Thomas Calculus 13th [Solutions]

13. Assuming Earth has a circular orbit with radius
6
2
150 10 km
365.256 days
9 2
2.24 10 km /sec.
Section 13.6 Velocity and Acceleration in Polar Coordinates 961
6
150 10 km, the rate of change of area is
14. Solving Keplers third law for T we find
T
4
GM
2
10
3
77.8 10 m 11.857 years.
15. Solving Keplers third law for the mass M of the body around which Io is orbiting we find
M
10
3
3
2 a 0.042 10 m
2 27
2 2
4 4 1.876 10 kg.
T G (1.769 days) G
16. To solve this we need a value for the mass of Earth, which is approximately
Keplers third law for the orbital radius we get
24
M 5.972 10 kg. Solving
1/3 1/3 1/3
1/3
2 2 2 6 24 5
a 4 T GM 4 2.36055 10 sec G 5.972 10 kg 3.831 10 km.
Since Earths radius is about 6371, the orbit of the moon is about 383,143 6371 376.772 km from
the surface, assuming a circular orbit for the moon.
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