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. $\frac{2m{g}_0{R}^2}{R+h}$
• B. $-\frac{2m{g}_0{R}^2}{R+h}$
• C. $\frac{m{g}_0{R}^2}{2(R+h)}$
• D. $-\frac{m{g}_0{R}^2}{2(R+h)}$

Answer 1

The correct option is D $-\frac{m{g}_0{R}^2}{2(R+h)}$

We know that the total energy of a satellite is given by $TE=-\frac{1}{2}\frac{GMm}{(R+h)}.$ Here the symbols have their usual meanings

We also know that the acceleration due to gravity on the earth-surface is $g_{{}_0}$ given by $g_{{}_0}=\frac{GM}{R^2}$

$\therefore GM=g_{{}_0}{R}^2$

$TE=-\frac{1}{2}\frac{GMm}{(R+h)}$

$\therefore TE=-\frac{m{g}_{{}_0}{R}^2}{2(R+h)}$