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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 :<div class=

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 :

Option: 1

\sqrt{\frac{Gm}{a^{3}}}


Option: 2

\sqrt{\frac{Gm}{4a^{3}}}


Option: 3

\sqrt{\frac{Gm}{2a^{3}}}


Option: 4

\sqrt{\frac{Gm}{8a^{3}}}


Answers (1)

best_answer

\begin{aligned} & \mathrm{F}_{\mathrm{C}}=m \omega^2 \mathrm{a} \\ & \frac{\mathrm{Gm}^2}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\ & \omega=\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{a}^3}} \end{aligned}                 

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