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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=2GR⋅43πR3ρ=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 (ve∝Rv_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
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