![Chapter 11224
Using Kepler’s third law, relate the
period of the Sun T to its mean
distance R0 from the center of the
galaxy:
3
0
G
2
2 4
R
GM
T
π
= ⇒ 2
S
3
0
2
S
G 4
TGM
R
M
M π
=
Substitute numerical values and evaluate MG/Ms:
( )
s
11
galaxy
11
2
7630
2
2
11
315
42
s
G
101.1101.1
y
s
10156.3y10250kg1099.1
kg
mN
106742.6
y
m10461.9
y103.004
MM
c
c
M
M
×≈⇒×≈
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
××××⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×
×⋅×
=
−
π
Kepler’s Laws
25 •• [SSM] One of the so-called ″Kirkwood gaps″ in the asteroid belt
occurs at an orbital radius at which the period of the orbit is half that of Jupiter’s.
The reason there is a gap for orbits of this radius is because of the periodic pulling
(by Jupiter) that an asteroid experiences at the same place in its orbit every other
orbit around the sun. Repeated tugs from Jupiter of this kind would eventually
change the orbit of such an asteroid. Therefore, all asteroids that would otherwise
have orbited at this radius have presumably been cleared away from the area due
to this resonance phenomenon. How far from the Sun is this particular 2:1
resonance ″Kirkwood″ gap?
Picture the Problem The period of an orbit is related to its semi-major axis (for
circular orbits this distance is the orbital radius). Because we know the orbital
periods of Jupiter and a hypothetical asteroid in the Kirkwood gap, we can use
Kepler’s third law to set up a proportion relating the orbital periods and average
distances of Jupiter and the asteroid from the Sun from which we can obtain an
expression for the orbital radius of an asteroid in the Kirkwood gap.
Use Kepler’s third law to relate
Jupiter’s orbital period to its mean
distance from the Sun:
3
Jupiter
Sun
2
2
Jupiter
4
r
GM
T
π
=](https://image.slidesharecdn.com/ch11ssm-181218172926/85/Ch11-ssm-4-320.jpg)
![Chapter 11236
(b) Apply Newton’s second law to
your body to obtain:
2
N ωmr
R
r
mgF −=−
where the net force ( )2
ωmr− , directed
toward the center of Earth, is the
centripetal force acting on your body.
Solving for FN yields: 2
N ωmrr
R
mg
F −⎟
⎠
⎞
⎜
⎝
⎛
=
Note that this equation tells us that your effective weight increases linearly with
distance from the center of Earth. However, due just to the effect of rotation, as
you approach the center the centripetal force decreases linearly and, doing so,
increases your effective weight.
77 •• A solid sphere of radius R has its center at the origin. It has a uniform
mass density ρ0, except that there is a spherical cavity in it of radius Rr 2
1
=
centered at Rx 2
1
= as in Figure 11-27. Find the gravitational field at points on the
x axis for Rx > . Hint: The cavity may be thought of as a sphere of mass m =
(4/3)π r3
ρ0 plus a sphere of ″negative″ mass –m.
Picture the Problem We can use the hint to find the gravitational field along the
x axis.
Using the hint, express :( )xg ( ) spherehollowspheresolid ggxg +=
Substitute for and and simplify to obtain:spheresolidg spherehollowg
( )
( )
( ) ( )[ ]
( )
( ) ⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
−
−
+=
−
+=
2
2
12
3
0
2
2
1
3
2
1
3
4
0
2
3
3
4
0
2
2
1
spherehollow
2
spheresolid
8
11
3
4
Rxx
R
G
Rx
RG
x
RG
Rx
GM
x
GM
xg
πρ
πρπρ
81 ••• A small diameter hole is drilled into the sphere of Problem 80 toward
the center of the sphere to a depth of 2.0 m 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.
Picture the Problem We can use conservation of energy to relate the work done
by the gravitational field to the speed of the small object as it strikes the bottom of
the hole. Because we’re given the mass of the sphere, we can find C by](https://image.slidesharecdn.com/ch11ssm-181218172926/85/Ch11-ssm-16-320.jpg)
![Chapter 11242
Because MN = 17.1ME and MS = 333,000ME:
232
U
N5
232
U
N
E
5
E
-SUg,
U-Ng,
11033.3
1.17
11033.3
1.17
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
T
T
T
T
M
M
F
F
Substitute numerical values and evaluate
S-Ug,
U-Ng,
F
F
:
4
232
5
-SUg,
U-Ng,
102
1
y0.84
y8.164
1033.3
1.17 −
×≈
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
F
F
Because this ratio is so small, during the time at which Neptune is closest to
Uranus, the force exerted on Uranus by Neptune is much less than the force
exerted on Uranus by the Sun.
97 •• [SSM] Four identical planets are arranged in a square as shown in
Figure 11-29. If the mass of each planet is M and the edge length of the square is
a, what must be their speed if they are to orbit their common center under the
influence of their mutual attraction?
