QJEEM23Ph9284: JEE Main 2023 Physics - Class 11 JEEM23Ph9284 Solution: A planet having mass and radius , where and are mass and radius …

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:

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

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:

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