A planet having mass $9 \mathrm{Me}$ and radius $4 \mathrm{R}_{\mathrm{e}}$, where $\mathrm{Me}$ and $\mathrm{Re}$ are mass and radius of earth respectively, has escape velocity in $\mathrm{km} / \mathrm{s}$ given by:

(Given escape velocity on earth $\mathrm{V}_{\mathrm{e}}=11.2 \times 10^{3} \mathrm{~m} / \mathrm{s}$ )

The escape velocity on a planet is given by the following formula: <br/><br/> $v_{esc} = \sqrt{\frac{2GM}{R}}$ <br/><br/> where $v_{esc}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet. <br/><br/> For Earth, we are given that $v_e = 11.2 \times 10^3$ m/s. We know that the mass of the planet in question is $9M_e$ and its radius is $4R_e$. Let's find the escape velocity of this planet: <br/><br/> $v_{esc} = \sqrt{\frac{2G(9M_e)}{4R_e}}$ <br/><br/> Divide both sides by the Earth's escape velocity formula: <br/><br/> $$\frac{v_{esc}}{v_e} = \frac{\sqrt{\frac{2G(9M_e)}{4R_e}}}{\sqrt{\frac{2GM_e}{R_e}}}$$ <br/><br/> Simplify: <br/><br/> $\frac{v_{esc}}{11.2 \times 10^3 \text{ m/s}} = {\sqrt{\frac{9}{4}}}$ <br/><br/> $\frac{v_{esc}}{11.2 \times 10^3 \text{ m/s}} = \frac{3}{2}$ <br/><br/> Now, solve for $v_{esc}$: <br/><br/> $v_{esc} = 11.2 \times 10^3 \text{ m/s} \times \frac{3}{2}$ <br/><br/> $v_{esc} = 16.8 \times 10^3 \text{ m/s}$ <br/><br/> Converting this to km/s, we get: <br/><br/> $v_{esc} = 16.8 \text{ km/s}$