Classically, the escape velocity is independent of the direction of emission, because the gravitational force is potential (unlike, say, magnetism or friction). In GR the situation is more complicated because of the potential capture by an event horizon.
In a real world, the escape speed from the system (Earth)-(any black hole) is heavily dependent from the direction you choose to escape. In the direction from Earth to the black hole, it is greater then c. While in the opposite direction it is only the well known 11+ km per second.
What bothers me further. For an observer, far away on the other side of the super massive black hole in the Center, we are sometimes behind the event horizon, sometimes we are not. True?
In a real world, the escape speed from the system (Earth)-(any black hole) is heavily dependent from the direction you choose to escape. In the direction from Earth to the black hole, it is greater then c. While in the opposite direction it is only the well known 11+ km per second.
That cannot be right. For example, in the Earth-Sun system the escape velocity from the Earth's surface is about 11.2km/s (to escape the Earth), but this only gets you to an orbit around the Sun. You need to accelerate to about 42.1 km/s to escape the solar system (neglecting ...
As mister shminux mentioned somewhere, he is happy and qualified to answer questions in the field of the Relativity. Here is mine:
A long rod (a cylinder) could have a large escape velocity in the direction of its main axe. From its end, to the "infinity". Larger than the speed of light. While the perpendicular escape velocity is lesser than the speed of light.
Is this rod then an asymmetric black hole?