Question

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

Solution

The gravitational force on each mass due to other $$3$$ masses provides the necessary centripetal forceConsider the particle $$A$$. Net force on particle $$A$$ is given by$$\vec {F}_{A} = \vec {F}_{AB} + \vec {F}_{AD} + \vec {F}_{AC}$$$$\Rightarrow F_{AY} = \dfrac {GM^{2}}{(AB)^{2}} \cos 45^{\circ} + \dfrac {GM^{2}}{(AD)^{2}} \cos 45^{\circ} + \dfrac {GM^{2}}{(AC)^{2}}$$$$= \dfrac {GM^{2}}{(\sqrt {2}R)^{2}} \dfrac {1}{\sqrt {2}} + \dfrac {GM^{2}}{(\sqrt {2}R)^{2}} \dfrac {1}{\sqrt {2}} + \dfrac {GM^{2}}{(2R)^{2}} = \dfrac {2GM^{2}}{2\sqrt {2}R^{2}} + \dfrac {GM^{2}}{4R^{2}} = \dfrac {GM^{2}}{R^{2}}\left [\dfrac {1}{\sqrt {2}} + \dfrac {1}{4}\right ]$$ along $$AO$$This must provide the necessary centripetal forceHence, $$\dfrac {GM^{2}}{R^{2}}\left [\dfrac {1}{\sqrt {2}} + \dfrac {1}{4}\right ] =\dfrac {mv^{2}}{R}$$$$\Rightarrow v = \sqrt {\dfrac {GM}{R} \left [\dfrac {2\sqrt {2} + 1}{4}\right ]}$$.Physics

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