Explanation

(c) If $m$ is the mass and $v$ is the orbital velocity of the satellite, then kinetic energy.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{1}{2}{mv}^{2}$
$or\ \ \ \ \ \ \ \ \ \ \ Em = \frac{1}{2}m^{2}v^{2}$
$or\ \ \ \ \ \ \ m^{2}v^{2} = 2Em$
$or\ \ \ \ \ \ \ \ \ \ \ \ mv = \sqrt{2Em}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$\ $If $r$ is the radius of the orbit of the satellite, then its angular momentum
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = mvr$
Using Eq. (i),
$\ \ \ \ \ \ \ \ \ L = \left( \sqrt{2Em} \right)r = \sqrt{{2Emr}^{2}}$