Explanation

(c) The maximum velocity with which a body must be projected in the atmosphere, so as to enable it to just overcome the gravitational pull, is known as escape velocity.
Escape velocity from earth’s surface is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{es} = \sqrt{\frac{2GM_{e}}{R_{e}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{\frac{2G \bullet \frac{4}{3}\pi R_{e}^{3}d_{e}}{R_{e}}}\ \ \ \ \ \ \ \ \left( \therefore M = \frac{4}{3}\pi R_{e}^{3}d_{e} \right)$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{es} \propto \sqrt{d_{e}} \times R_{e}\ \ \ \ldots.(i)$
similarly, for a planet
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {v^{'}}_{es} \propto \sqrt{d_{p}} \times R_{p}\ \ \ \ \ \ \ldots(ii)$
$So,\ \ \ \ \ \ \ \ \ \ \ \frac{v_{es}}{{v^{'}}_{es}} = \left( \frac{d_{e}}{d_{p}} \right)^{1/2} \times \frac{R_{e}}{R_{p}}$
$Given,\ \ d_{p} = \frac{1}{4}d_{e},\ R_{p} = {2R}_{e}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v_{es}}{v_{es}} = \left( \frac{\frac{d_{e}}{d_{e}}}{4} \right)^{\frac{1}{21}} \times \frac{R_{e}}{{2R}_{e\ }}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (4)^{1/2} \times \frac{1}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 \times \frac{1}{2} = 1$
$So,\ \ \ \ \ \ \ \ \ \frac{v_{es}}{{v'}_{es}} = 1:1$