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MHT CET 2026 (13 Apr, Shift 2) · Escape Velocity

Escape Velocity question from MHT CET 2026 (13 Apr, Shift 2)

Q3MHT CET 2026 · 13 APR · SHIFT 2Medium

Escape Velocity

If earth has a mass nine times and radius twice to the planet P. Then $\frac{v_e}{3}\sqrt{x}$ ms $-1$ 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: (a)

$v_e = \sqrt{\frac{2GM}{R}}$ for earth; planet P has $M_P = \frac{M}{9}$ , $R_P = \frac{R}{2}$ .
$v_P = \sqrt{\frac{2G(M/9)}{R/2}} = v_e\sqrt{\frac{2}{9}} = \frac{v_e}{3}\sqrt{2}$ .
Comparing with $\frac{v_e}{3}\sqrt{x}$ gives $x = 2$ .
So the answer is (a) $2$ .

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