JEE Physics 2023 Question with Solution

Given: Mass of the planet is $16$ times the mass of Earth and radius of the planet is $4$ times the radius of Earth.

Find: The ratio of escape velocity of the planet to that of Earth.

Escape velocity is given by

$$v_e = \sqrt{\frac{2GM}{R}}$$

Let the escape velocities of the planet and Earth be $v_{ep}$ and $v_{ee}$ respectively.

Then

$$v_{ee} = \sqrt{\frac{2GM_e}{R_e}}$$

and

$$v_{ep} = \sqrt{\frac{2GM_p}{R_p}}$$

Therefore,

$$\frac{v_{ep}}{v_{ee}} = \frac{\sqrt{\frac{2GM_p}{R_p}}}{\sqrt{\frac{2GM_e}{R_e}}} = \sqrt{\frac{M_p R_e}{M_e R_p}}$$

Using $M_p = 16M_e$ and $R_p = 4R_e$,

$$\frac{v_{ep}}{v_{ee}} = \sqrt{\frac{16M_e R_e}{M_e \cdot 4R_e}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$$

Therefore, the ratio of escape velocity of the planet to the escape velocity of Earth is $2:1$. The correct option is B.

Given: $M_p = 16M_e$ and $R_p = 4R_e$.

Find: $\frac{v_{ep}}{v_{ee}}$.

Since escape velocity varies as

$$v_e \propto \sqrt{\frac{M}{R}}$$

we can directly write

$$\frac{v_{ep}}{v_{ee}} = \sqrt{\frac{M_p/R_p}{M_e/R_e}} = \sqrt{\frac{16/4}{1}} = \sqrt{4} = 2$$

Hence, the required ratio is $2:1$. The correct option is B.