Question

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

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

Answer 1

The correct option is B $\frac{1}{2}\sqrt{\frac{G}{R^3}}$

Due to mutual gravitational attraction, the particles will be at the position shown below.


From FBD we have
$F_g=F_c$
$\Rightarrow \frac{G{m}^2}{{\left(2R\right)}^2}=mR{\omega }^2$
$\Rightarrow \omega =\frac{1}{2}\sqrt{\frac{G}{R^3}}$