A thin rod of length L is bent to form a semi | vedclass

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Question

Gravitational potential of the centre $=\frac{-\mathrm{GM}}{\mathrm{r}}$

since, $\pi r=L \Rightarrow r=\frac{L}{\pi}$

Gravitational potential $=

Vedclass · Accepted Answer

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

Vedclass · Answer

Gravitational potential of the centre $=\frac{-\mathrm{GM}}{\mathrm{r}}$

since, $\pi r=L \Rightarrow r=\frac{L}{\pi}$

Gravitational potential $=\frac{-\mathrm{GM}}{(\mathrm{L} / \pi)}$

$=\frac{-\pi G M}{L}$

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

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