Question:
Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is:
$\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$
$\sqrt{2 \sqrt{2} \frac{G M}{R}}$
$\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$
$\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$
Question of from chapter.
JEE Main Previous Year 2014
Correct Option: 4
Solution:
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