Radius of planet A is twice that of planet B and the density of A is one third that of B. The ratio of the acceleration due to gravity at the surface of A to that at the surface of B is

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4 : 3

3 : 2

3 : 4

2 : 3

answer is A.

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Detailed Solution

Given:

$R_{A} = 2 R_{B}$

$ρ_{A} = \frac{ρ_{B}}{3}$

Acceleration due to gravity is given as:
$g = \frac{G M}{R^{2}}$
$M a s s = V o l u m e \times D e n s i t y$
$M = \frac{4}{3} {πR}^{3} ρ$
$g = \frac{G \frac{4}{3} {πR}^{3} ρ}{R^{2}}$

$g \propto R ρ$
Substituting the values

$\Rightarrow \frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} \frac{ρ_{A}}{ρ_{B}}$

$\Rightarrow \frac{g_{A}}{g_{B}} = \frac{2}{3}$

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