I know that for basic shapes like a solid cube or a solid sphere the proof is simple. But if we had a homogeneous rigid body with an arbitrary shape, can we prove that the body's "Center of mass" is exactly at its "Geometric center"?
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It follows from the definition. Given a volume $V$ with mass $M$ and density $\rho(\textbf{r})$ the position of the center of mass is:
$$\textbf{r}_{CM}={1\over M}\int_V \rho(\textbf{r}) \textbf{r} dV$$
However if $\rho(\textbf{r}) = \rho_0={M\over V}$ i.e. the system is homogeneous
$$\textbf{r}_{CM}={1\over M}{M\over V}\int_V \textbf{r} dV={1\over V}\int_V \textbf{r} dV$$ which now is the definition of geometric center.
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$\begingroup$ In general relativity theory this is not so clear anymore., see e.g. nsf.gov/awardsearch/showAward?AWD_ID=1005560 $\endgroup$ Commented Nov 6, 2016 at 18:30
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$\begingroup$ nice, I will look into it. Yet I think OP did not have that in mind (: $\endgroup$– JalfredPCommented Nov 6, 2016 at 18:37
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1$\begingroup$ OP did not list general relativity as a topic. @JalfredP's answer is perfectly good. BTW, it is a simple result from first year analytical mechanics (classical). $\endgroup$– Bob BeeCommented Nov 7, 2016 at 0:34