Four 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:

Step-by-step solution

F 1 = G m m 2 R 2 = G m 2 4 R 2   F 2 = G m m 2 R 2 = G m 2 2 R 2   F 3 = G m m 2 R 2 = G m 2 2 R 2 ⇒ F net = F 1 + F 2 cos 45 ° + F 3 cos 45 ° = G m 2 4 R 2 + G m 2 2 R 2 1 2 + G m 2 2 R 2 1 2 = G m 2 R 2 1 4 + 1 2 = G m 2 4 R 2 1 + 2 2 F net = G m 2 4 R 2 1 + 2 2 = m v 2 R ⇒ v = G 1 + 2 2 2