Question

a satellite going around earth in an elliotic orbit has a speed of 10 km per second at the perigee which is at a distance if 227 km from the surface of the earth. calculate the apogee distance and its speed at that point

Answer 1

For a satellite or planet in an elliptical orbit due to gravitational attraction, the angular momentum is constant at all points. This is because there is no external torque. Force is radial.
    So  m v1 r1 = m v2 r2     =>  v1 r1 = v2 r2

Also the total energy is constant.
    1/2 m v1² - GMe m / r1 = 1/2 m v2² - GMe m / r2

Solving these equations we get:
     v1² = 2 G Me r2 / [ r1 (r1+r2) ] 
     v2² = 2 G Me r1 / [ r2 (r1+r2) ]
 
For Earth, GMe = 4 *10¹⁴ units,    Re = 6,375,000 m
   Given   r1 = Re+227 km = 6, 602,000 m
               v1 = 10,000 m/s
     (10⁴)² = 2 * 4*10¹⁴ * r2 / [ 6,602,000 (6,602,000+ r2) ]

So at Apogee:
      r2 = 24, 575.68 km

Altitude of the satellite = r2 - Re = 18, 200.68 km  at Apogee.

Velocity v2 = v1 * r1 / r2 =  10,000 m/s * 6,602 km/ 24,575.68 km
                   = 2,686.4 m/s