| Four identical particles of mass M are located at the corners of square of side 'a'. What should be their speed if each of them revolves under the influence of others gravitational field in a circular orbit circumscribing the square? |
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A) \[1.16\sqrt{\frac{GM}{a}}\]
B) \[1.21\sqrt{\frac{GM}{a}}\]
C) \[1.35\sqrt{\frac{GM}{a}}\]
D) \[1.41\sqrt{\frac{GM}{a}}\]
Correct Answer: A
Solution :
\[\frac{G{{M}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)=\frac{M{{V}^{2}}}{a/\sqrt{2}}\] \[\Rightarrow \] \[V=\sqrt{\frac{Gm}{a}}\sqrt{1+\frac{1}{2\sqrt{2}}}=1.16\sqrt{\frac{Gm}{a}}\]
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