Question

If escape velocity on earth surface is 11.1 kmh^-1 then find the escape velocity on moon surface. If mass of moon is $\frac{1}{81}$ times of mass of earth and radius of moon is $\frac{1}{4}$ times radius of earth.

Moderate

Solution

Given, escape velocity on the surface of earth,
$v_{e} = 11.1 km / h$
i.e.,    $v_{e} = \sqrt{2 g R_{e}}$
where, R_e is radius of earth.
As,    $v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$    $(∵ g = \frac{G M_{e}}{R_{e}^{2}})$
where, M_e is mass of earth.
Mass of moon,    $M_{m} = \frac{M_{e}}{81}$
Radius of moon, $R_{m} = \frac{R_{e}}{4}$
$∴$ Escape velocity on the surface of moon
$v_{m} = \sqrt{\frac{2 G M_{m}}{R_{m}}} = \sqrt{\frac{2 G \frac{M_{e}}{81}}{\frac{R_{e}}{4}}}$
$= \frac{2 \sqrt{2}}{9} \sqrt{\frac{G M_{e}}{R_{e}}} = \frac{2}{9} \sqrt{\frac{2 G M_{e}}{R_{e}}} = \frac{2}{9} v_{e}$
$= \frac{2}{9} \times 11.1 = 2.46 km h^{- 1}$

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