CHAPTER
11
Gravity
1* · True or false: (a) Kepler’s law of equal areas implies that gravity varies inversely with the square of the distance.
(b) The planet closest to the sun, on the average, has the shortest period of revolution about the sun.
(a) False (b) True
2
· If the mass of a satellite is doubled, the radius of its orbit can remain constant if the speed of the satellite (a)
increases by a factor of 8. (b) increases by a factor of 2. (c) does not change. (d) is reduced by a factor of 8. (e) is
reduced by a factor of 2.
(c)
3
· One night, Lucy picked up a strange message on her ham radio. “Help! We ran away from Earth to live in peace
and serenity, and we got disoriented. All we know is that we are orbiting the sun with a period of 5 years. Where are
we?” Lucy did some calculations and told the travelers their mean distance from the sun. What is it?
Use Equ. 11-2; R = RES (5/1) 2/3; RES = 1.5 × 1011 m
R = (1.5 × 1011)(5) 2/3 m = 4.39 × 1011 m
4
· Halley’s comet has a period of about 76 y. What is its mean distance from the sun?
Rmean = (1 AU)(76) 2/3 (see Problem 3)
Rmean = 1.5 × 1011 × 762/3 m = 26.9 × 1011 m
5* · A comet has a period estimated to be about 4210 y. What is its mean distance from the sun? (4210 y was the estimated period of the comet Hale –Bopp, which was seen in the Northern Hemisphere in early 1997. Gravitational
interactions with the major planets that occurred during this apparition of the comet greatly changed its period, which is
now expected to be about 2380 y.)
Rmean = (1 AU)(4210) 2/3 (see Problem 4)
Rmean = 1.5 × 1011 × 42102/3 m = 3.91 × 1013 m
6
· The radius of the earth’s orbit is 1.496 × 1011 m and that of Uranus is 2.87 × 1012 m. What is the period of
Uranus?
Use Equ. 11-2; TU = (1 y)(RU/1 AU)3/2
TU = (2.87 × 1012/1.5 × 1011)3/2 y = 83.7 y
7
· The asteroid Hektor, discovered in 1907, is in a nearly circular orbit of radius 5.16 AU about the sun. Determine
the period of this asteroid.
Chapter 11
TH = (1 y)(5.16/1) 3/2 (see Problem 6)
8
Gravity
TH = 11.7 y
·· The asteroid Icarus, discovered in 1949, was so named because its highly eccentric elliptical orbit brings it close to
the sun at perihelion. The eccentricity e of an ellipse is defined by the relation d p = a(1 - e), where d p is the perihelion
distance and a is the semimajor axis. Icarus has an eccentricity of 0.83. The period of Icarus is 1.1 years. (a)
Determine the semimajor axis of the orbit of Icarus. (b) Find the perihelion and aphelion distances of the orbit of
Icarus.
(a) Use Kepler’s third law; a = (1 AU)(TI)2/3
a = 1.5 × 1011 × 1.12/3 m = 1.6 × 1011 m
(b) d p = a(1 - e); d a + d p = 2a; d a = 2a - d p
d p = (1.6 × 1011 × 0.17) m = 2.72 × 1010 m;
d a = 2.93 × 1011 m
9* · Why don’t you feel the gravitational attraction of a large building when you walk near it?
The mass of the building is insignificant compared to the mass of the earth.
10 · Astronauts orbiting in a satellite 300 km above the surface of the earth feel weightless. Why? Is the force of
gravity exerted by the earth on them negligible at this height?
They feel weightless because the force of gravity provides just their centripetal acceleration. The force of gravity
is not negligible. (see Example 11-2)
11 ·· The distance from the center of the earth to a point where the acceleration due to gravity is g/4 is (a) RE.
(b) 4RE. (c) RE/2. (d) 2RE. (e) none of the above.
(d) g ∝ 1/R2.
12 ·· At the surface of the moon, the acceleration due to the gravity of the moon is a. At a distance from the center of
the moon equal to four times the radius of the moon, the acceleration due to the gravity of the moon is (a) 16a. (b)
a/4. (c) a/3. (d) a/16. (e) none of the above.
(d) a ∝ 1/R2.
13* · One of Jupiter’s moons, Io, has a mean orbital radius of 4.22 × 108 m and a period of 1.53 × 105 s. (a) Find the
mean orbital radius of another of Jupiter’s moons, Callisto, whose period is 1.44 × 106 s. (b) Use the known value of G
to compute the mass of Jupiter.
(a) Use Equ. 11-2; RC = RI (TC/TI)2/3
RC = (4.22 × 108)(14.4/1.53) 2/3 m = 18.8 × 108 m
(b) Use Equ. 11-15; M J = 4π 2RI3/GTI2
M J = 4π 2(4.22 × 108)3/[6.67 × 10-11(1.53 × 105)2]
= 1.9 × 1027 kg
14 · The mass of Saturn is 5.69 × 1026 kg. (a) Find the period of its moon Mimas, whose mean orbital radius is
1.86 × 108 m. (b) Find the mean orbital radius of its moon Titan, whose period is 1.38 × 106 s.
(a) Use Equ. 11-15; T = (4π 2R3/GMS)1/2
TM = [4π 2(1.86 × 108)3/(6.67 × 10-11 × 5.68 × 1026)]1/2 s
= 8.18 × 104 s
(b) Use Equ. 11-2
RT = 1.86 × 108(138/8.18) 2/3 m = 12.2 × 108 m
15 ·
Calculate the mass of the earth from the period of the moon T = 27.3 d, its mean orbital radius
Chapter 11
Gravity
rm = 3.84 × 108 m, and the known value of G.
Convert T to s: T = 2.36 × 106 s; Use Equ. 11-15
M E = 4π 2(3.84 × 108)3/G(2.36 × 106)2 kg
= 6.02 × 1024 kg
Note: This neglects the mass of the moon; consequently, the mass calculated here is slightly too great.
16 · Use the period of the earth (1 y), its mean orbital radius (1.496 × 1011 m), and the value of G to calcula te the mass
of the sun.
Proceed as in Problem 15. 1 y = 3.156 × 107 s; substituting in the numerical values yields M S = 1.99 × 1030 kg.
17* · An object is dropped from a height of 6.37 × 106 m above the surface of the earth. What is its initial accelera-tion?
Since R = 2RE, a = g E/4 = (9.81/4) m/s2 = 2.45 m/s2.
18 · Suppose you leave the solar system and arrive at a planet that has the same mass per unit volume as the earth but
has 10 times the earth’s radius. What would you weigh on this planet compared with what you weigh on earth?
1. Find M P, mass of the planet, in units of M E
For same ρ , M ∝ R3; M = 103M E
2. Find g P in units of g E
Since g ∝ M/R2, g P = g E × 103/102 = 10g E
3. Weight on planet = 10 × weight on earth
19 · Suppose that the earth retained its present mass but was somehow compressed to half its present radius. What
would be the value of g, the acceleration due to gravity, at the surface of this new, compact planet?
Since g is proportional to 1/R2, g on planet would be 4 × 9.81 m/s2 = 39.24 m/s2.
20 · A planet moves around a massive sun with constant angular momentum. When the planet is at perihelion, it has a
speed of 5 × 104 m/s and is 1.0 × 1015 m from the sun. The orbital radius increases to 2.2 × 1015 m at aphelion. What is
the planet’s speed at aphelion?
