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 :

• A. $\sqrt{\frac{G}{2 R^{3}}}$
• B. $\frac{1}{2} \sqrt{\frac{G}{R^{3}}}$
• C. $\frac{1}{2 R} \sqrt{\frac{1}{G}}$
• D. $\sqrt{\frac{2 G}{R^{3}}}$

Answer 1

Answer: (B)

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