gi0s3cierMar
contestada

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.. .

Respuesta :

meerkat18
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

Otras preguntas