The net gravitational force at the centre of | vedclass

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(A) Let the distance from the centre to each corner be $r$. Let a test mass $m_{0}$ be placed at the centre.In the initial configuration,the forces ex

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$2$

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(A) Let the distance from the centre to each corner be $r$. Let a test mass $m_{0}$ be placed at the centre.In the initial configuration,the forces exerted by the masses at the corners on $m_{0}$ are directed towards the corners. Let $k = \frac{Gm_{0}}{r^{2}}$.The forces are $F_{M} = kM$,$F_{2M} = 2kM$,$F_{3M} = 3kM$,and $F_{4M} = 4kM$.Opposite pairs are $(M, 3M)$ and $(2M, 4M)$.The net force along the diagonal with $M$ and $3M$ is $F_{diag1} = (3M - M)k = 2kM$ (towards $3M$).The net force along the diagonal with $2M$ and $4M$ is $F_{diag2} = (4M - 2M)k = 2kM$ (towards $4M$).Since the diagonals are perpendicular,$F_{1} = \sqrt{(2kM)^{2} + (2kM)^{2}} = 2\sqrt{2}kM$.In the new configuration,$3M$ and $4M$ are interchanged. The corners now have $M, 2M, 4M,$ and $3M$ in order.Opposite pairs are $(M, 4M)$ and $(2M, 3M)$.The net force along the diagonal with $M$ and $4M$ is $F'_{diag1} = (4M - M)k = 3kM$ (towards $4M$).The net force along the diagonal with $2M$ and $3M$ is $F'_{diag2} = (3M - 2M)k = kM$ (towards $3M$).$F_{2} = \sqrt{(3kM)^{2} + (kM)^{2}} = \sqrt{9+1}kM = \sqrt{10}kM$.Ratio $\frac{F_{1}}{F_{2}} = \frac{2\sqrt{2}kM}{\sqrt{10}kM} = \frac{2\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{2}{\sqrt{5}}$.Comparing with $\frac{\alpha}{\sqrt{5}}$,we get $\alpha = 2$.

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