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NEET PHYSICSGRAVITATIONMedium

Question

The ratio of escape velocity at the Earth (vev_e) to the escape velocity at a planet (vpv_p) whose radius and mean density are twice that of the Earth is:

A

1 : 2\sqrt{2}

B

1 : 4

C

1 : \sqrt{2}

D

1 : 2

Step-by-Step Solution

  1. Escape Velocity Formula: The escape velocity vev_e from a spherical body of mass MM and radius RR is given by ve=2GMRv_e = \sqrt{\frac{2GM}{R}} .
  2. In Terms of Density: To relate velocity to density (ρ\rho), express mass as M=Volume×Density=43πR3ρM = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 \rho. Substituting this into the escape velocity formula: v=2GR(43πR3ρ)=8πGρR23v = \sqrt{\frac{2G}{R} \left( \frac{4}{3}\pi R^3 \rho \right)} = \sqrt{\frac{8\pi G \rho R^2}{3}} From this, we see that vRρv \propto R\sqrt{\rho}.
  3. Ratio Calculation: Let Earth's radius be RR and density be ρ\rho. Then the planet has radius R=2RR' = 2R and density ρ=2ρ\rho' = 2\rho. vevp=RρRρ=Rρ(2R)2ρ\frac{v_e}{v_p} = \frac{R\sqrt{\rho}}{R'\sqrt{\rho'}} = \frac{R\sqrt{\rho}}{(2R)\sqrt{2\rho}} vevp=122\frac{v_e}{v_p} = \frac{1}{2\sqrt{2}} The ratio is 1:221 : 2\sqrt{2}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from GRAVITATION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSGRAVITATIONescapevelocityescapevelocityplanet

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