Explanation

(d) The gravitational force exerted on satellite at a height $x$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F_{G} = \frac{GM_{e}m}{(R + x)^{2}}$
where $M_{e} = mass\ of\ earth.$
Since, gravitational force provides the necessary centripetal force, so,
$\ \ \ \ \ \ \frac{GM_{e}m}{(R + x)^{2}} = \frac{mv_{o}^{2}}{(R + x)}$
where $v_{o}$ is orbital speed of satellite.
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{GM_{e}m}{(R + x)} = mv_{o}^{2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{gR^{2}m}{(R + x)} = mv_{o}^{2}\ \ \ \ \ \ \ \ \ \left( \therefore g = \frac{GM_{e}}{R^{2}} \right)$
$\Longrightarrow \ \ v_{o} = \sqrt{\left\lbrack \frac{gR^{2}}{(R + x)} \right\rbrack} = \left\lbrack \frac{gR^{2}}{(R + x)} \right\rbrack^{1/2}$