Physics - Gravitation Question with Solution | TestHub

The orbital radius of the satellite is $r=R+h$.

Given $h=3R$, we have $r=R+3R=4R$.

The orbital velocity of the satellite is $v=\sqrt{\frac{GM}{r}}$.

The kinetic energy of the satellite is $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$.

Using the relation $g=\frac{GM}{R^2}$, we get $GM=g{R}^2$.

Substituting $GM=g{R}^2$ and $r=4R$ into the expression for kinetic energy:

$K=\frac{g{R}^{2}m}{2(4R)}=\frac{mgR}{8}$

Answer: $\frac{mgR}{8}$