A particle is projected vertically upwards from the surface of the Earth such that it reaches a maximum height equal to three times the radius of the Earth ( 3R ). If V_e is the escape velocity on the surface of the Earth, the initial velocity of projection of the particle is

Step-by-step solution

Let the mass of the Earth be M , radius be R , and mass of the particle be m . By conservation of mechanical energy between the surface of the Earth and the maximum height: K_i + U_i = K_f + U_f At the surface, the distance from the center of the Earth is R . At the maximum height h = 3R , the distance from the center is r = R + 3R = 4R . 1 2 mv^2 - GMm R = 0 - GMm 4R 1 2 mv^2 = GMm R - GMm 4R = 3GMm 4R We know that the escape velocity V_e = 2GM R , which gives GM R = V_e^2 2 . Substituting this into the energy equ