Chapter 11 Gravity
Chapter 11 Gravity
Chapter 11 Gravity
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<strong>Chapter</strong> <strong>11</strong><br />
<strong>Gravity</strong><br />
Conceptual Problems<br />
1 • True or false:<br />
(a) For Kepler’s law of equal areas to be valid, the force of gravity must vary<br />
inversely with the square of the distance between a given planet and the<br />
Sun.<br />
(b) The planet closest to the Sun has the shortest orbital period.<br />
(c) Venus’s orbital speed is larger than the orbital speed of Earth.<br />
(d) The orbital period of a planet allows accurate determination of that planet’s<br />
mass.<br />
(a) False. Kepler’s law of equal areas is a consequence of the fact that the<br />
gravitational force acts along the line joining two bodies but is independent<br />
of the manner in which the force varies with distance.<br />
(b) True. The periods of the planets vary with the three-halves power of their<br />
distances from the Sun. So the shorter the distance from the Sun, the shorter the<br />
period of the planet’s motion.<br />
(c) True. Setting up a proportion involving the orbital speeds of the two planets in<br />
terms of their orbital periods and mean distances from the Sun (see Table <strong>11</strong>-1)<br />
shows that v = 1. 17v<br />
.<br />
Venus<br />
Earth<br />
(d) False. The orbital period of a planet is independent of the planet’s mass.<br />
3 • During what season in the northern hemisphere does Earth attain its<br />
maximum orbital speed about the Sun? What season is related to its minimum<br />
orbital speed? Hint: Earth is at perihelion in early January.<br />
Determine the Concept Earth is closest to the Sun during winter in the northern<br />
hemisphere. This is the time of fastest orbital speed. Summer would be the time<br />
for minimum orbital speed.<br />
7 • At the surface of the moon, the acceleration due to the gravity of the<br />
moon is a. At a distance from the center of the moon equal to four times the<br />
radius of the moon, the acceleration due to the gravity of the moon is (a) 16a, (b)<br />
a/4, (c) a/3, (d) a/16, (e) None of the above.<br />
Picture the Problem The acceleration due to gravity varies inversely with the<br />
square of the distance from the center of the moon.<br />
221
222<br />
<strong>Chapter</strong> <strong>11</strong><br />
Express the dependence of the<br />
acceleration due to the gravity of the<br />
moon on the distance from its center:<br />
Express the dependence of the<br />
acceleration due to the gravity of the<br />
moon at its surface on its radius:<br />
Divide the first of these expressions<br />
by the second to obtain:<br />
Solving for a′ and simplifying yields:<br />
1<br />
a' ∝ 2<br />
r<br />
1<br />
a ∝<br />
R<br />
a'<br />
=<br />
a<br />
2<br />
M<br />
R<br />
r<br />
R<br />
=<br />
r<br />
2<br />
M<br />
2<br />
R<br />
a' a a 16<br />
( R )<br />
1<br />
2<br />
2<br />
M<br />
M = =<br />
2<br />
2<br />
4 M<br />
and (d) is correct.<br />
<strong>11</strong> •• Suppose the escape speed from a planet was only slightly larger than<br />
the escape speed from Earth, yet the planet is considerably larger than Earth. How<br />
would the planet’s (average) density compare to Earth’s (average) density? (a) It<br />
must be denser. (b) It must be less dense. (c) It must be the same density. (d) You<br />
cannot determine the answer based on the data given.<br />
Picture the Problem The densities of the planets are related to the escape speeds<br />
from their surfaces through v = 2GM<br />
R .<br />
The escape speed from the planet is<br />
given by:<br />
The escape speed from Earth is given<br />
by:<br />
Expressing the ratio of the escape<br />
speed from the planet to the escape<br />
speed from Earth and simplifying<br />
yields:<br />
Because vplanet ≈ vEarth:<br />
e<br />
v =<br />
planet<br />
v =<br />
v<br />
v<br />
Earth<br />
planet<br />
Earth<br />
1 ≈<br />
=<br />
R<br />
R<br />
Earth<br />
planet<br />
2GM<br />
R<br />
2GM<br />
R<br />
planet<br />
Earth<br />
2GM<br />
R<br />
planet<br />
Earth<br />
planet<br />
2GM<br />
R<br />
M<br />
M<br />
Earth<br />
planet<br />
Earth<br />
Earth<br />
planet<br />
=<br />
a<br />
R<br />
R<br />
Earth<br />
planet<br />
M<br />
M<br />
planet<br />
Earth
Squaring both sides of the equation<br />
yields:<br />
R<br />
1 ≈<br />
R<br />
Earth<br />
planet<br />
M<br />
M<br />
planet<br />
Earth<br />
<strong>Gravity</strong><br />
Express Mplanet and MEarth in terms of their densities and simplify to obtain:<br />
R<br />
1 ≈<br />
R<br />
Earth<br />
planet<br />
ρ<br />
ρ<br />
planet<br />
Earth<br />
V<br />
V<br />
planet<br />
Earth<br />
R<br />
=<br />
R<br />
Earth<br />
planet<br />
planet<br />
Earth<br />
Solving for the ratio of the densities<br />
yields:<br />
Because the planet is considerably<br />
larger than Earth:<br />
ρ<br />
ρ<br />
V<br />
V<br />
planet<br />
Earth<br />
R<br />
=<br />
R<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
planet<br />
Earth<br />
Earth<br />
planet<br />
ρ<br />
ρ<br />
R<br />
≈<br />
R<br />
planet
224<br />
<strong>Chapter</strong> <strong>11</strong><br />
Using Kepler’s third law, relate the<br />
period of the Sun T to its mean<br />
distance R0 from the center of the<br />
galaxy:<br />
Substitute numerical values and evaluate MG/Ms:<br />
M<br />
M<br />
G<br />
s<br />
galaxy<br />
T<br />
4π<br />
M 4π<br />
R<br />
2<br />
2 3<br />
2<br />
=<br />
GM G<br />
3 G R0<br />
⇒<br />
M S<br />
0 =<br />
2<br />
GM ST<br />
15<br />
2⎛<br />
4 9.<br />
461×<br />
10 m ⎞<br />
4π<br />
⎜<br />
⎜3.00×<br />
10 c ⋅ y×<br />
c y ⎟<br />
⎝<br />
⋅<br />
=<br />
⎠<br />
2<br />
⎛ −<strong>11</strong><br />
N ⋅ m ⎞<br />
30 ⎛<br />
6<br />
⎜<br />
⎜6.<br />
6742×<br />
10 ( 1.<br />
99×<br />
10 kg)<br />
⎜250×<br />
10 y×<br />
3.<br />
156×<br />
10<br />
2<br />
kg ⎟<br />
⎝<br />
⎠<br />
⎝<br />
<strong>11</strong><br />
≈ 1.<br />
1×<br />
10 ⇒ M ≈<br />
<strong>11</strong><br />
1.<br />
1×<br />
10 M<br />
Kepler’s Laws<br />
25 •• [SSM] One of the so-called ″Kirkwood gaps″ in the asteroid belt<br />
occurs at an orbital radius at which the period of the orbit is half that of Jupiter’s.<br />
The reason there is a gap for orbits of this radius is because of the periodic pulling<br />
(by Jupiter) that an asteroid experiences at the same place in its orbit every other<br />
orbit around the sun. Repeated tugs from Jupiter of this kind would eventually<br />
