The radius of a planet is nine times of the radius of earth. Both have almost equal average mass-densities. If $V_P$ and $V_E$ are the escape velocities of the planet and the earth respectively, then

The correct option is D $V_P=9V_E$

Let the mass densities of both planet and earth be $\rho$.

As we know that the escape velocity near the surface of any planet is given by

$V_e=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2G(\frac{4\pi }{3}{R}^3\rho )}{R}}$

$[\because M=\frac{4}{3}\pi R^{3}p]$

$\Rightarrow V_e=\sqrt{\frac{8}{3}\pi G{R}^2\rho }$

$\Rightarrow V_e\propto R\sqrt{\rho }$

Given: $(R_P=9R_E;{\rho }_P={\rho }_E)$

$\therefore \frac{V_P}{V_E}=\frac{R_P}{R_E}\times \frac{\sqrt{{\rho }_P}}{\sqrt{{\rho }_E}}=\frac{9}{1}\times \frac{1}{1}=9$

$\Rightarrow V_P=9V_E$

Hence, option (d) is the correct answer.