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Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is:

Option 1 -

√( (1/2) * (G/R³) )

Option 2 -

√( (1/2) * (G/R³) )

Option 3 -

√( (1/2R) * (1/G) )

Option 4 -

√(2G/R³)

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    F = Gm²/ (2R)² = mω²R Given m = 1kg
    ⇒ ω² = Gm/ (4R³)
    ⇒ ω = (1/2)√ (G/R³)

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