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JEE Main 2023 (1 Feb, Shift 1) · Escape Velocity

Escape Velocity question from JEE Main 2023 (1 Feb, Shift 1)

Q29JEE Main 2023 · 1 FEB · SHIFT 1Easy

Escape Velocity

If earth has a mass nine times and radius twice to that of a planet P. Then $\frac{v_e}{3}\sqrt{x}\text{ m s}^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$ , where $v_e$ is escape velocity on earth. The value of $x$ is

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Answer: (c)

$v \propto \sqrt{\frac{M}{R}}$ ; planet P has $\frac{M}{9}$ and $\frac{R}{2}$
$v_P = v_e\sqrt{\frac{1/9}{1/2}} = v_e\sqrt{\frac{2}{9}} = \frac{v_e}{3}\sqrt{2}$
So $x = 2$ , answer (c).

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