Explanation

(b) Let a satellite is revolving around earth with orbital velocity $v$. The gravitational potential energy of satellite is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U = - \frac{GM_{e}m}{R_{e}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
The kinetic energy of satellite is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ K = \frac{1}{2}\frac{GM_{e}m}{R_{e}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$\therefore\ $Total energy of satellite
$\ \ \ \ \ \ \ \ \ \ \ \ E = U + K$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{GM_{e}m}{R_{e}} + \frac{1}{2}\frac{GM_{e}m}{R_{e}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2}\frac{GM_{e}m}{R_{e}}$ …(iii)
But we know that necessary centripetal force to the satellite is provided by the gravitational force. $ie,$
$\frac{mv^{2}}{R_{e}} = \frac{GM_{e}m}{R_{e}^{2}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ mv^{2} = \frac{GM_{e}m}{R_{e}}\ \ \ \ \ \ \ \ \ \ \ \ \ldots(iv)$
Hence, from Eqs. (iii) and (iv), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = - \frac{1}{2}mv^{2}$