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JEE Mains · Physics · STD 11 - 7. gravitation

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 :

  1. A \(\frac{1}{2 R} \sqrt{\frac{1}{G}}\)
  2. B \(\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)
  3. C \(\sqrt{\frac{2 G}{R^{3}}}\)
  4. D \(\sqrt{\frac{G}{2 R^{3}}}\)
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 \(F=\frac{G m^{2}}{(2 R)^{2}}=m R \omega^{2}\) \(\omega=\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)
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