change the orbit of such an asteroid. Therefore, all asteroids that would otherwise<br />
have orbited at this radius have presumably been cleared away from the area due<br />
to this resonance phenomenon. How far from the Sun is this particular 2:1<br />
resonance ″Kirkwood″ gap?<br />
Picture the Problem The period of an orbit is related to its semi-major axis (for<br />
circular orbits this distance is the orbital radius). Because we know the orbital<br />
periods of Jupiter and a hypothetical asteroid in the Kirkwood gap, we can use<br />
Kepler’s third law to set up a proportion relating the orbital periods and average<br />
distances of Jupiter and the asteroid from the Sun from which we can obtain an<br />
expression for the orbital radius of an asteroid in the Kirkwood gap.<br />
Use Kepler’s third law to relate<br />
Jupiter’s orbital period to its mean<br />
distance from the Sun:<br />
T<br />
s<br />
2<br />
Jupiter<br />
2<br />
4π<br />
=<br />
GM<br />
Sun<br />
r<br />
3<br />
3<br />
Jupiter<br />
7<br />
s<br />
y<br />
⎞<br />
⎟<br />
⎠<br />
2
Use Kepler’s third law to relate the<br />
orbital period of an asteroid in the<br />
Kirkwood gap to its mean distance<br />
from the Sun:<br />
Dividing the second of these<br />
equations by the first and simplifying<br />
yields:<br />
Solving for rKirkwood yields:<br />
Because the period of the orbit of an<br />
asteroid in the Kirkwood gap is half<br />
that of Jupiter’s:<br />
T<br />
2<br />
Kirkwood<br />
T<br />
2<br />
Kirkwood<br />
2<br />
TJupiter<br />
2<br />
4π<br />
=<br />
GM<br />
Sun<br />
2<br />
4π<br />
GM<br />
=<br />
4π<br />
GM<br />
Sun<br />
2<br />
r<br />
Sun<br />
3<br />
Kirkwood<br />
r<br />
3<br />
Kirkwood<br />
r<br />
3<br />
Jupiter<br />
<strong>Gravity</strong><br />
r<br />
=<br />
2 3<br />
Kirkwood<br />
Kirkwood Jupiter ⎟<br />
Jupiter<br />
⎟<br />
⎛ ⎞<br />
⎜<br />
T<br />
= r<br />
⎜ T<br />
r<br />
Kirkwood<br />
=<br />
=<br />
⎝<br />
10 ( 77.<br />
8×<br />
10 m)<br />
4.<br />
90×<br />
10<br />
<strong>11</strong><br />
⎠<br />
⎛ 1<br />
2 T<br />
⎜<br />
⎝ T<br />
m<br />
3<br />
Kirkwood<br />
3<br />
rJupiter<br />
Jupiter<br />
Jupiter<br />
⎞<br />
⎟<br />
⎠<br />
<strong>11</strong> 1AU<br />
= 4.<br />
90×<br />
10 m×<br />
1.50×<br />
10<br />
= 3.<br />
27 AU<br />
Remarks: There are also significant Kirkwood gaps at 3:1, 5:2, and 7:3 and<br />
resonances at 2.5 AU, 2.82 AU, and 2.95 AU.<br />
29 •• Kepler determined distances in the Solar System from his data. For<br />
example, he found the relative distance from the Sun to Venus (as compared to<br />
the distance from the Sun to Earth) as follows. Because Venus’s orbit is closer to<br />
the Sun than is Earth’s orbit, Venus is a morning or evening star—its position in<br />
the sky is never very far from the Sun (Figure <strong>11</strong>-24). If we suppose the orbit of<br />
Venus is a perfect circle, then consider the relative orientation of Venus, Earth,<br />
and the Sun at maximum extension, that is when Venus is farthest from the Sun in<br />
the sky. (a) Under this condition, show that angle b in Figure <strong>11</strong>-24 is 90º. (b) If<br />
the maximum elongation angle a between Venus and the Sun is 47º, what is the<br />
distance between Venus and the Sun in AU? (c) Use this result to estimate the<br />
length of a Venusian ″year.″<br />
Picture the Problem We can use a property of lines tangent to a circle and radii<br />
drawn to the point of contact to show that b = 90°. Once we’ve established that b<br />
is a right angle we can use the definition of the sine function to relate the distance<br />
from the Sun to Venus to the distance from the Sun to Earth.<br />
r<br />
2 3<br />
<strong>11</strong><br />
m<br />
225
226<br />
<strong>Chapter</strong> <strong>11</strong><br />
(a) The line from Earth to Venus' orbit is tangent to the orbit of Venus at the point<br />
of maximum extension. Venus will appear closer to the sun in Earth’s sky when it<br />
passes the line drawn from Earth and tangent to its orbit. Hence b = 90°<br />
(b) Using trigonometry, relate the<br />
distance from the sun to Venus d<br />
to the angle a:<br />
Substitute numerical values and<br />
evaluate dSV<br />
:<br />
(c) Use Kepler’s third law to relate<br />
Venus’s orbital period to its mean<br />
distance from the Sun:<br />
Use Kepler’s third law to relate<br />
Earth’s orbital period to its mean<br />
distance from the Sun:<br />
Dividing the first of these equations<br />
by the second and simplifying yields:<br />
Solving for TVenus<br />
yields:<br />
SV<br />
Using the result from Part (b) yields:<br />
dSV<br />
a = ⇒ dSV<br />
= d sin a<br />
d<br />
sin SE<br />
d<br />
T<br />
T<br />
SV<br />
=<br />
=<br />
2<br />
Venus<br />
2<br />
Earth<br />
T<br />
T<br />
T<br />
2<br />
Venus<br />
2<br />
Earth<br />
SE<br />
( 1.<br />
00AU)<br />
0.<br />
73AU<br />
2<br />
4π<br />
=<br />
GM<br />
2<br />
4π<br />
=<br />
GM<br />
Sun<br />
Sun<br />
2<br />
4π<br />
GM<br />
=<br />
4π<br />
GM<br />
sin 47°<br />
=<br />
r<br />
r<br />
Sun<br />
2<br />
Sun<br />
3<br />
Venus<br />
3<br />
Earth<br />
r<br />
3<br />
Venus<br />
r<br />
3<br />
Earth<br />
3 2<br />
Venus<br />
Venus Earth ⎟<br />
Earth<br />
⎟<br />
⎛ r ⎞<br />
= T ⎜<br />
r<br />
Venus<br />
=<br />
⎝<br />
( 1.<br />
00 y)<br />
0.<br />
63 y<br />
⎠<br />
r<br />
=<br />
r<br />
0.<br />
731AU<br />
3<br />
Venus<br />
3<br />
Earth<br />
⎛ 0.<br />
731AU<br />
⎞<br />
⎜ ⎟<br />
⎝ 1.<br />
00 AU ⎠<br />
Remarks: The correct distance from the sun to Venus is closer to 0.723 AU.<br />
Newton’s Law of <strong>Gravity</strong><br />
31 • Jupiter’s satellite Europa orbits Jupiter with a period of 3.55 d at an<br />
average orbital radius of 6.71 × 10 8 m. (a) Assuming that the orbit is circular,<br />
determine the mass of Jupiter from the data given. (b) Another satellite of Jupiter,<br />
Callisto, orbits at an average radius of 18.8 × 10 8 m with an orbital period of<br />
T<br />
=<br />
3 2
<strong>Gravity</strong><br />
16.7 d. Show that these data are consistent with an inverse square force law for<br />
gravity (Note: DO NOT use the value of G anywhere in Part (b)).<br />
Picture the Problem While we could apply Newton’s law of gravitation and<br />
second law of motion to solve this problem from first principles, we’ll use<br />
Kepler’s third law (derived from these laws) to find the mass of Jupiter in Part (a).<br />
