The escape velocity is $11 k m / s e c$ at the surface ofthe earth. If the planet have radius twice of the earth and equal density as earth then the escape velocity at the surface of the planet is.

11 k m / s e c

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22 k m / s e c

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55 k m / s e c

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225 k m / s e c

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Solution

The correct option is B $22 k m / s e c$
Escape velocity , $v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$

As $M = ρ \frac{4 π}{3} R_{e}^{3}$ where $ρ$ is the density of earth.

$⟹ v_{e} = \sqrt{\frac{8 π}{3} ρ G R_{e}^{2}} = 11 k m / s$

Now for $R^{'} = 2 R_{e} ρ^{'} = ρ$

$v_{e}^{'} = \sqrt{\frac{8 π}{3} ρ G (2 R_{e})^{2}} = 2 v_{e} = 22 k m / s$

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The escape velocity is 11km/sec at the surface ofthe earth. If the planet have radius twice of the earth and equal density as earth then the escape velocity at the surface of the planet is.

The correct option is B 22km/secEscape velocity , ve=√2GMeReAs M=ρ4π3R3e where ρ is the density of earth.⟹ve=√8π3ρGR2e=11km/s Now for R′=2Reρ′=ρv′e=√8π3ρG(2Re)2=2ve=22km/s

The escape velocity is $11 k m / s e c$ at the surface ofthe earth. If the planet have radius twice of the earth and equal density as earth then the escape velocity at the surface of the planet is.