Some time ago, I asked a question about Black holes and the velocity of light, which was kindly answered, and similar to another query I have.
The escape velocity of the Earth is 11.2 km/s, which I thought defined the speed, that an unpowered ballistic object requires, to break free from Earth's gravitational pull and reach an orbit at infinity. If I shoot an object skyward at just below escape velocity, it may still leave Earth, just not reach infinity... but this could be a significant distance from Earth.
I understood that a black hole's event horizon was a surface at which its escape velocity equalled the speed of light. Since light is travelling at the escape velocity, at the event horizon, won't it be able to leave its surface, and reach an "orbit at infinity".
In comparing the use of "escape velocity" with regard to Earth and black holes, it seems that the former allows objects to leave the surface, whereas a black hole event horizon is described as an impenetrable surface.
One solution is that an observer at infinity would see something different to an observer at twice the event horizon distance. But even if a photon can extend a tiny bit above the event horizon, then it could extend further. But can someone clarify my dilemma?