The lengths of seconds pendulums on the surface of the earth and at an altitude ' h ' from the surface of the earth are l_s and l_h respectively. The radius of the earth is

Step-by-step solution

The time period of a simple pendulum is given by T = 2 l g . For a seconds pendulum, the time period is constant ( T = 2 s) at both locations. Therefore, T_s = T_h 2 l_s g_s = 2 l_h g_h . This gives l_s l_h = g_s g_h . The acceleration due to gravity on the surface of the earth is g_s = GM R^2 and at an altitude h is g_h = GM (R+h)^2 . Substituting these values, we get: l_s l_h = GM R^2 GM (R+h)^2 = (R+h)^2 R^2 = (1 + h R )^2 Taking the square root on both sides: l_s l_h = 1 + h R h R = l_s l_h - 1 = l_s - l_h l_h