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

Answer 1

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


$K{E}_1=\frac{1}{2}m{V}^2=\frac{|P.E.|}{2}$
$P.E_1=-\frac{GMm}{r}$
Finally, $KE+P{E}_r=0$
$KE=-P{E}_r$
$=+\frac{GMm}{r}=m{V}^2$