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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
$\frac{{2m{g_0}{R^2}}}{{R + h}}$
$-$$\;\frac{{2m{g_0}{R^2}}}{{R + h}}$
$\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}$
$-$$\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}$
Solution
Total energy of satellite at height $h$ from the earth surface,
$E=PE+KE$
$ = – \frac{{GMm}}{{\left( {R + h} \right)}} + \frac{1}{2}m{v^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)$
Also, $\frac{{m{v^2}}}{{\left( {R + h} \right)}} = \frac{{GMm}}{{\left( {R + {h^2}} \right)}}$
or, ${v^2} = \frac{{GM}}{{R + h}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)$
From eqns, $(i)$ and $(ii)$,
$E = – \frac{{GMm}}{{\left( {R + h} \right)}} + \frac{1}{2}\frac{{GMm}}{{\left( {R + h} \right)}} = – \frac{1}{2}\frac{{GMm}}{{\left( {R + h} \right)}}$
$ = – \frac{1}{2}\frac{{GM}}{{{R^2}}} \times \frac{{m{R^2}}}{{\left( {R + h} \right)}}$
$ = – \frac{{m{g_0}{R^2}}}{{2\left( {R + h} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\left( {{g_0} = \frac{{GM}}{{{R^2}}}} \right)$
