JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Gravitation

If earth has a mass nine times and radius twice to the of a planet P. Then

\frac{v _{e}}{3} x ms^{- 1}

will be the minimum velocity required by a rocket to pull out of gravitational force of P, where

v_{e}

is escape velocity on earth. The value of

x

(A)2✓ Correct
(B)3
(C)18
(D)1

Answer: (A)

Solution

The escape speed from a body of mass

M

and radius

R

v = \frac{2 GM}{R}

, so for the earth

v_{e} = \frac{2 G M _{e}}{R _{e}}

. For planet P, using

M_{P} = \frac{M _{e}}{9}

and

R_{P} = \frac{R _{e}}{2}

v_{P} = \frac{2 G M _{P}}{R _{P}} = \frac{2 G ( M _{e} /9 )}{R _{e} /2} = \frac{4 G M _{e}}{9 R _{e}} = \frac{2}{9} \frac{2 G M _{e}}{R _{e}} = \frac{2}{3} v_{e} .

Comparing with

\frac{v _{e}}{3} x

, the quantity under the root is

x = 2

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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