Consider a spacecraft in an elliptical orbit around the earth. at the low point, or perigee, of its orbit, it is 500 above the earth's surface; at the high point, or apogee, it is 4000 above the earth's surface.. using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee..

We know that semimajor axis is a = (1/2)[(Re + hp)+(Re +ha)]
                                                  = 9.13*106 m
 Re = radius of earth= 6380 km = 6380000 m
 hp = height of perigee = 500 km = 500000 m
ha = height of apogee = 5000 km = 5000000 m
 M = mass of the earth = 5.98*1024 kg
 G = 6.67*10-11 Nm2/kg2
we know that time period of revolution of the spacecraft is
T= 2πa3/2/√GM
 Then T = 2π*(9.13*106 )3/2/√6.67*10-11 *5.98*1024 = 8.67*103 sec

mvprp =mvara
 vp = spacecraft's speed at perigee
va=spacecraft's speed at apogee

Then vp /va =ra/rp = (Re+ha)/(Re +hp) = 6380+5000/6380+500
 =1.654

Is this a question of an answer