship between x and y, so there is a linear relationship between
log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u will be a
straight line.
(c) The slope of the straight line represented by the equation
=
y (q/p ) x + log C/p is q/p. Thus, the slope of the log vversus-log u graph will be q/p.
(d) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating
these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is=
M M sun .
 4π 2
1
1
log C = log 
 GM
2
2
sun

M sun

4π 2
1
=10−18.528 = 18.528
 =−9.264 ⇒
GM
10
sun

2
 4π  18.528
)=
1.996 × 1030 kg
=

 (10
 G 
The tabulated value, to three significant figures, is M sum =1.99 × 1030 kg. We have used the orbits of the planets to
“weigh the sun!”
13.52. Solve: (a) Dividing the circumference of the orbit by the period gives
v=
2π Rs 2π (1.0 × 104 m)
=
= 6.3 × 104 m/s
T
1.0 s
(b) Using the formula for the acceleration at the surface,
gsurface =
GM s
Rs2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
(1.0 × 104 m) 2
= 1.3 × 1012 m/s 2
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Newton’s Theory of Gravity
13-25
(c) The mass of an object on the earth will be the same as its mass on the star. The gravitational force is
( FG )star = mg surface = 1.3 × 1012 N
(d) The radius of the orbit of the satellite is r = 1 × 104 m + 1.0 × 103 m = 1.1 × 104 m. The period is
4π 2 r 3
4π 2 (1.1 × 104 m)3
T2 =
=
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
⇒ T =6.29 × 10−4 s
This means there are 1589 revolutions per second or 9.5 × 104 orbits per minute.
(e) Applying Equation 13.25 for a geosynchronous orbit,
r3 =
GM s
4π 2
T2 =
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(1.0 s)
4π 2
⇒ r = 1.5 × 106 m
13.53. Model: Assume the solar system is a point particle.
Solve: (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means
r = 25,000 light years = 2500(3.0 × 108 m/s)(365 d/y)(24 h/d)(3600 s/h)(1 y) = 2.36 × 1020 m
T=
2π r 2π (2.36 × 1020 m)
=
= 6.64 × 1015 s = 2.1 × 108 years
v
2.30 × 105 m/s
5.0 × 109 years
= 24 orbits.
2.05 × 108 years
(c) Applying Newton’s second law yields
(b) The number of=
orbits
GM g center mss
r
2
=
mssv 2
r
⇒ M g center =
v 2 r (2.30 × 105 m/s) 2 (2.36 × 1020 m)
=
= 1.9 × 1041 kg
G
6.67 × 10−11 N ⋅ m 2 /kg 2
(d) The number of stars in the center of the galaxy is
1.87 × 1041 kg
1.99 × 1030 kg
= 9.4 × 1010
13.54. Model: Assume the three stars are spherical masses.
Visualize:
The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three stars.
The gravitational force between any two stars is the same. On a given star the two forces from the other stars make an
angle of 60°.
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13-26
Chapter 13
Solve: The value of r can be found as follows:
L/2
L
1.0 × 1012 m
=cos(30°) ⇒ r =
=
=0.577 × 1012 m
r
2cos(30°)
2cos(30°)
The gravitational force between any two stars is
GM 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2
Fg = 2 =
=2.64 × 1026 N
(1.0 × 1012 m) 2
L
The component of this force toward the center is
Fc = Fg cos(30°) = (2.64 × 1026 N)cos(30°) = 2.29 × 1026 N
The net force on a star toward the center is twice this force, and that force equals MRω 2 . This means
 2π 
=
2 × 2.29 × 1026
N MR=
ω 2 MR  
 T 
T=
4π 2 MR
4.58 × 10
26
N
=
4π 2 (1.99 × 1030 kg)(0.577 × 1012 m)
4.58 × 1026 N
2
=3.15 × 108 s =10 years
13.55. Model: Angular momentum is conserved for a particle following a trajectory of any shape.
Visualize:
For a particle in an elliptical orbit, the particle’s angular momentum is
=
L mrv
=
t mrv sin β , where v is the velocity


tangent to the trajectory and β is the angle between r and v.
Solve: At the distance of closest approach (rmin ) and also at the most distant point, β= 90°. Since there is no tangential force (the only force being the radial force) there is no torque, so angular momentum must be conserved:
mPlutov1rmin = mPlutov2 rmax
v2 =v1 (rmin /rmax ) =(6.12 × 103 m/s)((4.43 × 1012 m)/(7.30 × 1012 m) =3.71 km/s
13.56. Model: Angular momentum is conserved for a particle following a trajectory of any shape.
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Newton’s Theory of Gravity
13-27
Visualize:
For a particle in an elliptical orbit, the particle’s angular momentum is=
L mrv
=
t mrv sin β , where v is the velocity


