A meteor of mass m is initially at a distance of 4R from the centre of the Earth, where R is the radius of the Earth. As it falls towards the Earth, what is the loss in its gravitational potential energy when it reaches the Earth's surface? (Assume g is the acceleration due to gravity on the surface of the Earth)

Step-by-step solution

Initial gravitational potential energy of the meteor is U_i = - GMm 4R Final gravitational potential energy of the meteor on the Earth's surface is U_f = - GMm R Loss in gravitational potential energy is U = U_i - U_f U = - GMm 4R - (- GMm R ) = 3GMm 4R Acceleration due to gravity on the Earth's surface is g = GM R^2 GM = gR^2 Substituting the value of GM , we get U = 3(gR^2)m 4R = 3 4 mgR Answer: 3 4 mgR