Gravitation

JEE Main level, single correct, medium

7 of 223

Question 7 of 223

Two identical particles each of mass mm go round a circle of radius aa under the action of their mutual gravitational attraction. The angular speed of each particle will be :

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Answer: (C) Gm4a3\sqrt{\dfrac{Gm}{4{a}^{3}}}

The gravitational force (FG)\left({F}_{G}\right) between the particles is given by
FG=Gm2(2a)2=Gm24a2...(1)\begin{matrix}{F}_{G} & = & \dfrac{G{m}^{2}}{{\left(2a\right)}^{2}} \\ & = & \dfrac{G{m}^{2}}{4{a}^{2}}...\left(1\right)\end{matrix}
The centripetal force (FC)\left({F}_{C}\right) of each particle can be written as
FC=mω2a...(2){F}_{C}=m{\omega }^{2}a...\left(2\right)
Under balanced condition, equate equation (1) and equation (2) and simplify to obtain the angular speed of each particle.
mω2a=Gm24a2⇒ω2=Gm4a3⇒ω=Gm4a3\begin{matrix}m{\omega }^{2}a & = & \dfrac{G{m}^{2}}{4{a}^{2}} \\ & \Rightarrow & {\omega }^{2}=\dfrac{Gm}{4{a}^{3}} \\ & \Rightarrow & \omega =\sqrt{\dfrac{Gm}{4{a}^{3}}}\end{matrix}

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