## Black holes & escape velocity

What is the reason why something, once it enters the event horizon of a black hole, it cannot escape? I've often heard the answer given that at the event horizon the escape velocity equals the speed of light, and from within the horizon a superluminal velocity is required, which is impossible. The problem I have with this explanation is that escape velocity refers to the inertial velocity an object must have in order to escape a gravitational field, with no forces acting on the object apart from gravity. But if we have other forces acting on the object (like thrust from a rocket engine), it is possible to escape from a gravitational field travelling at a velocity less than the escape velocity, by providing a force counteracting that of gravity.

For instance, suppose we have a very big black hole with mass of 10^50 kg. The Schwartzchild radius will be 2GM/c^2=2(6.66x10^-11x10^50)/9x10^16=1.48x10^23 metres. The gravitational field strength at the event horizon will hence be GM/r^2 = (6.66x10^-11x10^50)/(1.48x10^23)^2= 3x10^-7 N/kg, which is almost trivial. For an object, say, a one kilogram mass one metre past the event horizon, to escape it must only overcome a measly gravitational force of 3x10^-7 N for enough time to traverse one metre, which is hardly difficult. Indeed, it should be possible to play a game of table tennis across the event horizon without noticing anything amiss.

So, what is the actual reason for why the event horizon acts as a "point of no return"?

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 Quote by Pyrovus What is the reason why something, once it enters the event horizon of a black hole, it cannot escape? I've often heard the answer given that at the event horizon the escape velocity equals the speed of light, and from within the horizon a superluminal velocity is required, which is impossible. The problem I have with this explanation is that escape velocity refers to the inertial velocity an object must have in order to escape a gravitational field, with no forces acting on the object apart from gravity. But if we have other forces acting on the object (like thrust from a rocket engine), it is possible to escape from a gravitational field travelling at a velocity less than the escape velocity, by providing a force counteracting that of gravity. For instance, suppose we have a very big black hole with mass of 10^50 kg. The Schwartzchild radius will be 2GM/c^2=2(6.66x10^-11x10^50)/9x10^16=1.48x10^23 metres. The gravitational field strength at the event horizon will hence be GM/r^2 = (6.66x10^-11x10^50)/(1.48x10^23)^2= 3x10^-7 N/kg, which is almost trivial. For an object, say, a one kilogram mass one metre past the event horizon, to escape it must only overcome a measly gravitational force of 3x10^-7 N for enough time to traverse one metre, which is hardly difficult. Indeed, it should be possible to play a game of table tennis across the event horizon without noticing anything amiss. So, what is the actual reason for why the event horizon acts as a "point of no return"?

IIRC the spacetime geometry within the even horizon is very odd. You are no longer free to travel through space as you choose, all spatial paths lead to the singularity, but there is no longer a natural flow of time either.

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## Black holes & escape velocity

 it is possible to escape from a gravitational field travelling at a velocity less than the escape velocity
Wrong. It's called the escape velocity because you must have attain at least that velocity at some point in your trajectory to be able to escape to infinite distance. If you don't have at least that velocity, you will remain gravitationally bound and will eventually fall back onto the body.

In other words, you have to have more kinetic energy than gravitational potential energy, and the only way to do that is to reach escape velocity. When you're inside a black hole, you will need a velocity greater than c to be able to escape. As has been mentioned, if you solve the equations (not just hand wave about table tennis), you'll see that all worldlines inside the event horizon eventually end at the singularity. There's no fighting it.

- Warren

 Recognitions: Gold Member Science Advisor I believe Franznietzsche's answer is a closest to the truth. You see, the thing about a black hole is not just the fact that escape velocity is greater than lightspeed, but that space is so curved as to make this condition exist. The advent of space elevators (if they are ever actually build) will disapprove the idea that escape velocity has to be reached in order for an object to leave the gravity well of another object.
 Recognitions: Gold Member Science Advisor Staff Emeritus LURCH, Of course not. As you climb the space elevator, it will push on you and you will be gaining tangential velocity, thus enabling you to orbit. In the end, you better have 17,000 mph for low earth orbit, or you won't be in orbit at all, regardless of how you get there. Besides, space elevators won't allow you to actually leave the Earth's gravitational influence altogether -- you'll need 11 km/s for that, no matter how you choose to get it. - Warren

 it is possible to escape from a gravitational field travelling at a velocity less than the escape velocity, by providing a force counteracting that of gravity.
What I was thinking of was Newton's first law - i.e. that a body in motion remains in motion so long as no net force acts on it. For instance, shouldn't a rocket travelling at, say, 20m/s be able to escape the earth if the engines provide a thrust of 9.8 N/kg (of course less as the rocket reaches higher altitudes), so that the net force acting on the rocket is zero?

