Two identical particles of mass 1 , kg each g | vedclass

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(B) The two particles of mass $m = 1 \, kg$ are separated by a distance $d = 2R$.The gravitational force between them provides the necessary centripet

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$\frac{1}{2} \sqrt{\frac{G}{R^{3}}}$

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(B) The two particles of mass $m = 1 \, kg$ are separated by a distance $d = 2R$.The gravitational force between them provides the necessary centripetal force for circular motion.The gravitational force is given by $F = \frac{G m^2}{(2R)^2}$.The centripetal force required for a particle of mass $m$ moving in a circle of radius $R$ with angular speed $\omega$ is $F_c = m R \omega^2$.Equating the two forces: $\frac{G m^2}{4 R^2} = m R \omega^2$.Simplifying for $\omega^2$: $\omega^2 = \frac{G m}{4 R^3}$.Given $m = 1 \, kg$,we get $\omega^2 = \frac{G}{4 R^3}$.Taking the square root,$\omega = \frac{1}{2} \sqrt{\frac{G}{R^3}}$.

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

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:

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