Question

A satellite is moving with a constant speed $v$ in circular orbit around the earth. An object of mass ${}^{\prime }{m}^{\prime }$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is

• A. $\frac{3}{2}m{v}^2$
• B. $m{v}^2$
• C. $2m{v}^2$
• D. $\frac{1}{2}m{v}^2$

Answer 1

The correct option is B $m{v}^2$

At height $r$ from center of earth. orbital velocity
$=\sqrt{\frac{GM}{r}}$
$\therefore$ By energy conservation
$KE$ of ${}^{\prime }{m}^{\prime }+(-\frac{GMm}{r})=0+0$
(At infinity, $PE=KE=0)$
$\Rightarrow KE$ of ${}^{\prime }{m}^{\prime }=\frac{GMm}{r}={\left(\sqrt{\frac{GM}{r}}\right)}^{2}m=m{v}^2$.