A 6 kg satellite is
travelling in free flight along an elliptical orbit such that at A, where
r_{A} = 10 Mm, it has a speed V_{A} = 20 Mm/h. What is
the speed of the satillite when it reaches point B, where r_{B}
= 40 Mm. Also assume M_{e} = 5.976 (10^{24}) and also
that
G = 66.73 (10^{-12}) m^{3}/(kg-s^{2}).

1) 7981.6 m/s
2) 5243.5 m/s
3) 4231.5 m/s
4) 7234.5 m/s

Given that the velocity
of A is V_{A} = 20 Mm/h and
V_{A} = 40 Mm/h = 11111.1 m/s
Since V = -
(G Me m)/r
Applying the conservation of energy principle we have:
ST_{1} + S
V_{1} = ST_{2} + S
V_{2
1/2(60)(11111.1)2 - 66.73 (10)-12 (5.976)1024
(60)/107 = 1/2(60)VB2 - 66.73 (10)-12
(5.976)1024 (60)/40(10)6} _{VB2 } = 6.37 10-7 m^{2}/s^{2}
V^{B} = 7981.63 m/s

So the velocity of
the satillite at B V^{B} = 7981.6 m/s.