JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2021
- Medium
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)
Step-by-step Solution
Detailed explanation
Using newtons law of gravitation
\(F=\frac{G m^{2}}{(2 R)^{2}}=m R \omega^{2}\)
\(\omega=\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)
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