tangent to the trajectory, and β is the angle between r and v.
Solve: At the distance of closest approach (rmin ) and also at the most distant point, β= 90°. Since there is no tangential force (the only force being the radial force) there is no torque, so angular momentum must be conserved:
mMercuryv1rmin = mMercuryv2 rmax
(6.99 × 1010 m)(38.8 km/s)
rmin =rmax v2 /v1 =
=4.60 × 1010 m
59.0 km/s
13.57. Model: For the sun + comet system, the mechanical energy is conserved.
Visualize:
Solve: The conservation of energy equation K f + U f = Ki + U i gives
1
GM s M c 1
GM s M c
M cv22 −
= M cv12 −
r2
r1
2
2
Using G =6.67 × 10−11 Nm 2 /kg 2 , M s =1.99 × 1030 kg, r1 =8.79 × 1010 m, r2 =4.50 × 1012 m, and v=
1 54.6 km/s,
we get v2 =
4.49 km/s.
13.58. Model: Model the planet (p) as a spherical mass and the spaceship (s) as a point mass.
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13-28
Chapter 13
Visualize:
Solve: (a) For the circular motion of the spaceship around the planet,
GM p ms
r02
=
mv02
r0
⇒ v0=
GM p
r0
Immediately after the rockets were fired v1 = v0 /2 and r1= r0 . Therefore,
v1 =
1 GM p
r0
2
(b) The spaceship’s maximum distance is rmax= r0 . Its minimum distance occurs at the other end of the ellipse. The
energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is,
GM p ms 1
GM p ms
1
msv12 −
= msv22 −
r1
r2
2
2
Since the angular momentum at these two ends is conserved, we have
mv1=
r1 mv2r2
⇒ v=
2 v1 ( r1/r2 )
With this expression for v2 , the energy equation simplifies to
GM p
1 2 GM p 1 2
v1 − =
v1 (r1/r2 ) 2 −
2
2
r1
r2
Using r1 = r0 and=
v1 v=
0 /2
1 GM p
,
r0
2
1  1 GM p  GM p 1  1 GM p  r02 GM p
= 

−
 −
r0
r2
2  4 r0 
2  4 r0  r22
⇒
1
1
r
1
− = 0 −
8r0 r0 8r22 r2
 7  2
r
r0
7
1
+ 02 − = 0 ⇒ 
 r2 − r2 + = 0
r
8r0 8r2 r2
8
8
 0
The solutions are r2 = r0 (the initial distance) and =
r2 r0 /7. Thus the minimum distance is rmin
= r0 /7.
13.59. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on the
surface of the earth?
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Newton’s Theory of Gravity
13-29
(b)
(c) The distance is 6.21 × 107 m. This is 1.06RSaturn
13.60. Solve: (a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the orbit?
(b)
(c) The radius of the orbit is
GM E
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
r=
=
=1.00 × 108 m
2
2
(vpayload )
(1997 m/s)
13.61. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moon’s radius. At
what speed does it impact the moon’s surface?
(b)
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13-30
Chapter 13
(c) The speed
is v2 1680 m/s.
=
13.62. Solve: (a) Kepler’s third law for circular orbits is
 4π 2  3
T2 
r ⇒
T
=
=
 GM 


Letting a =
4π 2 32
r
GM
3
4π 2
, the first satellite obeys T = ar 2 . For the second satellite, which orbits the same mass,
GM
3
 ∆r  2
T + ∆=
T a (r + ∆r )= ar 1 +

r 

3
2
Since
3
2
∆r
 1, we can use the approximation (1 ± x) n ≈ 1 ± nx.
r
Thus
3 
3 ∆r 
T + ∆T ≈ ar 2 1 +

