JEE Main 2025 — Gravitation Question with Solution

The escape velocity ($v_{\text{esc}}$) for a planet with mass $M$ and radius $R$ is given by the formula:

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

where $G$ is the gravitational constant.

For Earth, let's denote its mass as $M_E$ and its radius as $R_E$. The escape velocity is given as 11.2 km/s. Therefore:

$$v_{\text{esc, Earth}}=\sqrt{\frac{2G{M}_E}{R_E}}=11.2\text{ km/s}$$

For the planet in question, its mass $M_P$ is $M_E/8$ and its radius $R_P$ is $R_E/2$. The escape velocity from the planet is:

$$v_{\text{esc, Planet}}=\sqrt{\frac{2G{M}_P}{R_P}}=\sqrt{\frac{2G(M_E/8)}{R_E/2}}$$

Simplifying the expression:

$$v_{\text{esc, Planet}}=\sqrt{\frac{2G(M_E/8)}{R_E/2}}=\sqrt{\frac{2G{M}_E\cdot 2}{8R_E}}=\sqrt{\frac{G{M}_E}{2R_E}}$$

Now, relate it to the escape velocity from Earth:

$$v_{\text{esc, Planet}}=\frac{1}{\sqrt{4}}\cdot v_{\text{esc, Earth}}$$

Calculating further:

$$v_{\text{esc, Planet}}=\frac{1}{2}\cdot 11.2\text{ km/s}=5.6\text{ km/s}$$

Thus, the escape velocity from the planet is 5.6 km/s.

Option C: 5.6 km/s is the correct answer.