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NEET PHYSICSGRAVITATIONMedium

Question

The escape velocity from the Earth's surface is vv. The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is:

A

3v3v

B

4v4v

C

vv

D

2v2v

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: Since the planet has the same mass density (ρ\rho), we 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: ve=2GR43πR3ρ=8πGρR23=R8πGρ3v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 \rho} = \sqrt{\frac{8\pi G \rho R^2}{3}} = R \sqrt{\frac{8\pi G \rho}{3}}
  3. Proportionality: From the derived formula, if density ρ\rho is constant, the escape velocity is directly proportional to the radius RR (veRv_e \propto R).
  4. Calculation: Given that the radius of the planet is four times that of Earth (Rp=4RER_p = 4R_E): vpvE=RpRE=4RERE=4\frac{v_p}{v_E} = \frac{R_p}{R_E} = \frac{4R_E}{R_E} = 4 vp=4vE=4vv_p = 4 v_E = 4v

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.

PHYSICSGRAVITATIONescapevelocityearthssurfaceescape

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