In Part (b) we can compare the ratio of the centripetal accelerations of Europa and<br />
Callisto to show that these data are consistent with an inverse square law for<br />
gravity.<br />
(a) Assuming a circular orbit, apply<br />
Kepler’s third law to the motion of<br />
Europa to obtain:<br />
Substitute numerical values and evaluate MJ:<br />
2<br />
8<br />
π ( 6.71×<br />
10 m)<br />
−<strong>11</strong><br />
2 2 ⎛<br />
( 6.<br />
673×<br />
10 N ⋅ m /kg ) ⎜3.55d<br />
T<br />
2<br />
E<br />
2<br />
2<br />
4π<br />
3 4π<br />
= RE<br />
⇒ M J = R<br />
GM<br />
GT<br />
3<br />
4 27<br />
M J =<br />
= 1.<br />
90 × 10<br />
2<br />
24 h 3600s<br />
⎞<br />
× × ⎟<br />
⎝ d h ⎠<br />
Note that this result is in excellent agreement with the accepted value of<br />
1.902×10 27 kg.<br />
(b) Express the centripetal<br />
acceleration of both of the moons to<br />
obtain:<br />
Using this result, express the<br />
centripetal accelerations of Europa<br />
and Callisto:<br />
Divide the first of these equations by<br />
the second and simplify to obtain:<br />
Substitute for the periods of Callisto<br />
and Europa using Kepler’s third<br />
law to obtain:<br />
J<br />
2<br />
2<br />
E<br />
3<br />
E<br />
kg<br />
227<br />
⎛ 2πR<br />
⎞<br />
2 ⎜ ⎟ 2<br />
v T 4π<br />
R<br />
acentripetal<br />
= =<br />
⎝ ⎠<br />
= 2<br />
R R T<br />
where R and T are the radii and periods<br />
of their motion.<br />
a<br />
4π<br />
R<br />
4π<br />
R<br />
2<br />
2<br />
E = 2<br />
TE<br />
E and aC<br />
= C<br />
2<br />
TC<br />
a<br />
a<br />
a<br />
a<br />
E<br />
C<br />
E<br />
C<br />
4π<br />
R<br />
T<br />
=<br />
4π<br />
T<br />
CR<br />
=<br />
CR<br />
2<br />
E<br />
2<br />
E<br />
2<br />
RC<br />
2<br />
C<br />
3<br />
C<br />
3<br />
E<br />
R<br />
R<br />
E<br />
C<br />
T<br />
=<br />
T<br />
2<br />
C<br />
2<br />
E<br />
R<br />
=<br />
R<br />
2<br />
C<br />
2<br />
E<br />
R<br />
R<br />
E<br />
C
228<br />
<strong>Chapter</strong> <strong>11</strong><br />
This result, together with the fact that the gravitational force is directly<br />
proportional to the acceleration of the moons, demonstrates that the gravitational<br />
force varies inversely with the square of the distance.<br />
33 • The mass of Saturn is 5.69 × 10 26 kg. (a) Find the period of its moon<br />
Mimas, whose mean orbital radius is 1.86 × 10 8 m. (b) Find the mean orbital<br />
radius of its moon Titan, whose period is 1.38 × 10 6 s.<br />
Picture the Problem While we could apply Newton’s law of gravitation and<br />
second law of motion to solve this problem from first principles, we’ll use<br />
Kepler’s third law (derived from these laws) to find the period of Mimas and to<br />
relate the periods of the moons of Saturn to their mean distances from its center.<br />
(a) Using Kepler’s third law, relate<br />
the period of Mimas to its mean<br />
distance from the center of Saturn:<br />
Substitute numerical values and evaluate TM:<br />
T<br />
T<br />
2<br />
M<br />
2<br />
8 3<br />
π ( 1.<br />
86×<br />
10 m)<br />
<strong>11</strong> 2 2<br />
26<br />
( 6.<br />
6726×<br />
10 N ⋅ m /kg )( 5.<br />
69×<br />
10 kg)<br />
2<br />
4π<br />
3<br />
= rM<br />
⇒ T<br />
GM<br />
4 4<br />
M = = 8.<br />
18×<br />
10<br />
−<br />
(b) Using Kepler’s third law, relate<br />
the period of Titan to its mean<br />
distance from the center of Saturn:<br />
Substitute numerical values and evaluate rT:<br />
r<br />
T<br />
=<br />
3<br />
T<br />
4π<br />
S<br />
M<br />
2<br />
2<br />
3<br />
T = rT<br />
⇒<br />
GM S<br />
3 r T =<br />
=<br />
s ≈<br />
2<br />
4π<br />
GM<br />
2<br />
TT<br />
GM<br />
2<br />
4π<br />
S<br />
r<br />
3<br />
M<br />
22.7 h<br />
6 2<br />
−<strong>11</strong><br />
2 2<br />
26<br />
( 1. 38×<br />
10 s)<br />
( 6.<br />
6726×<br />
10 N ⋅ m /kg )( 5.<br />
69 × 10 kg)<br />
9<br />
= 1.<br />
22 × 10 m<br />
2<br />
4π<br />
41 •• A superconducting gravity meter can measure changes in gravity of<br />
the order Δg/g = 1.00 × 10 –<strong>11</strong> . (a) You are hiding behind a tree holding the meter,<br />
and your 80-kg friend approaches the tree from the other side. How close to you<br />
can your friend get before the meter detects a change in g due to his presence? (b)<br />
You are in a hot air balloon and are using the meter to determine the rate of ascent<br />
(assume the balloon has constant acceleration). What is the smallest change in<br />
altitude that results in a detectable change in the gravitational field of Earth?<br />
S
<strong>Gravity</strong><br />
Picture the Problem We can determine the maximum range at which an object<br />
with a given mass can be detected by substituting the equation for the<br />
gravitational field in the expression for the resolution of the meter and solving for<br />
the distance. Differentiating g(r) with respect to r, separating variables to obtain<br />
dg/g, and approximating Δr with dr will allow us to determine the vertical change<br />
in the position of the gravity meter in Earth’s gravitational field is detectable.<br />
(a) Earth’s gravitational field is<br />
given by:<br />
Express the gravitational field due to<br />
the mass m (assumed to be a point<br />
mass) of your friend and relate it to<br />
the resolution of the meter:<br />
Solving for r yields:<br />
GM<br />
g E =<br />
R<br />
g<br />
() r<br />
r = R<br />
E<br />
2<br />
E<br />
Gm<br />
= = 1.<br />
00×<br />
10<br />
2<br />
r<br />
−<strong>11</strong><br />
GM<br />
= 1.<br />
00×<br />
10<br />
R<br />
E<br />
1.<br />
00×<br />
10<br />
M<br />
Substitute numerical values and<br />
<strong>11</strong><br />
6 1.<br />
00×<br />
10<br />
evaluate r: ( ) ( 80kg)<br />
r = 6.37× 10 m<br />
24<br />
(b) Differentiate g(r) and simplify to<br />
obtain:<br />
=<br />
7.<br />
37m<br />
E<br />
<strong>11</strong><br />
m<br />
−<strong>11</strong><br />
E<br />
2<br />
E<br />
g<br />
E<br />
5.98×<br />
10<br />
kg<br />
dg − 2Gm<br />
2 ⎛ Gm ⎞ 2<br />
= = − = − g<br />
3 ⎜ 2 ⎟<br />
dr r r ⎝ r ⎠ r<br />
Separate variables to obtain: dg dr −<strong>11</strong><br />
= −2<br />
= 10<br />
g r<br />
Approximating dr with Δr, evaluate<br />
Δr with r = RE:<br />
Δr<br />
= −<br />
1<br />
2<br />
−<strong>11</strong><br />
6<br />
( 1.<br />
00×<br />
10 )( 6.<br />
37×<br />
10 m)<br />
= 31.<br />
9μm<br />
43 •• Earth’s radius is 6370 km and the moon’s radius is<br />
1738 km. The acceleration of gravity at the surface of the moon is 1.62 m/s 2 .<br />
