Download Physics 140 HOMEWORK Chapter 13B

Physics 140
HOMEWORK Chapter 13B
My question about Pr 13-4 on set 13A: The Sun exerts more than twice the gravitational force on the
Moon than Earth does, so why doesn’t the Sun snatch the Moon away from us? The Sun exerts the
same gravitational acceleration (F/m) on the Earth that it does on the Moon. So the Sun pulls equally
on both objects, and the pair orbits the Sun.
Q5. Figure 13-24 shows three situations involving a point particle P with mass m and a spherical shell
with a uniformly distributed mass M . The radii of the shells are given. Rank the situations according
to the magnitude of the gravitational force on particle P due to the shell, greatest first.
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b = c > a = 0. As long as P is outside, the shell acts as though it were a point mass of equal mass at
the center. If P is inside a spherical shell, the shell exerts zero force on it.
Q8. Figure 13-26 gives the gravitational acceleration ag for four planets as a function of the radial
distance r from the center of the planet, starting at the surface of the planet (at radius R1 , R2 , R3 ,
or R4 ). Plots 1 and 2 coincide for r ≥ R2 ; plots 3 and 4 coincide for r ≥ R4 . Rank the four planets
according to (a) mass and (b) mass per unit volume, greatest first.
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(a) M1 = M2 > M3 = M4 . The value of ag at a constant radius is directly proportional to the mass
of the planet.
(b) ρ1 > ρ2 > ρ3 > ρ4 . Comparing Planet 1 and Planet 2, both have the same mass, and Planet 1 is
clearly smaller, so has the higher density. Planets 2 and 3 have the same size, but Planet 2 has more
mass, so it’s denser than Planet 3. Planet 4 is ranked the same way Planets 1 and 2 were.
P45. The Martian satellite Phobos travels in an approximately circular orbit of radius 9.4 × 106 m
with a period of 7 h 39 min. Calculate the mass of Mars from this information.
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We have Kepler’s Third Law (modern statement), Eq. 13-34:
T 2 = 4πr3 /(GM ), where M is the mass of the planet. Solve for M :
M = 4π 2 r3 /(T 2 ) = 39.478·(9.4×106 m)3 /[(6.67×10−11 N m2 /kg2 )(2.754×105 s)2 ] = 6.48 × 1023 kg.
Note that it is not necessary to know the mass of Phobos.
P46. The first known collision between space debris and a functioning satellite occurred in 1996: At an
altitude of 700 km, a year-old French spy satellite was hit by a piece of an Ariane rocket. A stabilizing
boom on the satellite was demolished, and the satellite was sent spinning out of control. Just before
the collision and in kilometers per hour, what was the speed of the rocket piece relative to the satellite
if both were in circular orbits and the collision was (a) head-on and (b) along perpendicular paths?
[and added by Skelton, (c) crossing at 1◦ ].
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We first needp
the speed for
p circular orbit, which was derived in class.
vcirc orbit = GM/r = (6.67 × 10−11 5.98 × 1024 /(7.1 × 106 m) = 7495 m/s.
(a) vrel = 2 vcirc orbit = 15.0 km/s = 54, 000 km/h.
(b) ~vrel =√vcirc orbit ı̂ + vcirc orbit ̂
vrel = 2 vcirc orbit = 10.6 km/s = 37, 000 km/h.
(c) vrel = 2 vcirc orbit sin 0.5◦ = 131 m/s = 470 km/h. This is about 300 mph. I added this part to show
that collisions in space are very seldom gentle.
P50. An orbiting satellite stays over a certain spot on the equator of (rotating) Earth. What is the
altitude of the orbit (called a geosynchronous orbit)?
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Kepler’s Third Law (modern statement), Eq. 13-34, again:
T 2 = 4π 2 r3 /(GM ), where M is the mass of the planet. Solve for r:
h
i1/3
r = GM T 2 /(4π 2 )
= 42.25 × 106 m. Since this is measured from the center of the Earth, the
altitude is:
h = r − REarth = 42.25 × 106 m − 6.4 × 106 m = 35.75 × 106 m.
The actual altitude is slightly lower since the sidereal day (86164.1 s) should be used instead of 24
hours (86400 s).
P53. A 20 kg satellite has a circular orbit with a period of 2.4 h and a radius of 8.0 × 106 m around a
planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet
is 8.0 m/s2 , what is the radius of the planet?
