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Question:

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:

12√GMR(1+2√2)

√GMR(1+2√2)

√GMR

√2√2GMR

Solution:

Net force on any one particle
GM²/ (2R)² + GM²/ (R√2)² cos45° + GM²/ (R√2)² cos45° = GM²/R²[1/4 + 1/√2]
This force will be equal to centripetal force so
Mv²/R = GM²/R²[1 + 2√2/4]
v = √(GM/4R)[1 + 2√2] = 1/2√GMR(2√2+1)