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Three small identical bodies each of mass \mathrm{m} are moving in circular orbit around a fixed point with same angular velocity under their gravitational interaction. If the separation between any two bodies is \mathrm{R}, the total energy possessed by the system is given by
 

Option: 1

\mathrm{-\frac{3 G M^2}{2 R}}


 


Option: 2

\mathrm{-\frac{3 G M^2}{4 R}}
 


Option: 3

\mathrm{-\frac{3 G M^2}{2 R \cos 30^{\circ}}}
 


Option: 4

\mathrm{-\frac{3 G M^2}{R}}


Answers (1)

best_answer

In the figure,\mathrm{ O A=\frac{2}{3} R \cos 30^{\circ}}

Net force towards \mathrm{ O} on the mass \mathrm{ A}

\mathrm{ 2\left(\frac{G M^2}{R^2}\right) \cos 30^{\circ}=M \omega^2\left(\frac{2}{3} R \cos 30^{\circ}\right) }

\mathrm{\Rightarrow \omega^2=\frac{3 G M}{R^3}}

\mathrm{\text { Total energy } 3[K E+P E]=3\left[\frac{1}{2} M^2+\left(\frac{-G M \cdot M}{R}\right)\right] }

\mathrm{=3\left[\frac{1}{2} M\left(\frac{2}{3} R \cos 30\right)^2 \frac{3 G M}{R^3}-\frac{G M^2}{R}\right]=-\frac{3}{2} \frac{G M^2}{R} }

Hence option 1 is correct.



 





 

Posted by

manish painkra

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