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 $m{V}^2$

$V$ is the orbital velocity and $V_e$ is the escape velocity

then $V_e=\sqrt{2V}$

The kinetic energy at the time of ejection

$KE=\frac{1}{2}m{V_e}^2=$ $\frac{1}{2}m(\sqrt{2V}{)}^2=m{V}^2$