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(D) The gravitational potential $V$ at a distance $r$ from a point mass $m$ is given by $V = -\frac{Gm}{r}$.For a continuous distribution of mass,the

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$-\frac{\pi GM}{L}$

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(D) The gravitational potential $V$ at a distance $r$ from a point mass $m$ is given by $V = -\frac{Gm}{r}$.For a continuous distribution of mass,the potential at the center of a circular arc is the same for every point on the arc because every point is at the same distance $r$ from the center.Given that the length of the rod is $L$ and it is bent into a semicircle,the radius $r$ of the semicircle is related to the length by the formula for the circumference of a semicircle: $\pi r = L$.Therefore,the radius is $r = \frac{L}{\pi}$.Since the total mass $M$ is distributed uniformly along the arc,the gravitational potential at the center is $V = -\frac{GM}{r}$.Substituting the value of $r$,we get $V = -\frac{GM}{(L/\pi)}$.Thus,the gravitational potential at the center is $V = -\frac{\pi GM}{L}$.

$A$ thin rod of length $L$ is bent to form a semicircle. The mass of the rod is $M$. What will be the gravitational potential at the center of the circle?

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