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JEE Main202315 Apr 2023Morning ShiftPhysicsGravitationActual

Two identical particles each of mass m go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be :

Options

  1. AG m a 3
  2. BG m 8 a 3
  3. CG m 4 a 3
  4. DG m 2 a 3

Correct answer

C. G m 4 a 3

Step-by-step solution

The gravitational force F G between the particles is given by F G = G m 2 2 a 2 =   G m 2 4 a 2       . . . 1 The centripetal force F C of each particle can be written as F C = m ω 2 a       . . . 2 Under balanced condition, equate equation (1) and equation (2) and simplify to obtain the angular speed of each particle. m ω 2 a = G m 2 4 a 2 ⇒ ω 2 = G m 4 a 3 ⇒ ω = G m 4 a 3

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