Picture the Problem Note that, due to the symmetrical arrangement of the
planets, each experiences the same centripetal force. We can apply Newton’s
second law to any of the four planets to relate its orbital speed to this net
(centripetal) force acting on it.
a
a
F
F
F
1
2
θ
θ
M M
MM
1
Applying radialradial aF
rr
m=∑ to one of
the planets gives:
21c cos2 FFF += θ](https://image.slidesharecdn.com/ch11ssm-181218172926/85/Ch11-ssm-22-320.jpg)
Ch11 ssm
- 1. Chapter 11 Gravity Conceptual Problems 1 • True or false: (a) For Kepler’s law of equal areas to be valid, the force of gravity must vary inversely with the square of the distance between a given planet and the Sun. (b) The planet closest to the Sun has the shortest orbital period. (c) Venus’s orbital speed is larger than the orbital speed of Earth. (d) The orbital period of a planet allows accurate determination of that planet’s mass. (a) False. Kepler’s law of equal areas is a consequence of the fact that the gravitational force acts along the line joining two bodies but is independent of the manner in which the force varies with distance. (b) True. The periods of the planets vary with the three-halves power of their distances from the Sun. So the shorter the distance from the Sun, the shorter the period of the planet’s motion. (c) True. Setting up a proportion involving the orbital speeds of the two planets in terms of their orbital periods and mean distances from the Sun (see Table 11-1) shows that .EarthVenus 17.1 vv = (d) False. The orbital period of a planet is independent of the planet’s mass. 3 • During what season in the northern hemisphere does Earth attain its maximum orbital speed about the Sun? What season is related to its minimum orbital speed? Hint: Earth is at perihelion in early January. Determine the Concept Earth is closest to the Sun during winter in the northern hemisphere. This is the time of fastest orbital speed. Summer would be the time for minimum orbital speed. 7 • 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. Picture the Problem The acceleration due to gravity varies inversely with the square of the distance from the center of the moon. 221
- 2. Chapter 11222 Express the dependence of the acceleration due to the gravity of the moon on the distance from its center: 2 1 r a' ∝ Express the dependence of the acceleration due to the gravity of the moon at its surface on its radius: 2 M 1 R a ∝ Divide the first of these expressions by the second to obtain: 2 2 M r R a a' = Solving for a′ and simplifying yields: ( ) aa R R a r R a' 16 1 2 M 2 M 2 2 M 4 === and )(d is correct. 11 •• Suppose the escape speed from a planet was only slightly larger than the escape speed from Earth, yet the planet is considerably larger than Earth. How would the planet’s (average) density compare to Earth’s (average) density? (a) It must be denser. (b) It must be less dense. (c) It must be the same density. (d) You cannot determine the answer based on the data given. Picture the Problem The densities of the planets are related to the escape speeds from their surfaces through RGMv 2e = . The escape speed from the planet is given by: planet planet planet 2 R GM v = The escape speed from Earth is given by: Earth Earth Earth 2 R GM v = Expressing the ratio of the escape speed from the planet to the escape speed from Earth and simplifying yields: Earth planet planet Earth Earth Earth planet planet Earth planet 2 2 M M R R R GM R GM v v == Because vplanet ≈ vEarth: Earth planet planet Earth 1 M M R R ≈
- 3. Gravity 223 Squaring both sides of the equation yields: Earth planet planet Earth 1 M M R R ≈ Express Mplanet and MEarth in terms of their densities and simplify to obtain: 2 EarthEarth 2 planetplanet 3 Earth3 4 Earth 3 planet3 4 planet planet Earth EarthEarth planetplanet planet Earth EarthEarth planetplanet planet Earth 1 R R R R R R V V R R V V R R ρ ρ πρ πρ ρ ρ ρ ρ ===≈ Solving for the ratio of the densities yields: 2 planet 2 Earth Earth planet R R ≈ ρ ρ Because the planet is considerably larger than Earth: 1 Earth planet << ρ ρ and ( )b is correct. 13 •• Near the end of their useful lives, several large Earth-orbiting satellites have been maneuvered so they burn up as they enter Earth’s atmosphere. These maneuvers have to be done carefully so large fragments do not impact populated land areas. You are in charge of such a project. Assuming the satellite of interest has on-board propulsion, in what direction would you fire the rockets for a short burn time to start this downward spiral? What would happen to the kinetic energy, gravitational potential energy and total mechanical energy following the burn as the satellite came closer and closer to Earth? Determine the Concept You should fire the rocket in a direction to oppose the orbital motion of the satellite. As the satellite gets closer to Earth after the burn, the potential energy will decrease. However, the total mechanical energy will decrease due to the frictional drag forces transforming mechanical energy into thermal energy. The kinetic energy will increase until the satellite enters the atmosphere where the drag forces slow its motion. Estimation and Approximation 17 • Estimate the mass of our galaxy (the Milky Way) if the Sun orbits the center of the galaxy with a period of 250 million years at a mean distance of 30,000 c⋅y. Express the mass in terms of multiples of the solar mass MS. (Neglect the mass farther from the center than the Sun, and assume that the mass closer to the center than the Sun exerts the same force on the Sun as would a point particle of the same mass located at the center of the galaxy.) Picture the Problem To approximate the mass of the galaxy we’ll assume the galactic center to be a point mass with the Sun in orbit about it and apply Kepler’s third law. Let MG represent the mass of the galaxy.
