> to be 1G at its "surface"

This doesn't make any sense to me. By definition at the event horizon gravity is enough to keep light from escaping. Light can easily escape from 1G.

Well, as we all know, on earth's surface, the gravity is 1G, and rockets do escape it. So if the gravity at the event horizon was about 1G, then why couldn't you accelerate temporarily, from inside to outside, until at least your fuel ran out?

Without thinking too deeply, I suspect the issue is that you could never be stopped at that location in the first place, unlike Earth's surface.

It can escape from a local 1G gravity if the black hole were as small as the Earth. As soon as you move 6400 km away, that 1G gravity turns into 0.25G.

But the event horizon of such a black hole would be far larger than that of the Earth. Move 6400 km away, and you'll still have a gravity of ~1G.

Rinse and repeat, and you'll find that even light won't be able to escape.

The handy calculator says 1G at the event horizon implies a radius of half a light-year.

Which means it'll be at 1/4G after you travel a further half light year.

That's a hell of a lot of energy. (If I'm doing the math right, the potential energy difference is on the order of ~20% of the rest mass at the event horizon)

Yeah either the calculator is wrong or I don’t know what I’m talking about. Gravity at the surface of the black hole or the event horizon may well be 1G but all world lines point towards the singularity. I think what is missing is a sufficiently large black hole would have a really low average density at the event horizon thus you wouldn’t get speghettified instantly due to the huge discrepancy of gravity on different parts of your body.

Yeah, black hole "radius" (not volume) grows linearly with it's mass. So very heavy black holes aren't that "dense".

Light doesn't really accelerate, so if you put it in any gravity well that's wide enough, it will eventually fall in.