What is the ratio of the average density of the moon to that of Earth?<br />
Picture the Problem We can use the definitions of the gravitational fields at the<br />
surfaces of Earth and the moon to express the accelerations due to gravity at these<br />
locations in terms of the average densities of Earth and the moon. Expressing the<br />
ratio of these accelerations will lead us to the ratio of the densities.<br />
229
230<br />
<strong>Chapter</strong> <strong>11</strong><br />
Express the acceleration due to<br />
gravity at the surface of Earth in<br />
terms of Earth’s average density:<br />
The acceleration due to gravity at the<br />
surface of the moon in terms of the<br />
moon’s average density is:<br />
g<br />
E<br />
M<br />
GM<br />
=<br />
R<br />
=<br />
4<br />
3<br />
E<br />
2<br />
E<br />
Gρ<br />
π R<br />
4<br />
3<br />
E<br />
Gρ<br />
V<br />
=<br />
R<br />
M<br />
E<br />
g = Gρ<br />
π R<br />
E E<br />
2<br />
E<br />
M<br />
Gρ<br />
=<br />
Divide the second of these equations g M ρ M RM<br />
ρM<br />
gM<br />
RE<br />
= ⇒ =<br />
by the first to obtain: g E ρ E RE<br />
ρE<br />
gE<br />
RM<br />
Substitute numerical values and<br />
ρ M<br />
evaluate :<br />
ρ<br />
E<br />
Gravitational Potential Energy<br />
ρM<br />
ρ<br />
E<br />
=<br />
=<br />
4<br />
E 3<br />
2<br />
RE<br />
π R<br />
2<br />
6<br />
( 1.62m/s<br />
)( 6.37×<br />
10 m)<br />
2<br />
6<br />
( 9.81m/s<br />
)( 1.738×<br />
10 m)<br />
0.<br />
605<br />
47 • Knowing that the acceleration of gravity on the moon is 0.166 times<br />
that on Earth and that the moon’s radius is 0.273RE, find the escape speed for a<br />
projectile leaving the surface of the moon.<br />
Picture the Problem The escape speed from the moon is given by<br />
v e, m = 2GM m Rm<br />
, where Mm and Rm represent the mass and radius of the moon,<br />
respectively.<br />
Express the escape speed from the<br />
moon:<br />
Because g m = 0. 166gE<br />
and<br />
R = 0. 273R<br />
:<br />
m<br />
E<br />
2GM<br />
m<br />
v e.m = =<br />
Rm<br />
v =<br />
Substitute numerical values and evaluate ve,m:<br />
v<br />
e.m<br />
=<br />
2<br />
e.m<br />
2g<br />
m<br />
R<br />
m<br />
( 0.<br />
166g<br />
)( 0.<br />
273 )<br />
2 R<br />
2<br />
6<br />
( 0.<br />
166)(<br />
9.<br />
81m/s<br />
)( 0.<br />
273)(<br />
6.<br />
371×<br />
10 m)<br />
= 2.<br />
38 km/s<br />
51 •• An object is dropped from rest from a height of 4.0 × 10 6 m above the<br />
surface of Earth. If there is no air resistance, what is its speed when it strikes<br />
Earth?<br />
E<br />
E<br />
3<br />
E
<strong>Gravity</strong><br />
Picture the Problem Let the zero of gravitational potential energy be at infinity<br />
and let m represent the mass of the object. We’ll use conservation of energy to<br />
relate the initial potential energy of the object-Earth system to the final potential<br />
and kinetic energies.<br />
Use conservation of energy to relate<br />
the initial potential energy of the<br />
system to its energy as the object is<br />
about to strike Earth:<br />
Express the potential energy of the<br />
object-Earth system when the object<br />
is at a distance r from the surface of<br />
Earth:<br />
Substitute in equation (1) to obtain:<br />
Solving for v yields:<br />
Substitute numerical values and evaluate v:<br />
v =<br />
Gravitational Orbits<br />
2<br />
231<br />
K f − Ki<br />
+ U f −U<br />
i = 0<br />
or, because Ki = 0,<br />
K ( RE<br />
) + U ( RE<br />
) −U<br />
( RE<br />
+ h)<br />
= 0 (1)<br />
where h is the initial height above<br />
Earth’s surface.<br />
U<br />
() r<br />
mv<br />
GM<br />
= −<br />
r<br />
GM<br />
E<br />
m<br />
m<br />
GM<br />
m<br />
1 2 E<br />
E<br />
2 − + =<br />
RE<br />
RE<br />
+ h<br />
v =<br />
=<br />
⎛ GM<br />
2 ⎜<br />
⎝ RE<br />
E<br />
GM ⎞ E − ⎟<br />
RE<br />
+ h ⎠<br />
⎛ h ⎞<br />
2gR<br />
⎜<br />
⎟<br />
E<br />
⎝ RE<br />
+ h ⎠<br />
2<br />
6<br />
6<br />
( 9.81m/s<br />
)( 6.37×<br />
10 m)(<br />
4.<br />
0×<br />
10 m)<br />
= 6.<br />
9km/s<br />
6.37 × 10<br />
6<br />
m + 4.<br />
0×<br />
10<br />
59 •• Many satellites orbit Earth with maximum altitudes of<br />
1000 km or less. Geosynchronous satellites, however, orbit at an altitude of<br />
35 790 km above Earth’s surface. How much more energy is required to launch a<br />
500-kg satellite into a geosynchronous orbit than into an orbit 1000 km above the<br />
surface of Earth?<br />
Picture the Problem We can express the energy difference between these two<br />
orbits in terms of the total energy of a satellite at each elevation. The application<br />
of Newton’s second law to the force acting on a satellite will allow us to express<br />
6<br />
m<br />
0
232<br />
<strong>Chapter</strong> <strong>11</strong><br />
the total energy of each satellite as a function of its mass, the radius of Earth, and<br />
its orbital radius.<br />
Express the energy difference: Δ E = Egeo<br />
− E1000<br />
(1)<br />
Express the total energy of an<br />
orbiting satellite:<br />
Apply Newton’s second law to a<br />
satellite to relate the gravitational<br />
force to the orbital speed:<br />
Solving for v 2 yields:<br />
Substitute in equation (2) to obtain:<br />
Etot<br />
= K + U<br />
2 GM Em<br />
1 = 2 mv −<br />
R<br />
where R is the orbital radius.<br />
F<br />
v<br />
E<br />
radial<br />
2 =<br />
tot<br />
=<br />
GM<br />
=<br />
R<br />
gR<br />
R<br />
Substituting in equation (1) and simplifying yields:<br />
1<br />
2<br />
2<br />
E<br />
E<br />
2<br />
gR<br />
m<br />
R<br />
2<br />
m v<br />
= m<br />
R<br />
2<br />
2<br />
2<br />
mgR<br />
⎛ ⎞<br />
E mgRE<br />
mgRE<br />
⎜<br />
1 1<br />
ΔE<br />
= − + =<br />
− ⎟<br />
2R<br />
⎜ ⎟<br />
geo 2R1000<br />
2 ⎝ R1000<br />
Rgeo<br />
⎠<br />
2<br />
mgR ⎛ 1<br />
1 ⎞<br />
E = ⎜<br />
−<br />
⎟<br />
2 ⎝ RE<br />
+ 1000 km RE<br />
+ 35 790 km ⎠<br />
Substitute numerical values and evaluate ΔE:<br />
ΔE =<br />
2<br />
E<br />
(2)<br />
2<br />
gREm<br />
mgR<br />
− = −<br />
R 2R<br />
6 2<br />
( 500 kg)(<br />
9.81N<br />
/ kg)(<br />
6.37 × 10 m)<br />
⎛ 1<br />
1 ⎞<br />
⎜<br />
−<br />
⎟ = <strong>11</strong>.<br />
1GJ<br />
The Gravitational Field ( g r )<br />
2<br />
⎜<br />
⎝ 7.<br />
37 × 10<br />
6<br />
m<br />
4.<br />
22×<br />
107 m ⎟<br />
⎠<br />
63 •• A point particle of mass m is on the x axis at x = L and an identical<br />
point particle is on the y axis at y = L. (a) What is the direction of the<br />