L = constant = mvprp = mvara; va = vprp/ra
va = (5 × 104 × 1015/2.2 × 1015) m/s = 2.27 × 104 m/s
21* · A comet orbits the sun with constant angular momentum. It has a maximum radius of 150 AU, and at aphelion its
speed is 7 × 103 m/s. The comet’s closest approach to the sun is 0.4 AU. What is its speed at perihelion?
vp = vara/rp (see Problem 20)
vp = 7 × 103 × 150/0.4 m/s = 2625 km/s
22 ·· The speed of an asteroid is 20 km/s at perihelion and 14 km/s at aphelion. Determine the ratio of the aphelion to
perihelion distance.
ra/rp = vp/va (see Problem 20)
ra/rp = 20/14 = 1.43
23 ·· A satellite with a mass of 300 kg moves in a circular orbit 5 × 107 m above the earth’s surface. (a) What is the
gravitational force on the satellite? (b) What is the speed of the satellite? (c) What is the period of the satellite?
(a) Fg = Mg(R) = Mg × [RE/(RE + h)]2; RE
Fg = 300 × 9.81 × (6.37/56.37) 2 N = 37.58 N
= 6.37 × 106 m
(b) Mv2/R = Fg; v = (FgR/M)1/2
v = (37.58 × 56.37 × 106/300) 1/2 m/s = 2.657 km/s
(c) T = 2π R/v
T = 2π × 56.37 × 106/2657 s = 1.333 × 105 s = 37 h
Chapter 11
Gravity
24 ·· At the airport, a physics student weighs 800 N. The student boards a jet plane that rises to an altitude of 9500 m.
What is the student’s loss in weight?
1. Find w at 9.5 km; w(h) = wE × [RE/(RE + h)]2
w(h) = 800(6370/6379.5) 2 N = 797.6 N
2. ∆w = wE - w(h)
∆w = 2.4 N
25* ·· Suppose that Kepler had found that the period of a planet’s circular orbit is proportional to the square of the orbit
radius. What conclusion would Newton have drawn concerning the dependence of the gravitational attraction on
distance between two masses?
Take F = CRn, where C is a constant. Then, for a stable circular orbit, v2/R = F = CRn. The period of the orbit is given
by T = 2π R/v, and so T = 2π R/C1/2R(n+1)/2. Therefore, if T ∝ R2, 1 - (n + 1)/2 = 2, n = -3, and F ∝ 1/R3.
26 ·· A superconducting gravity meter can measure changes in gravity of the order ∆g/g = 10-11. (a) Estimate the
maximum range at which an 80-kg person can be detected by this gravity meter. Assume that the gravity meter is
stationary, and that the person’s mass can be considered to be concentrated at his or her center of gravity. (b) What
vertical change in the position of the gravity meter in the earth’s gravitational field is detectable?
(a) Find the gravitational field, g(r) due to mass m
g(r) = Gm/r2; given: g(r) = 10-11g E = 10-11(GME/RE2)
Solve for r; r = RE(1011m/M E)1/2
r = 6.37 × 106(8 × 1012/6 × 2024)1/2 m = 7.36 m
(b) Write g(r) = C/r2 and differentiate
dg/dr = -2C/r3; dg/g = -2 dr/r = 10-11
-11
6
-5
Find dr = ∆r for r = RE
∆r = 1/2 × 10 × 6.37 × 10 m ≈ 3.2 × 10 m
= 0.032 mm
27 ·· During a solar eclipse, when the moon is between the earth and the sun, the gravitational pull of the moon and the
sun on a student are in the same direction. (a) If the pull of the earth on the student is 800 N, what is the force of the
moon on the student? (b) What is the force of the sun on the student? (c) What percentage correction due to the sun
and moon when they are directly overhead should be applied to the reading of a very accurate scale to obtain the
student’s weight?
(a) F due to moon is FM = GmM M/REM2. Write FM
FM = 800(M M/M E)(RE/REM)2 N
in terms of FE = GmM E/RE2 = 800 N and
FM = 800(7.35/598)(6.37/384.4) 2 N = 0.0027 N
evaluate FM.
(b) Likewise, FS = 800(M S/M E)(RE/RES )2
FS = 800(1.99 × 106/5.98)(6.37/1.496 × 105)2 = 0.483 N
(c) Percentage correction = (-0.485/800) × 100
correction = -0.061%
28 ·· Suppose that the attractive interaction between a star of mass M and a planet of mass m << M were of the form
F = KMm/r, where K is the gravitational constant. What would be the relation between the radius of the planet’s
circular orbit and its period?
KMm/r = mv2/r; so v2 = KM, independent of r
T = 2π r/v = 2π r/(KM)1/2 , i.e., proportional to r
29* ·· The mass of the earth is 5.97 × 1024 kg and its radius is 6370 km. The radius of the moon is 1738 km. The
acceleration of gravity at the surface of the moon is 1.62 m/s2. What is the ratio of the average density of the moon to
that of the earth?
Chapter 11
Gravity
1. Write g E and g M in terms of ρ E and ρ M
g E = G(4πρ ERE3/3)/RE2; g M = G(4πρ MRM3/3)/RM2
2. Find g M/g E and solve for and evaluate ρ M/ρ E
g M/g E = ρ MRM/ρ ERE; ρ M/ρ E = (1.62/9.81)(6.37/1.738)
= 0.605
30 ··· A plumb bob near a large mountain is slightly deflected from the vertical by the gravitational attraction of the
mountain. Estimate the order of magnitude of the angle of deflection using any assumptions you like.
First, we note that the two force fields are approximately at right angles. Their magnitudes are GME/RE2 and
Gmm/Dm2, where M m is the mass of the mountain and Dm the distance to its center. Thus the angle of deflection is
given by θ = tan-1(M mRE2/M ED2). Assume that the density of the mountain is the average density of the earth. Let the
“mountain” be a cylinder 10 km in diameter and 4 km high. Its mass is ρ Eπ r2h. We consider the plumb bob to be
adjacent to the mountain cylinder so that D = r = 5 km. Now M m/M E = π r2h/(4π RE3/3) ≈ 3 × 10-10 and RE2/D2 ≈ 1.6 ×
106. The angle of deflection is then approximately θ = tan-1(5 × 10-4) = 5 × 10-4 rad ≈ 0.03o.
31 · Why is G so difficult to measure?
Measurement of G is difficult because masses accessible in the laboratory are very small compared to the mass of the
earth.
32 · The masses in a Cavendish apparatus are m1 = 10 kg and m2 = 10 g, the separation of their centers is 6 cm, and
the rod separating the two small masses is 20 cm long. (a) What is the force of attraction between the large and small
masses? (b) What torque must be exerted by the suspension to balance these forces?
(a) Use Equs. 11-3 and 11-4
F = (6.67 × 10-11 × 10 × 10-2/36 × 10-4) N
= 1.85 × 10-9 N
-10
(b) τ = 2Fr, r = 0.1 m
τ = 3.7 × 10 N.m
33* · The masses in a Cavendish apparatus are m1 = 12 kg and m2 = 15 g, and the separation of their centers is
7 cm. (a) What is the force of attraction between these two masses? (b) If the rod separating the two small masses
is 18 cm long, what torque must be exerted by the suspension to balance the torque exerted by gravity?
(a) See Problem 32
F = (6.67 × 10-11 × 12 × 1.5 × 10-2/49 × 10-4) N
= 2.45 × 10-9 N
-10
(b) τ = 2Fr, r = 0.09 m
τ = 4.41 × 10 N.m
34 ·· How would everyday life change if gravitational and inertial mass were not identical?
The force required to accelerate an object (car, person, etc.) would not be proportional to its weight. Also, the period
of a pendulum clock would depend on the mass of the bob.
35 ·· If gravitational and inertial mass were not identical, what would change for (a) an offensive lineman on a football
team? (b) a car? (c) a paperweight?
