Three identical bodies of mass M are located at the vertices of an equilateral triangle with side L. At what speed must they move if they all revolve under the influence of one another's gravity in a circular orbit circumscribing the triangle while still preserving the equilateral triangle
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The radius of the circle is L3√
The net gravitational force acting on any particle
F=Gm2L2(cos300+cos300)=3√Gm2L2
This force must be equal to the centripetal force on the particle. so,
3√Gm2L2=mv2L3√/⇒v2=GmL⇒v=GmL‾‾‾√
So the speed of the particle is GmL‾‾‾√
mark me brainlist !!!!
The radius of the circle is L3√
The net gravitational force acting on any particle
F=Gm2L2(cos300+cos300)=3√Gm2L2
This force must be equal to the centripetal force on the particle. so,
3√Gm2L2=mv2L3√/⇒v2=GmL⇒v=GmL‾‾‾√
So the speed of the particle is GmL‾‾‾√
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