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.

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3.46 $k m h^{- 1}$

2.46 $k m h^{- 1}$

4.4 $k m h^{- 1}$

None of these

answer is A.

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Detailed 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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