Question

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

Answer 1

Answer: (B)

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