Q. The orbital velocity of a satellite at a height $h$ above the surface of earth is $v$. The value of escape velocity from the same location is given by:
J & K CETJ & K CET 2000
Solution:
The orbital velocity $ (v_{o}) $ of a satellite at a height h above the surface of earth is
$v_{o}=\sqrt{\frac{G M_{e}}{R_{e}+h} \ldots}(i) $
The escape velocity $\left(v_{e}\right)$ is
$v_{e}=\sqrt{\frac{2 G M_{e}}{R_{e}+h}} \ldots (ii) $
From Eqs. (i) and (ii), we get
$v_{e}=\sqrt{2} . v_{o}$
Given, $ v_{o}=v $
$\therefore v_{e}=\sqrt{2} v$
