Q. 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
A) $\frac{1}{2}{mv}^{2}$
B) ${mv}^{2}$
C) $\frac{3}{2}{mv}^{2}$
D) $2{mv}^{2}$
Explanation
(b) In circular orbit of a satellite of potential energy
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 2 \times (kinetic\ energy)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 2 \times \frac{1}{2}m^{v} = - {mv}^{2}$
Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore , its kinetic energy should be $+ {mv}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.