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The Schwarzschild Radius is a theoretical radius, that if the mass of the object was compressed into that certain radius, the escape speed from the surface of the compressed object would equal the speed of light, therefore creating a black hole. But I wanted to know, under Earth's gravitational pull, how dense would an object have to be to reach its Schwarzschild Radius?

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I'm not quite sure how the density of the object plays into it. The Earth would have to become a black hole for it to be possible for an object to reach the Schwarzschild radius. – HDE 226868 Aug 12 '14 at 12:33
up vote 2 down vote accepted

Earth's gravity is quite weak. It manages to compress air (luckily for us), but that's about it. Since the Earth is not a black hole it can never compress something else to be a black hole.

That being said, lets consider some object and assume something compresses it to a black hole.

The Schwarzschild radius is defined as $$ R_s = \frac{2GM}{c^2}. $$ Now lets say the object is a sphere of radius $R_s$, then its uniform density is $$ \rho = \frac{M}{ \frac{4\pi}{3} R_s^3 }. $$ Using this to get rid of the mass term leaves us with $$ R_s = \frac{2G \rho \frac{4\pi}{3} R_s^3}{c^2} $$ Inverting this expression gives the black hole density for a given radius $$ \rho = \frac{3 c^2}{8 \pi G R_s^2}. $$

Notice that if you want a really small black hole, say $R=1$ m, the density has to be about $1.6\times10^{26}$ kg/m$^3$. That's about 20 times the mass of Earth stuffed in a cubic meter, so quite a lot.

But if the black hole can be as large as we please, then the density can get extremely low. A great example of this are supermassive black holes, which have about the same density as water (that's $10^3$ kg/m$^3$).

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