Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:

Gravitation PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q61 with Solution

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Correct answer: C

Use the escape velocity formula

The escape velocity from the surface of a planet is $v_e = \sqrt{\frac{2GM}{R}}$

where $M$ is the mass and $R$ is the radius.
Relate Earth and the planet

Let the planet have mass $M$ and radius $R$ .

Then Earth has: $M_E = 8M, \qquad R_E = 2R$
Write escape velocities

For Earth: $v_E = \sqrt{\frac{2G(8M)}{2R}} = \sqrt{\frac{8GM}{R}}$

For the planet: $v_P = \sqrt{\frac{2GM}{R}}$
Find the ratio

$\frac{v_E}{v_P} = \sqrt{\frac{8GM/R}{2GM/R}} = \sqrt{4} = 2$

So, $v_E = 2v_P$
Substitute the given value

Given: $v_E = 11.2\text{ km/s}$

Therefore, $v_P = \frac{11.2}{2} = 5.6\text{ km/s}$
Check options
- A: $8.4$ ❌
- B: $11.2$ ❌
- C: $5.6$ ✅
- D: $2.8$ ❌

Therefore, the correct answer is Option C.