Question

# Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Solution

## $$\begin{array}{l}f = \dfrac{{Gm.m}}{{{{\left( {2R} \right)}^2}}} = \dfrac{{G{m^2}}}{{\left( {4R} \right)}}\\f = \dfrac{{m{v^2}}}{R}\\\dfrac{{G{m^2}}}{{\left( {4R} \right)}} = \dfrac{{m{v^2}}}{R}\\v = \dfrac{1}{2}\sqrt {\dfrac{{Gm}}{R}} \end{array}$$Physics

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