Prove that the orbital velocity of a satellite at a height h from the earth's surface is equal to R e g ( R e + h ) , where R e is the radius of earth and g is the acceleration due to gravity. Find out the time period of revolution of the satellite.

If the mass of earth is M e . The gravitational force exerted by the earth on the satellite of mass m is F = G M e m r 2 where G is the gravitational constant. The gravitational force provides the required centripetal force, we have G M e m r 2 = m v o 2 r GM e = v o 2 r v o 2 = GM e r v o = GM e r . . . . . ( 1 ) Its R e be the radius of earth and h the height of the satellite above the earth's surface. So the distance of the satellite from the center of the earth will be r = R e + h Substituting the value of r in above equation, we have v o = G M e R e + h . . . . . . ( 2 ) If the acceleration due to gravity on the earth's surface g , then g = G M e R e 2 or GM e = gR e 2 Substituting the value of GM e in equation (1) we get v o = R e g R e + h . It is the evident from the above equations that the speed of satellite depend only upon its height above the earth. The time period of the revolving satellite is given by: T = 2 πr V = 2 πr G M r = 2 πr 3 2 G M = 2 π ( R e + h ) 3 2 G M using equation 1 here the letters have usual meaning