Question

A satellite is moving with a constant speed $v$ in circular orbit around 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 ejection, the kinetic energy of the object is

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

Answer 1

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

At a distance $r$ from the centre of earth, orbital velocity is,
$v=\sqrt{\frac{GM}{r}}$
According to principle of energy conservation of energy,
$\text{KE of m}+(-\frac{GMm}{r})=0+0$
($\because$ At infinity, PE = KE = 0)
$\therefore \text{KE of m}=\frac{GMm}{r}$

$={\left(\sqrt{\frac{GM}{r}}\right)}^{2}m=m{v}^2$

Hence, $\left(B\right)$ is the correct answer.