Question 182 of 223

Two identical particles of mass $1kg$ each go round a circle of radius $R$, under the action of their mutual gravitational attraction. The angular speed of each particle is:

A$\sqrt{\dfrac{G}{2{R}^{3}}}$B$\dfrac{1}{2}\sqrt{\dfrac{G}{{R}^{3}}}$C$\dfrac{1}{2R}\sqrt{\dfrac{1}{G}}$D$\sqrt{\dfrac{2G}{{R}^{3}}}$

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Answer: (B) $\dfrac{1}{2}\sqrt{\dfrac{G}{{R}^{3}}}$

$F=\dfrac{G{m}^{2}}{(2R{)}^{2}}=mR{\omega }^{2}$
$\omega =\dfrac{1}{2}\sqrt{\dfrac{G}{{R}^{3}}}$