Four particles,each of mass M and equidistant | vedclass

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(C) Consider one particle of mass $M$. It is acted upon by three other particles. Let the distance between adjacent particles be $a = R\sqrt{2}$ and t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Consider one particle of mass $M$. It is acted upon by three other particles. Let the distance between adjacent particles be $a = R\sqrt{2}$ and the distance between opposite particles be $d = 2R$.The gravitational force exerted by the two adjacent particles is $F = \frac{GM^2}{a^2} = \frac{GM^2}{2R^2}$. The components of these forces directed towards the center are $F \cos(45^{\circ})$ each.The gravitational force exerted by the diagonally opposite particle is $F' = \frac{GM^2}{d^2} = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}$.The net centripetal force required for circular motion is provided by the sum of these forces directed towards the center:$F_{net} = 2F \cos(45^{\circ}) + F' = \frac{Mv^2}{R}$Substituting the values:$2 \left( \frac{GM^2}{2R^2} \right) \frac{1}{\sqrt{2}} + \frac{GM^2}{4R^2} = \frac{Mv^2}{R}$$\frac{GM^2}{R^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Mv^2}{R}$$v^2 = \frac{GM}{R} \left( \frac{4 + \sqrt{2}}{4\sqrt{2}} \right) = \frac{GM}{R} \left( \frac{2\sqrt{2} + 1}{4} \right)$$v = \frac{1}{2} \sqrt{\frac{GM}{R} (1 + 2\sqrt{2})}$

Four particles,each of mass $M$ and equidistant from each other,move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

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