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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.