Three point masses,each of mass m,are kept at | vedclass

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(C) Consider a mass $m$ at corner $A$. The gravitational forces exerted on it by the other two masses at $B$ and $C$ are $F_1$ and $F_2$,where $F_1 =

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$L^{3/2}$

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(C) Consider a mass $m$ at corner $A$. The gravitational forces exerted on it by the other two masses at $B$ and $C$ are $F_1$ and $F_2$,where $F_1 = F_2 = G \frac{m^2}{L^2}$.The angle between these two forces is $60^{\circ}$. The resultant force $F$ is given by:$F = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos 60^{\circ}} = \sqrt{3} F_1 = \sqrt{3} G \frac{m^2}{L^2}$.The distance of each mass from the center of the triangle is $r = \frac{L}{\sqrt{3}}$.For uniform circular motion,the gravitational force provides the necessary centripetal force:$mr \omega^2 = F$$m \left( \frac{L}{\sqrt{3}} \right) \omega^2 = \sqrt{3} G \frac{m^2}{L^2}$$\omega^2 = 3 G \frac{m}{L^3}$Since $\omega = \frac{2\pi}{T}$,we have:$\left( \frac{2\pi}{T} \right)^2 = \frac{3Gm}{L^3}$$T^2 = \frac{4\pi^2 L^3}{3Gm}$$T \propto L^{3/2}$.

Three point masses,each of mass $m$,are kept at the corners of an equilateral triangle of side $L$. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to

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