An asteroid, initially at rest, falls towards a planet from a distance of 4R from the center of the planet, where R is the radius of the planet. If v_e is the escape velocity from the surface of the planet, what will be the speed of the asteroid just before it strikes the planet's surface? (Assume the planet is a uniform sphere and neglect any atmospheric friction)

Step-by-step solution

Escape velocity from the surface of the planet is given by v_e = 2GM R . By conservation of mechanical energy between the initial position and the surface of the planet: K_i + U_i = K_f + U_f 0 - GMm 4R = 1 2 mv^2 - GMm R 1 2 mv^2 = GMm R - GMm 4R 1 2 mv^2 = 3GMm 4R v^2 = 3GM 2R Since GM R = v_e^2 2 , substituting this into the equation for v^2 gives: v^2 = 3 2 ( v_e^2 2 ) = 3 4 v_e^2 v = 3 2 v_e Answer: 3 2 v_e