(a) His effectiveness would depend on his mass rather than his weight. (b) The power requirement would not be
determined by the car’s weight, but by its mass. (c) There would be no significant effect.
36 · A standard object defined as having a mass of exactly 1 kg is given an acceleration of 2.6587 m/s2 when a certain
force is applied to it. A second object of unknown mass acquires an acceleration of 1.1705 m/s2 when the same force
Chapter 11
Gravity
is applied to it. (a) What is the mass of the second object? (b) Is the mass that you determined in
part (a) gravitational or inertial mass?
(a) From F = ma, m2 = (2.6587/1.1705) kg = 2.2714 kg. (b) It is the inertial mass of m2.
37* · The weight of a standard object defined as having a mass of exactly 1 kg is measured to be 9.81 N. In the same
laboratory, a second object weighs 56.6 N. (a) What is the mass of the second object? (b) Is the mass you determined
in part (a) gravitational or inertial mass?
(a) As in Problem 36, m2 = (56.6/9.81) kg = 5.77 kg. (b) This is the gravitational mass of m2—determined by the
effect on m2 of the earth’s gravitational field.
38 · (a) Taking the potential energy to be zero at infinite separation, find the potential energy of a 100-kg object at the
surface of the earth. (Use 6.37 × 106 m for the earth’s radius.) (b) Find the potential energy of the same object at a
height above the earth’s surface equal to the earth’s radius. (c) Find the escape speed for a body projected from this
height.
(a) Use Equ. 11-18; U(RE) = -GMEm/RE = -gmR E
U(RE) = -(9.81 × 100 × 6.37 × 106) J = -6.25 × 109 J
(b) G, ME, and m are unchanged, U(2RE) = U(RE)/2
U(2RE) = -3.12 × 109 J
(c) From (b), Kesc(2RE) = 1/2Kesc(RE); K ∝ v2
vesc(2RE) = vesc(RE)/ 2 = (11.2/ 2 0) km/s = 7.92 km/s
39 · A point mass m0 is initially at the surface of a large sphere of mass M and radius R. How much work is needed to
remove it to a very large distance away from the large sphere?
W = ∆U = Uf - Ui. Uf = 0, Ui = -GMm0/R. So W = GMm0/R.
40 · Suppose that in space there is a duplicate earth, except that it has no atmosphere, is not rotating, and is not in
motion around any sun. What initial velocity must a spacecraft on its surface have to travel vertically upward a
distance above the surface of the planet equal to one earth radius?
From the results of Problem 38, we know that the change in gravitational potential is exactly half the
gravitational potential difference between R = RE and R = ∞. Consequently, v = vesc/ 2 = 7.92 km/s.
41* ·· An object is dropped from rest from a height of 4 × 106 m above the surface of the earth. If there is no air
resistance, what is its speed when it strikes the earth?
2
14
6
6
1. Use 1/2mv2 = -∆U = U(4 × 106 + RE) - U(RE)
1/2v = (3.99×10 )[(1/6.37×10 ) - (1/10.37×10 )] J/kg
v = 6.95 km/s
2. Solve for v
42 ·· An object is projected upward from the surface of the earth with an initial speed of 4 km/s. Find the maximum
height it reaches.
1. Uf = Ki + Ui; divide both sides by m
-GME/(RE + h) = 1/2vi2 - GME/RE
2. Solve for and evaluate h
h = (Rv i2)/[(2GME/RE) - vi2] = 9.3 × 105 m
43 ·· A spherical shell has a radius R and a mass M. (a) Write expressions for the force exerted by the shell on a point
mass m0 when m0 is outside the shell and when it is inside the shell. (b) What is the potential-energy function U(r) for
this system when the mass m0 is at a distance r (r ≥ R) if U = 0 at r = ∞? Evaluate this function at r = R. (c) Using
the general relation for dU = -F⋅d r = -Fr dr, show that U is constant everywhere inside the shell. (d) Using the fact
Chapter 11
Gravity
that U is continuous everywhere, including at r = R, find the value of the constant U inside the shell. (e) Sketch U(r)
versus r for all possible values of r.
(a) F = m0g ; use Equ. 11-24
Outside: F = GMm0/r2, radially in; inside: F = 0
r
U(r) = -GMm0/r; U(R) = -Gmm0/R
(b) U(r) = - ∫ F r dr for r > R
∞
(c), (d) F = 0 for r < R, dU/dr = 0, U = constant.
Since U is continuous, then for r < R, U(r)
= U(R) = -Gmm0/R.
(e) A sketch of U(r) is shown
44 ··· Our galaxy can be considered to be a large disk of radius R and mass M of approximately uniform mass density.
(a) Consider a ring element of radius r and thickness dr of such a disk. Find the gravitational potential energy of a
1-kg mass on the axis of this element a distance x from its center. (b) Integrate your result for part
(a) to find the total gravitational potential energy of a 1-kg mass at a distance x due to the disk. (c) From
Fx = - dU/dx and your result for part (b), find the gravitational field g x on the axis of the disk.
Let σ = mass/unit area = M/π R2. Let d =
x2 + r 2 0 be the distance from a point at a radius r from the center of
the disk to the point a distance x along its axis.
(a) Write an expression for dU
R
(b) U = -
∫
0
2πrσ G dr
2
x + r
(c) Fx = -dU(x)/dx
2
= - 2πσ G ( x 2 + R2 - x)
dU = -
U(x) = -
Fx = -
2πrσ G dr
x2 + r 2
2G M
R
2
( x 2 + R 2 - x)
2GM 
12
R 

 = g(x)
2
2 
x + R 
x
45* ··· The assumption of uniform mass density in Problem 44 is rather unrealistic. For most galaxies, the mass density
increases greatly toward the center of the galaxy. Repeat Problem 44 using a surface mass density of the form σ(r) =
C/r, where σ(r) is the mass per unit area of the disk at a distance r from the center. First determine the constant C in
terms of R and M; then proceed as in Problem 44.
Chapter 11
R
(a) M =
∫
R
∫
2πrσ dr = 2π C dr = 2π C R
0
(b) dU = -
Gravity
0
GM dr
C = M/2π R
U(x) = −
R x2 + r 2
(c) Fx = -dU/dx = g(x)
Fx =
GM
R
R
∫
0
GM
x x 2 + R2
dr
x2 + r2
=−
GM  R + x 2 + R 2
ln
R 
x





= g(x)
46 · What is the effect of air resistance on the escape speed near the earth’s surface?
It causes an increase in the escape speed.
47 · Would it be possible in principle for the earth to escape from the solar system?
Yes, if it interacted with a huge comet.
48 · If the mass of a planet is doubled with no increase in its size, the escape speed for that planet will be (a) increased
by a factor of 1.4. (b) increased by a factor of 2. (c) unchanged. (d) reduced by a factor of 1.4. (e) reduced by a
factor of 2.
(a) See Equ. 11-19.
49* · The planet Saturn has a mass 95.2 times that of the earth and a radius 9.47 times that of the earth. Find the escape
speed for objects near the surface of Saturn.
ve(Sat) = (2GMS/RS)1/2 = ve(earth)[(M S/M E)(RE/RS)]1/2
ve(Sat) = 11.2(95.2/9.47) 1/2 km/s = 35.5 km/s
50 · Find the escape speed for a rocket leaving the moon. The acceleration of gravity on the moon is 0.166 times that
on earth, and the moon’s radius is 0.273RE.
ve(moon) = (2g mRm)1/2 = ve(earth)[g mRm/g ERE]1/2
ve(moon) = 11.2(0.166 × 0.273) 1/2 km/s = 2.38 km/s
51 ·· A particle is projected from the surface of the earth with a speed equal to twice the escape speed. When it is very
far from the earth, what is its speed?
