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Schwarzschild radius of black hole is proportional to its mass. From here we can deduce that black hole density getting lower as black hole grows in pace that is inverse to the square of mass. If it grows big enough - density of the hole becomes less than density of surrounding vacuum (background radiation + intergalactic gas, etc...).

So can we expect that black hole can start experiencing "Archimedes force" and we can observe its movements against surrounding gravitational potential?

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The only problem with your hypothetical ultralight black hole is that it would be big. Very big.

First, note that if you had a black hole with mass $M$ and corresponding volume $V$ such that $M/V < \rho_\mathrm{space}$, then a volume $V$ of a typical patch of space would have more than enough mass to be a black hole. This should worry you if you thought you'd find black holes like this in abundance, since it would mean most every thing would be a large black hole.

Let's look at some numbers. The average density of ordinary matter in the universe is $0.2$ particles per cubic meter, which we'll take to be hydrogen. That gives us $\rho = 3.3\times10^{-28}\ \mathrm{kg/m^3}$. A black hole of mass $M$ has Schwarzschild radius $R = 2GM/c^2$, and "volume" $V = (4\pi/3) R^3$. Putting everything together, we have the density formula $\rho = M/V = (3/8\pi) c^2/GR^2$, or $$ R = \sqrt{\frac{3c^2}{8\pi G\rho}} = 7\times10^{26}\ \mathrm{m} = 7\times10^{10}\ \mathrm{ly}. $$ Thus, such a black hole would be about the size of the observable universe.

This is not a coincidence either. Our universe's matter density is not too far from the critical density that would, in the absence of dark energy, determine whether it expanded forever or eventually contracted.

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  • $\begingroup$ Thank you for bringing numbers and show connection with the size of the observable universe. $\endgroup$ – hOff Mar 23 '15 at 2:14

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