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:
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 - <p>√( (1/2) * (G/R³) )</p>
Option 2 - <p>√( (1/2) * (G/R³) )</p>
Option 3 - <p>√( (1/2R) * (1/G) )<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>√(2G/R³)</p>
2 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
V
Answered by
7 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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