JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Gravitation

The escape velocity of a body on a planet ' A ' is

12 kms^{- 1}

. The escape velocity of the body on another planet ' B ', whose density is four times and radius is half of the planet ' A ', is :

Answer: (A)

Solution

Escape velocity is given by

v_{escape} = \frac{2 GM}{R}

. For a sphere,

M = ρ \cdot \frac{4}{3} π R^{3}

, so:

v_{escape} = \frac{2 Gρ \cdot \frac{4}{3} π R ^{3}}{R} = \frac{8 π Gρ R ^{2}}{3}

Therefore,

v_{escape} \propto ρ R^{2}

.

For planet B: density is 4 times (

ρ_{B} = 4 ρ_{A}

) and radius is half (

R_{B} = \frac{R _{A}}{2}

\frac{v _{B}}{v _{A}} = \frac{ρ _{B} R _{B}^{2}}{ρ _{A} R _{A}^{2}} = \frac{4 ρ _{A} \cdot ( R _{A} /2 ) ^{2}}{ρ _{A} R _{A}^{2}} = \frac{4 \cdot R _{A}^{2} /4}{R _{A}^{2}} = 1

Therefore,

v_{B} = v_{A} = 12 kms^{- 1}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option