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We can combine Kepler’s Third Law (Eq 13-34) with the equation for g = GM/R2 to get:
p
p
R = 2π r3 /T 2 g = 6.28 · 8.6 × 106 m)3 /[(8640 s)2 )(8.0 m/s2 )] = 5.82 × 106 m.
P71. Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed of material
that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a
homogeneous thin ring of mass M and outer radius R (Fig. 13-51). (a) What gravitational attraction
does it exert on a particle of mass m located on the ring’s central axis a distance x from the ring center?
(b) Suppose the particle falls from rest as a result of the attraction of the ring of matter. What is the
speed with which it passes through the center of the ring?
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You could work part (a) with the force equation by taking only the component along the axis, but part
(b) would be more difficult this way. Taking Skelton’s hint:
Ug = −GM m/(R2 + x2 )1/2 . Note that all of the mass in the ring is at the same distance. The
potential is a scalar (not a vector) so we don’t have to worry about components.
To get F , use the general formula that relates the force to the potential:
F = −dU/dx = −GMmx/(R2 + x2 )3/2 .
(b) Use Conservation of Energy, with the above formula for Ug :
Kf + Ug,f = K0 + Ug,0 , or
(1/2)mvf2 + (−GM m/R) = 0 + (−GM m/(R2 + x2 )( 1/2). Solve for vf :
p
vf = (2GM)[1/R − 1/(R2 + x2 )(1/2) ]
P80 (optional). The fastest possible rate of rotation of a planet is that for which the gravitational
force on material at the equator just barely provides the centripetal force needed for the rotation.
(Why?) (a) Show that the corresponding shortest period of rotation is
T =
r
3π
Gρ
where ρ is the uniform density (mass per unit volume) of the spherical planet. (b) Calculate the
rotation period assuming a density of 3.0 g/cm3 , typical of many planets, satellites, and asteroids. No
astronomical object has ever been found to be spinning with a period shorter than that determined by
this analysis.
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Why: if the gravitational were less than the required (centripetal) force, a rock on the surface would
go into orbit. And it doesn’t matter how big the rock is, so the entire surface would be in orbit.
(a) FBD for rock on surface: mg down; n up. ~a = v 2 /R toward center. As the speed limit is approached,
n goes to zero, so:
mg = mv 2 /R. Use the equations for g, v, and M :
g = GM/R2 ;
v = 2πR/T ;
M = (4/3)πR3 ρ. (Volume of a sphere and density.))
Solve for T with a little algebra to get the stated result.
(b) Remember
to convert 3 g/cm3 to 3000 kg/m3 :
p
T = 9.42/[(6.67 × 10−11 N m2 /kg2 )(3000 kg/m3 )] = 6860 m/s.
This also the orbital speed for a satellite right above the surface (provided the planet is perfectly
smooth, with no atmosphere). Interesting that the size doesn’t matter.
P87 (optional). (a) If the legendary apple of Newton could be released from rest at a height of 2 m
from the surface of a neutron star with a mass 1.5 times that of our Sun and a radius of 20 km, what
would be the apple’s speed when it reached the surface of the star? (b) If the apple could rest on the
surface of the star, what would be the approximate difference between the gravitational acceleration at
the top and at the bottom of the apple? (Choose a reasonable size for an apple; the answer indicates
that an apple would never survive near a neutron star.)
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Let’s calculate g for a neutron star. (Suppress units)
g = GM/r2 = (6.67 × 10−11 )(1.5 · 1.99 × 1030 )/(2 × 104 )2 = 4.98 × 1011 m/s2 .
v 2 − v02 = 2a∆y ⇒
p
√
v = 2a∆y = 2 · (−4.98 × 1011 N m2 /kg2 )(−2 m) = 1.41 × 106 m/s.
(b) Take the diameter of an apple as 0.1 m. Then, by direct calculation,
gsurface = GM/r2 = 4.9774875 × 1011 m/s2 .
gsurface+0.1 m = 4.9774377 × 1011 m/s2 .
A difference of 5 × 106 m/s2 , or 5 × 105 gEarth .
That is to say, the force separating the top from the bottom is 5 × 105 times the weight of the apple
on Earth, or about 50 tons; this tidal force, or tidal stress stretches the apple while it falls. It gets
squished 5000X harder than that when it hits!
Alternative way to get ∆g is via a derivative:
∆g = (dg/dr)∆r = (−2GM/r3 )(∆r) = (−4.977 × 107 )(−0.1) = 5 × 106 m/s2 .