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Two identical particles each of mass ' $m$ ' go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be:
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Verified Answer
The correct answer is:
$\sqrt{\frac{G m}{4 a^3}}$
Radius of circle, $\mathrm{r}=\mathrm{a}$
Centrifugal force, $F=m \omega^2 r$
$\begin{aligned}
& \Rightarrow \frac{\mathrm{Gmm}}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\
& \left(\because \mathrm{F}=\frac{\mathrm{G} \mathrm{M} \mathrm{M}_2}{\mathrm{~d}}\right)
\end{aligned}$
Here $M_1=m ; M_2=m$ and $d=2 a$
$\Rightarrow$ angular speed, $\omega=\sqrt{\frac{G m}{4 a^3}}$
Centrifugal force, $F=m \omega^2 r$
$\begin{aligned}
& \Rightarrow \frac{\mathrm{Gmm}}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\
& \left(\because \mathrm{F}=\frac{\mathrm{G} \mathrm{M} \mathrm{M}_2}{\mathrm{~d}}\right)
\end{aligned}$
Here $M_1=m ; M_2=m$ and $d=2 a$
$\Rightarrow$ angular speed, $\omega=\sqrt{\frac{G m}{4 a^3}}$
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