- 4. Chapter 11224 Using Kepler’s third law, relate the period of the Sun T to its mean distance R0 from the center of the galaxy: 3 0 G 2 2 4 R GM T π = ⇒ 2 S 3 0 2 S G 4 TGM R M M π = Substitute numerical values and evaluate MG/Ms: ( ) s 11 galaxy 11 2 7630 2 2 11 315 42 s G 101.1101.1 y s 10156.3y10250kg1099.1 kg mN 106742.6 y m10461.9 y103.004 MM c c M M ×≈⇒×≈ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ××××⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ × ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ × ×⋅× = − π Kepler’s Laws 25 •• [SSM] One of the so-called ″Kirkwood gaps″ in the asteroid belt occurs at an orbital radius at which the period of the orbit is half that of Jupiter’s. The reason there is a gap for orbits of this radius is because of the periodic pulling (by Jupiter) that an asteroid experiences at the same place in its orbit every other orbit around the sun. Repeated tugs from Jupiter of this kind would eventually change the orbit of such an asteroid. Therefore, all asteroids that would otherwise have orbited at this radius have presumably been cleared away from the area due to this resonance phenomenon. How far from the Sun is this particular 2:1 resonance ″Kirkwood″ gap? Picture the Problem The period of an orbit is related to its semi-major axis (for circular orbits this distance is the orbital radius). Because we know the orbital periods of Jupiter and a hypothetical asteroid in the Kirkwood gap, we can use Kepler’s third law to set up a proportion relating the orbital periods and average distances of Jupiter and the asteroid from the Sun from which we can obtain an expression for the orbital radius of an asteroid in the Kirkwood gap. Use Kepler’s third law to relate Jupiter’s orbital period to its mean distance from the Sun: 3 Jupiter Sun 2 2 Jupiter 4 r GM T π =
- 5. Gravity 225 Use Kepler’s third law to relate the orbital period of an asteroid in the Kirkwood gap to its mean distance from the Sun: 3 Kirkwood Sun 2 2 Kirkwood 4 r GM T π = Dividing the second of these equations by the first and simplifying yields: 3 Jupiter 3 Kirkwood 3 Jupiter Sun 2 3 Kirkwood Sun 2 2 Jupiter 2 Kirkwood 4 4 r r r GM r GM T T == π π Solving for rKirkwood yields: 32 Jupiter Kirkwood JupiterKirkwood ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = T T rr Because the period of the orbit of an asteroid in the Kirkwood gap is half that of Jupiter’s: ( ) AU27.3 m101.50 AU1 m1090.4 m1090.4 m108.77 11 11 11 32 Jupiter Jupiter2 1 10 Kirkwood = × ××= ×= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ×= T T r Remarks: There are also significant Kirkwood gaps at 3:1, 5:2, and 7:3 and resonances at 2.5 AU, 2.82 AU, and 2.95 AU. 29 •• Kepler determined distances in the Solar System from his data. For example, he found the relative distance from the Sun to Venus (as compared to the distance from the Sun to Earth) as follows. Because Venus’s orbit is closer to the Sun than is Earth’s orbit, Venus is a morning or evening star—its position in the sky is never very far from the Sun (Figure 11-24). If we suppose the orbit of Venus is a perfect circle, then consider the relative orientation of Venus, Earth, and the Sun at maximum extension, that is when Venus is farthest from the Sun in the sky. (a) Under this condition, show that angle b in Figure 11-24 is 90º. (b) If the maximum elongation angle a between Venus and the Sun is 47º, what is the distance between Venus and the Sun in AU? (c) Use this result to estimate the length of a Venusian ″year.″ Picture the Problem We can use a property of lines tangent to a circle and radii drawn to the point of contact to show that b = 90°. Once we’ve established that b is a right angle we can use the definition of the sine function to relate the distance from the Sun to Venus to the distance from the Sun to Earth.
- 6. Chapter 11226 (a) The line from Earth to Venus' orbit is tangent to the orbit of Venus at the point of maximum extension. Venus will appear closer to the sun in Earth’s sky when it passes the line drawn from Earth and tangent to its orbit. Hence °= 90b (b) Using trigonometry, relate the distance from the sun to Venus to the angle a: SVd add d d a sinsin SESV SE SV =⇒= Substitute numerical values and evaluate :SVd ( ) AU73.0 AU731.074sinAU00.1SV = =°=d (c) Use Kepler’s third law to relate Venus’s orbital period to its mean distance from the Sun: 3 Venus Sun 2 2 Venus 4 r GM T π = Use Kepler’s third law to relate Earth’s orbital period to its mean distance from the Sun: 3 Earth Sun 2 2 Earth 4 r GM T π = Dividing the first of these equations by the second and simplifying yields: 3 Earth 3 Venus 3 Earth Sun 2 3 Venus Sun 2 2 Earth 2 Venus 4 4 r r r GM r GM T T == π π Solving for yields:VenusT 23 Earth Venus EarthVenus ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = r r TT Using the result from Part (b) yields: ( ) y63.0 AU00.1 AU731.0 y00.1 23 Venus = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =T Remarks: The correct distance from the sun to Venus is closer to 0.723 AU. Newton’s Law of Gravity 31 • Jupiter’s satellite Europa orbits Jupiter with a period of 3.55 d at an average orbital radius of 6.71 × 108 m. (a) Assuming that the orbit is circular, determine the mass of Jupiter from the data given. (b) Another satellite of Jupiter, Callisto, orbits at an average radius of 18.8 × 108 m with an orbital period of