Since its speed is twice the escape speed, its initial kinetic energy is four times the energy needed to escape.
Therefore, when it has escaped, it has an energy that is three times the escape energy and its speed is ve 3 =
19.4 km/s.
52 ·· What initial speed should a partic le be given if it is to have a final speed when it is very far from the earth equal to
its escape speed?
Using the same reasoning as in Problem 51, one finds that vi = ve 2 = 15.8 km/s.
53* ·· A space probe launched from the earth with an initial speed vi is to have a speed of 60 km/s when it is very far
from the earth. What is vi?
Chapter 11
1/2mvi
2
= 1/2mvf2 + 1/2mve2; vi = (vf2 + ve2)1/2
Gravity
vi = (602 + 11.22)1/2 km/s = 61.04 km/s
54 ·· (a) Calculate the energy in joules necessary to launch a 1-kg mass from the earth at escape speed. (b) Convert
this energy to kilowatt-hours. (c) If energy can be obtained at 10 cents per kilowatt-hour, what is the minimum cost of
giving an 80-kg astronaut enough energy to escape the earth’s gravitational field?
(a) E =1/2mve2
E = 1/2(11.2 × 103)2 J = 62.72 MJ
(b) 1 kW.h = 3.6 MJ
E = (62.72/3.6) kW.h = 17.42 kW.h
(c) Cost = $ (M × 0.1 × E)
55 ··
Cost = $ 139.36
Show that the escape speed from a planet is related to the speed of a circular orbit just above the surface of the
planet by ve = 2 vc, where vc is the speed of the object in the circular orbit.
From Example 11-6, the kinetic energy of a mass m in a circular orbit is K = 1/2U. Note that this result is true for
any circular orbit of a mass m about a massive center. But U is the escape energy of 1/2mve2. Consequently, K =
2
1/2mvc
= 1/2(1/2mve2) and vc = ve/ 2 .
56 ·· Find the speed of the earth vc as it orbits the sun, assuming a circular orbit. Use this and the result of Problem 55
to calculate the speed veS needed by the earth to escape from the sun.
1. vc = 2π RES /T; RES = 1.5 × 1011 m,
vc = (2π × 1.5 × 1011/3.156 × 107) m/s = 29.9 km/s
T = 3.156 × 107 s
ve = 42.2 km/s
2. ve = vc 2
57* ·· If an object has just enough energy to escape from the earth, it will not escape from the solar system because of
the attraction of the sun. Use Equation 11-19 with M S replacing M E and the distance to the sun rS replacing RE to
calculate the speed veS needed to escape from the sun’s gravitational field for an object at the surface of the earth.
Neglect the attraction of the earth. Compare your answer with that in Problem 56. Show that if ve is the speed needed
to escape from the earth, neglecting the sun, then the speed of an object at the earth’s surface needed to escape from
the solar system is given by ve,solar2 = ve2 + veS 2, and calculate ve,solar.
From Equ. 11-19, veS = (2GMS/rS)1/2 = [2 × (6.67 × 10-11) × (2 × 1030)/1.5 × 1011]1/2 = 42.2 km/s; this is the same speed
calculated in Problem 56, as it should be. The energy needed to escape the solar system, starting from the surface of
the earth, is the sum of the energy needed to escape the earth’s gravity plus that required to escape from the sun,
starting at the earth’s orbit radius. These energies are proportional to the squares of the corresponding escape
velocities. Therefore, ve,solar2 = ve2 + veS 2; ve,solar = (11.22 + 42.22)1/2 km/s = 43.7 km/s.
58 ·· Why is it reasonable to neglect the other planets in calculating the speed needed to escape from the solar system?
Would you expect the actual value of this speed to be greater or less than that calculated in Problem 57?
The masses of the planets are only about 0.1% of the sun’s mass. The actual value of ve,solar is slightly greater because
the total mass of the solar system is slightly larger than that of the sun.
59 ·· An object is projected vertically from the surface of the earth. Show that the maximum height reached by the
object is H = REH′/(RE - H′), where H′ is the height that it would reach if the gravitational field were constant.
H′ = v2/2g. To determine H use energy conservation: 1/2mv2 = -∆U = GMm[1/RE - 1/(RE+H)] or
v2 = 2gRE2[1/RE - 1/(RE+H)]. So H′ = REH/(RE + H) and, solving for H, H = REH′/(RE - H′).
Chapter 11
Gravity
60 ·· An object (say, a newly discovered comet) enters the solar system and makes a pass around the sun. How can we
tell if the object will return many years later, or if it will never return?
If the path is an ellipse, it will return; if its path is hyperbolic or parabolic, it will not return.
61* ·· A spacecraft of 100 kg mass is in a circular orbit about the earth at a height h = 2RE. (a) What is the period of the
spacecraft’s orbit about the earth? (b) What is the spacecraft’s kinetic energy? (c) Express the angular momentum L
of the spacecraft about the earth in terms of its kinetic energy K and find its numerical value.
We will use the result of Problem 55, i.e., vc = ve/ 2 , where vc is the speed of a circular orbit just above the
surface.
(a) 1. K(3RE) = K(RE)/3; v = vc/ 3 = ve/ 6
v = 11.2/ 6 km/s = 4.57 km/s
2. T = 2π (3RE)/v
(b) K = 1/2mv2
(c) K = L2/2I; L = (2KI)1/2; I = m(3RE)2
T = 2π × 3 × 6370/(4.57 × 3600) h = 7.3 h
K = 50 × (4.57 × 103)2 = 1.04 × 109 J
L = 3RE(2Km)1/2 = 3REmv = 8.73 × 1010 J.s
62 ·· Many satellites orbit the earth about 1000 km above the earth’s surface. Geosynchronous satellites orbit at a
distance of 4.22 × 107 m from the center of the earth. How much more energy is required to launch a 500-kg satellite
into a geosynchronous orbit than into an orbit 1000 km above the surface of the earth?
Let EO be the total energy of a satellite in orbit. Then, from Problem 55, EO = KO + UO = 1/2UO.
7
6
∆E = Egeo - E1000 = 1/2(Ugeo - U1000)
∆E = -1/2GMEm(1/4.22 × 10 - 1/7.73 × 10 )
= 1.05 × 1010 J
63 ·· It is theoretically possible to place a satellite at a position between the earth and the sun on the line joining them,
where the gravitational forces of the sun and the earth on the satellite combine in such a way that the satellite will
execute a circular orbit around the sun that is synchronous with the earth’s orbit around the sun.
(In other words, the satellite and the earth have the same orbital period about the sun, even though they are at
different distances from the sun. The satellite always remains on the line joining the earth and the sun.) Write an
expression that relates the appropriate circular orbital speed v of a satellite in such a situation to its distance r
from the sun. Your expression may also contain quantities shown in Figure 11-20 plus the gravitational
constant G.
Write Fnet = Fcentrip = mv2/r = GmM S/r2 - GmM E/(D - r)2; v =
G M S /r - G M E r/(D - r )2 .
64 · A 3-kg mass experiences a gravitational force of 12 N i at some point P. What is the gravitational field at that
point?
Use Equ. 11-21
g = (12/3) i N/kg = 4 i N/kg
65* · The gravitational field at some point is given by g = 2.5 × 10-6 N/kg j. What is the gravitational force on a mass of
4 g at that point?
-3
-6
-8
Use Equ. 11-21
F = mg = (4 × 10 × 2.5 × 10 ) j N = 10 j N
Chapter 11
Gravity
66 ·· A point mass m is on the x axis at x = L and a second equal point mass m is on the y axis at y = L. (a) Find the
gravitational field at the origin. (b) What is the magnitude of this field?
