[allendale]

OK but explain a singularity. Always thought that all objects had a “gravitational” pull that is proportional to its mass. If an object such as a star “collapses” unto itself, its mass would be a bit more dense but overall less as some of that contracting mass dissipates as energy. Then why is the gravity field so much stronger?

[FredZarguna]

Well, remember, gravitational attraction depends only on mass, so theoretically a black hole can have any mass as long as it's small enough. [This isn't really true because the Uncertainty principle limits how small a black hole can be -- but in General Relativity or Newtonian gravity there's no lower size boundary] gravity The fact that some of the star mass is lost doesn't matter all that much. There is also mass loss due to binding energy when the Fermions start collapsing into themselves, and that also doesn't make a whole lot of difference.

The solution of Einstein Field equations is beyond the scope of this reply. However, you can get a very good estimate of the size (it actually turns out to be too large by a factor of 1/2) by using the Virial Theorem of classical physics, and Newtons Law:

Virial Therorem for central forces:
Average Kinetic energy = 1/2 Average Potential Energy.
1/2 mv² = GMm/2r
r = GM/v²
At Schwarzchild Radius, r = r_s, v = c;
r_s = GM/c²
Couple examples:
for solar mass [the sun won't become a black hole because it doesn't have enough mass to become a neutron star, but just an example] ~2 x 10³⁰ kg
r_s = 1.989E30 kg x 6.67384E-11 m3 kg-1 s-2 / (3.0E8 ms-1) ~ 1500 m. So a black hole with one solar mass would have r_s of 1.5 kilometers.

A black Hole with a mass the same as earth: 5.972E24 kg would have r_s of about 0.4 cm .

Both of these answers are off by 2, because Newton's gravity doesn't hold for strong gravitational fields.