A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be ( g = acceleration due to gravity on earth's surface) (R - radius of earth)

Step-by-step solution

Initial potential energy of the body on the surface of the earth is U_i = - GMm R Final potential energy of the body at height h = 2R is U_f = - GMm R+h = - GMm R+2R = - GMm 3R Increase in potential energy is U = U_f - U_i = - GMm 3R - (- GMm R ) = 2GMm 3R Using the relation g = GM R^2 , we have GM = gR^2 Substituting this in the expression for U , we get U = 2(gR^2)m 3R = 2 3 mgR Answer: 2 3 mgR