If the radius of a planet is doubled keeping density constant, what will be the value of escape velocity

Escape velocity remains same

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Escape velocity doubles

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Escape velocity becomes halved

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Escape velocity becomes one fourth

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Solution

The correct option is B Escape velocity doubles
Doubling radius and keeping density constant $⟹ M^{'} = V ρ = \frac{4 π (2 R)^{3} ρ}{3} = 8 M$ , M is the mass of the planet with R as the radius
Thus escape velocity $v_{e}^{'} = \sqrt{(} 2 G . (8 M / 2 R)) = 2 v_{e}$
Thus, escape velocity doubles

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