Three particles of equal mass m are situated | vedclass

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(C) The distance of each particle from the center of the equilateral triangle is $r = \frac{l}{\sqrt{3}}$.The gravitational force exerted on one parti

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$l^{3/2}$

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(C) The distance of each particle from the center of the equilateral triangle is $r = \frac{l}{\sqrt{3}}$.The gravitational force exerted on one particle by the other two is the vector sum of the two forces,each of magnitude $F = \frac{Gm^2}{l^2}$.The resultant force towards the center is $F_{net} = 2F \cos(30^\circ) = 2 \left( \frac{Gm^2}{l^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3}Gm^2}{l^2}$.This force provides the necessary centripetal force for circular motion: $\frac{mv^2}{r} = F_{net}$.Substituting $r = \frac{l}{\sqrt{3}}$,we get $\frac{mv^2}{l/\sqrt{3}} = \frac{\sqrt{3}Gm^2}{l^2}$,which simplifies to $v^2 = \frac{Gm}{l}$,or $v = \sqrt{\frac{Gm}{l}}$.The time period $T$ is given by $T = \frac{2\pi r}{v} = \frac{2\pi (l/\sqrt{3})}{\sqrt{Gm/l}} = \frac{2\pi}{\sqrt{3G m}} l^{3/2}$.Thus,$T \propto l^{3/2}$.

Three particles of equal mass $m$ are situated at the vertices of an equilateral triangle of side $l$. They are moving in a circle under the influence of their mutual gravitational interaction. Then their time period of revolution is directly proportional to

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