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 :
- A \(\frac{1}{2 R} \sqrt{\frac{1}{G}}\)
- B \(\frac{1}{2} \sqrt{\frac{G}{R^{3}}}\)
- C \(\sqrt{\frac{2 G}{R^{3}}}\)
- D \(\sqrt{\frac{G}{2 R^{3}}}\)
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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