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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 ve3x\frac{v_e}{3}\sqrt{x} ms−1-1 will be the minimum velocity required by a rocket to pull out of gravitational force of P, where vev_e is escape velocity on earth. The value of xx is

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Answer: (a)

  1. ve=2GMRv_e = \sqrt{\frac{2GM}{R}} for earth; planet P has MP=M9M_P = \frac{M}{9}, RP=R2R_P = \frac{R}{2}.
  2. vP=2G(M/9)R/2=ve29=ve32v_P = \sqrt{\frac{2G(M/9)}{R/2}} = v_e\sqrt{\frac{2}{9}} = \frac{v_e}{3}\sqrt{2}.
  3. Comparing with ve3x\frac{v_e}{3}\sqrt{x} gives x=2x = 2.
  4. So the answer is (a) 22.
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