The escape velocity of a projectile on the earth's surface is 11.2 ~km / s . A body is projected out with thrice this speed. The speed of the body far away from the earth will be

Step-by-step solution

Let v_e be the escape velocity on the earth's surface, where v_e = 11.2 km/s . The energy conservation principle states that the total energy at the surface equals the total energy at infinity. The initial velocity of the body is v = 3v_e . The total energy at the surface is E_i = 1 2 mv^2 - GMm R = 1 2 m(3v_e)^2 - GMm R . Since v_e = 2GM R , we have GM R = v_e^2 2 . Substituting this into the energy equation: E_i = 1 2 m(9v_e^2) - m ( v_e^2 2 ) = 9 2 mv_e^2 - 1 2 mv_e^2 = 4mv_e^2 . At a large distance from the ear