2
2
(a), (b) Use Equ. 11-24 and vector addition
g = (Gm/L )(i + j); g = 2 Gm/L
67 ·· Five equal masses M are equally spaced on the arc of a semicircle of radius R as in Figure 11-21. A mass m is
located at the center of curvature of the arc. (a) If M is 3 kg, m is 2 kg, and R is 10 cm, what is the force on m due to
the five masses? (b) If m is removed, what is the gravitational field at the center of curvature of the arc?
(a) By symmetry, g x = 0; write expression for Fy
Fy = (GMm/R2)(2 sin 45o + 1)
-8
Substitute numerical values
F = 9.66 × 10 N j
-8
(b) g = F/m
g = 4.83 × 10 N/kg j
68 ·· A point mass m1 = 2 kg is at the origin and a second point mass m2 = 4 kg is on the x axis at x = 6 m. Find the
gravitational field at (a) x = 2 m, and (b) x = 12 m. (c) Find the point on the x axis for which g = 0.
The configuration is shown below.
(a) g = g 1 + g 2; g 1 = -Gm1/4 i; g 2 = Gm2/16 i
(b) g = (-Gm1/144 - Gm2/36) i
(c) g = 0 when 2/x2 = 4/(6-x)2; solve for x
-11
g = G(1/4 - 1/2) i = -1.67 × 10 i N/kg
-12
g = -G(1/72 + 1/9) i = -G/8 i = -8.34 × 10 i N/kg
x2 + 72x - 36 = 0; x = 2.484 m, -8.48 m. From the diagram
it is clear that only at x = 2.484 m is g = 0.
69* ·· (a) Show that the gravitational field of a ring of uniform mass is zero at the center of the ring. (b) Figure 11-22
shows a point P in the plane of the ring but not at its center. Consider two elements of the ring of length s1 and s2 at
distances of r1 and r2, respectively. 1. What is the ratio of the masses of these elements? 2. Which produces the
greater gravitational field at point P? 3. What is the direction of the field at point P due to these elements? (c) What is
the direction of the gravitational field at point P due to the entire ring? (d) Suppose that the gravitational field varied as
1/r rather than 1/r2. What would be the net gravitational field at point P due to the two elements? (e) How would your
answers to parts (b) and (c) differ if point P were inside a spherical shell of uniform mass rather than inside a plane
circular ring?
Let λ = mass per unit length of the ring.
(a) g of opposite elements of mass Rλ d θ cancel
By symmetry g = 0 at center
(b) 1. m1 = r1λ d θ; m2 = r2λ d θ
m1/m2 = r1/r2
2
2
2. g = Gm/r = Grλ d θ/r = Gλ d θ/r
r1 < r2, therefore g 1 > g 2
3. By symmetry, g points along OP
g points toward m1, i.e., in direction of OP
(c) Take x along OP
By symmetry, g y = 0; g points in the direction OP
(d) For g ∝ 1/r, g 1 = g 2 ∝ λ d θ
g 1 = -g 2; g = 0
Chapter 11
(e) Now m1 and m2 ∝ r2, so g 1 = g 2
Gravity
g 1 = -g 2; g = 0; note: g = 0 everywhere inside the shell
70 ·· Show that the maximum value of g x for the field of Example 11-7 occurs at the points x = ±a/ 2 .
From Example 11-7, g x = -2GMx/(x2 + a 2)3/2. To find maximum, differentiate with respect to x and equate to 0.
dg x/dx = -2GM[(x2 + a 2)-3/2 - 3x2(x2 + a 2)-5/2] = 0. Solve for x: x = ±a/ 2 .
71 ·· A nonuniform stick of length L lies on the x axis with one end at the origin. Its mass density λ (mass per unit
length) varies as λ = Cx, where C is a constant. (Thus, an ele ment of the stick has mass dm = λ dx.) (a) What is the
total mass of the stick? (b) Find the gravitational field due to the stick at a point x0 > L.
L
L
1
2
(a) M = λ dx = C x dx = C L .
2
0
0
∫
∫
x dx
2G M   x 0   L 
=
ln 
2
 - 
 i.
( - x)
L2   x 0 - L   x0 - L 
0 x0
L
(b) d g = -G dm/(x0 - x) i; g = - G C ∫
2
72 ··· A uniform rod of mass M and length L lies along the x axis with its center at the origin. Consider an element of
length dx at a distance x from the origin. (a) Show that this element produces a gravitational field at a point x0 on the x
axis (x0 > 1/2L) given by
dg x = -
GM
dx
2
L( x 0 - x )
(b) Integrate this result over the length of the rod to find the total gravitational field at the point x0 due to the rod. (c)
What is the force on an object of mass m0 at x0? (d) Show that for x0 >> L, the field is approximately equal to that of a
point mass M.
(a), (b) See Example 11-8.
(c) F = m0g = -GMm0/(x02 - L2/4) i N.
(d) For x0 >> L, g x = -GM/x02, which is the result for a point mass M at the origin.
73* ·· Explain why the gravitational field increases with r rather than decreasing as 1/r2 as one moves out from the
center inside a solid sphere of uniform mass.
g is proportional to the mass within the sphere and inversely proportional to the radius, i.e., proportional to
r3/r2 = r.
74 · A spherical shell has a radius of 2 m and a mass of 300 kg. What is the gravitational field at the following
distances from the center of the shell: (a) 0.5 m; (b) 1.9 m; (c) 2.5 m?
(a) g = 0 [see Problem 69(e)]. (b) g = 0. (c) g = GM/r2 = 3.2 × 10-9 N/kg.
75 · A spherical shell has a radius of 2 m and a mass of 300 kg, and its center is located at the origin of a coordinate
system. Another spherical shell with a radius of 1 m and mass 150 kg is inside the larger shell with its center at 0.6 m
on the x axis. What is the gravitational force of attraction between the two shells?
Chapter 11
Gravity
The gravitational attraction is zero. The gravitational field inside the 2 m shell due to that shell is zero; therefore,
it exerts no force on the 1 m shell, and, by Newton’s third law, that shell exerts no force on the larger shell.
76 · Two spheres, S 1 and S 2, have equal radii R and equal masses M. The density of sphere S 1 is constant, whereas
that of sphere S 2 depends on the radial distance accordingρ (tor ) = C / r
. If the acceleration of gravity at the
surface of sphere S 1 is g 1, what is the acceleration of gravity at the surface of sphere S 2?
The accelerations of gravity are the same for both spheres; they depend only on M and R.
77* ·· Two homogeneous spheres, S 1 and S 2, have equal masses but different radii, R1 and R2. If the acceleration of
gravity on the surface of sphere S 1 is g 1, what is the acceleration of gravity on the surface of sphere S 2?
g ∝ M/r2
g 2 = g 1(R12/R22)
78 ·· Two concentric uniform spherical shells have masses M 1 and M 2 and radii a and 2a as in Figure 11-23. What is
the magnitude of the gravitational force on a point mass m located (a) a distance 3a from the center of the shells? (b)
a distance 1.9a from the center of the shells? (c) a distance 0.9a from the center of the shells?
(a) At r = 3a, both masses contribute to g
F = Gm(M 1 + M 2)/9a 2
(b) At r = 1.9a, g due to M 2 = 0
F = GmM 1/3.61a 2
(c) At r = 0.9a, g = 0
F=0
79 ·· The inner spherical shell in Problem 78 is shifted such that its center is now at x = 0.8a. The points 3a, 1.9a, and
0.9a lie along the same radial line from the center of the larger spherical shell. (a) What is the force on m at
x = 3a? (b) What is the force on m at x = 1.9a? (c) What is the force on m at x = 0.9a?
