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(D) Let the vertices of the equilateral triangle be $P, Q,$ and $R$. The mass $m_2$ is placed at point $S$,which is the midpoint of side $QR$.The grav

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$\frac{12 Gm_1 m_2}{L^2}$

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(D) Let the vertices of the equilateral triangle be $P, Q,$ and $R$. The mass $m_2$ is placed at point $S$,which is the midpoint of side $QR$.The gravitational forces exerted by the masses at $Q$ and $R$ on the mass $m_2$ at $S$ are equal in magnitude and opposite in direction because $S$ is the midpoint of $QR$. Thus,these two forces cancel each other out.The net force on $m_2$ is only due to the mass $m_1$ located at vertex $P$.In $	riangle PQS$,the side $PQ = \frac{L}{3}$ and $\angle PQS = 60^{\circ}$. The distance $h$ (the altitude from $P$ to $QR$) is given by $h = PQ \sin 60^{\circ} = \frac{L}{3} \cdot \frac{\sqrt{3}}{2} = \frac{L\sqrt{3}}{6}$.The gravitational force $F$ on $m_2$ due to $m_1$ at $P$ is:$F = \frac{G m_1 m_2}{h^2} = \frac{G m_1 m_2}{(\frac{L\sqrt{3}}{6})^2} = \frac{G m_1 m_2}{\frac{3L^2}{36}} = \frac{G m_1 m_2}{\frac{L^2}{12}} = \frac{12 G m_1 m_2}{L^2}$.

$A$ system consists of three particles each of mass $m_1$ placed at the corners of an equilateral triangle of side $\frac{L}{3}$. $A$ particle of mass $m_2$ is placed at the midpoint of any one side of the triangle. Due to the system of particles,the force acting on $m_2$ is

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