Question

Three identical bodies (each mass M) are placed at
vertices of an equilateral triangle of arm L, keeping
the triangle as such by which angular speed the
bodies should be rotated in their gravitational fields
so that the triangle moves along circumference of
circular orbit :
3GM
(1)
L​

Answer 1

$\huge\pink{\overbrace{\purple{\underbrace{\blue{\mathbf{:Answer:}}}}}}$

• √(3GM/L³) ---> option A

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$\huge\pink{\overbrace{\purple{\underbrace{\blue{\mathbf{Explanation}}}}}}$

As shown in figure, Fc= Centripetal force and Fg= Gravitational force. We take upper body for calculation.Let angular speed of bodies are ω.

Given side of equilateral triangle is L so,

radius of orbit =L/√3

===>Fc = (ML/√3) x ω²

===>Fg = GM²/L²

∴ Net gravitational force acting towards the center of the triangle Fnet= 2Fgcos(30°)

===>Fnet = ✓3GM²/L²,

===>Fnet = Fc ===> MLω²/√3 =√3GM²/L²

===>ω= √(3GM/L³) [ whole root]

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