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75. (a) Using Kepler’s law of periods, we obtain
2
34
4
2.15
10 s .
Tr
GM
π
§·
==
×
¨¸
©¹
(b) The speed is constant (before she fires the thrusters), so
v
o
= 2
π
r/T
= 1.23
×
10
4
m/s.
(c) A two percent reduction in the previous value gives
v
= 0.98
v
o
= 1.20
×
10
4
m/s.
(d) The kinetic energy is
K
= ½
mv
2
= 2.17
×
10
11
J.
(e) The potential energy is
U
=
−
GmM/r
=
−
4.53
×
10
11
J.
(f) Adding these two results gives
E
=
K
+
U
=
−
2.35
×
10
11
J.
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 Spring '08
 Any
 Physics

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