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प्रश्न
Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.
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उत्तर
Assume that three particles are at points A, B and C on the circumference of a circle.
BC = CD = \[\sqrt{2}a\]
The force on the particle at C due to gravitational attraction of the particle at B is
\[\overrightarrow{F}_{CB} = \frac{G M^2}{2 R^2} \hat j\] .
The force on the particle at C due to gravitational attraction of the particle at D is
\[\overrightarrow{F}_{CD} = - \frac{G M^2}{2 R^2} \hat i\].
Now, force on the particle at C due to gravitational attraction of the particle at A is given by
\[\overrightarrow{F}_{CA} = - \frac{G M^2}{4 R^2}\cos 45 \hat i + \frac{G M^2}{4 R^2}\sin 45 \hat j\]
\[ \therefore \overrightarrow{F}_C = \overrightarrow{F}_{CA} + \overrightarrow{F}_{CB} + \overrightarrow{F}_{CD} \]
\[ = \frac{- G M^2}{4 R^2}\left( 2 + \frac{1}{\sqrt{2}} \right) \hat i + \frac{G M^2}{4 R^2}\left( 2 + \frac{1}{\sqrt{2}} \right) \hat j\]
So, the resultant gravitational force on C is \[F_C = \frac{G m^2}{4 R^2}\sqrt{2\sqrt{2} + 1}\]
Let v be the velocity with which the particle is moving.
Centripetal force on the particle is given by
\[F = \frac{m v^2}{R}\]
\[ \Rightarrow v = \sqrt{\frac{GM}{R}\left( \frac{2\sqrt{2} + 1}{4} \right)}\]
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