Q. The escape velocity from the earth is $11\ \ kms^{- 1}$. The escape velocity from a planet having twice the radius and the same mean density as the earth would be
A) $5.5\ \ kms^{- 1}$
B) $11\ \ kms^{- 1}$
C) $15.5\ kms^{- 1}$
D) $22\ kms^{- 1}$
Explanation
(d) Escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$
$v_{e} = \sqrt{\frac{2G\ \frac{4}{3}\pi R^{3} \times d}{R}}$
$\sqrt{2G\ \frac{4}{3}\pi R^{3} \times d} = R\sqrt{\frac{8}{3}\pi Gd}$
where $d =$ mean density of earth
$\because\ \ v_{e} \propto R\sqrt{d}$
$\therefore\ \ \ \frac{v_{e}}{v_{p}} = \frac{R_{e}}{R_{p}}\sqrt{\frac{d_{e}}{d_{p}}}$
$= \frac{R_{e}}{2R_{e}}\sqrt{\frac{d_{e}}{d_{e}}}$
$= v_{p} = 2v_{e}$
$= 2 \times 11 = 22kms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.