The gravitational force of attraction between two bodies is F₁. If the mass of each body is doubled and the distance between them is halved, then the gravitational force between them is F₂. Then

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F₂ = 8F₁

F₂ = 16F₁

F₁ = F₂

F₂ = 4F₁

answer is D.

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Detailed Solution

$F_{1} = \frac{G \times m_{1} \times m_{1}}{R^{2}}$

Doubling both masses and halving the distance

$F_{2} = G \times \frac{2 m_{1} \times 2 m_{2}}{{(\frac{R}{2})}^{2}}$

$∴ F_{2} = \frac{G \times 4 \times m_{1} \times m_{2}}{\frac{1}{4} \times R^{2}}$

$∴ F_{2} = \frac{G \times m_{1} \times m_{1}}{R^{2}} \times 16$

$∴ F_{2} = 16 F_{1}$

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