ExamBro
ExamBro
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.
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)

Step-by-step Solution

Detailed explanation

Using newtons law of gravitation

\(F=\frac{G m^{2}}{(2 R)^{2}}=m R \omega^{2}\)

\(\omega=\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)

Same subject
Explore more questions on app