A satellite is traveling in a free - flight elliptical orbit  around the earth. Its perigee is 622 Km above the  surface of earth and  its  velocity  at  this  point  is  9245 m /s . Find the eccentricity of its orbital path.( The universal gravitational constant ,
G = 66.73 ( 10 ^-12 ) , the mass of the earth is
5.976 ( 10 ²⁴ ) Kg and the radius of the Earth is     6378 Km . )

 a ) 0.50
 b ) 0.67
 c ) 0.43
 d ) 0.77

Let R_p be the radius of the orbit at perigee ; e , be the eccentricity of the orbit ; M _e the mass of the earth ; v_p , be its velocity when it is at its perigee ; C , the reciprocal of the distance between the focus and the directrix of the orbital path and G , the universal gravitational constant

   v _p = 9245 m / s
   R _p = radius of earth + height above the surface .
        = 6378 Km + 622 Km = 7 ( 10 ⁶ ) m
Then we have ,


Hence the eccentricity of the elliptical orbit is 0.50