JEE Main 2023 — Gravitation Question with Solution

The escape velocity on a planet is given by the following formula:

$$v_{esc}=\sqrt{\frac{2GM}{R}}$$

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.

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:

$$v_{esc}=\sqrt{\frac{2G(9M_e)}{4R_e}}$$

Divide both sides by the Earth's escape velocity formula:

$$\frac{v_{esc}}{v_e}=\frac{\sqrt{\frac{2G(9M_e)}{4R_e}}}{\sqrt{\frac{2G{M}_e}{R_e}}}$$

Simplify:

$$\frac{v_{esc}}{11.2\times 10^3\text{ m/s}}=\sqrt{\frac{9}{4}}$$

$$\frac{v_{esc}}{11.2\times 10^3\text{ m/s}}=\frac{3}{2}$$

Now, solve for $v_{esc}$:

$$v_{esc}=11.2\times 10^3\text{ m/s}\times \frac{3}{2}$$

$$v_{esc}=16.8\times 10^3\text{ m/s}$$

Converting this to km/s, we get:

$$v_{esc}=16.8\text{ km/s}$$