Question

A 400 kg satellite revolves in a circular orbit of radius $2R_e$ about the earth; calculate the kinetic energy, potential energy and total energy of the satellite ?
Given : $R_e=6.4\times {10}^{6}m$

Answer 1

$M_1=6\times {10}^{24}kg{M}_2=400kg{R}_e=6.4\times {10}^{6}mG=6.67\times {10}^{-4}{Nm}^2/kg\frac{G{M}_1{M}_2}{2\left(2R_e\right)}=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 400}{2\times 12.8\times {10}^6}=6.25\times {10}^{9}J\therefore K.E=6.25\times {10}^{9}JP.E=-2KE=-12.5\times {10}^{9}J\therefore T.E=P.E+K.E=6.25\times {10}^{9}J$