Explanation

(c) The orbital velocity of satellite close to the earth is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{0} = \sqrt{gR_{e}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
where $R_{e}$ is radius of the earth. The escape velocity for a body thrown from the earth’s surface is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{gR_{e}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
$Thus,\ \ \frac{v_{o}}{v_{e}} = \frac{\sqrt{gR_{e}}}{\sqrt{2gR_{e}}} = \frac{1}{\sqrt{2}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{2}v_{o}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{o} = \frac{v_{e}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\ {kms}^{- 1}$