Two particles of equal mass ' m ' move in a circle of radius ' r ' under the action of their mutual gravitational attraction. The speed of each particle will be ( G = Universal gravitational constant)

Step-by-step solution

Two particles of equal mass m orbit each other in circular paths of radius r under mutual gravitational attraction. The center of mass lies midway between them, so the separation distance is d = 2r . The gravitational force providing the centripetal acceleration is F_g = G m^2 (2r)^2 = G m^2 4r^2 . For circular motion with speed v , the required centripetal force is F_c = m v^2 r . Equating these gives G m^2 4r^2 = m v^2 r , which simplifies to v^2 = G m 4r . Each particle's speed is therefore v = G m 4r , correspo