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Now, ##g = \frac{Gm}{r^2}## in this case and substituting gives ##u_g = \frac{-GM^2}{8 \pi r^4}##. Integrating this over volume will give ##U = \iiint \frac{GM^2}{8 \pi r^4} dv = \frac{-GM^2}{2R}##.

Is my solution correct?

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- Thread starter Ananthan9470
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- #1

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Now, ##g = \frac{Gm}{r^2}## in this case and substituting gives ##u_g = \frac{-GM^2}{8 \pi r^4}##. Integrating this over volume will give ##U = \iiint \frac{GM^2}{8 \pi r^4} dv = \frac{-GM^2}{2R}##.

Is my solution correct?

- #2

TSny

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Yes, that looks correct.

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