Q. Gravitational acceleration on the surface of a planet is$\ \frac{\sqrt{6}}{11}\ g,\ $where g is the gravitational$\ $ acceleration on the surface of earth. The average mass density of the planet is$\ \ \ \frac{2}{\ 3}\ $times that of the earth. If the escape speed on the surface of the earth is taken on be $11\ {kms}^{- 1},$ the escape speed on the surface of the planet in $k{ms}^{- 1}\ $will be
A) 5
B) 7
C) 3
D) 11
Explanation
(c) $\ \ \ \ g = \frac{GM}{R^{2}} = \frac{G\left( \frac{4}{3}\pi R^{3} \right)\rho}{R^{2}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ g \propto \rho R$
$or\ \ \ \ \ \ \ \ \ \ \ \ R \propto \frac{g}{\rho}$
Now escape velocity, $v_{e} = \sqrt{2gR}$
$or\ \ \ \ \ \ \ \ v_{e} \propto \sqrt{gR}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^{2}}{\rho}}$
$\therefore(v_{e})_{panet} = ({11\ ms}^{- 1})\sqrt{\frac{6}{121} \times \frac{3}{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3\ {km\ s}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.