# 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

$\begin{array}{1 1}(A)\;\sqrt{\frac{GM}{R} (1+2 \sqrt 2)} \\ (B)\;\frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt 2)} \\(C)\;\sqrt{\frac{GM}{R} } \\(D)\;\sqrt{2 \sqrt 2 \frac{GM}{R} }\end{array}$

The speed of each particle is $\frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt 2)}$
Hence B is the correct answer.

Net force acting on any one particle M=GM²/(2R)²+GM²/(R√2)²Cos45°+GM²/(R√2)²Cos45° M=GM²/R² [1/4+1/√2] This force will equal to centripetal force So,. Mv²/R=GM²/R²[1/4+1/√2] V=√[GM²/4R(1+2√2)] V=1/2√[GM/R(2√2+1)]