- 7. Gravity 227 16.7 d. Show that these data are consistent with an inverse square force law for gravity (Note: DO NOT use the value of G anywhere in Part (b)). Picture the Problem While we could apply Newton’s law of gravitation and second law of motion to solve this problem from first principles, we’ll use Kepler’s third law (derived from these laws) to find the mass of Jupiter in Part (a). In Part (b) we can compare the ratio of the centripetal accelerations of Europa and Callisto to show that these data are consistent with an inverse square law for gravity. (a) Assuming a circular orbit, apply Kepler’s third law to the motion of Europa to obtain: 3 E J 2 2 E 4 R GM T π = ⇒ 3 E2 E 2 J 4 R GT M π = Substitute numerical values and evaluate MJ: ( ) ( ) kg1090.1 h s3600 d h24 d3.55/kgmN10673.6 m106.714 27 2 2211 382 J ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ××⋅× × = − π M Note that this result is in excellent agreement with the accepted value of 1.902×1027 kg. (b) Express the centripetal acceleration of both of the moons to obtain: 2 2 2 2 lcentripeta 4 2 T R R T R R v a π π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ == where R and T are the radii and periods of their motion. Using this result, express the centripetal accelerations of Europa and Callisto: 2 E E 2 E 4 T R a π = and 2 C C 2 C 4 T R a π = Divide the first of these equations by the second and simplify to obtain: C E 2 E 2 C 2 C C 2 2 E E 2 C E 4 4 R R T T T R T R a a == π π Substitute for the periods of Callisto and Europa using Kepler’s third law to obtain: 2 E 2 C C E 3 E 3 C C E R R R R CR CR a a ==
- 8. Chapter 11228 This result, together with the fact that the gravitational force is directly proportional to the acceleration of the moons, demonstrates that the gravitational force varies inversely with the square of the distance. 33 • 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. Picture the Problem While we could apply Newton’s law of gravitation and second law of motion to solve this problem from first principles, we’ll use Kepler’s third law (derived from these laws) to find the period of Mimas and to relate the periods of the moons of Saturn to their mean distances from its center. (a) Using Kepler’s third law, relate the period of Mimas to its mean distance from the center of Saturn: 3 M S 2 2 M 4 r GM T π = ⇒ 3 M S 2 M 4 r GM T π = Substitute numerical values and evaluate TM: ( ) ( )( ) h22.7s1018.8 kg1069.5/kgmN106726.6 m1086.14 4 262211 382 M ≈×= ×⋅× × = − π T (b) Using Kepler’s third law, relate the period of Titan to its mean distance from the center of Saturn: 3 T S 2 2 T 4 r GM T π = ⇒ 3 2 S 2 T T 4π GMT r = Substitute numerical values and evaluate rT: ( ) ( )( ) m1022.1 4 kg1069.5/kgmN106726.6s1038.1 93 2 26221126 T ×= ×⋅×× = − π r 41 •• A superconducting gravity meter can measure changes in gravity of the order Δg/g = 1.00 × 10–11 . (a) You are hiding behind a tree holding the meter, and your 80-kg friend approaches the tree from the other side. How close to you can your friend get before the meter detects a change in g due to his presence? (b) You are in a hot air balloon and are using the meter to determine the rate of ascent (assume the balloon has constant acceleration). What is the smallest change in altitude that results in a detectable change in the gravitational field of Earth?
- 9. Gravity 229 Picture the Problem We can determine the maximum range at which an object with a given mass can be detected by substituting the equation for the gravitational field in the expression for the resolution of the meter and solving for the distance. Differentiating g(r) with respect to r, separating variables to obtain dg/g, and approximating Δr with dr will allow us to determine the vertical change in the position of the gravity meter in Earth’s gravitational field is detectable. (a) Earth’s gravitational field is given by: 2 E E E R GM g = Express the gravitational field due to the mass m (assumed to be a point mass) of your friend and relate it to the resolution of the meter: ( ) 2 E E11 E 11 2 1000.1 1000.1 R GM g r Gm rg − − ×= ×== Solving for r yields: E 11 E 1000.1 M m Rr × = Substitute numerical values and evaluate r: ( ) ( ) m37.7 kg105.98 kg801000.1 m106.37 24 11 6 = × × ×=r (b) Differentiate g(r) and simplify to obtain: g rr Gm rr Gm dr dg 222 23 −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= − = Separate variables to obtain: 11 102 − =−= r dr g dg Approximating dr with Δr, evaluate Δr with r = RE: ( )( ) m9.31 m1037.61000.1Δ 611 2 1 μ= ××−= − r 43 •• Earth’s radius is 6370 km and the moon’s radius 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 Earth? Picture the Problem We can use the definitions of the gravitational fields at the surfaces of Earth and the moon to express the accelerations due to gravity at these locations in terms of the average densities of Earth and the moon. Expressing the ratio of these accelerations will lead us to the ratio of the densities.
