Q. The binding energy of a satellite of mass $m$ in a orbit of radius $r$ is $(R = radius\ of\ earth,\ g = acceleration\ due\ to\ gravity)$
A) $\frac{{mgR}^{2}}{r}$
B) $\frac{{mgR}^{2}}{2r}$
C) $- \frac{{mgR}^{2}}{r}$
D) $- \frac{{mgR}^{2}}{2r}$
Explanation
(b) The energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satellite. It is equal to negative of total mechanical energy of satellite in its orbit.
$Thus,\ binding\ \ energy = - E = \frac{GMm}{2r}$
$but,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ GM = {gR}^{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ BE = \frac{gmR^{2}}{2r}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.