Q. The escape velocity from the earth is ${11\ kms}^{- 1}$. The escape velocity from a planet having twice the radius and same mean density as that of earth is
A) $5.5\ {kms}^{- 1}$
B) $11\ {kms}^{- 1}$
C) $22\ {kms}^{- 1}$
D) None of these
Explanation
(c) Escape velocity,
$\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ }v_{ascape}\mathbf{=}\sqrt{\frac{2\ GM}{R}}\mathbf{\ \ }$
$\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =}R\sqrt{\frac{8}{3}\pi G\rho}$
$\therefore v_{e} \propto R\ if\ \rho = constant.$
Since the planet is having double radius in comparision to earth, therefore escape velocity becomes twice $ie,$ ${22\ kms}^{- 1}$ .
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.