Question:
Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be
(a) $1 / 4$ times
(b) 4 times
(c) $1 / 2$ times
(d) unchanged.
Solution:
(b) Explanation: $\mathrm{F}=\frac{\mathrm{G} m_{1} m_{2}}{r^{2}}, \mathrm{~F}^{\prime}=\frac{\mathrm{G}\left(2 m_{1}\right)\left(2 m_{2}\right)}{r^{2}}=4 \mathrm{~F}$.
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