[lutusp]

> A huge black hole may have pretty weak gravitation at its horizon.

Not really. The amount of spacetime curvature at the event horizon is fixed in the theory -- it's always the same. It's how the event horizon is defined. Near the event horizon, light orbits endlessly (in principle), and (again in principle) if you were located at an event horizon and there was sufficient illumination, anywhere you turned you would see the back of your own head.

> Thus no-escape is valid only in the sense that escaping object just would never reach "a point at infinity" from the black hole and the light would get red-shifted into full oblivion.

From a more distant frame of reference, yes, but not at the event horizon itself.

[dragonwriter]

> The amount of spacetime curvature at the event horizon is fixed in the theory -- it's always the same.

No, its greater the smaller the horizon is (so, equivalently, less the bigger the horizon is).

> Near the event horizon, light orbits endlessly (in principle)

Sure, at the event horizon, light orbits endlessly. But the amount of curvature needed to do that is less the further across the horizon is.

[lutusp]

>> The amount of spacetime curvature at the event horizon is fixed in the theory -- it's always the same.

> No, its greater the smaller the horizon is (so, equivalently, less the bigger the horizon is).

We're using different meanings of "curvature". For all black holes regardless of their properties, the event horizon has the same spacetime curvature -- that required to produce orbiting photons that cannot escape. Outside the event horizon, photons can escape. Inside, they cannot (and those photons don't orbit either, but cross the horizon). All the same.

> But the amount of curvature needed to do that is less the further across the horizon is.

No, it's the same. Same curvature, different large-scale geometry. For a sufficiently small zone near the event horizon, the conditions are identical.

Again, this is about the meaning of "curvature". The circumference of the horizon is greater for a large mass than a small one, but the spacetime curvature at the horizon is the same -- it's the specific value that causes photons to orbit perpetually, and interestingly from a local perspective, the entire horizon surface appears as a plane of infinite extent, sort of like two facing mirrors but with more dimensions.

From the perspective of a hypothetical observer at the horizon, he wouldn't be able to judge the size of the black hole using local observations -- there would be a plane of infinite extent for any black hole regardless of size.