PhysicsHard159×since 2002Q7261Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :AG2(1+22)\sqrt {{G \over 2}(1 + 2\sqrt 2 )}2G(1+22)BG2(22−1)\sqrt {{G \over 2}(2\sqrt 2 - 1)}2G(22−1)CG(1+22)\sqrt {G(1 + 2\sqrt 2 )}G(1+22)D12G(1+22){1\over2}\sqrt {G(1 + 2\sqrt 2 )}21G(1+22)Check answerSkip