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Question

Because of their mutual gravitational attraction, four identical planets each of mass m are orbiting in a circular path of radius r in the same sense (angular direction). The magnitude of the velocity of each planet is :

Options

A.[(Gm/r)((1 + 2√2)/4)]1/2
B.3√(Gm/r)
C.√[(Gm/r)(1 + 2√2)]
D.[(1/2)(Gm/r)((1 + √2)/2)]1/2

Solution

[Key Concept: Each planet's centripetal force is the resultant of the gravitational pulls of the other three planets at the square's vertices.]

Net force on planet A is :

$\begin{matrix} & F = F_{1}\sqrt{2} + F_{0} \\ & F = \frac{Gm^{2}}{(r\sqrt{2})^{2}}\sqrt{2} + \frac{Gm^{2}}{(2r)^{2}} \\ & F = \frac{Gm^{2}}{r^{2}}\left\lbrack \frac{1}{\sqrt{2}} + \frac{1}{4} \right\rbrack = \frac{mv^{2}}{r} \\ & v = \sqrt{\frac{Gm}{r}\frac{(1 + 2\sqrt{2})}{4}} \end{matrix}$.

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