Explanation

(d) $Given\ that,\ the\ orbital\ velocity\ of\ satellite = \frac{escape\ velocity}{2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{o} = \frac{v_{e}}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$\ $But we know that,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{o} = \sqrt{\frac{{gR}^{2}}{R + h}}\ and\ v_{e} = \sqrt{2gR}$
On putting these values in Eq. (i)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{\frac{{gR}^{2}}{R + h}} = \frac{\sqrt{2gR}}{2}$
On squaring both sides, we obtain
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{gR}^{2}}{R + h} = \frac{2gR}{4}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {2gR}^{2} = gR(R + h)$
$or\ \ \ \ \ \ \ \ \ 2R = R + h\ \ or\ \ \ R = h\ \ $
$or\ \ \ \ \ \ \ \ \ \ \ h = R = 6400\ km$