Solveeit Logo

Question

Physics Question on Gravitation

Escape velocity from Earth is about 11 km/s. The escape velocity from a planet, 4 times the size of the Earth with same mean density is

A

44 km/s

B

22 km/s

C

16.5 km/s

D

88 km/s

Answer

44 km/s

Explanation

Solution

Escape velocity of a planet
v=2GMR=2G×43πR3×ρRv=\sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \times\frac{4}{3} \pi R^{3} \times\rho}{R}}
v=R8π3Gρv=R\sqrt{\frac{8\pi}{3} G\rho}
For same ρ,vR\rho, v \propto R
vEvP=RERP\therefore\:\: \frac{v_{E}}{v_{P}} = \frac{R_{E}}{R_{P}}
Here, vE=11Kms1,RE=R,Rp=4R,vP=?v_{E} = 11 Km s^{-1} , R_{E} = R, R_{p} = 4R, v_{P } = ?
11vP=R4R\therefore\:\:\: \frac{11}{v_{P} } = \frac{R}{4R}
or, vP=44kms1 v_{P} = 44 km s^{-1}