Are you feeling lucky, punk?

Let's say we want to know how big the earth would be if it were compressed down so far that it became a black hole. We don't really know much about black holes, but we do know something about escape velocity. Stand on the surface of the earth (which had radius r) and fire a projectile upward, and it will escape if the kinetic energy we gave it at launch is equal or greater than the gravitational potential energy. But at a black hole surface, we the escape velocity is the speed of light. Why not plug that fact into the equation and see what we get?

We start off with the potential energy on the left and the kinetic energy on the right:

Solve for r:

For a black hole, the escape velocity v is equal to the speed of light c:

Plugging in the mass of the earth, we see that the radius would be about 8 millimeters - the size of your average breath mint. As a black hole, the whole sun would be less than 3 kilometers in radius*. Black holes are pretty compact. A theorem due to Newton but also true in general relativity asserts that gravitational fields created by spherically symmetric bodies behave just as if the mass were all concentrated at the center. If the earth were replaced by an earth-mass black hole at its center, you standing on a platform hovering exactly where you are sitting now would experience the same gravity as you do now. Black holes don't suck things in like a vacuum cleaner, they have regular gravity just like everything else. Just a lot more as you get close to the event horizon.

But is our equation right? Not even close. The gravitational potential energy term is only good for Newtonian gravity, and is totally wrong for the extreme fields near a black hole. It's only a good approximation for weak gravity. The kinetic energy term isn't right either, as it's only the leading-order term in the special relativity expression for kinetic energy.

However through blind luck the equations happen to be wrong in offsetting ways and the result that we got for the radius of a black hole is in fact correct. If we had gone through and used the much more difficult full-blown general relativity equations we'd have gotten the same answer.

Blind luck has actually saved the bacon of theoretical physics more than once. In his famous alpha particle scattering experiment, Rutherford used a formula for the classical scattering off of a 1/r potential and found that it was in great agreement with his experiment for a particular tiny r, the radius of the nucleus. But quantum mechanically speaking that classical scattering equation is not even remotely correct. But in the lucky case of a 1/r potential you just happen to get the same answer. Had this not been the case, the model of an atom as containing a tiny massive nucleus at the center might have been delayed by many years.

*You may notice that this seems a little big compared to the 8 millimeter earth-mass black hole. But it's true, escape velocity scales linearly with the mass, not as the cube root of the mass we'd expect if black holes all had the same density. But they don't, and in fact supermassive black holes can have densities in the gram per cubic meter range. "Density" is a bit of an odd word to use for black holes, so don't take it too literally. I just mean it in terms of mass divided by volume, where volume is just defined in the classical (4/3) pi r^3 sense.