The configuration is shown on the right. The centers of
the spheres are indicated by the center-lines. The
locations of the mass m for parts (a), (b), and (c) are
shown by small dots along the x axis.
(a) At x = 3a, g 1 = GM1/(2.2a)2 and g 2 = GM2/(3a)2
(b) At x = 1.9a, g 1 = GM1/(1.1a)2 and g 2 = 0
(c) At x = 0.9a, g 1 = g 2 = 0
F = (Gm/a 2)(M 1/4.84 + M 2/9)
F = GmM 1/1.21a 2
F=0
80 ·· Suppose the earth were a sphere of uniform mass. If there were a deep elevator shaft going 15,000 m into the
earth, what would be the loss in weight at the bottom of this deep shaft for a student who weighs 800 N at the surface
of the earth?
1. Write an expression for M inside R = RE - 15 km
M = M E[(RE - 15)/RE]3, where RE is in km
2. Write an expression for g at R = RE - 15 km
g = GM/R2 = GME(RE -15)3/[(RE -15)2RE3]
Chapter 11
3. w(R) = mg(R) = w(RE)[(RE - 15)/RE]
Gravity
= g E(RE -15)/RE
∆w = (800 N)(15/RE) = (800 × 15/6370) N = 1.88 N
81* ·· A sphere of radius R has its center at the origin. It has a uniform mass density ρ 0, except that there is a spherical
cavity in it of radius r = 1/2R centered at x = 1/2R as in Figure 11-24. Find the gravitational field at points on the x axis
for x > R. (Hint: The cavity may be thought of as a sphere of mass m = (4/3) π r3ρ 0 plus a sphere of mass -m.)
Write g(x) using the hint. That is, find the sum of g of
3
the solid sphere plus the field of a sphere of radius 1/2R of
 4π ρ 0  R3
R / 8 
g(x)= G 
 2 2
negative mass centered at x = 1/2R.
 3  x ( x - ½ R ) 
82 ··· For the sphere with the cavity in Problem 81, show that the gravitational field inside the cavity is uniform, and find
its magnitude and direction.
1. Find the x and y components of g 1, where g 1 is the field due to a solid sphere of radius R and density ρ 0.
2. Find the x and y components of g 2, where g 2 is a sphere of radius 1/2R and negative density ρ 0 centered at 1/2R.
3. Add the x and y components to obtain the components of the field g = g 1 + g 2.
1. g 1 = 4πρ 0Gr3/3r2 = 4πρ 0Gr/3
g 1x = -g 1 cos θ = -g 1(x/r) = -4πρ 0Gx/3;
g 1y = -4πρ 0Gy/3 (the negative sign because the field
points inward)
3
2
2. g 2 = 4πρ 0Gr2 /3r2 = 4πρ 0Gr2/3, where
g 2x = g 2[(x - 1/2R)/r2] = 4πρ 0G(x - 1/2R)/3;
2
2 1/2
r2 = [(x - 1/2R) + y ]
g 2y = g 2(y/r2) = 4πρ 0Gy/3
3. g x = g 1x + g 2x; g y = g 1y + g 2y
g x = -2πρ 0GR/3; g y = 0. g = g x, a constant
83 ··· A straight, smooth tunnel is dug through a spherical planet whose mass density ρ 0 is constant. The tunnel passes
through the center of the planet and is perpendicular to the planet’s axis of rotation, which is fixed in space. The planet
rotates with an angular velocity ω such that objects in the tunnel have no acceleration relative to the tunnel. Find ω.
2
Fg = 4πρ 0Gmr/3 (see Problem 82); set Fr = mrω2
ω = 4πρ 0G/3; ω = (4πρ 0G/3)1/2
84 ··· The density of a sphere is given by p(r) = C/r. The sphere has a radius of 5 m and a mass of 1011 kg. (a)
Determine the constant C. (b) Obtain expressions for the gravitational field for (1) r > 5 m, and (2) r < 5 m.
5
(a)
∫
0
5
∫
4π pr 2 dr = 4πC r dr = M
0
g = 6.67 × 10-11 × 1011/r2 = 6.74 × 10-8/r2 N/kg
(b) 1. For r > 5 m, g = GM/r2
2. For r < 5 m, g = G
r
∫04π
2π C × 25 = 1011 kg; C = 6.436 kg/m2
Cr dr
r2
g = 2π GC = 2.7 × 10-9 N/kg
(Note that g is continuous at r = 5 m)
Chapter 11
Gravity
85* ··· A hole is drilled into the sphere of Problem 84 toward the center of the sphere to a depth of 2 km below the
sphere’s surface. A small mass is dropped from the surface into the hole. Determine the speed of the small mass as it
strikes the bottom of the hole.
3
1. Write the work per kg done by g between
E = ∫ g dr = 2πGC (5 - 3) = 0 5.4 × 10-9 J = 1/2v2
r = 5 m and r = 3 m; note that g points inward
5
2. Evaluate v
v = 0.104 mm/s
86 ··· The solid surface of the earth has a density of about 3000 kg/m3. A spherical deposit of heavy metals with a
density of 8000 kg/m3 and radius of 1000 m is centered 2000 m below the surface. Find ∆g/g at the surface directly
above this deposit, where ∆g is the increase in the gravitational field due to the deposit.
1. Determine g ′ at r = 2000 m due to a sphere of
M = 4π∆ρ R3/3 = 2.09 × 1013 kg; g ′ = GM/r2
radius 1000 m with density ∆ρ = 5000 kg/m3
g ′ = ∆g = 3.49 × 10-4 N/kg
-5
2. Evaluate ∆g/g were g = 9.81 N/kg
∆g/g = 3.56 × 10
87 ··· Two identical spherical hollows are made in a lead sphere of radius R. The hollows have a radius R/2. They touch
the outside surface of the sphere and its center as in Figure 11-25. The mass of the lead sphere before hollowing was
M. (a) Find the force of attraction of a small sphere of mass m to the lead sphere at the position shown in the figure.
(b) What is the attractive force if m is located right at the surface of the lead sphere?
1. First find the force FS due to the solid lead sphere. 2. Find the force FC due to each sphere of negative mass.
3. Add the forces.
(a) 1. Write the force due to the solid sphere
FS = Gmm/d 2; FS acts in the negative x direction
2. Write FC
FC = G(M/8)m/(d 2 + R2/4)
3. Find the x and y component of the two
2FCx = 2FC cos θ = 2FC d/(d 2 + R2/4)1/2 in the positive x
forces FC
direction; y components add to 0 by symmetry
2
3
2
2
3/2
4. Find F = Fx i = (-FS + 2FCx) i
F = -(GMm/d )[1 - (d /4)/(d + R /4) ] i
2
(b) Set d = R
F = -0.821(GMm/R ) i
88 · If K is the kinetic energy of the moon in its orbit around the earth, and U is the potential energy of the earth–moon
system, what is the relationship between K and U?
K = -1/2U. (see Example 11-6)
89* ·· A woman whose weight on earth is 500 N is lifted to a height two earth radii above the surface of the earth. Her
weight will (a) decrease to one-half of the original amount. (b) decrease to one-quarter of the original amount. (c)
decrease to one-third of the original amount. (d) decrease to one-ninth of the original amount.
(d) g depends on 1/r2.
90 · The mean distance of Pluto from the sun is 39.5 AU. Find the period of Pluto.
