Question

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.

Answer 1

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ANSWER::

Force on M at C due to gravitational attraction

Force in CB = Gm²j / 2R²

Force in CD = -Gm²i / 4R²

Force in CB = [-Gm² cos45° j / 4R²] + [GM² sin45° j / 4R²]

Resultant force on C ,

Force in C = Force in CA + Force in CB + Force in CD

Force in C = (-GM²/4R²)(2+ 1/√2)i + (GM²/4R²)(2+ 1/√2)j

Force in C = (GM²/4R²)(2√2+1)

For moving along circle ,

Force , F = mv²/R or (GM²/4R²)(2√2+1) = MV²/R or V

= √[(GM/R){(2√2+1)/4}]

Hope it helps!