A satellite is moving with a constant speed $v$ in a circular orbit about the earth. An object of mass $m$…
A satellite is moving with a constant speed $v$ in a circular orbit about the earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is
$\frac{1}{2} m v^2$
$m v^2$
$\frac{3}{2} m v^2$
$2 m v^2$
Solution
In circular orbit of a satellite, potential energy
$
\begin{aligned}
& =-2 \times(\text { kinetic energy) } \\
& =-2 \times \frac{1}{2} m v^2=-m v^2
\end{aligned}
$
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 $U=-\frac{G M m}{r}$ and $K=\frac{G M m}{2 r}$ or $U=-2 K$
Here, $U$ is potential energy and $K$ is kinetic energy