WombatGambit

WombatGambit t1_irqjoxv wrote

What is meant by the event horizon? It's the distance at which anything moving slower than c, the speed of light, will not escape the object's gravitational field. In other words, the object has an escape velocity, v_e, that is greater than c. And v_e depends on the mass and radius of the object. If you have enough mass contained within a small enough radius, then v_e > c. That's why they're called "black" holes, because light cannot reflect off them and come back to our sensors (eye, telescope, etc.) But the only event that can bring that much mass within that small a radius is a (massive) collapsing star. Once the nuclear fuel burns out, there is no more outward force, and the star's massive gravitational field starts pulling everything inward. If the mass is high enough, protons and electrons are smashed together to form neutrons, a supernova occurs, and you have a neutron star left (itself, extremely dense, but still no event horizon - light can still escape from even something as dense as a neutron star. There just isn't enough mass within a small enough space to make the escape velocity high enough.) But if the star's original mass is large enough, even the "quantum degeneracy pressure" of the neutrons isn't enough to keep all that mass at a certain distance. The star continues to collapse. So we end up with a huge amount of mass in a very tiny space --> enormous gravitational field --> so big that c < v_e --> light cannot escape --> black hole.

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WombatGambit t1_irqhdpq wrote

This would be something of a miraculous "perpetual motion machine," being able to provide a force like that forever. I think it's a great question, though. To clear up the confusion, we need to ask what is meant by escape velocity, which I'll denote v_e.

This term is relevant because we can only provide a push for some finite time. Whether it's a rocket or a cannon ball, you get an impulse and then the projectile is off and moving. If we can't get it moving more than v_e once we stop pushing, then it will move in a parabola and fall back to earth. It might go *really* far, but it will still eventually come back down. If we provide v_e, then it will go up and maintain a stable orbit. It is "falling to earth at the same rate that earth is curving away from it." If, when we stop pushing, we exceed v_e, then it will be moving so fast that, even as earth's gravitational field decelerates it, the rocket has to be infinitely far away before that deceleration alone brings it to v = 0. So it will never come back to earth and has "escaped."

As a sidenote, keep in mind that the force of gravity decreases as you get farther away. So if our magic force generator truly provided a constant force, then "just barely moving" at earth's surface would translate into "rocket speeds" as it gets further away. This thing would just keep moving faster and faster. 1N fighting 0.9N isn't much, but 1N fighting 0.009N is totally different!

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