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What would the size of the universe be if it were physically possible to remove all of the empty space, leaving only matter?

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closed as off topic by mbq Jan 26 '11 at 21:48

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I think this question may be indeterminate. Consider the difference between an electron degenerate gas (as in a white dwarf star) and a neucleon degenerate gas (as in a heavy nucleus or a neutron star). Each is arguable matter with all the space taken out, but the real density is monstrously different. And of course, there is an upper limit to the stability with respect to gravity of matter at nuclear densities. –  dmckee Jan 26 '11 at 20:51
    
Way too basic, closed. –  mbq Jan 26 '11 at 21:51
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1 Answer 1

Yes, well making some assumptions, you would simply get a black hole with a radius $2GM/c^2$, so plug in values. You would have to decide if the radius is to the horizon of the observable universe. Wiki entry "observable universe" estimates the combined mass of normal and dark matter to be around $1.5 \times 10^{53}$ kg. The size of the "entire" universe cannot be estimated. And "empty space" is hard to characterize--as Frank Wilczek (via Sidney Coleman) says, "Nothing is unstable."

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or a FRW metric....question is not really possible to answer properly. –  Gordon Jan 26 '11 at 22:09
    
Except of course, if you do the calculation you suggest, you'll find that $2GM/c^2 =2G(1.5 \times 10^{53})kg/c^2 = 2 \times 10^{26}m $, just about the radius of the observable universe. The universe is already just about the same density as a black hole of equivalent radius. this doesn't make the universe a black hole but it does make this question a bit intractable. –  Mark Beadles Jan 5 '12 at 15:56
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