 2 r 
3
Subtracting the equation T = ar 2 for the first satellite from this,
3 3  ∆r 
∆T = ar 2  
2
 r 
Dividing this by the equation for the first satellite,
∆T 3 ∆r
=
T
2 r
(b) The satellites orbit the earth. The fractional difference in their periods is
∆T 3 1 km
=
=2.24 × 10−4
T
2 6700 km
After
1
2.24 × 10−4
= 4467 periods they will meet again. For the inner satellite,
T
=
4π 2
G (5.98 × 10
3
24
kg)
(6.700 × 106 m) 2
= 5456 s = 1.52 hrs
So the satellites will meet again in 4467 × 1.52
=
hrs 6770
=
hrs 282 days.
Assess: A communications satellite has an orbital period of around 1.5 h. The surprising length of time between the
two satellites meeting is due to the small differences in their periods.
13.63. Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the satellite’s
distance from the earth is very small compared to the earth’s (and satellite’s) distance to the sun.
Visualize:
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Newton’s Theory of Gravity
13-31
Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational force
toward the earth. This net force is responsible for circular motion around the sun. We want to chose the distance d to
make the period T match the period Te with which the earth orbits the sun. The earth’s orbital period is given by
=
Te2 (4π 2 /GM s ) Re3 . Thus
2
GM m GM e m
mv 2 m  2pp
r
4 2
GM
Fnet = 2s −
ma
mr
mr 3 s
=
=
=
=
=


centripetal
2
2
r
r  Te 
r
d
Te
Re
Using =
r Re − d and canceling the Gm term gives
Ms
( Re − d ) 2
Me Ms
=
( Re − d )
d2
Re3
−
This equation can’t be solved exactly, but we can make use of the fact that d  Re to use the binomial approximation. Factor the Re out of the expressions Re − d to get
Ms
2
Re (1 − d/Re ) 2
−
Me
d
2
=
Ms
Re2
(1 − d/Re )
If we think of d/Re = x  1, we can simplify the first term by using (1 ± x) n ≈ 1 ± nx. Here n = −2, so we get
Ms
Re2
Me Ms
[1 −−−
( 2)d/Re ] =
(1 − d/Re ) ⇒
d 2 Re2
Me
M
=
3 2s
2
d
Re
Thus d =[ M e /(3M s )]1/3 Re =1.50 × 109 m.
Assess: d/Re = 0.010, so our assumption that d/Re  1 is justified.
13.64. Model: The earth and sun are spherical masses. The earth is in a circular orbit around the sun. The projectile is a particle. The effect of the earth’s rotation on the projectile’s velocity is ignored. The earth and sun are so
massive compared to the rocket that they are unaffected by the rocket’s motion (no recoil). The rocket escapes from
the influence of the earth’s gravitation in a distance small in comparison with the distance from the earth to the sun,
so that the change in solar potential of the rocket as it escapes the earth is negligible.
Visualize:
Solve: To escape the sun’s gravitational pull at the earth’s distance from the sun, a projectile must have speed
vescape . The energy conservation equation for the projectile is
M m
1 2
mvescape - G s =
0J
Re-s
2
vescape =2G
Ms
2(6.67 × 10-11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
4 207 × 104 m/s
=
=.
Re-s
(1.50 × 1011 m)
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13-32
Chapter 13
The earth’s speed in its orbit is found by considering Equation 13.22:
ve =
GM s
=.
2 975 × 104 m/s
Re-s
The projectile must also escape from the earth’s gravitational pull. Far from the earth, the projectile’s total speed is
the sum of its speed relative to the earth far from the influence of earth’s gravity, vp , and the earth’s speed.
vescape= ve + vp
vp = vescape − ve = 4.207 × 104 m/s − 2.975 × 104 m/s = 1.232 × 104 m/s.
To obtain the speed vp far from the earth, the projectile must be launched from the earth with speed vlaunch . Using
energy conservation,
M m 1
1 2
mvlaunch − G e = mvp2
Re
2
2
⇒ vlaunch =
2(6.67 × 102 11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
(6.37 × 106 m)
2G
Me
+ vp2
Re
+ (1.232 × 104 m/s) 2
=1.66 × 104 m/s
Assess: The projectile must attain a speed of 16.6 km/s if launched in the direction of earth’s motion. This is about a
factor of 7 times less than what is required from rest. Note that the speed of the earth in its orbit about the sun ve
differs from the escape speed from the sun vescape by a factor of
2.
13.65. Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a spherical
mass. Momentum is conserved during the inelastic collision of the two satellites.
Visualize:
Solve: For the given orbit, r0 = Re + 1 × 106 m = 7.37 × 106 m. The speed of a satellite in this orbit is
=
v0
GM e
=
7357 m/s
r0
The two satellites collide, stick together, and move with velocity v1. The equation for momentum conservation for
the perfectly inelastic collision gives
=
(400 kg + 100
kg)v1 (400 kg)(7357 m/s) − (100 kg)(7357 m/s)
v1 = 4414 m/s
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Newton’s Theory of Gravity
13-33
The new satellite’s radius immediately after the collision is still r1 =r0 =7.37 × 106 m. Now it is moving in an elliptical orbit. We need to determine if the minimum distance r2 is larger or smaller than the earth’s radius
Re =6.37 × 106 m.
The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is
L = mrv sin β (see Equation 13.26) and is conserved in a trajectory of any shape. The angle β is 90° when
v1 = 4414 m/s and when the satellite is at its closest approach to the earth. From conservation of angular momentum,
we have
r1v1 =
r2v2 =
(4414 m/s) =.
3 253 × 1010 m 2 /s
r2 =
3.253 × 1010 m 2 /s
v2
Using the conservation of energy equation at positions 1 and 2,
1
GM e (500 kg) 1
GM e (500 kg)
(500 kg)(4414 m/s) 2 −
= (500 kg)v22 −
2
r2
7.37 × 106 m 2
Using the above expression for r2 , we can simplify the energy equation to
v22 − (2.452 × 104 m/s)v2 + (8.876 × 107 m 2 /s 2 ) =
0 m 2 /s 2
v2 = 20,107 m/s and 4414 m/s
A velocity of 20,107 m/s for v2 yields
3.253 × 1010 m 2 /s
r2 =
=1.62 × 106 m
20,107 m/s
Since r2 < Re =6.37 × 106 m, the combined mass of the two satellites will crash into the earth.
13.66. Model: Planet Physics is a spherical mass. The cruise ship is in a circular orbit.
Solve: (a) At the surface, the free-fall acceleration=
is g GM/R 2 . From kinematics,
y f = yi + v0∆t − 2 g (∆t ) 2
⇒ 0 m = 0 m + (11 m/s)(2.5 s) − 2 g (2.5 s) 2
g = 2.20 m/s 2
The period of the cruise ship’s orbit is 230 ×=
60 13,800 s. For the circular orbit of the cruise ship,
 4π 2 
(2 R )3 ⇒
T2 = 
 GM 


