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Two particles of equal mass $m$ go round a circle of radius $\mathrm{R}$ under the action of their mutual gravitational attraction. The speed of each particle is
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Verified Answer
The correct answer is:
$\frac{1}{2} \sqrt{\frac{G m}{R}}$
From given condition
$\frac{\mathrm{Gmm}}{(2 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}, \mathrm{v}=\frac{\mathrm{Gm}}{4 \mathrm{R}}=\frac{1}{2} \sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$
$\frac{\mathrm{Gmm}}{(2 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}, \mathrm{v}=\frac{\mathrm{Gm}}{4 \mathrm{R}}=\frac{1}{2} \sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$
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