At the surface of earth, the acceleration due to gravity is g. If an object of mass m is raised from the surface of earth to a height equal to radius of earth (R ). The potential energy gained by the object is

Options

A. mgR
B. `(mgR^(2))/(2)`
C. `(+mgR)/(2)`
D. `(-mgR)/(2)`

Answer

C

Explanation

On surface of earth, the gravitational potential energy
`U_(2) = -(GMm)/(R )`
At a height R from surface of earth, the gravitational potential energy
`U_(2) = -(GMm)/((R+R)) = - (GMm)/(2R)`
`Delta U = U_(2) - U_(1) = -(GMm)/(2R) - (-(GMm)/(R ))`
`= (GMm)/(2R)`
`= (gR^(2)m)/(2R) " "(because g = (GM)/(R^(2)),(gR^(2) = GM))`
Therefore, `U_(2) - U_(1) = (MgR)/(2)`