A body of mass m is released from rest from a height h = 2R above the surface of the Earth, where R is the radius of the Earth and M is its mass. The kinetic energy of the body just before it strikes the Earth's surface is:

Step-by-step solution

According to the principle of conservation of mechanical energy, the total energy of the body remains constant. Initial kinetic energy, K_i = 0 . Initial distance from the center of the Earth, r₁ = R + 2R = 3R . Initial potential energy, U_i = - GMm 3R . Final distance from the center of the Earth (at the surface), r₂ = R . Final potential energy, U_f = - GMm R . Let the final kinetic energy be K_f . Applying conservation of energy: K_i + U_i = K_f + U_f 0 - GMm 3R = K_f - GMm R K_f = GMm R - GMm 3R = 2GMm 3R . Ans