Question

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

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

Answer 1

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

Total energy of the satellite $T=\frac{-GMm}{2r}$ where $r=R+h$

$\therefore$ $T=\frac{-GMm}{2(R+h)}$

Acceleration due to gravity at earth's surface $g_o=\frac{GM}{R^2}$ $⟹GM=g_o{R}^2$

$⟹$ $T=\frac{-m{g}_o{R}^2}{2(R+h)}$