gravitational field at the origin? (b) What is the magnitude of this field?<br />
2<br />
E
<strong>Gravity</strong><br />
Picture the Problem We can use the definition of the gravitational field due to a<br />
point mass to find the x and y components of the field at the origin and then add<br />
these components to find the resultant field. We can find the magnitude of the<br />
field from its components using the Pythagorean theorem.<br />
(a) The gravitational field at the<br />
origin is the sum of its x and y<br />
components:<br />
Express the gravitational field due to<br />
the point mass at x = L:<br />
Express the gravitational field due to<br />
the point mass at y = L:<br />
Substitute in equation (1) to obtain:<br />
r r r<br />
g = g + g<br />
(1)<br />
r<br />
g<br />
r<br />
g<br />
x =<br />
y =<br />
x<br />
y<br />
Gm<br />
iˆ<br />
2<br />
L<br />
Gm<br />
2<br />
L<br />
(b) The magnitude of g r is given by: 2 2<br />
= g x + g y<br />
Substitute for gx and gy and simplify<br />
ˆj<br />
r r r Gm<br />
g g g iˆ<br />
Gm<br />
=<br />
ˆ<br />
x + y = + j ⇒ the<br />
2 2<br />
L L<br />
direction of the gravitational field is<br />
along a line at 45° above the +x axis.<br />
⎛ Gm ⎞ ⎛ Gm ⎞ Gm<br />
to obtain: g = ⎜<br />
= 2 2 ⎟ + ⎜ 2 ⎟<br />
2<br />
⎝ L ⎠ ⎝ L ⎠ L<br />
r<br />
67 ••• A nonuniform thin rod of length L lies on the x axis. One end of the<br />
rod is at the origin, and the other end is at x = L. The rod’s mass per unit length λ<br />
varies as λ = Cx, where C is a constant. (Thus, an element of the rod has mass<br />
dm = λ dx.) (a) What is the total mass of the rod? (b) Find the gravitational field<br />
due to the rod on the x axis at x = x0, where x0 > L.<br />
Picture the Problem We can find the mass of the rod by integrating dm over its<br />
length. The gravitational field at x0 > L can be found by integrating g at x<br />
r<br />
d 0 over<br />
the length of the rod.<br />
(a) The total mass of the stick is<br />
given by:<br />
Substitute for λ and evaluate the<br />
integral to obtain:<br />
g r<br />
M =<br />
L<br />
∫<br />
0<br />
λ dx<br />
L<br />
0<br />
2<br />
M = C∫<br />
xdx =<br />
1<br />
2<br />
CL<br />
2<br />
2<br />
233
234<br />
<strong>Chapter</strong> <strong>11</strong><br />
(b) Express the gravitational field<br />
due to an element of the stick of<br />
mass dm:<br />
Integrate this expression over the<br />
r<br />
dg<br />
= −<br />
= −<br />
Gdm<br />
2 ( x − x)<br />
( x − x)<br />
0<br />
GCxdx<br />
( ) i 2<br />
x − x<br />
0<br />
L<br />
r<br />
g = −GC∫<br />
iˆ<br />
= −<br />
length of the stick to obtain: ( x − x)<br />
=<br />
0<br />
0<br />
ˆ<br />
xdx<br />
2<br />
iˆ<br />
Gλ<br />
dx<br />
0<br />
2<br />
iˆ<br />
2GM<br />
⎡ ⎛ x0<br />
⎞ ⎛ L ⎞⎤<br />
ln<br />
iˆ<br />
2 ⎢ ⎜<br />
⎟ −<br />
⎜<br />
⎟<br />
⎟⎥<br />
L ⎣ ⎝ x0<br />
− L ⎠ ⎝ x0<br />
− L ⎠⎦<br />
The Gravitational Field ( g r ) due to Spherical Objects<br />
71 •• Two widely separated solid spheres, S1 and S2, each have radius R and<br />
mass M. Sphere S1 is uniform, whereas the density of sphere S2 is given by<br />
ρ(r) = C/r, where r is the distance from its center. If the gravitational field<br />
strength at the surface of S1 is g1, what is the gravitational field strength at the<br />
surface of S2?<br />
Picture the Problem The gravitational field strength at the surface of a sphere is<br />
2<br />
given by g = GM R , where R is the radius of the sphere and M is its mass.<br />
Express the gravitational field<br />
strength on the surface of S1:<br />
Express the gravitational field<br />
strength on the surface of S2:<br />
GM<br />
g 1 = 2<br />
R<br />
GM<br />
g 2 = 2<br />
R<br />
Divide the second of these equations GM<br />
by the first and simplify to obtain: g 2<br />
2 = R<br />
g GM<br />
1<br />
2<br />
R<br />
= 1<br />
g = g<br />
⇒ 1 2<br />
75 •• Suppose you are standing on a spring scale in an elevator that is<br />
descending at constant speed in a mine shaft located on the equator. Model Earth<br />
as a homogeneous sphere. (a) Show that the force on you due to Earth’s gravity<br />
alone is proportional to your distance from the center of the planet.<br />
(b) Assume that the mine shaft located on the equator and is vertical. Do not<br />
neglect Earth’s rotational motion. Show that the reading on the spring scale is<br />
proportional to your distance from the center of the planet.
<strong>Gravity</strong><br />
Picture the Problem There are two forces acting on you as you descend in the<br />
elevator and are at a distance r from the center of Earth; an upward normal force<br />
(FN) exerted by the scale, and a downward gravitational force (mg) exerted by<br />
Earth. Because you are in equilibrium (you are descending at constant speed)<br />
under the influence of these forces, the normal force exerted by the scale is equal<br />
in magnitude to the gravitational force acting on you. We can use Newton’s law<br />
of gravity to express this gravitational force.<br />
(a) Express the force of gravity<br />
acting on you when you are a<br />
distance r from the center of Earth:<br />
Using the definition of density,<br />
express the density of Earth between<br />
you and the center of Earth and the<br />
density of Earth as a whole:<br />
The density of Earth is also given by:<br />
Equating these two expressions for ρ<br />
and solving for M(r) yields:<br />
Substitute for M(r) in equation (1)<br />
and simplify to obtain:<br />
Apply Newton’s law of gravity<br />
to yourself at the surface of Earth<br />
to obtain:<br />
Substitute for g in equation (2) to<br />
obtain:<br />
GM ( r)<br />
m<br />
Fg = (1)<br />
2<br />
r<br />
where M(r) is the mass of Earth<br />
enclosed within the radius r: that is,<br />
loser to the center of Earth than your<br />
position.<br />
( r)<br />
() r<br />
M<br />
ρ = =<br />
V<br />
M<br />
E ρ = =<br />
4 VE<br />
3<br />
M<br />
() r<br />
= M<br />
E<br />
M<br />
( r)<br />
3<br />
4<br />
3 π r<br />
M<br />
π R<br />
⎛ r<br />
⎜<br />
⎝ R<br />
E<br />
3<br />
⎞<br />
⎟<br />
⎠<br />
3<br />
⎛ r ⎞<br />
GM E ⎜ ⎟ m<br />
R GM Em<br />
r<br />
=<br />
⎝ ⎠<br />
=<br />
(2)<br />
2<br />
r R R<br />
F g 2<br />
3<br />
GM Em<br />
GM E<br />
mg = ⇒g =<br />
2<br />
2<br />
R R<br />
where g is the magnitude of the<br />
gravitational field at the surface of<br />
Earth.<br />
235<br />
⎛ mg ⎞<br />
Fg = ⎜ ⎟r ⎝ R ⎠<br />
That is, the force of gravity on you is<br />
proportional to your distance from the<br />
center of Earth.