- 10. Chapter 11230 Express the acceleration due to gravity at the surface of Earth in terms of Earth’s average density: EE3 4 2 E 3 E3 4 E 2 E EE 2 E E E RG R RG R VG R GM g πρ πρρ = === The acceleration due to gravity at the surface of the moon in terms of the moon’s average density is: MM3 4 M RGg πρ= Divide the second of these equations by the first to obtain: EE MM E M R R g g ρ ρ = ⇒ ME EM E M Rg Rg = ρ ρ Substitute numerical values and evaluate E M ρ ρ : ( )( ) ( )( ) 605.0 m101.738m/s9.81 m106.37m/s1.62 62 62 E M = × × = ρ ρ Gravitational Potential Energy 47 • Knowing that the acceleration of gravity on the moon is 0.166 times that on Earth and that the moon’s radius is 0.273RE, find the escape speed for a projectile leaving the surface of the moon. Picture the Problem The escape speed from the moon is given by mmme, 2 RGMv = , where Mm and Rm represent the mass and radius of the moon, respectively. Express the escape speed from the moon: mm m m e.m 2 2 Rg R GM v == Because and : Em 166.0 gg = Em 273.0 RR = ( )( )EEe.m 273.0166.02 Rgv = Substitute numerical values and evaluate ve,m: ( )( )( )( ) km/s38.2m10371.6273.0m/s81.9166.02 62 e.m =×=v 51 •• An object is dropped from rest from a height of 4.0 × 106 m above the surface of Earth. If there is no air resistance, what is its speed when it strikes Earth?
- 11. Gravity 231 Picture the Problem Let the zero of gravitational potential energy be at infinity and let m represent the mass of the object. We’ll use conservation of energy to relate the initial potential energy of the object-Earth system to the final potential and kinetic energies. Use conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike Earth: 0ifif =−+− UUKK or, because Ki = 0, ( ) ( ) ( ) 0EEE =+−+ hRURURK (1) where h is the initial height above Earth’s surface. Express the potential energy of the object-Earth system when the object is at a distance r from the surface of Earth: ( ) r mGM rU E −= Substitute in equation (1) to obtain: 0 E E E E2 2 1 = + +− hR mGM R mGM mv Solving for v yields: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= hR h gR hR GM R GM v E E E E E E 2 2 Substitute numerical values and evaluate v: ( )( )( ) km/s9.6 m100.4m106.37 m100.4m106.37m/s9.812 66 662 = ×+× ×× =v Gravitational Orbits 59 •• Many satellites orbit Earth with maximum altitudes of 1000 km or less. Geosynchronous satellites, however, orbit at an altitude of 35 790 km above Earth’s surface. 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 Earth? Picture the Problem We can express the energy difference between these two orbits in terms of the total energy of a satellite at each elevation. The application of Newton’s second law to the force acting on a satellite will allow us to express
- 12. Chapter 11232 the total energy of each satellite as a function of its mass, the radius of Earth, and its orbital radius. Express the energy difference: 1000geo EEE −=Δ (1) Express the total energy of an orbiting satellite: R mGM mv UKE E2 2 1 tot −= += (2) where R is the orbital radius. Apply Newton’s second law to a satellite to relate the gravitational force to the orbital speed: R v m R mGM F 2 2 E radial == Solving for v2 yields: R gR v 2 E2 = Substitute in equation (2) to obtain: R mgR R mgR R gR mE 2 2 E 2 E 2 E 2 1 tot −=−= Substituting in equation (1) and simplifying yields: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −=+−=Δ km79035 1 km1000 1 2 11 222 EE 2 E geo1000 2 E 1000 2 E geo 2 E RR mgR RR mgR R mgR R mgR E Substitute numerical values and evaluate ΔE: ( )( )( ) GJ1.11 m10722.4 1 m1037.7 1 2 m106.37kg/N9.81kg500 6 26 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × − × × =ΔE The Gravitational Field ( g r ) 63 •• A point particle of mass m is on the x axis at x = L and an identical point particle is on the y axis at y = L. (a) What is the direction of the gravitational field at the origin? (b) What is the magnitude of this field?
- 13. Gravity 233 Picture the Problem We can use the definition of the gravitational field due to a point mass to find the x and y components of the field at the origin and then add these components to find the resultant field. We can find the magnitude of the field from its components using the Pythagorean theorem. (a) The gravitational field at the origin is the sum of its x and y components: yx ggg rrr += (1) Express the gravitational field due to the point mass at x = L: ig ˆ 2 L Gm x = r Express the gravitational field due to the point mass at y = L: jg ˆ 2 L Gm y = r Substitute in equation (1) to obtain: jiggg ˆˆ 22 L Gm L Gm yx +=+= rrr ⇒ the direction of the gravitational field is along a line at 45° above the +x axis. (b) The magnitude of g r is given by: 22 yx gg +=g r Substitute for gx and gy and simplify to obtain: 2 2 2 2 2 2 L Gm L Gm L Gm =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =g r 67 ••• A nonuniform thin rod of length L lies on the x axis. One end of the rod is at the origin, and the other end is at x = L. The rod’s mass per unit length λ varies as λ = Cx, where C is a constant. (Thus, an element of the rod has mass dm = λ dx.) (a) What is the total mass of the rod? (b) Find the gravitational field due to the rod on the x axis at x = x0, where x0 > L. Picture the Problem We can find the mass of the rod by integrating dm over its length. The gravitational field at x0 > L can be found by integrating at xg r d 0 over the length of the rod. (a) The total mass of the stick is given by: ∫= L dxM 0 λ Substitute for λ and evaluate the integral to obtain: 2 2 1 0 CLxdxCM L == ∫
- 14. Chapter 11234 (b) Express the gravitational field due to an element of the stick of mass dm: ( ) ( ) ( ) i iig ˆ ˆˆ 2 0 2 0 2 0 xx GCxdx xx dxG xx Gdm d − −= − −= − −= λr Integrate this expression over the length of the stick to obtain: ( ) i ig ˆln 2 ˆ 00 0 2 0 2 0 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = − −= ∫ Lx L Lx x L GM xx xdx GC L r The Gravitational Field ( g r ) due to Spherical Objects 71 •• Two widely separated solid spheres, S1 and S2, each have radius R and mass M. Sphere S1 is uniform, whereas the density of sphere S2 is given by ρ(r) = C/r, where r is the distance from its center. If the gravitational field strength at the surface of S1 is g1, what is the gravitational field strength at the surface of S2? Picture the Problem The gravitational field strength at the surface of a sphere is given by ,2 RGMg = where R is the radius of the sphere and M is its mass. Express the gravitational field strength on the surface of S1: 21 R GM g = Express the gravitational field strength on the surface of S2: 22 R GM g = Divide the second of these equations by the first and simplify to obtain: 1 2 2 1 2 == R GM R GM g g ⇒ 21 gg = 75 •• Suppose you are standing on a spring scale in an elevator that is descending at constant speed in a mine shaft located on the equator. Model Earth as a homogeneous sphere. (a) Show that the force on you due to Earth’s gravity alone is proportional to your distance from the center of the planet. (b) Assume that the mine shaft located on the equator and is vertical. Do not neglect Earth’s rotational motion. Show that the reading on the spring scale is proportional to your distance from the center of the planet.
