QJEEM25Phe4bb: JEE Main 2025 Physics - Class 11 JEEM25Phe4bb Solution: Earth has mass 8 times and radius 2 times that of a planet. If…

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.