 R2
= R

32π
 GM
T2
2

1
 = R  
g

(2.20 m/s 2 )(13,800 s) 2
=
R=
1.327 × 106 m.
32π 2
The mass is thus M = ( R 2 /G ) g = 5.8 × 1022 kg.
(b) From part (a), R = 1.3 × 106 m.
13.67. Model: The moon is a spherical mass. The moon lander is originally in a circular orbit.
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13-34
Chapter 13
Visualize:
Solve: Energy and momentum are conserved between points 1 and 2 in the elliptical orbit. Also, at both points
Let h 1000 km. The original speed of the lander
is v0
β= 90°, so =
L mrv sin=
β mrv.=
=
Conservation of angular momentum requires mr1v1 =
mr2v2
⇒
GM m
=
1338 m/s.
Rm + h
v2 r1 Rm + h
= =
. The energy conservation equav1 r2
Rm
tion gives
1
M m 1
M m
mv12 − G m =mv22 − G m
Rm + h 2
Rm
2
R +h
Substituting v2 =  m
 v1,
 Rm 
2
Mm
Mm
1 2
1 2  Rm + h 
v1 − G =
v1 
 −G
Rm + h 2  Rm 
Rm
2
  R + h 2 


 2GM m  1 − 1 
v12 1 −  m
=
  Rm  
 Rm + h Rm 


2GM m h
1
2GM m  Rm 
=
v12 =
1.392 × 106 m 2 /s 2

=
Rm ( Rm + h) [( Rm + h)/Rm ]2 − 1 ( Rm + h)  2 Rm + h 
v1 = 1180 m/s
The fractional change in speed required to just graze the moon at point 2 is
1338 m/s − 1180 m/s
= 11.8%
1338 m/s
Assess: A reduction in speed by almost 12% is reasonable.
13.68. Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated system, so
mechanical energy is conserved. Also, the angular momentum of the satellite is conserved.
Visualize: Please refer to Figure CP13.68.
Solve: (a) Angular momentum
is L mrv sin β . The angle β= 90° at points 1 and 2, so conservation of angular
=
momentum requires
mr1v=
1′ mr2v2′
 r2 
⇒ v=
 v2′
1′ 
 r1 
The energy conservation equation is
GMm 1
GMm
1
m(v2′ ) 2 −
m(v1′ ) 2 −
=
r2
r1
2
2
 1 1
2
⇒ (v2′ ) 2 − (v1′ )=
2GM  − 
 r2 r1 
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Newton’s Theory of Gravity
13-35
Using the angular momentum result for v1′ gives
2
2
r 
 r1 −−−
 r1 r2 
r2 
r22 
2  r1
=
(v2′ ) 2 − (v2′ ) 2  2=
 2GM 
 2GM 
 ⇒ (v2′ ) 

