Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.5 m.

Sol. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is also zero. So (10 kg)v base 1 = (20 kg) v base 2 Or v base 1 = v base 2 …(1) Since P.E. is conserved Initial P.E. = -6.67 * 10^-11 * 10 *20/1 = -13.34*10^-19 J When separation is 0.5 m, -13.34 * 10^-9 + 0 = -13.34 * 10^-9/(1/2) + (1/2) * 10 v base^2 + (1/2) * 20 v base 2^2 …(2) ⇒ - 13.34 * 10^-19 = -26.68 * 10^-9 + 5v base 1^2 + 10 v base 2^2 ⇒ - 13.34 * 10^-19 = -26.68 * 10^-9 + 30 v base 2^2 ⇒ v base 2^2 = 13.34 * 10^-9/30 = 4.44 * 10^-10 ⇒ v base 2 = 2.1 * 10^-15 m/s. So, v base 1 = 4.2 × 10^–5 m/s.