Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} \mathrm{~g}$, where $g$ is the…
Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} \mathrm{~g}$, where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be 11 $\mathrm{kms}^{-1}$, the escape speed on the surface of the planet in $\mathrm{kms}^{-1}$ will be
Solution
$g=\frac{G M}{R^2}=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho}{R^2}$
or $\quad g \propto \rho R$
or $\quad R \propto \frac{g}{\rho}$
Now escape velocity, $v_e=\sqrt{2 g R}$
or $\quad v_e \propto \sqrt{g R}$
or $\quad v_e \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^2}{\rho}}$
$\therefore\left(v_e\right)_{\text {planet }}=\left(11 \mathrm{~km}-\mathrm{s}^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}}$
$\therefore$ The correct answer is 3 .