Proceed as in Problem 7
Tp = (39.5) 3/2 y = 248 y
Chapter 11
Gravity
91 · The semimajor axis of Ganymede, a moon of Jupiter discovered by Galileo, is 1.07 × 106 km, and its period is 7.155
days. Determine the mass of Jupiter.
Use the same procedure as for Problem 15
M J = 4π 2(1.07 × 109)3/G(6.18 × 105)2 = 1.9 × 1027 kg
92 · Calculate the mass of the earth using the known values of G, g, and RE.
Use Equ. 11-27; M E = gRE2/G
M E = [9.81(6.37 × 106)2/6.67 × 10-11] kg
= 5.97 × 1024 kg
93* · Uranus has a moon, Umbriel, whose mean orbital radius is 2.67 × 108 m and whose period is 3.58 × 105 s. (a) Find
the period of another of Uranus’s moons, Oberon, whose mean orbital radius is 5.86 × 108 m. (b) Use the known value
of G to find the mass of Uranus.
(a) Use Equ. 11-2; TO = TU(RO/RU)3/2
TO = (3.58 × 105 s)(5.86/2.67) 3/2 = 1.16 × 106 s
(b) M = 4π 2R3/GT 2
M U = 4π 2(2.67 × 108)3/G(3.58 × 105)2 kg
= 8.79 × 1025 kg
94 ·· Joe and Sally learn that there is a point between the earth and the moon where the gravitational effects of the two
bodies balance each other. Being of a New Age bent, they decide to try to conceive a child free from the bondage of
gravity, so they book an earth-to-moon trip. How far from the center of the earth should they try to conceive Zerog,
the first zero-gravity baby?
1. If M E/r2 = M m /(REM - r)2, g = 0; solve for r
r = βREM/(1 + β), where β = (M E/M m)1/2
2. Evaluate r for M E/M m = 81.36,
r = 3.46 × 108 m
REM = 3.844 × 108 m
95 ·· The force exerted by the earth on a particle of mass m a distance r from the center of the earth has the magnitude
GMEm/r2 = mgRE2/r2. (a) Calculate the work you must do against gravity to move the particle from a distance r1 to r2.
(b) Show that when r1 = RE and r2 = RE + h, the result can be written W = mgRE2[(1/RE) - 1/(RE + h)]. (c) Show that
when h << RE, the work is given approximately by W = mgh.
(a) W = −
r2
∫
r1
Fg dr = GM E m
r2
∫
r1
1 1
1 1 
dr
= −GM E m −  = GM E m −  .
r
 r2 r2 
 r1 r2 
(b) In the above expression, replace GMEm by mgRE2, r1 by RE, and r2 by RE + h to obtain the result given.
(c) [(1/RE) - 1/(RE + h)] = h/[RE(RE + h)]; if h << RE, the denominator ≈ RE2 and W = mgh.
96 ·· Suppose that the gravitational force of attraction depended not on 1/r2 but was proportional to the distance
between the two masses, like the force of a spring. In a planetary system like the solar system, what would then be the
relation between the period of a planet and its orbit radius, assuming all orbits were circular?
If F = Cr, where C is a constant, then rω2 ∝ r. Thus ω = constant, and T is constant, independent of r.
97* ·· A uniform sphere of radius 100 m and density 2000 kg/m3 is in free space far from other massive objects. (a) Find
the gravitational field outside of the sphere as a function of r. (b) Find the gravitational field inside the sphere as a
Chapter 11
function of r.
(a) g = -GM/r2; M = 4πρ R3/3
(b) Use Equ. 11-27
Gravity
M = 8.38 × 109 kg; g = -0.559/r2
g = -5.59 × 10-7r
98 ·· Two spherical planets have identical mass densities. Planet P1 has a radius R1, and planet P2 has a radius R2. If the
acceleration of gravity at the surface of planet P1 is g 1, what is the acceleration of gravity at the surface of planet P2?
g ∝ M/R2 ∝ R3/R2 = R
g 2 = g 1R2/R1
99 ·· Jupiter has a mass 320 times that of Earth and a volume 1320 times that of Earth. A “day” on Jupiter is 9 h 50 min
long. Find the height h above Jupiter at which a satellite must be revolving to have a period equal to 9 h 50 min.
1. Use Equ. 11-15 with M J replacing M S
R = (T 2GMJ/4π 2)1/3; M J = 1.91 × 1027 kg
2. Determine RJ = RE(1320) 1/3
RJ = 6.37 × 106(1320) 1/3 m = 6.99 × 107 m
3. Find R for T = 35400 s; h = R - RJ
R = 1.59 × 108 m; h = R - RJ = 8.94 × 107 m
100 ·· The average density of the moon is ρ = 3340 kg/m3. Find the minimum possible period T of a spacecraft orbiting
the moon.
The minimum period is when the orbit radius equals the object’s radius, i.e., orbit just above the surface of the moon.
1. Write the condition for a stable orbit, Ro = Rm = R
mRω2 = mMG/R2 = 4πρ R3mG/3R2 = 4πρ mGR/3
2. Solve for T 2 = 4π 2/ω2 and T
T 2 = 3π /ρ G; T = (3π /ρ G)1/2
3. Evaluate T = Tmin
Tmin = (3π /3340 × 6.67 × 10-11)1/2 s = 6500 s = 1h 48 min
101* ·· A satellite is circling around the moon (radius 1700 km) close to the surface at a speed v. A projectile is launched
from the moon vertically up at the same initial speed v. How high will it rise?
2
1. From Problem 55, v2 = ve2/2; 1/2mve2 = GmM m/Rm
1/2v = 1/2GMm/Rm = GMm[1/Rm - 1/(Rm + h)]
2. Solve for h
h = Rm = 1700 m
102 ·· Two space colonies of equal mass orbit a star (Figure 11-26). The Yangs in m1 move in a circular orbit of radius
1011 m with a period of 2 y. The Yins in m2 move in an elliptical orbit with a closest distance r1 = 1011 m and a farthest
distance r2 = 1.8 × 1011 m. (a) Using the fact that the mean radius of an elliptical orbit is the length of the semimajor
axis, find the length of the Yin year. (b) What is the mass of the star? (c) Which colony moves faster at point P in
Figure 11-26? (d) Which colony has the greater total energy? (e) How does the speed of the Yins at point P compare
with their speed at point A?
(a) 1. Find R2, the semi-major axis of Yins
R2 = 1/2 × 2.8 × 1011 m = 1.4 × 1011 m
2. Use Equ. 11-2; TYin = T2 = TYang(R2/r1)3/2
T2 = 2(1.4) 3/2 y = 3.31 y
(b) From Equ. 11-15, M = 4π 2r3/GT 2
M = 4π 2 × 1033/[6.67 × 10-11(6.31 × 107)2] kg
= 1.49 × 1029 kg
(c) Since R2 > r1, E2 > E1; U2P = U1; E = K + U
K2 > K1; v2 > v1
(d) See above
E2 > E1
(e) L is conserved; r2vA = r1vP
vP = 1.8vA
Chapter 11
Gravity
103 ·· In a binary star system, two stars orbit about their common center of mass. If the stars have masses m1 and m2
and are separated by a distance r, show that the period of rotation is related to r by T 2 = 4π r3/[G(m1 + m2)].
Take the coordinate origin at the center of mass. Then r1m1 = r2m2 and r = r1 + r2. The force holding m2 in orbit is
Gm1m2/(r1 + r2)2 = m2r2ω2. ω2 = Gm1/r2(r1 + r2)2. Now r2 = rm1/(m1 + m2), so ω2 = 4π 2/T 2 = G(m1 + m2)/r3 and
T 2 = 4π 2r3/G(m1 + m2).
