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. 
 

Given, escape velocity on the surface of earth, 

$v_e=11.1 \mathrm{km}/\mathrm{h}$

i.e.,     $v_e=\sqrt{2g{R}_e}$

where, R_e is radius of earth.

As,     $v_e=\sqrt{\frac{2G{M}_e}{R_e}}$                          $\left(\because g=\frac{G{M}_e}{R_e^2}\right)$

where, M_e is mass of earth.

Mass of moon,   $M_m=\frac{M_e}{81}$

Radius of moon,  $R_m=\frac{R_e}{4}$

$\therefore$  Escape velocity on the surface of moon

$v_m=\sqrt{\frac{2G{M}_m}{R_m}}=\sqrt{\frac{2G\frac{M_e}{81}}{\frac{R_e}{4}}}$

$=\frac{2\sqrt{2}}{9}\sqrt{\frac{G{M}_e}{R_e}}=\frac{2}{9}\sqrt{\frac{2G{M}_e}{R_e}}=\frac{2}{9}{v}_e$

$=\frac{2}{9}\times 11.1=2.46 \mathrm{km}{ \mathrm{h}}^{-1}$