236<br />
<strong>Chapter</strong> <strong>11</strong><br />
(b) Apply Newton’s second law to<br />
your body to obtain:<br />
r<br />
2<br />
FN − mg = −mrω<br />
R<br />
2<br />
where the net force ( − mrω<br />
) , directed<br />
toward the center of Earth, is the<br />
centripetal force acting on your body.<br />
Solving for FN yields: ⎛ mg ⎞<br />
2<br />
FN = ⎜ ⎟r<br />
− mrω<br />
Note that this equation tells us that your effective weight increases linearly with<br />
distance from the center of Earth. However, due just to the effect of rotation, as<br />
you approach the center the centripetal force decreases linearly and, doing so,<br />
increases your effective weight.<br />
77 •• A solid sphere of radius R has its center at the origin. It has a uniform<br />
1<br />
mass density ρ0, except that there is a spherical cavity in it of radius r = 2 R<br />
1<br />
centered at x = 2 R as in Figure <strong>11</strong>-27. Find the gravitational field at points on the<br />
x axis for x > R . Hint: The cavity may be thought of as a sphere of mass m =<br />
(4/3)π r 3 ρ0 plus a sphere of ″negative″ mass –m.<br />
Picture the Problem We can use the hint to find the gravitational field along the<br />
x axis.<br />
Using the hint, express x :<br />
g ( )<br />
g ( x)<br />
= gsolid<br />
sphere + g hollow sphere<br />
Substitute for and g and simplify to obtain:<br />
g<br />
( x)<br />
=<br />
gsolid sphere hollow sphere<br />
GM<br />
=<br />
x<br />
solid sphere<br />
2<br />
⎛ 4πρ0R<br />
G<br />
⎜<br />
⎝ 3<br />
3<br />
GM<br />
+<br />
1 ( x − R)<br />
⎞⎡<br />
1<br />
⎟<br />
⎟⎢<br />
− 2<br />
⎠⎢⎣<br />
x 8<br />
hollow sphere<br />
2<br />
2<br />
1<br />
⎤<br />
( ) ⎥ ⎥ 1 2<br />
x − 2 R ⎦<br />
⎝<br />
Gρ0<br />
3 =<br />
x<br />
R<br />
⎠<br />
3 [ ]<br />
4 3<br />
4 1<br />
( π R ) Gρ<br />
− π ( R)<br />
2<br />
+<br />
0<br />
3<br />
2<br />
2<br />
1 ( x − R)<br />
81 ••• A small diameter hole is drilled into the sphere of Problem 80 toward<br />
the center of the sphere to a depth of 2.0 m below the sphere’s surface. A small<br />
mass is dropped from the surface into the hole. Determine the speed of the small<br />
mass as it strikes the bottom of the hole.<br />
Picture the Problem We can use conservation of energy to relate the work done<br />
by the gravitational field to the speed of the small object as it strikes the bottom of<br />
the hole. Because we’re given the mass of the sphere, we can find C by<br />
2
<strong>Gravity</strong><br />
expressing the mass of the sphere in terms of C. We can then use the definition of<br />
the gravitational field to find the gravitational field of the sphere inside its surface.<br />
The work done by the field equals the negative of the change in the potential<br />
energy of the system as the small object falls in the hole.<br />
Use conservation of energy to relate<br />
the work done by the gravitational<br />
field to the speed of the small object<br />
as it strikes the bottom of the hole:<br />
Express the mass of a differential<br />
element of the sphere:<br />
Integrate to express the mass of<br />
K f − Ki<br />
+ ΔU<br />
= 0<br />
or, because Ki = 0 and W = −ΔU,<br />
1 2 2W<br />
W = mv 2 ⇒ v = (1)<br />
m<br />
where v is the speed with which the<br />
object strikes the bottom of the hole<br />
and W is the work done by the<br />
gravitational field.<br />
dm = ρ dV = ρ<br />
2 ( 4π r dr)<br />
2<br />
M 4 π C rdr = ( 50m<br />
) π C<br />
the sphere in terms of C:<br />
Solving for C yields:<br />
Substitute numerical values and<br />
5.<br />
0<br />
= ∫<br />
0<br />
m<br />
M<br />
C = 2 ( 50m<br />
)π<br />
1.<br />
0×<br />
10<br />
C =<br />
evaluate C: 2 ( 50m<br />
)<br />
<strong>11</strong><br />
kg<br />
8<br />
= 6.<br />
37×<br />
10 kg/m<br />
π<br />
Use its definition to express the gravitational field of the sphere at a distance from<br />
its center less than its radius:<br />
2<br />
4π<br />
r ρ dr<br />
Fg<br />
GM 0<br />
g = = = G = G<br />
2<br />
2<br />
m r r<br />
Express the work done on the small<br />
object by the gravitational force<br />
acting on it:<br />
Substitute in equation (1) and<br />
simplify to obtain:<br />
r<br />
∫<br />
r<br />
∫<br />
0<br />
4π<br />
r<br />
r<br />
2<br />
2<br />
r<br />
C<br />
dr 4π<br />
C r dr<br />
r ∫<br />
0<br />
= G = 2π<br />
GC<br />
2<br />
r<br />
3.0m<br />
W = − ∫ mgdr = ( 2m)mg<br />
v =<br />
=<br />
5.<br />
0<br />
2<br />
m<br />
( 2.<br />
0m)<br />
m(<br />
2π<br />
GC)<br />
m<br />
( 8.<br />
0m)<br />
π GC<br />
2<br />
237
238<br />
<strong>Chapter</strong> <strong>11</strong><br />
Substitute numerical values and evaluate v:<br />
v =<br />
−<strong>11</strong><br />
2 2<br />
8 2<br />
( 8.<br />
0m)<br />
π ( 6.<br />
673×<br />
10 N ⋅ m /kg )( 6.<br />
37 × 10 kg/m ) = 1.<br />
0m/s<br />
83 ••• Two identical spherical cavities are made in a lead sphere of radius R.<br />
The cavities each have a radius R/2. They touch the outside surface of the sphere<br />
and its center as in Figure <strong>11</strong>-28. The mass of a solid uniform lead sphere of<br />
radius R is M. Find the force of attraction on a point particle of mass m located at<br />
a distance d from the center of the lead sphere.<br />
Picture the Problem The force of attraction of the small sphere of mass m to the<br />
lead sphere of mass M is the sum of the forces due to the solid sphere ( FS ) and<br />
the cavities ( ) of negative mass.<br />
r<br />
F r<br />
C<br />
Express the force of attraction: S C F F F<br />
r r r<br />
= +<br />
(1)<br />
Use the law of gravity to express the<br />
force due to the solid sphere:<br />
Express the magnitude of the force<br />
acting on the small sphere due to one<br />
cavity:<br />
Relate the negative mass of a cavity<br />
to the mass of the sphere before<br />
hollowing:<br />
Letting θ be the angle between the x<br />
axis and the line joining the center of<br />
the small sphere to the center of<br />
either cavity, use the law of gravity<br />
to express the force due to the two<br />
cavities:<br />
Use the figure to express cosθ :<br />
r<br />
F<br />
S<br />
GMm<br />
= − iˆ<br />
2<br />
d<br />
C<br />
2<br />
2<br />
2 ⎟ GM'm<br />
F =<br />
⎛ R ⎞<br />
d + ⎜<br />
⎝ ⎠<br />
where M′ is the negative mass of a<br />
cavity.<br />
3 ⎡ R 4 ⎛ ⎞ ⎤<br />
M' = −ρ<br />
V = −ρ<br />
