Question:
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 :
Correct Option: , 2
Solution:
$F=\frac{\mathrm{Gm}^{2}}{(2 \mathrm{R})^{2}}=m R \omega^{2}$
$\omega=\frac{1}{2} \sqrt{\frac{G}{R^{3}}}$
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