Four particles,each of mass M,move along a ci | vedclass

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(A) Consider one particle of mass $M$. The gravitational forces acting on it due to the other three particles are:$1$. Two forces $F = \frac{GM^2}{(\s

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$\frac{1}{2} \sqrt{\frac{GM}{R}(2\sqrt{2}+1)}$

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(A) Consider one particle of mass $M$. The gravitational forces acting on it due to the other three particles are:$1$. Two forces $F = \frac{GM^2}{(\sqrt{2}R)^2}$ acting along the sides of the square (at $90^\circ$ to each other).$2$. One force $F_1 = \frac{GM^2}{(2R)^2}$ acting along the diagonal.The net gravitational force towards the center of the circle is:$F_{	ext{net}} = \sqrt{F^2 + F^2} + F_1 = \sqrt{2}F + F_1$$F_{	ext{net}} = \sqrt{2} \left( \frac{GM^2}{2R^2} ight) + \frac{GM^2}{4R^2} = \frac{GM^2}{R^2} \left( \frac{\sqrt{2}}{2} + \frac{1}{4} ight) = \frac{GM^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} ight)$This net force provides the necessary centripetal force for circular motion:$F_{	ext{net}} = \frac{MV^2}{R}$$\frac{MV^2}{R} = \frac{GM^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} ight)$$V^2 = \frac{GM}{R} \left( \frac{2\sqrt{2} + 1}{4} ight)$$V = \frac{1}{2} \sqrt{\frac{GM}{R}(2\sqrt{2} + 1)}$

Four particles,each of mass $M$,move along a circle of radius $R$ under the action of their mutual gravitational attraction as shown in the figure. The speed of each particle is:

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