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Planet X is twice the radius of planet Y and density of material of X is twice that of density of material of Y then the ratio of acceleration due to gravity at the surface of X to that at the surface Y is

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1 : 4

4 : 1

1 : 2

2 : 1

answer is D.

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Detailed Solution

We know,

$g = \frac{G M}{R^{2}}$

$M = ρ X V$ , where $ρ$ is the density and $V$ is the volume.

$g = \frac{G \frac{4}{3} {πR}^{3} ρ}{R^{2}}$

$g = \frac{4}{3} π G ρ R$

$g \propto ρ R$

Given that

$R_{x} = 2 R_{y}$

$ρ_{x} = 2 ρ_{y}$

$\frac{g_{x}}{g_{y}} = \frac{ρ_{x} R_{x}}{ρ_{y} R_{y}}$

$\frac{g_{x}}{g_{y}} = \frac{4 ρ_{y} R_{y}}{ρ_{y} R_{y}}$

$g_{x} : g_{y} = 4 : 1$

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