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# Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

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#### Force on M at C due to gravitational attraction.$$\vec { F_ { CB} } = \dfrac{Gm^2}{2R^2} \hat{i}$$$$\vec {F_{CD}} = \dfrac{GM^2}{4R^2} cos 45 \hat{j} + \dfrac{GM^2}{4R^2} sin 45\hat{j}$$So resultant force on C, $$\therefore \vec { F_ c } =\quad \vec { F_ { CA} } \quad +\quad \vec { F_ { CB} } +\quad \vec {F_{CD}}$$ $$= - \dfrac{GM^2}{4R^2} \left( 2+\dfrac { 1 }{ \sqrt { 2 } } \right) \hat { i } +\dfrac { GM^2 }{ 4R^2 } \left( 2+\dfrac { 1 }{ \sqrt { 2 } } \right) \$$ $$\therefore F_c \dfrac{GM^2}{4R^2}(2\sqrt{2}+1)$$For moving along the circle , $$\vec { F } = \dfrac{mv^2}{R}$$ or $$\dfrac{GM^2}{4R^2}(2\sqrt{2} +1) = \dfrac{MV^2}{R} or V = \sqrt { \dfrac { GM }{ R } \left( \dfrac { 2\sqrt { 2 } +1 }{ 4 } \right) }$$

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