An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the surface of earth. (Radius of earth `= 6400 km`)
(a) Dentermine the height of the satellite above the earth's surface.
(b) If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, find the speed with which it hits the surface of earth.

Explanation

(a) Let height above the earth's surface = h then
`v_("orbital")= sqrt((GM)/(R+h))=1/2 v_(e)=1/2 sqrt((2GM)/R)=sqrt((GM)/(2R)) implies R+h=2R implies h=R`
(b) If the satellite is stopped suddenly then it total energy `E_(1)= - (GMm)/(2R)`
Let its speed be v when it hits the earth's surface then its total energy on earth surface `E_(2)=- (GMm)/R +1/2 mv^(2)`
conservation law for mechanical energy yields `E_(1)=E_(2)implies (-GMm)/(2R) = - (GMm)/R +1/2 mv^(2) implies v=sqrt((GM)/R)`