A satellite A revolves around a planet of radius R at a height R from its surface. Its time period of revolution is T . Another satellite B revolves around the same planet with a time period of 8T . What is the height of satellite B from the surface of the planet?

Step-by-step solution

According to Kepler's third law, the square of the time period is proportional to the cube of the orbital radius: T^2 r^3 , which gives T r^ 3/2 . Let r_A and r_B be the orbital radii of satellites A and B respectively. The orbital radius of satellite A is r_A = R + h_A = R + R = 2R . Taking the ratio of their time periods: T_B T_A = ( r_B r_A )^ 3/2 Given T_A = T and T_B = 8T : 8T T = ( r_B 2R )^ 3/2 8 = ( r_B 2R )^ 3/2 Raising both sides to the power of 2/3 : r_B 2R = (8)^ 2/3 = 4 r_B = 4 2R = 8R The height of sa