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Physics

General

Easy

Question

Two particle of equal mass go round a circle of radius r. Under the action of their mutual gravitational force. The speed of each particle is …….

v= $\frac{1}{2 r} \sqrt{\frac{1}{G m}}$
v= $\sqrt{\frac{G m}{2 r}}$
v= $\frac{1}{2} \sqrt{\frac{G m}{r}}$
v= $\sqrt{\frac{4 G m}{r}}$

The correct answer is: v= $1 half square root of fraction numerator G m over denominator r end fraction end root$

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