- 15. Gravity 235 Picture the Problem There are two forces acting on you as you descend in the elevator and are at a distance r from the center of Earth; an upward normal force (FN) exerted by the scale, and a downward gravitational force (mg) exerted by Earth. Because you are in equilibrium (you are descending at constant speed) under the influence of these forces, the normal force exerted by the scale is equal in magnitude to the gravitational force acting on you. We can use Newton’s law of gravity to express this gravitational force. (a) Express the force of gravity acting on you when you are a distance r from the center of Earth: 2 )( r mrGM Fg = (1) where M(r) is the mass of Earth enclosed within the radius r: that is, loser to the center of Earth than your position. Using the definition of density, express the density of Earth between you and the center of Earth and the density of Earth as a whole: ( ) ( ) ( ) 3 3 4 r rM rV rM π ρ == The density of Earth is also given by: 3 3 4 E E E R M V M π ρ == Equating these two expressions for ρ and solving for M(r) yields: ( ) 3 E ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = R r MrM Substitute for M(r) in equation (1) and simplify to obtain: R r R mGM r m R r GM Fg 2 E 2 3 E = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = (2) Apply Newton’s law of gravity to yourself at the surface of Earth to obtain: 2 E R mGM mg = ⇒ 2 E R GM g = where g is the magnitude of the gravitational field at the surface of Earth. Substitute for g in equation (2) to obtain: r R mg Fg ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = That is, the force of gravity on you is proportional to your distance from the center of Earth.
- 16. Chapter 11236 (b) Apply Newton’s second law to your body to obtain: 2 N ωmr R r mgF −=− where the net force ( )2 ωmr− , directed toward the center of Earth, is the centripetal force acting on your body. Solving for FN yields: 2 N ωmrr R mg F −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Note that this equation tells us that your effective weight increases linearly with distance from the center of Earth. However, due just to the effect of rotation, as you approach the center the centripetal force decreases linearly and, doing so, increases your effective weight. 77 •• A solid sphere of radius R has its center at the origin. It has a uniform mass density ρ0, except that there is a spherical cavity in it of radius Rr 2 1 = centered at Rx 2 1 = as in Figure 11-27. Find the gravitational field at points on the x axis for Rx > . Hint: The cavity may be thought of as a sphere of mass m = (4/3)π r3 ρ0 plus a sphere of ″negative″ mass –m. Picture the Problem We can use the hint to find the gravitational field along the x axis. Using the hint, express :( )xg ( ) spherehollowspheresolid ggxg += Substitute for and and simplify to obtain:spheresolidg spherehollowg ( ) ( ) ( ) ( )[ ] ( ) ( ) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − − += − += 2 2 12 3 0 2 2 1 3 2 1 3 4 0 2 3 3 4 0 2 2 1 spherehollow 2 spheresolid 8 11 3 4 Rxx R G Rx RG x RG Rx GM x GM xg πρ πρπρ 81 ••• A small diameter hole is drilled into the sphere of Problem 80 toward the center of the sphere to a depth of 2.0 m 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. Picture the Problem We can use conservation of energy to relate the work done by the gravitational field to the speed of the small object as it strikes the bottom of the hole. Because we’re given the mass of the sphere, we can find C by
- 17. Gravity 237 expressing the mass of the sphere in terms of C. We can then use the definition of the gravitational field to find the gravitational field of the sphere inside its surface. The work done by the field equals the negative of the change in the potential energy of the system as the small object falls in the hole. Use conservation of energy to relate the work done by the gravitational field to the speed of the small object as it strikes the bottom of the hole: 0if =Δ+− UKK or, because Ki = 0 and W = −ΔU, 2 2 1 mvW = ⇒ m W v 2 = (1) where v is the speed with which the object strikes the bottom of the hole and W is the work done by the gravitational field. Express the mass of a differential element of the sphere: ( )drrdVdm 2 4πρρ == Integrate to express the mass of the sphere in terms of C: ( ) CrdrCM ππ 2 m0.5 0 m504 == ∫ Solving for C yields: ( )π2 m50 M C = Substitute numerical values and evaluate C: ( ) 28 2 11 kg/m1037.6 m50 kg100.1 ×= × = π C Use its definition to express the gravitational field of the sphere at a distance from its center less than its radius: GC r drrC G r dr r C r G r drr G r GM m F g rrr π ππρπ 2 444 2 0 2 0 2 2 0 2 2 g ====== ∫∫∫ Express the work done on the small object by the gravitational force acting on it: ( )mgmgdrW m2 m3.0 m0.5 =−= ∫ Substitute in equation (1) and simplify to obtain: ( ) ( ) ( ) GC m GCm v π π m0.8 2m0.22 = =