 Recognitions: Gold Member Science Advisor Staff Emeritus Pyrovus, Imagine if you did this by launching the rocket straight "up," leaving the earth along a radial line. At every point along the trajectory, the rocket would only be moving at 20 m/s away from the earth. (Assume that the rocket produces thrust that matches the earth's gravitational pull at all times -- as it gets further away, the force of gravity is smaller, so the rocket throttles back.) Now, what happens when the rocket is only 100 feet off the ground? If it were to turn off its engines, it's 20 m/s of velocity would be gone in only two seconds (since gravity accelerates it downward at roughly 10 m/s) and it would begin falling back to earth. When it's 10,000 feet off the ground, the same thing would happen. It would take a little longer than two seconds, because g is a little less than 9.81 m/s there. When it's 10,000,000 feet off the ground, the same thing would happen. It would take longer, but it would still happen. When it's 10,000 light years away, the same thing would happen. It would take longer, but it would still happen. And so on -- as you can see, if you turn off your rocket at any point along this trajectory, even when you're very very far away, you'll begin falling back to earth. Only when you get to infinite distance (which isn't possible) will falling back take infinite time and thus not happen. - Warren
 Understood. But what's so special about the event horizon of a black hole that effectively makes it a one-way barrier? In my example in my first post, why can't I cross back to outside the event horizon by continually applying that force of 3x10^-7 N/kg, once I'm moving away from the black hole, even if I can't escape to infinity. Escaping from the event horizon isn't the same as escaping to infinity.
 Recognitions: Gold Member Science Advisor Chroot - one thing that's not quite true, if a rocket acheives a velocity (radially away from the earth) of 20 m/s at a distnce greater than approximately 1/5,000 of a light year it will escape to infinity. Escape velocity is depednent on distance and it's the velocity that's needed only if your not being acted on by another force.
 Recognitions: Gold Member Science Advisor Pyrovus the problem is your trying to apply Newtonian physics to a black hole. What you have said would bve true of Michell's "dark stars", but blcak hole are governed by general relativity which can't be ignored; the fact that the event horizon is at where in Newtonian physics Vesc = c is almost coincidental i.e. the event horizon doesn't come from Newtonian physics. You find that force that any real force applied to an object within the event horizon will not halt or reverse it's advance to the singularity, this is a consequence of GR.
 I must be reading some posts wrong. I do not see why you can not get to space from earth at a meer 20 m/s.

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 Quote by chroot LURCH, Of course not. As you climb the space elevator, it will push on you and you will be gaining tangential velocity, thus enabling you to orbit. In the end, you better have 17,000 mph for low earth orbit, or you won't be in orbit at all, regardless of how you get there. Besides, space elevators won't allow you to actually leave the Earth's gravitational influence altogether -- you'll need 11 km/s for that, no matter how you choose to get it. - Warren
But the point is, an object can move away from the center of gravitational pull at speeds far less than escape velocity. Therefore, the mere fact that at the event horizon, escape velocity is V>c does not provide an adequate explanation as to why escape from the event horizon is impossible.

 Think about it differently: If you require a certain escape velocity, that velocity would correspond to the energy required to gain enough positive potential energy to escape the gravitation of the object. Inside the event horizon, an escape velocity of >c is required. If we convert this to energy terms, we find that we have a potential energy inside the event horizon of greater than -infinity, relative to being infinitely distant. No matter how long we continue to deliver force, or how much force we do, we still have the same potential barrier, and we will not be able to do enough work against gravitation to escape the black hole, or reach a position where it is possible.

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 Quote by jcsd Chroot - one thing that's not quite true, if a rocket acheives a velocity (radially away from the earth) of 20 m/s at a distnce greater than approximately 1/5,000 of a light year it will escape to infinity. Escape velocity is depednent on distance and it's the velocity that's needed only if your not being acted on by another force.
An excellent point jcsd, thanks!

- Warren

 Quote by FZ+ Think about it differently: If you require a certain escape velocity, that velocity would correspond to the energy required to gain enough positive potential energy to escape the gravitation of the object. Inside the event horizon, an escape velocity of >c is required. If we convert this to energy terms, we find that we have a potential energy inside the event horizon of greater than -infinity, relative to being infinitely distant. No matter how long we continue to deliver force, or how much force we do, we still have the same potential barrier, and we will not be able to do enough work against gravitation to escape the black hole, or reach a position where it is possible.
Does this mean, that by conservation of energy, an object inside a black hole will have infinite kinetic energy and hence travel at the speed of light towards the singularity?