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When the distance between two bodies of masses $m_{1}$ and $m_{2}$ is tripled. Then the gravitational force between them becomes

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\frac{1}{6}

times

\frac{1}{3}

times

\frac{1}{8}

times

\frac{1}{9}

times

answer is D.

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Detailed Solution

Given that two masses

m_{1}, m_{2}

separated by a distance r then.

Gravitational force $F_{1} = \frac{{Gm}_{1} m_{2}}{r^{2}}$

Now $r^{|} = 3 r$ then

Gravitational force $F_{2} = \frac{{Gm}_{1} m_{2}}{{(3 r)}^{2}}$

$\Rightarrow \frac{F_{1}}{F_{2}} = \frac{{(3r)}^{2}}{r^{2}} = \frac{{9r}^{2}}{r^{2}} = F_{2} = \frac{F_{1}}{9}$

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When the distance between two bodies of masses $m_{1}$ and $m_{2}$ is tripled. Then the gravitational force between them becomes