An earth satelites S has orbit radius which is 4 times that of communication satellite C . The period of revolution of S will be

Step-by-step solution

Larger the distance of planet from the sun, larger will be its period of revolution around the sun. Kepler's third low planetary motion the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit T^2 R^3 T_s T_c = ( R_s R_c )^ 3 / 2 given R_s=4 R_c aligned & T_s T_c = ( 4 R_c R_c )^ 3 / 2 =8 & for T_c=1 day & T_s=8 days. aligned