JEE Main 2023 · 15 April · morning · Single correct · +4 / −1
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 :
- (a) √((G m)/(2 a³))
- (b) √((G m)/a³)
- (c) √((G m)/(8 a³))
- (d) √((G m)/(4 a³))
Answer and solution
Answer (d) √((G m)/(4 a³))
Solution
Physics sir · 4 steps
The two particles sit at opposite ends of a diameter, so they are 2a apart while each circles at radius a.
- Picture the motion. The only force on each particle is the pull of the other one. That pull must point at the centre of the circle, so the two particles stay at opposite ends of a diameter. Their separation is 2a, but each one moves on a circle of radius a.
- Gravity gives centripetal force. The gravitational force uses the separation 2a. The centripetal force uses the radius of the circle, a.
(G m²)/(2a)² = m ω² a
- Solve for omega. Cancel one m: (G m)/4a² = ω² a, so ω² = (G m)/4a³.
- So. ω = √((G m)/(4 a³)), which is option (d).
The idea to keep. In a two-body orbit, keep two distances apart: the separation goes into Newton's law, and each body's own radius about the centre goes into mω² r.
Answer from the NTA final answer key. Solution by PrepWiser.