Quote by Bernie G
“How can the gravitational acceleration at an event horizon be smaller than at the surface of a neutron star ?”
Because gravitational acceleration varies as the inverse of r squared. One of us is making a mistake. I was under the impression that distant supermassive black holes (10 billion solar masses) “disappeared” because the gravitational acceleration at the event horizon is so small (and the curvature so large) that infalling material doesn’t even radiate until it is well within the black hole. Hence I volunteer to sit on the ring and bravely stick my toes inside the event horizon of a trillion solar mass black hole, where the gravity (gulp) should be about as strong as in California.
To challange the staus quo even further, here in a nutshell is my minority viewpoint about the size of a star composed of relativistic material inside a black hole:
The gravitational energy could be as low as (4GM^2)/(5R) for a typical density profile, or possibly as high as (GM^2)/R (unlikely) if the star has a high density core. The total energy creating pressure would be (Mc^2)/3. Using the viral theorem (the energy creating pressure equals half the gravitational energy), a nonrotating star of relativistic material would have a radius as small as (1.2GM)/(c^2) or as large as (1.5GM)/(c^2), or between 60  75% of the Schwarzchild radius.
If this model is true, it could be verified someday by the observation of the merger of two approximately equal mass black holes: a massive ejection from the relativistic stars would occur.

I don't really get what you are saying; the event horizon is a boundary beyond which photons cannot escape the gravitational pull of the BH. Its radius is only dependent on the total mass of the BH. As neutrons stars are stable and do not collapse gravitationally, the gravitational acceleration at the event horizon for a BH of equal mass must be much stronger than at the surface of the neutron star ?! If it was the other way around all neutron stars would immediately collapse...