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 $11km{s}^{-1}$ the escape speed on the surface of the planet in $km{s}^{-1}$ will be

Answer 1

Answer: 3

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