p14 082

background image

82.

(a) Kepler’s law of periods is

T

2

=



4π

2

GM



r

3

.

Thus, with M = 6.0

× 10

30

kg and T = 300(86400) = 2.6

× 10

7

s, we obtain r = 1.9

× 10

11

m.

(b) That its orbit is circular suggests that its speed is constant, so

v =

2πr

T

= 4.6

× 10

4

m/s .


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