Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :

Solution

Physics sir · 3 steps

Add the pulls of the other three masses along the line to the centre. That sum is the centripetal force.

1. Geometry first. The four masses sit at the corners of a square inscribed in the circle. Each side is √2R and each diagonal is 2R. By symmetry the net force on any one mass points straight at the centre.

   Pull from	Distance	Force	Part toward centre
   Adjacent mass 1	√2R	GM²/2R²	GM²/2R²cos 45° = GM²/2√2R²
   Adjacent mass 2	√2R	GM²/2R²	GM²/2√2R²
   Opposite mass	2R	GM²/4R²	GM²/4R² (already along the diameter)

2. Net inward force. The sideways parts of the two adjacent pulls cancel. The inward parts add: F = GM²/R²(1/√2 + 1/4) = GM²/4R²(2√2+1).

   Mv²/R = GM²/4R²(2√2+1) ⇒ v² = GM/4R(2√2+1)

3. So. v = 1/2√(GM/R(2√2+1)), which is option (b).

The idea to keep. For bodies orbiting under their mutual pull, resolve every force along the radius and add. That net inward force equals Mv²/R. Every other mass pulls inward, so no term is subtracted.

Answer from the NTA final answer key. Solution by PrepWiser.