PhysicsMedium159×since 2002Q7280Two identical particles each of mass ' mmm ' go round a circle of radius aaa under the action of their mutual gravitational attraction. The angular speed of each particle will be :AGm2a3\sqrt{\frac{G m}{2 a^{3}}}2a3GmBGma3\sqrt{\frac{G m}{a^{3}}}a3GmCGm8a3\sqrt{\frac{G m}{8 a^{3}}}8a3GmDGm4a3\sqrt{\frac{G m}{4 a^{3}}}4a3GmCheck answerSkip