The gravitational pull of the moon is ( 1 6 )^ th of the earth and mass of moon is ( 1 8 )^ th of the earth. This implies that the

Step-by-step solution

The acceleration due to gravity at a celestial body's surface is governed by g = GM R^2 , where M is mass and R is radius. Given that lunar gravity is one-sixth of Earth's, g_m = 1 6 g_e , and lunar mass is one-eighth of Earth's, M_m = 1 8 M_e . Substituting into the gravity formula yields GM_m R_m^2 = 1 6 GM_e R_e^2 . After canceling G and M_e using the mass relation, 1 8R_m^2 = 1 6R_e^2 . Cross-multiplying gives 6R_e^2 = 8R_m^2 , simplifying to 3R_e^2 = 4R_m^2 . Solving for the radii, R_e = 4 3 R_m , confirming t