Three particles each of mass m = 1 kg , are located at the vertices of an equilateral triangle of side a = 4 m . The speed at which they must move to revolve in a circular orbit, under the influence o

Question

Three particles each of mass m = 1 kg , are located at the vertices of an equilateral triangle of side a = 4 m . The speed at which they must move to revolve in a circular orbit, under the influence of mutual gravitational force, circumscribing and preserving the equilateral triangle is v = n G , then the value of n is

TestHub · Accepted Answer

The net gravitational force on any particle due to the other two is
$F_{g} = \frac{2 G m^{2}}{a^{2}} \cos (30 °) = \frac{\sqrt{3} G m^{2}}{a^{2}}$
Now, $r = \frac{a}{\sqrt{3}}$

For circular motion,

$F_{g} = \frac{\sqrt{3} G m^{2}}{a^{2}} = \frac{m v^{2}}{r}$

Solving it we obtain,
$v = 0.50 \sqrt{G}$

Physics - Gravitation Question with Solution | TestHub

Doubts & Discussion