Four particles of equal masses ‘M’ move along circle of radius ‘R’ under the action of their mutual gravitational attraction. Find the speed of each particle.

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22−1M×GMR12

22+14×GMR12

22+14×GMR2

22−1M×GMR2

answer is B.

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Detailed Solution

The force on M at ‘A’ Force of attraction between two masses=Gmmr2; universal gravitation constant=G; mass=m; distance=r=d=2R=AB=ADand AC=2R; The force on M at ‘A’ are F1,F2,F3 F1=F2=GM22R2 resultant of F1,F2is F1=F12+F22 here F1=F2=F F1=2F F1=GM22R2 F3=GM24R2 Net force on mass at A is net force=Fnet=F1+F3Fnet=2F+F3Fnet=GM2R214+12 along AOThis force provide necessary centripetal force F=Mv2rMv2r=Fnetsubstitute∴Mv2r=GM2R214+12V=22+14×GMR12= speed of each particle

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