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How small would the earth have to be squashed so that it would become a black hole?

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You can define a Schwarzschild Radius based solely on total mass, i.e. $R_s = \frac{2GM}{c^2}$. If you plug in the mass of the earth, the radius is about 9 mm --- which is how small you would have to compress it to make a black-hole.

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  • $\begingroup$ why is it not necessary to account for relativistic effects? $\endgroup$ – CognisMantis Jul 5 '15 at 16:24
  • $\begingroup$ @CognisMantis The Schwarzschild Radius is a relativistic effect. What else might one account for? $\endgroup$ – DilithiumMatrix Jul 5 '15 at 19:23
  • $\begingroup$ oh, sorry. I had the impression that the derivation was to have mv^2/2=GMm/r, which magically gets the right answer. Why is it not justified to use mc^2(L-1)=GMm/r, but when I do that, I get that the radius is zero. Why must we use the field equations? $\endgroup$ – CognisMantis Jul 6 '15 at 12:58
  • $\begingroup$ @CognisMantis The Einstein field equations in vacuum is satisfied by the Schwarzschild black hole solution. $\endgroup$ – JamalS Dec 8 '16 at 16:38

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