Physics - Gravitation Question with Solution | TestHub

Kinetic energy of the satellite $\mathrm{KE}=\frac{1}{2}m{v}_0^2$ ...(1) where $v_0=\sqrt{\frac{GM}{R}}$ Now putting the value of $v_0$ is eq. (1), we get $\mathrm{KE}=\frac{1}{2}m{\left(\sqrt{\frac{GM}{R}}\right)}^2=\frac{1}{2}\frac{mGM}{R}$ Hence, $\mathrm{KE}\propto \frac{1}{R}$