220
15. Newton’s law of gravity depends on the inverse
square of the distance, so if the two planets’ masses
had been equal, then the factor of 0.83/0.059=14 in
distance would have caused the force on planet c to
be 14
2
=2.0x10
2
times weaker. However, planet c’s
mass is 3.0 times greater, so the force on it is only
smaller by a factor of 2.0x10
2
/3.0=65.
16. The reasoning is reminiscent of section 10.2.
From Newton’s second law we have F=ma=mv
2
/r =
m(2
p
r/T)
2
/r = 4
p
2
mr/T
2
,and Newton’s law of gravity
gives F=GMm/r
2
, where M is the mass of the
earth.Setting these expressions equal to each other,
we have
4
p
2
mr/T
2
= GMm/r
2
,
which gives
r=
GMT
2
4
p
2
3
=4.22x10
4
km .
This is the distance from the center of the earth, so to
find the altitude, we need to subtract the radius of the
earth. The altitude is 3.58x10
4
km.
17. Any fractional change in r results in double that
amount of fractional change in 1/r
2
. For example,
raising r by 1% causes 1/r
2
to go down by very nearly
2%. The fractional change in 1/r
2
is actually
2
×
(
1
/
27
)
cm
3.84
×
10
5
km
×
1km
10
5
cm
= 2
×
10
–12
19. (a) The asteroid’s mass depends on the cube of
its radius, and for a given mass the surface gravity
depends on r
–2
. The result is that surface gravity is
directly proportional to radius. Half the gravity means
half the radius, or one eighth the mass. (b) To agree
with a, Earth’s mass would have to be 1/8 Jupiter’s.
We assumed spherical shapes and equal density.
Both planets are at least roughly spherical, so the
only way out of the contradiction is if Jupiter’s density
is significantly less than Earth’s.