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CGPET 2015 Solved Paper

Section: Physics

Question : 5 of 150

Marks: +1, -0

M

R

Solution:

Consider the movement of three bodies along a circle centered at as shown in the diagram alongside.

Gravitational force between A and B

F_{BA} = \frac{Gm^2}{(AB)^2}

Gravitational force between B and C

F_{BC} = \frac{Gm^2}{(BC)^2}

From geometry of the figure, we can write

BO \cos 30^{\circ} + CO \cos 30^{\circ} = BC

\Rightarrow AB = BC = CA = 2R \cos 30^{\circ} = \sqrt{3}R

Net force on B,

F_B = F_{BA} \cos 30^{\circ} + F_{BC} \cos 30^{\circ}

= \frac{2Gm^2}{(AB)^2} \times \frac{\sqrt{3}}{2} = \frac{2Gm^2}{(\sqrt{3}R)^2} \times \frac{\sqrt{3}}{2} = \frac{2Gm^2}{3R^2} \times \frac{\sqrt{3}}{2}

This force will provide centripetal force for circular motion of B

F_B = \frac{2Gm^2}{3R^2} \times \frac{\sqrt{3}}{2} = \frac{mv_B^2}{R} = \frac{G}{\sqrt{3}R} = v_B^2

\Rightarrow v_B = \sqrt{ \frac{G}{\sqrt{3}R} }

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