Q. What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is $(3.1)^{2}ms^{- 2}$ and whose radius is $8100\ km$
A) $2790\ km.s^{- 1}$
B) $27.9\ km.s^{- 1}$
C) $\frac{27.9}{\sqrt{5}}km.s^{- 1}$
D) $27.9\sqrt{5}km.s^{- 1}$
Explanation
(b) $v = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = \sqrt{2 \times (3.1)^{2} \times 8100 \times (10)^{3}}$
$= 27.9\ km/s{ec}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.