A 6 kg satellite is travelling in free flight along an elliptical orbit such that at A, where rA = 10 Mm, it has a speed VA = 20 Mm/h. What is the speed of the satillite when it reaches point B, where rB = 40 Mm. Also assume Me = 5.976 (1024) and also that
G = 66.73 (10-12) m3/(kg-s2)
.

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 VA = 20 Mm/h and
VA = 40 Mm/h = 11111.1 m/s
Since
V = - (G Me m)/r
Applying the conservation of energy principle we have:
ST1 + S V1 = ST2 + S V2
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 m2/s2
VB = 7981.63 m/s

So the velocity of the satillite at B VB = 7981.6 m/s.