104 ·· Two particles of mass m1 and m2 are released from rest with infinite separation. Find their speeds v1 and v2 when
their separation distance is r.
1. Use energy conservation
Gm1m2/r = 1/2m1v12 + 1/2m2v22
(1)
2. Use conservation of linear momentum
m1v1 = -m2v2; v1 = -v2(m2/m1)
(2)
2
2
3. Use (2) in (1)
v2 (m2 + m2 /m1) = 2Gm1m2/r
(3)
2
1/2
4. Solve for v2
v2 = [2Gm1 /r(m1 + m2)]
(4)
2
1/2
5. To find v1 permute subscripts in (4)
v1 = [2Gm2 /r(m1 + m2)]
105* ·· A hole is drilled from the surface of the earth to its center as in Figure 11-27. Ignore the earth’s rotation and air
resistance. (a) How much work is required to lift a particle of mass m from the center of the earth to the earth’s
surface? (b) If the particle is dropped from rest at the surface of the earth, what is its speed when it reaches the
center of the earth? (c) What is the escape speed for a particle projected from the center of the earth? Express your
answers in terms of m, g, and RE.
RE
∫F
RE
∫ rdr =
gmRE
2
(a) From Equ. 11-27, F = GmM Er/RE3 = gmr/RE
W=
(b) Use energy conservation; 1/2mv2 = W
v = gRE
(c) Eesc = W + 1/2mve2 = 1/2mvesc2
vesc2 = gRE + 2gRE; vesc =(3gRE)1/2 = 13.7 km/s
0
dr =
gm
RE
0
106 ·· A thick spherical shell of mass M and uniform density has an inner radius R1 and an outer radius R2. Find the
gravitational field g r as a function of r for all possible values of r. Sketch a graph of g r versus r.
1. For r < R1, g = 0 (see Equ. 11-24b). 2. For r > R2, g(r) is that of a mass M centered at the origin, i.e., g(r) =
GM/r2. 3. For R1 < r < R2, g(r) is determined by the mass within the shell of radius r. The mass density is
3
3
3
3
3
3
3
3
ρ = 3M/[4π (R2 - R1 )], and the mass within a radius r is given by 4πρ (r - R1 )/3 = M(r - R1 )/(R2 - R1 ).
So in this region g(r) =GM(r3 - R13)/r2(R23 - R13).
Chapter 11
Gravity
A graph of g(r) is shown alongside.
Here we have set R1 = 2, R2 = 3, and GM = 1.
107 ·· (a) Sketch a plot of the gravitational field g x versus x due to a uniform ring of mass M and radius R whose axis is
the x axis. (b) At what points is the magnitude of g x maximum?
(a) The geometry of the problem is shown at the left above. Consider an element of the ring of length dL. The
element of field at a point x is dg = Gλ dL/(x2 + R2). By symmetry, the y and z components of g vanish. The x
component of dg = dg cos θ = Gλ dL[x/(x2 + R2)3/2]. Since all parts of the ring contribute equally and 2πλR = M,
the field g(x) = Gmx/(x2 + R2)3/2. A plot of g(x) is shown to the right above. The curve is normalized for R = 1
and GM = 1.
(b) Differentiate g(x) and set dg/dx = 0; (x2 + R2)3/2 - 3x2(x2 + R2)1/2 = 0; x = ±R/ 2 . This agrees with the sketch
shown.
108 ··· In this problem, you are to find the gravitational potential energy of the stick in Example 11-8 and a point mass m0
that is on the x axis at x0. (a) Show that the potential energy of an element of the stick dm and m0 is given by
dU = -
G m0 dm
G M m0
=
dx
L ( x0 - x)
x0 - x
where U = 0 at x0 = ∞. (b) Integrate your result for part (a) over the length of the rod to find the total potential energy
for the system. Write your result as a general function U(x) by setting x0 equal to a general point x. (c) Compute the
force on m0 at a general point x from Fx = -dU/dx and compare your result with m0g, where g is the field at x0
calculated in Example 11-8.
(a) Let U = 0 at x = ∞. Let λ = M/L. Then the potential energy of the masses m0 at x0 and dm = λ dx at x is
Chapter 11
Gravity
given by dU = -Gm0 dm/r where r = x0 - x. Thus dU = -Gmm0 dx/[L(x0 - x)].
GMm0
(b) U = L
dx
 x + L/ 2
GMm0
0
 .
=
[ ln ( x 0 - L/2) - ln( x0 + L / 2)] = − GMm
ln  0
L
- x
L
 x0 − L / 2 
- L/2 x0
L/2
∫
(c) Since x0 is a general point along the x axis, F(x0) = -dU/dx0 = (Gmm0/L)[1/(x0 + L/2) - 1/(x0 - L/2)]; this simplifies
to F(x0) = -Gmm0/(x2 - L2/4) in agreement with the result of Example 11-8.
109* ··· A uniform sphere of mass M is located near a thin, uniform rod of mass m and length L as in Figure 11-28. Find
the gravitational force of attraction exerted by the sphere on the rod. (see Problem 72)
We shall determine the force exerted by the rod on the sphere and then use Newton’s third law. The sphere is
equivalent to a point mass m located at the sphere’s center.
1. Use the result of Problem 72; x0 = a + L/2
F = GmM/[(a + L/2)2 - L2/4] = GmM/[a(a + L)]
2. By Newton’s third law, this is the force on the rod
110 ··· A uniform rod of mass M = 20 kg and length L = 5 m is bent into a semicircle. What is the gravitational force
exerted by the rod on a point mass m = 0.1 kg located at the center of the circular arc?
The semicircular rod is shown in the figure. We shall use
an element of length R dθ = (L/π ) d θ of mass (M/π ) d θ.
By symmetry, Fy = 0. So we first find dFx and then
integrate over θ from -π /2 to π /2 to determine Fx.
1. Obtain an expression for dFx
2. Integrate from θ = -π /2 to π /2
3. Evaluate Fx
dFx = [GMm/π (L/π )2]d θ cos θ
Fx = 2π GMm/L2
Fx = 3.35 × 10-11 N
111 ··· Both the sun and the moon exert gravitational forces on the oceans of the earth, causing tides. (a) Show that the
ratio of the force exerted by the sun to that exerted by the moon is M Srm2/M mrS2, where M S and M m are the masses
of the sun and moon and rS and rm are the distances from the earth to the sun and to the moon. Evaluate this ratio. (b)
Even though the sun exerts a much greater force on the oceans than the moon does, the moon has a greater effect on
the tides because it is the difference in the force from one side of the earth to the other that is important. Differentiate
the expression F = Gm1m2/r2 to calculate the change in F due to a small change in r. Show that dF/F = (-2 dr)/r. (c)
During one full day, the rotation of the earth can cause the distance from the sun or moon to an ocean to change by at
most the diameter of the earth. Show that for a small change in distance, the change in the force exerted by the sun is
related to the change in the force exerted by the moon by ∆FS/∆Fm ≈ (M Srm3)/(M mrS3) and calculate this ratio.
(a) Force on a mass m is F = GMm/r2
FS/Fm = M Srm2/M mrS2 = 179 (see Problem 27)
(b) Find dF/dr and dF/F
dF/dr = -2Gm1m2/r3 = -2F/r; dF/F = -2 dr/r
Chapter 11
(c) ∆F = -2(F/r)∆r
Insert numerical values
Gravity
3
3
∆FS/∆Fm = (FS/Fm)(rm/rS) = (M Srm )/(M mrS )
∆ F S (1.99 × 10 30)(3.844 ×10 8 )
=
= 0.46
∆ Fm (7.35 × 10 22)(1.496 × 1011 )3
3