Question

Three particles each of mass m are located at the vertices of an equilateral triangle of side a. At what speed must they move, if they all revolve under the influence of their gravitational force of attraction in a circular orbit circumscribing the triangle while still preserving the equilateral triangle?

Moderate

Solution

From the figure
$F_{A} = F_{A B} + F_{A C} = 2 [\frac{G m^{2}}{a^{2}}] \cos 30 ° = [\frac{G m^{2}}{a^{2}} \cdot \sqrt{3}]$
Also, $r = \frac{a}{\sqrt{3}}$
$Now, \frac{m v^{2}}{r} = |F_{A}| or \frac{m v^{2} \sqrt{3}}{a} = \frac{G m^{2}}{a^{2}} \sqrt{3}$
$∴ v = \sqrt{\frac{G m}{a}}$

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