A satellite of mass $m$ is orbiting the earth (of radius $R$) at a height $h$ from its surface. The

Question

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

Professor PC Thomas Classes · Answer

(a) : The orbital velocity $v$ of a satellite at height $h$ is given by $\frac{mv^2}{(R+h)} = \frac{GMm}{(R+h)^2}$, which simplifies to $v^2 = \frac{GM}{R+h}$. The kinetic energy is $K = \frac{1}{2}mv^2 = \frac{GMm}{2(R+h)}$. We know that acceleration due to gravity at the surface is $g_0 = \frac{GM}{R^2}$, which means $GM = g_0 R^2$. Substituting this into the kinetic energy expression, we get $K = \frac{m(g_0 R^2)}{2(R+h)} = \frac{m g_0 R^2}{2(R+h)}$.