- 18. Chapter 11238 Substitute numerical values and evaluate v: ( ) ( )( ) m/s0.1kg/m1037.6/kgmN10673.6m0.8 282211 =×⋅×= − πv 83 ••• Two identical spherical cavities are made in a lead sphere of radius R. The cavities each have a radius R/2. They touch the outside surface of the sphere and its center as in Figure 11-28. The mass of a solid uniform lead sphere of radius R is M. Find the force of attraction on a point particle of mass m located at a distance d from the center of the lead sphere. Picture the Problem The force of attraction of the small sphere of mass m to the lead sphere of mass M is the sum of the forces due to the solid sphere ( ) and the cavities ( ) of negative mass. SF r CF r Express the force of attraction: CS FFF rrr += (1) Use the law of gravity to express the force due to the solid sphere: iF ˆ 2S d GMm −= r Express the magnitude of the force acting on the small sphere due to one cavity: 2 2 C 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = R d GM'm F where M′ is the negative mass of a cavity. Relate the negative mass of a cavity to the mass of the sphere before hollowing: ( ) MR R VM' 8 13 3 4 8 1 3 3 4 2 −=−= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=−= πρ πρρ Letting θ be the angle between the x axis and the line joining the center of the small sphere to the center of either cavity, use the law of gravity to express the force due to the two cavities: iF ˆcos 4 8 2 2 2 C θ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = R d GMmr because, by symmetry, the y components add to zero. Use the figure to express cosθ : 4 cos 2 2 R d d + =θ
- 19. Gravity 239 Substitute for cosθ and simplify to obtain: i iF ˆ 4 4 ˆ 44 4 2/32 2 2 2 2 2 C ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = R d GMmd R d d R d GMmr Substitute in equation (1) and simplify: i iiF ˆ 4 41 ˆ 4 4 ˆ 2/32 2 3 2 2/32 2 2 ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + −−= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + +−= R d d d GMm R d GMmd d GMmr General Problems 89 •• A neutron star is a highly condensed remnant of a massive star in the last phase of its evolution. It is composed of neutrons (hence the name) because the star’s gravitational force causes electrons and protons to ″coalesce″ into the neutrons. Suppose at the end of its current phase, the Sun collapsed into a neutron star (it can’t in actuality because it does not have enough mass) of radius 12.0 km, without losing any mass in the process. (a) Calculate the ratio of the gravitational acceleration at the surface of the Sun following its collapse compared to its value at the surface of the Sun today. (b) Calculate the ratio of the escape speed from the surface of the neutron-Sun to its value today. Picture the Problem We can apply Newton’s second law and the law of gravity to an object of mass m at the surface of the Sun and the neutron-Sun to find the ratio of the gravitational accelerations at their surfaces. Similarly, we can express the ratio of the corresponding expressions for the escape speeds from the two suns to determine their ratio. (a) Express the gravitational force acting on an object of mass m at the surface of the Sun: 2 Sun Sun gg R mGM maF == Solving for ag yields: 2 Sun Sun g R GM a = (1)
- 20. Chapter 11240 The gravitational force acting on an object of mass m at the surface of a neutron-Sun is: 2 -Sunneutron -Sunneutron gg ' R mGM maF == Solving for yields:g'a 2 -Sunneutron -Sunneutron g' R GM a = (2) Divide equation (2) by equation (1) to obtain: 2 Sun Sun 2 Sun-neutron Sun-neutron 2 Sun Sun 2 Sun-neutron Sun-neutron g g' R M R M R GM R GM a a == Because :Sun-Sunneutron MM = 2 Sun-neutron 2 Sun g g' R R a a = Substitute numerical values and evaluate the ratio :' gg aa 9 2 3 8 g g 1036.3 m100.12 m1096.6' ×=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × = a a (b) The escape speed from the neutron-Sun is given by: -Sunneutron -Sunneutron e' R GM v = The escape speed from the Sun is given by: Sun Sun e R GM v = Dividing the first of these equations by the second and simplifying yields: -Sunneutron Sun e e' R R v v = Substitute numerical values and evaluate :' ee vv 241 m100.12 m1096.6' 3 8 e e = × × = v v 95 •• Uranus, the seventh planet in the Solar System, was first observed in 1781 by William Herschel. Its orbit was then analyzed in terms of Kepler’s Laws. By the 1840s, observations of Uranus clearly indicated that its true orbit was different from the Keplerian calculation by an amount that could not be accounted for by observational uncertainty. The conclusion was that there must be another influence other than the Sun and the known planets lying inside Uranus’s orbit. This influence was hypothesized to be due to an eighth planet, whose predicted orbit was described independently in 1845 by two astronomers: John Adams (no relation to the former president of the United States) and Urbain LeVerrier. In September of 1846, John Galle, searching in the sky at the place predicted by
