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
a ) 0.50
Let Rp be the radius of the orbit at perigee ; e , be the eccentricity of the orbit ; M e the mass of the earth ; vp , 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