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)

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month 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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