2
 r1 
 r1r2 
 r1r2 
 r1 
=
v2′
2GM (r1/r2 )
r1 + r2
 r2 
=
and
v1′ =
 v2′
 r1 
2GM (r1/r2 )
r1 + r2
(b) For the circular orbit,
v1
=
GM
=
r1
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
= 7730 m/s
(6.37 × 106 m + 3 × 105 m)
For the elliptical orbit,
r1 =Re + 300 km =6.37 × 106 m + 3 × 105 m =6.67 × 106 m
r2 = Re + 35,900 km = 6.37 × 106 m + 3.59 × 107 m = 4.23 × 107 m
=
v1′
2GM (r1/r2 )
=
⇒ v1′ 10,160 m/s
r1 + r2
1
1
(c) From the work-kinetic energy theorem, W =
∆K = mv1′2 − mv12 =.
2 17 × 1010 J
2
2
(d) v2′ =
2GM (r1/r2 )
r1 + r2
Using the same values of r1 and r2 as in=
(b), v2′ 1600 m/s. For the circular orbit,
=
v2
GM
= 3070 m/s
r2
1
1
(e) W = mv22 − mv2′ 2 =3.43 × 109 J
2
2
(f) The total work done is 2.52 × 1010 J. This is the same as in Example 13.6, but here we’ve learned how the work
has to be divided between the two burns.
13.69. Model: The rod is thin and uniform.
Visualize:
Solve: (a) The rod is not spherical so must be divided into thin sections each dr wide and having mass dm. Since the
rod is uniform,
dm dr
=
M
L
⇒ dm =
M
dr
L
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13-36
Chapter 13
The width of the rod is small enough so that all of dm is distance r away from m. The gravitational potential energy
of dm and m is
m dm
m( M/L)dr
−G
=
−G
dU =
r
r
The total potential energy of the rod and mass m is found by adding the contributions dU from every point along the
rod in an integral:
x+
L
2
GMm
dr
GMm  x + L/2 
−
−
U=
ln 

∫ dU =
∫ r =
L
L
 x − L/2 
L
x−
2
Note r is increasing with the limits chosen as they are.
(b) The force on m when at x is
1 
dU GMm d  
L
L   GMm  1

−
=
−
ln  x +  −−
ln  x
F=
 = 


2
2 
dx
L dx  
L  x + L/2 x − L/2 

4
L


, x≥
=
−GMm  2
2
2
 4x − L 
Assess: The direction of the force is toward the − x direction, as expected. The force magnitude approaches ∞ as
the mass m approaches the end of the rod, but goes to zero like x −2 as x gets large. This is expected since from far
away the rod looks like a point mass.
13.70. Model: The ring is uniform and is so thin that every point on it may be considered to be the same distance r
from m.
Visualize:
Solve: (a) We must determine the gravitational potential (dU) between m and an arbitrary part of the ring dm, then
add using an integral all the contributions to U. Since the ring is uniform,
dm
dl
=
M 2π r
The distance from m to dm is r =
⇒ dm =
M
dl
2π r
x 2 + R 2 . The gravitational potential between m and dm is
m dm
mM
dU =
−G
=
−G
r
2π r
dl
x + R2
2
The total gravitational potential is
U=
∫ dU =
−
GmM
2π r x 2 + R 2
∫
dl
ring
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Newton’s Theory of Gravity
13-37
Note that x and R do not change for any location of dm. The integral is just the length of the ring.
∫
dl = 2π R
ring
Thus
U= −
GmM
x2 + R2
(b) The force on m when at x is
3
1
dU
d
 1
−−
−
=
Fx =
GmM [( x 2 + R 2 ) 2 ] =
GmM  −  ( x 2 + R 2 ) 2 (2 x)
dx
dx
 2
x
= −GmM
3
( x2 + R2 ) 2
Thus the magnitude of the force is
F = GmM
x
3
( x + R2 ) 2
2
.
Assess: The force is zero at the center of the ring. Elsewhere its direction is toward the origin. As x gets large, the
force decreases like x2 2 This is expected since from far away the ring looks like a point mass.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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