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 (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 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:
ST₁ + S V₁ = ST₂ + 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 m²/s²
V^B = 7981.63 m/s

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