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

Orbital velocity `=v_(0) = sqrt(( gR^(2))/( R + h )) `
Given that `v_(0) = (1)/( 2) v_(e )` where `v_(e )` is the escape velocity
But `v_(e ) = sqrt( 2g R )`
`(1)/(2) v_(e ) = sqrt(( gR^(2))/( R + h )) = ( 1)/( 2) sqrt( 2 gR )`
`(gR^(2))/( R + h ) = ( 2 gR)/( 4)`
Solving h =R, the radius of the earth.
(ii) Decrease in P.E. of the satellite, when it falls from a height `( R + h )` to R ( the surface of the earth) is coverted into K.E. of the satellite.
`(1)/(2) mv^(2) = - ( GMm)/( R ) + ( GMm)/( 2R) = - ( GMm)/( 2R) = - ( gR^(2) m)/( 2R)`
`v = sqrt( gR)`