Question

If $V_e$ is the escape velocity of a body from a planet of mass $M$ and radius $R$. Then, the velocity of satellite revolving at height $h$ from the surface of planet will be

• A. $V=V_e\sqrt{\frac{R}{(R+h)}}$
• B. $V=V_e\sqrt{\frac{2R}{(R+h)}}$
• C. $V=V_e\sqrt{\frac{(R+h)}{R}}$
• D. $V=V_e\sqrt{\frac{R}{2(R+h)}}$

Answer 1

Answer: (D)

$V=V_e\sqrt{\frac{R}{2(R+h)}}$

The expression of escape velocity is $V_e=\sqrt{\frac{2GM}{R}}..................................\left(1\right)$
Let $V$ be the velocity of satellite revolving at height $h$ from the surface of planet.
here, $\frac{m{V}^2}{(R+h)}=\frac{GMm}{(R+h{)}^2}$
or $V=\sqrt{\frac{GM}{R+h}}.......................\left(2\right)$
$\left(2\right)/\left(1\right),V=V_e\sqrt{\frac{R}{2(R+h)}}$