Physics - Gravitation Question with Solution | TestHub

In circular orbit of a satellite, potential energy $$\begin{array}{rl}& =-2\times (\text{ kinetic energy) }\\ & =-2\times \frac{1}{2}m{v}^2=-m{v}^2\end{array}$$ Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be $+m{v}^2$. $\therefore$ Correct answer is (b). Analysis of Question (i) Question is moderately difficult. (ii) In circular orbit of satellite $U=-\frac{GMm}{r}$ and $K=\frac{GMm}{2r}$ or $U=-2K$ Here, $U$ is potential energy and $K$ is kinetic energy.