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61. (a) We use the law of periods:
T
2
= (4
π
2
/
GM
)
r
3
, where
M
is the mass of the Sun (1.99
×
10
30
kg) and
r
is the radius of the orbit. The radius of the orbit is twice the radius of
Earth’s orbit:
r
= 2
r
e
= 2(150
×
10
9
m) = 300
×
10
9
m. Thus,
23
2
9
3
7
11
3
2
30
4
4
(300
10 m)
8.96
10 s.
(6.67
10
m /s kg)(1.99
10 kg)
r
T
GM
ππ
−
×
==
=
×
×⋅
×
Dividing by (365 d/y) (24 h/d) (60 min/h) (60 s/min), we obtain
T
= 2.8 y.
(b) The kinetic energy of any asteroid or planet in a circular orbit of radius
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Mass

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