# The escape velocity on earth’s surface is 11.2 km/sec. If the radius of planet is 1/3rd that of earth and mass 1/4th, what will be the escape velocity on the planet?

The escape velocity on earth’s surface is, $v_e=\sqrt{2gR}$

The escape velocity on the surface of another planet is, $v_e’=\sqrt{2g’R’}$

According to the question, $M’=\frac{M}{4}$ and $R’=\frac{R}{3}$.

$\therefore g’=\frac{GM’}{R’^2}=\frac{G(M/4)}{(R/3)^2}$ $=\frac{9}{4}\left(\frac{GM}{R^2}\right)=\frac{9}{4}g$

Now, $v_e’=\sqrt{2\frac{9}{4}g\frac{R}{3}}=\sqrt{\frac{3}{4}}\sqrt{2gR}=\sqrt{\frac{3}{4}}v_e$

Putting $v_e=11.2$ km/sec, we get, $v_e’=\sqrt{\frac{3}{4}}×11.2=9.7\;\text{km/sec}$