Question

A satellite of mass $m$ is orbiting the earth (of radius $R$) at a height $h$ from its surface. The total energy of the satellite in terms of $g_0$, the value of acceleration due to gravity at the earth’s surface, is -

• A. $\frac{mg_0 R^2}{2(R - h)}$
• B. $-\frac{mg_0 R^2}{2(R + h)}$
• C. $\frac{2mg_0 R^2}{2(R + h)}$
• D. $- \frac{2 mg_0 R^2}{(R + h)}$

Answer 1

Answer: (B)

$T.E = - \frac{GMm}{2r} $
$\therefore g_{0} = g_{surface} = \frac{GM}{R^{2}} $
$GM = g_{0}R^{2} $
$ T.E = - \frac{g_{0}R^{2}m}{2\left(R+h\right)} $