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Physics Question on Gravitation

Gravitational acceleration on the surface of a planet is 611g\frac{\sqrt6}{11}g where g, is the gravitational acceleration oil the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11kms−1,11kms^{-1}, the escape speed on the surface of the planet in kms−1kms^{-1} will be

A

3

B

6

C

9

D

12

Answer

3

Explanation

Solution

let the gravitational acceleration on the surface be g′g^{\prime} and density be ρ′\rho^{\prime} g′g=611;ρ′ρ=23\frac{ g ^{\prime}}{ g }=\frac{\sqrt{6}}{11} ; \frac{\rho^{\prime}}{\rho}=\frac{2}{3} Hence, R′R=3622\frac{ R ^{\prime}}{ R }=\frac{3 \sqrt{6}}{22} vesc′vesc∝R′2ρ′R2ρ=311 vesc′=3km/s\begin{array}{l} \frac{ v _{ esc }^{\prime}}{ v _{ esc }} \propto \sqrt{\frac{ R ^{\prime 2} \rho^{\prime}}{ R ^{2} \rho}}=\frac{3}{11} \\\ v _{ esc }^{\prime}=3 km / s \end{array}