- 21. Gravity 241 Adams and LeVerrier, made the first observation of Neptune. Uranus and Neptune are in orbit about the Sun with periods of 84.0 and 164.8 years, respectively. To see the effect that Neptune had on Uranus, determine the ratio of the gravitational force between Neptune and Uranus to that between Uranus and the Sun, when Neptune and Uranus are at their closest approach to one another (i.e. when aligned with the Sun). The masses of the Sun, Uranus, and Neptune are 333,000, 14.5 and 17.1 times that of Earth, respectively. Picture the Problem We can use the law of gravity and Kepler’s third law to express the ratio of the gravitational force between Neptune and Uranus to that between Uranus and the Sun, when Neptune and Uranus are at their closest approach to one another. The ratio of the gravitational force between Neptune and Uranus to that between Uranus and the Sun, when Neptune and Uranus are at their closest approach to one another is given by: ( ) ( )2 UNS 2 UN 2 U SU 2 UN UN S-Ug, U-Ng, rrM rM r MGM rr MGM F F − = − = (1) Applying Kepler’s third law to Uranus yields: 3 U 2 U CrT = (2) Applying Kepler’s third law to Neptune yields: 3 N 2 N CrT = (3) Divide equation (3) by equation (2) to obtain: 3 U 3 N 3 U 3 N 2 U 2 N r r Cr Cr T T == ⇒ 32 U N UN ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = T T rr Substitute for rN in equation (1) to obtain: 2 U 32 U N US 2 UN S-Ug, U-Ng, ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = r T T rM rM F F Simplifying this expression yields: 232 U N S N S-Ug, U-Ng, 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = T T M M F F
- 22. Chapter 11242 Because MN = 17.1ME and MS = 333,000ME: 232 U N5 232 U N E 5 E -SUg, U-Ng, 11033.3 1.17 11033.3 1.17 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = T T T T M M F F Substitute numerical values and evaluate S-Ug, U-Ng, F F : 4 232 5 -SUg, U-Ng, 102 1 y0.84 y8.164 1033.3 1.17 − ×≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = F F Because this ratio is so small, during the time at which Neptune is closest to Uranus, the force exerted on Uranus by Neptune is much less than the force exerted on Uranus by the Sun. 97 •• [SSM] Four identical planets are arranged in a square as shown in Figure 11-29. If the mass of each planet is M and the edge length of the square is a, what must be their speed if they are to orbit their common center under the influence of their mutual attraction? Picture the Problem Note that, due to the symmetrical arrangement of the planets, each experiences the same centripetal force. We can apply Newton’s second law to any of the four planets to relate its orbital speed to this net (centripetal) force acting on it. a a F F F 1 2 θ θ M M MM 1 Applying radialradial aF rr m=∑ to one of the planets gives: 21c cos2 FFF += θ
- 23. Gravity 243 Substituting for F1, F2, and θ and simplifying yields: ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += += +°= 2 1 2 22 12 2 45cos 2 2 2 2 2 2 2 2 2 2 2 c a GM a GM a GM a GM a GM F Because a Mv a Mv F 22 c 2 2 == ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += 2 1 2 2 2 22 a GM a Mv Solve for v to obtain: a GM a GM v 16.1 22 1 1 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += 103 ••• In this problem you are to find the gravitational potential energy of the thin rod in Example 11-8 and a point particle of mass m0 that is on the x axis at x = x0. (a) Show that the potential energy shared by an element of the rod of mass dm (shown in Figure 11-14) and the point particle of mass m0 located at Lx 2 1 0 ≥ is given by ( ) s s0 0 s0 0 dx xxL GMm xx dmGm dU − = − −= 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. Generalize your function U(x0) to any place on the x axis in the region x > L/2 by replacing x0 by a general coordinate x and write it as U(x). (c) Compute the force on m0 at a general point x using Fx = –dU/dx and compare your result with m0g, where g is the field at x0 calculated in Example 11-8. Picture the Problem Let U = 0 at x = ∞. The potential energy of an element of the stick dm and the point mass m0 is given by the definition of gravitational potential energy: rdmGmdU 0−= where r is the separation of dm and m0. (a) Express the potential energy of the masses m0 and dm: s0 0 xx dmGm dU − −= The mass dm is proportional to the size of the element :sdx sdxdm λ= where L M =λ . Substitute for dm and λ to express dU in terms of xs: ( )s0 s0 s0 s0 xxL dxGMm xx dxGm dU − −= − −= λ
- 24. Chapter 11244 (b) Integrate dU to find the total potential energy of the system: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + −= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= − −= ∫− 2 2 ln 2 ln 2 ln 0 00 00 0 2/ 2/ s0 s0 Lx Lx L GMm L x L x L GMm xx dx L GMm U L L (c) Because x0 is a general point along the x axis: ( ) ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − + =−= 2 1 2 1 00 0 0 0 L x L x L GMm dx dU xF Further simplification yields: ( ) 422 0 0 Lx Gmm xF − −= This answer and the answer given in Example 11-8 are the same.