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JEE Mains · Physics · STD 11 - 7. gravitation

Two identical particles each of mass \(m\) go round a circle of radius \(a\) under the action of their mutual gravitational attraction. The angular speed of each particle will be

  1. A \(\sqrt{\frac{G m}{2 a^3}}\)
  2. B \(\sqrt{\frac{G m}{8 a^3}}\)
  3. C \(\sqrt{\frac{G m}{4 a^3}}\)
  4. D \(\sqrt{\frac{G m}{a^3}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt{\frac{G m}{4 a^3}}\)

Step-by-step Solution

Detailed explanation

\(F = m \omega^2 r\) \(\Rightarrow \frac{G m m}{(2 a)^2}=m \omega^2 a\) \(\Rightarrow \omega=\sqrt{\frac{G m}{4 a^3}}\)
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