Question

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g,$ where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $1.5$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11kms^{- 1}$ the escape speed on the surface of the planet in $kms^{- 1}$ will be

Answer 1

Gravitation acceleration in terms of universal gravitational constant is given as $g=\frac{G M}{R^{2}}=\frac{G \left(\frac{4}{3} \pi R^{3}\right) \rho }{R^{2}}$
or $g \propto \rho R$
Thus, radius is related as $R \propto \frac{g}{\rho }...\left(1\right)$
Now escape velocity, $v_{e} \propto \sqrt{2 gR}$
or $v_{e} \propto \sqrt{g R}..\left(2\right)$
From above equations (1) and (2), we get, $v_{e} \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^{2}}{\rho}}$
$\therefore \quad\left(v_{e}\right)_{\text {planet }}=\left(11 kms ^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}}$
$=3 km s ^{-1}$