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(A) For a very long cylindrical mass distribution,the gravitational field $E$ at a distance $r$ $(r > R)$ from the axis is given by Gauss's Law for gr

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$T \propto r$

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(A) For a very long cylindrical mass distribution,the gravitational field $E$ at a distance $r$ $(r > R)$ from the axis is given by Gauss's Law for gravitation: $E = \frac{2GM}{Lr}$.The gravitational force acting on a star of mass $m$ is $F = mE = \frac{2GMm}{Lr}$.Since the star is orbiting in a circular path,this gravitational force provides the necessary centripetal force: $\frac{mv^2}{r} = \frac{2GMm}{Lr}$.Substituting $v = r\omega = r(\frac{2\pi}{T})$,we get: $m r (\frac{2\pi}{T})^2 = \frac{2GMm}{Lr}$.Simplifying the equation: $r \frac{4\pi^2}{T^2} = \frac{2GM}{Lr}$.Rearranging for $T^2$: $T^2 = \frac{4\pi^2 L}{2GM} r^2$.Therefore,$T^2 \propto r^2$,which implies $T \propto r$.

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