The gravitational force between two bodies is decreased by $36 %$ when the distance between them is increased by $3 m$ . The initial distance between them is:

6 m

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9 m

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12 m

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15 m

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Solution

The correct option is B $12 m$
$\frac{F_{1}}{F_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$
$F_{2} = F_{1} - (\frac{36}{100}) F_{1}; d_{2} = d_{1} + 3$
$\frac{F_{1}}{(64 / 100) F_{1}} = \frac{(d_{1} + 3)^{2}}{(d_{1})^{2}}$
$\frac{5}{4} = 1 + \frac{3}{d_{1}}$
$d_{1} = 12 m$

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