A particle is launched vertically upward from the surface of the earth and reaches a maximum height of 6400 km . Determine the initial speed of the particle.

Step-by-step solution

By the principle of conservation of mechanical energy: 1 2 mv^2 - GMm R = - GMm R+h Given that the maximum height h = 6400 km , which is equal to the radius of the earth R . 1 2 mv^2 - GMm R = - GMm 2R 1 2 mv^2 = GMm 2R v = GM R = gR We know that the escape velocity from the surface of the earth is v_e = 2gR 11.2 km/s . Therefore, v = v_e 2 = 11.2 1.414 7.9 km/s