⎢ 3 π ⎜ ⎟ ⎥<br />
⎢⎣<br />
⎝ 2 ⎠ ⎥⎦<br />
= −<br />
1<br />
8<br />
4 3 1<br />
( πρ R ) = − M<br />
3<br />
r GMm<br />
F 2<br />
cos iˆ<br />
C =<br />
θ<br />
2 ⎛ 2 R ⎞<br />
8 ⎜<br />
⎜d<br />
+<br />
4 ⎟<br />
⎝ ⎠<br />
because, by symmetry, the y<br />
components add to zero.<br />
cosθ<br />
=<br />
d<br />
2<br />
d<br />
R<br />
+<br />
4<br />
2<br />
8
Substitute for cosθ and simplify to<br />
obtain:<br />
Substitute in equation (1) and<br />
simplify:<br />
General Problems<br />
r<br />
F<br />
C<br />
GMm<br />
=<br />
2 ⎛ 2 R ⎞<br />
4 ⎜<br />
⎜d<br />
+<br />
4 ⎟<br />
⎝ ⎠<br />
GMmd<br />
=<br />
2 ⎛ 2 R ⎞<br />
4 ⎜<br />
⎜d<br />
+<br />
4 ⎟<br />
⎝ ⎠<br />
3 / 2<br />
d<br />
2<br />
iˆ<br />
d<br />
R<br />
+<br />
4<br />
r GMm<br />
F iˆ<br />
GMmd<br />
= − + 2<br />
d<br />
2 ⎛ 2 R ⎞<br />
4 ⎜<br />
⎜d<br />
+<br />
4 ⎟<br />
⎝ ⎠<br />
=<br />
⎡<br />
⎢<br />
GMm ⎢<br />
− 1 2 ⎢ −<br />
d<br />
⎢ ⎧<br />
⎢<br />
⎨d<br />
⎣ ⎩<br />
2<br />
<strong>Gravity</strong><br />
2<br />
iˆ<br />
3 / 2<br />
3<br />
d<br />
4<br />
2<br />
R ⎫<br />
+<br />
4<br />
89 •• A neutron star is a highly condensed remnant of a massive star in the<br />
last phase of its evolution. It is composed of neutrons (hence the name) because<br />
the star’s gravitational force causes electrons and protons to ″coalesce″ into the<br />
neutrons. Suppose at the end of its current phase, the Sun collapsed into a neutron<br />
star (it can’t in actuality because it does not have enough mass) of radius 12.0 km,<br />
without losing any mass in the process. (a) Calculate the ratio of the gravitational<br />
acceleration at the surface of the Sun following its collapse compared to its value<br />
at the surface of the Sun today. (b) Calculate the ratio of the escape speed from<br />
the surface of the neutron-Sun to its value today.<br />
Picture the Problem We can apply Newton’s second law and the law of gravity<br />
to an object of mass m at the surface of the Sun and the neutron-Sun to find the<br />
ratio of the gravitational accelerations at their surfaces. Similarly, we can express<br />
the ratio of the corresponding expressions for the escape speeds from the two suns<br />
to determine their ratio.<br />
(a) Express the gravitational force<br />
acting on an object of mass m at the<br />
surface of the Sun:<br />
Solving for ag yields:<br />
GM<br />
F g = mag<br />
=<br />
R<br />
Sun<br />
2<br />
Sun<br />
Sun<br />
2<br />
Sun<br />
m<br />
⎬<br />
⎭<br />
iˆ<br />
3 / 2<br />
⎤<br />
⎥<br />
⎥<br />
iˆ<br />
⎥<br />
⎥<br />
⎥<br />
⎦<br />
GM<br />
a g = (1)<br />
R<br />
239
240<br />
<strong>Chapter</strong> <strong>11</strong><br />
The gravitational force acting on an<br />
object of mass m at the surface of a<br />
neutron-Sun is:<br />
Solving for a'<br />
yields:<br />
g<br />
Divide equation (2) by equation (1)<br />
to obtain:<br />
Because M = M :<br />
neutron-Sun<br />
Sun<br />
Substitute numerical values and<br />
evaluate the ratio a :<br />
'g ag<br />
(b) The escape speed from the<br />
neutron-Sun is given by:<br />
The escape speed from the Sun is<br />
given by:<br />
Dividing the first of these equations<br />
by the second and simplifying yields:<br />
GM<br />
F g = ma'g<br />
=<br />
R<br />
neutron-Sun<br />
2<br />
neutron-Sun<br />
neutron-Sun<br />
2<br />
neutron-Sun<br />
m<br />
GM<br />
a 'g<br />
= (2)<br />
R<br />
a'<br />
a<br />
g<br />
g<br />
a 'g<br />
=<br />
a<br />
g<br />
a'<br />
a<br />
g<br />
g<br />
v ' =<br />
e<br />
e<br />
v =<br />
v 'e<br />
=<br />
v<br />
e<br />
GM<br />
R<br />
=<br />
GM<br />
R<br />
R<br />
neutron-Sun<br />
2<br />
neutron-Sun<br />
R<br />
Sun<br />
2<br />
Sun<br />
2<br />
Sun<br />
2<br />
neutron-Sun<br />
8 ⎛ 6.<br />
96×<br />
10 m ⎞<br />
= ⎜<br />
3<br />
12.<br />
0 10 m<br />
⎟<br />
⎝ × ⎠<br />
GM<br />
R<br />
GM<br />
R<br />
R<br />
neutron-Sun<br />
neutron-Sun<br />
Sun<br />
Sun<br />
R<br />
Sun<br />
neutron-Sun<br />
M<br />
R<br />
=<br />
M<br />
R<br />
2<br />
neutron-Sun<br />
2<br />
neutron-Sun<br />
Substitute numerical values and<br />
8<br />
v'e<br />
6.<br />
96×<br />
10 m<br />
evaluate v 'e ve<br />
:<br />
=<br />
= 241<br />
3<br />
v 12.<br />
0×<br />
10 m<br />
e<br />
=<br />
Sun<br />
2<br />
Sun<br />
9<br />
3.<br />
36×<br />
10<br />
95 •• Uranus, the seventh planet in the Solar System, was first observed in<br />
1781 by William Herschel. Its orbit was then analyzed in terms of Kepler’s Laws.<br />
By the 1840s, observations of Uranus clearly indicated that its true orbit was<br />
different from the Keplerian calculation by an amount that could not be accounted<br />
for by observational uncertainty. The conclusion was that there must be another<br />
influence other than the Sun and the known planets lying inside Uranus’s orbit.<br />
This influence was hypothesized to be due to an eighth planet, whose predicted<br />
orbit was described independently in 1845 by two astronomers: John Adams (no<br />
relation to the former president of the United States) and Urbain LeVerrier. In<br />
September of 1846, John Galle, searching in the sky at the place predicted by
<strong>Gravity</strong><br />
Adams and LeVerrier, made the first observation of Neptune. Uranus and<br />
Neptune are in orbit about the Sun with periods of 84.0 and 164.8 years,<br />
respectively. To see the effect that Neptune had on Uranus, determine the ratio of<br />
the gravitational force between Neptune and Uranus to that between Uranus and<br />
the Sun, when Neptune and Uranus are at their closest approach to one another<br />
(i.e. when aligned with the Sun). The masses of the Sun, Uranus, and Neptune are<br />
333,000, 14.5 and 17.1 times that of Earth, respectively.<br />
Picture the Problem We can use the law of gravity and Kepler’s third law to<br />
express the ratio of the gravitational force between Neptune and Uranus to that<br />
between Uranus and the Sun, when Neptune and Uranus are at their closest<br />
approach to one another.<br />
The ratio of the gravitational force<br />
between Neptune and Uranus to that<br />
between Uranus and the Sun, when<br />
Neptune and Uranus are at their<br />
closest approach to one another is<br />
given by:<br />
Applying Kepler’s third law to<br />
Uranus yields:<br />
Applying Kepler’s third law to<br />
Neptune yields:<br />
