question_answer

If three particle of equal mass \['m'\] is equidistance from each other and doing motion on circular path having radius r due to their gravitational interaction. then their speed v will be :

A)  \[\sqrt{\frac{Gm}{r}}\]                

B)  \[\sqrt{\frac{Gm}{3r}}\]

C)  \[\sqrt{\frac{Gm}{r\sqrt{3}}}\]                

D)  \[\sqrt{\frac{3Gm}{r}}\]

Correct Answer: C

Solution :

Net gravitation for on A \[=\frac{G{{m}^{2}}}{{{(2r\cos 30)}^{2}}}\times \cos 30\times 2=\frac{m{{v}^{2}}}{r}\] \[=\frac{2G{{m}^{2}}\cos 30}{4{{r}^{2}}\cos 30\times \cos 30}=\frac{m{{v}^{2}}}{r}\] \[v=\sqrt{\frac{Gm}{r\sqrt{3}}}\]