Divide equation (3) by equation (2)<br />
to obtain:<br />
Substitute for rN in equation (1) to<br />
obtain:<br />
Simplifying this expression yields:<br />
F<br />
F<br />
g, N-U<br />
g, U-S<br />
2<br />
U<br />
GM<br />
M<br />
241<br />
( )<br />
( ) 2<br />
N U<br />
2<br />
2<br />
rN<br />
− rU<br />
M NrU<br />
= =<br />
(1)<br />
GM UM<br />
S M S rN<br />
− rU<br />
2<br />
rU<br />
Cr T = (2)<br />
2<br />
N<br />
3<br />
U<br />
Cr T = (3)<br />
T<br />
T<br />
2<br />
N<br />
2<br />
U<br />
F<br />
3<br />
N<br />
3 3<br />
CrN<br />
rN<br />
N<br />
= = ⇒<br />
3 3 N U<br />
CrU<br />
r<br />
⎟<br />
U<br />
U<br />
⎟<br />
⎛ T ⎞<br />
r = r<br />
⎜<br />
⎝ T ⎠<br />
g, N-U<br />
F<br />
F<br />
g, U-S<br />
g, N-U<br />
F<br />
g, U-S<br />
M<br />
=<br />
⎛<br />
⎜ ⎛ T<br />
M S rU<br />
⎜ ⎜<br />
⎝ ⎝ T<br />
2<br />
N U<br />
N<br />
U<br />
M N<br />
=<br />
⎛<br />
⎜⎛<br />
T ⎞ N M S⎜<br />
⎜<br />
⎟<br />
U ⎝⎝<br />
T ⎠<br />
r<br />
⎞<br />
⎟<br />
⎠<br />
2 3<br />
2 3<br />
− r<br />
⎞<br />
−1⎟<br />
⎟<br />
⎠<br />
2<br />
U<br />
⎞<br />
⎟<br />
⎟<br />
⎠<br />
2<br />
2 3
242<br />
<strong>Chapter</strong> <strong>11</strong><br />
Because MN = 17.1ME and MS = 333,000ME:<br />
F<br />
F<br />
g, N-U<br />
g, U-S<br />
=<br />
5<br />
3.<br />
33×<br />
10 M<br />
17.<br />
1M<br />
E<br />
E<br />
⎛<br />
⎜⎛<br />
T<br />
⎜ ⎜<br />
⎝⎝<br />
T<br />
N<br />
U<br />
⎞<br />
⎟<br />
⎠<br />
2 3<br />
Substitute numerical values and evaluate<br />
F<br />
g, N-U<br />
F<br />
g, U-S<br />
⎞<br />
−1⎟<br />
⎟<br />
⎠<br />
F<br />
2<br />
g, N-U<br />
F<br />
g, U-S<br />
17.<br />
1<br />
=<br />
⎛<br />
5 N<br />
3.<br />
33 10 ⎜⎛<br />
T ⎞<br />
×<br />
⎜ ⎜<br />
⎟<br />
U ⎝⎝<br />
T ⎠<br />
17. 1<br />
−4<br />
=<br />
≈ 2×<br />
10<br />
2 3<br />
2<br />
⎛<br />
5 164.<br />
8 y ⎞<br />
3.<br />
33 10 ⎜⎛<br />
⎞<br />
× ⎜ ⎟ −1⎟<br />
⎜ 84.<br />
0 y ⎟<br />
⎝⎝<br />
⎠ ⎠<br />
:<br />
2 3<br />
⎞<br />
−1⎟<br />
⎟<br />
⎠<br />
Because this ratio is so small, during the time at which Neptune is closest to<br />
Uranus, the force exerted on Uranus by Neptune is much less than the force<br />
exerted on Uranus by the Sun.<br />
97 •• [SSM] Four identical planets are arranged in a square as shown in<br />
Figure <strong>11</strong>-29. If the mass of each planet is M and the edge length of the square is<br />
a, what must be their speed if they are to orbit their common center under the<br />
influence of their mutual attraction?<br />
Picture the Problem Note that, due to the symmetrical arrangement of the<br />
planets, each experiences the same centripetal force. We can apply Newton’s<br />
second law to any of the four planets to relate its orbital speed to this net<br />
(centripetal) force acting on it.<br />
M a<br />
M<br />
∑<br />
r<br />
F<br />
F<br />
r<br />
= ma<br />
1<br />
M<br />
Applying radial radial<br />
the planets gives:<br />
θ<br />
θ<br />
F<br />
2<br />
F<br />
1<br />
to one of<br />
c<br />
1<br />
a<br />
M<br />
F = 2F cosθ<br />
+ F<br />
2<br />
2
Substituting for F1, F2, and θ and<br />
2<br />
2GM<br />
GM<br />
F = cos45°<br />
+<br />
simplifying yields: c 2<br />
a<br />
( 2a)<br />
Because<br />
2GM<br />
= 2<br />
a<br />
GM<br />
= 2<br />
a<br />
2<br />
2<br />
⎛<br />
⎜<br />
⎝<br />
1 GM<br />
+<br />
2 2a<br />
1 ⎞<br />
2 + ⎟<br />
2 ⎠<br />
2<br />
2<br />
Mv 2Mv<br />
Fc<br />
a 2 a<br />
= = 2<br />
2<br />
2Mv<br />
GM ⎛ 1 ⎞<br />
= ⎜ 2 +<br />
2 ⎟<br />
a a ⎝ 2 ⎠<br />
Solve for v to obtain:<br />
GM ⎛ 1 ⎞<br />
v = ⎜1<br />
+ ⎟ = 1.<br />
16<br />
a ⎝ 2 2 ⎠<br />
2<br />
2<br />
<strong>Gravity</strong><br />
2<br />
2<br />
GM<br />
a<br />
103 ••• In this problem you are to find the gravitational potential energy of the<br />
thin rod in Example <strong>11</strong>-8 and a point particle of mass m0 that is on the x axis at x<br />
= x0. (a) Show that the potential energy shared by an element of the rod of mass<br />
1<br />
dm (shown in Figure <strong>11</strong>-14) and the point particle of mass m0 located at x0 ≥ 2 L<br />
is given by<br />
Gm0dm<br />
GMm0<br />
dU = − = dxs<br />
x0<br />
− xs<br />
L(<br />
x0<br />
− xs<br />
)<br />
where U = 0 at x0 = ∞. (b) Integrate your result for Part (a) over the length of the<br />
rod to find the total potential energy for the system. Generalize your function<br />
U(x0) to any place on the x axis in the region x > L/2 by replacing x0 by a general<br />
coordinate x and write it as U(x). (c) Compute the force on m0 at a general point x<br />
using Fx = –dU/dx and compare your result with m0g, where g is the field at x0<br />
calculated in Example <strong>11</strong>-8.<br />
Picture the Problem Let U = 0 at x = ∞. The potential energy of an element of<br />
the stick dm and the point mass m0 is given by the definition of gravitational<br />
potential energy: = −Gm<br />
dm r where r is the separation of dm and m0.<br />
dU 0<br />
(a) Express the potential energy of<br />
Gm0dm<br />
dU = −<br />
x − x<br />
the masses m0 and dm: 0 s<br />
The mass dm is proportional to the<br />
size of the element dxs<br />
:<br />
Substitute for dm and λ to express<br />
dm = λ dxs<br />
M<br />
where λ = .<br />
L<br />
Gm0λ<br />
dx<br />
dU = −<br />
dU in terms of xs: x − x L(<br />
x − x )<br />
0<br />
s<br />
s<br />
=<br />
GMm<br />
−<br />
0<br />
0<br />
dx<br />
s<br />
s<br />
243
244<br />
<strong>Chapter</strong> <strong>11</strong><br />
(b) Integrate dU to find the total potential energy of the system:<br />
L /<br />
GMm0<br />
U = −<br />
L ∫<br />
=<br />
GMm<br />
−<br />
L<br />
dxs<br />
x − x<br />
−L<br />
/ 2 0 s<br />
0<br />
2<br />
⎛ x<br />
ln<br />
⎜<br />
⎝ x<br />
0<br />
0<br />
+<br />
−<br />
GMm<br />
=<br />
L<br />
L 2 ⎞<br />
⎟<br />
L 2 ⎠<br />
0<br />
⎡ ⎛<br />
⎢ln⎜<br />
x<br />
⎣ ⎝<br />
(c) Because x0 is a general point along the x axis:<br />
F<br />
( x )<br />
0<br />
0<br />
L ⎞ ⎛<br />
− ⎟ − ln⎜<br />
x<br />
2 ⎠ ⎝<br />
⎡<br />
⎤<br />
dU GMm ⎢<br />
⎥<br />
0 1 1<br />
= − = ⎢ − ⎥<br />
dx0<br />
L ⎢<br />
L L<br />
x + − ⎥<br />
0 x0<br />
⎣ 2 2 ⎦<br />
Further simplification yields:<br />
F(<br />
x )<br />
0<br />
Gmm<br />
= − 2<br />
x − L<br />
0<br />
0<br />
2<br />
L ⎞⎤<br />
+ ⎟<br />
2<br />
⎥<br />
⎠⎦<br />
This answer and the answer given in Example <strong>11</strong>-8 are the same.<br />
4