How Can we Find the Escape velocity of a black hole

[Quds Akbar]

It depends on your distance from the black hole, once crossing the event horizon, the escape velocity exceeds the speed of light.

 

[Quds Akbar]

It depends on your distance from the black hole, once crossing the event horizon, the escape velocity exceeds the speed of light.

 

[Gaz1982]

If nothing can escape a Black Hole then where does Hawking radiation come from?

 

[phinds]

If nothing can escape a Black Hole then where does Hawking radiation come from?

From outside the event horizon, obviously.

The English-language analogy that Hawking used to describe it (and this is NOT really quite what happens, and I'm paraphrasing, not quoting directly) is "a virtual particle-pair pops into existence just outside the event horizon and one falls in and one escapes. The one falling in always contributes negative mass to the black hole"

 

[Gaz1982]

From outside the event horizon, obviously.

The English-language analogy that Hawking used to describe it (and this is NOT really quite what happens, and I'm paraphrasing, not quoting directly) is "a virtual particle-pair pops into existence just outside the event horizon and one falls in and one escapes. The one falling in always contributes negative mass to the black hole"

Contributes negative mass?

Help

 

[phinds]

Contributes negative mass?

Help

You have hijacked this thread. You should start a new thread asking for an expanation of Hawking Radiation, but before doing that I suggest you do a little research on your own. Google is your friend.

 

[Borg]

You have hijacked this thread. You should start a new thread asking for an expanation of Hawking Radiation, but before doing that I suggest you do a little research on your own. Google is your friend.

I would say that the thread is dead anyway. The OP never even bothered to come back after asking the question 4 months ago. We might as well be talking to ourselves.

 

[Chronos]

We have a talent for that.

 

[AgentSmith]

How Can we Find the Escape velocity of a black hole.

You can only calculate the escape velocity at the event horizon. Turns out to be c. Inside event horizon nothing can escape.

 

[Indi SUmmers]

I had a stupid question:

Gravity from Black Holes is, under several assumptions, from a collapsed star. Assuming gravity is not constant, wouldn't the speed of gravity be directly proportional to the mass of the star?

A class O star that collapsed would have a stronger gravitational pull than a class M - wouldn't it? So why is the equation for gravity treating the variable of gravity as a constant?

Just curious.

 

[phinds]

Gravity is gravity. It is immaterial what it comes from, just how MUCH stuff is there. A larger mass will generate a larger gravitational force (at the same distance from the center of mass) than a smaller one, but so what. The equation for the force is the same, you just plug in a different mass.

Also, there is no such thing as "the speed of gravity". There IS such a thing as the speed of changes in gravity, and that is c.

 

[Indi SUmmers]

I think I understand. We don't treat gravity as an object (ie - as a variable in an equation), but as a force to be calculated (the sum/difference/answer) we calculate. And as such, forces aren't constant. By plugging in the mass, it will automatically calculate any differences in the amount of gravity proportionate to the object. Is that what you're saying?

Velocity and gravity aren't interchangeable, though. Wouldn't velocity be the answer, sum, difference here? That would place gravity as a variable. We assume gravity is constant in this particular velocity equation, right? We're assuming gravity is G always. But how can we calculate v if G isn't constant, which I'd assume it isn't inside of the extreme conditions of a Blackhole as we approach the singularity.

I guess I'm confused, but that's nothing new. I always was under the assumption that a Back Hole's gravity would increase as you approach the singularity.

http://physics.stackexchange.com/questions/33916/what-is-the-escape-velocity-of-a-black-hole

 

[Borg]

I had a stupid question:

Gravity from Black Holes is, under several assumptions, from a collapsed star. Assuming gravity is not constant, wouldn't the speed of gravity be directly proportional to the mass of the star?

A class O star that collapsed would have a stronger gravitational pull than a class M - wouldn't it? So why is the equation for gravity treating the variable of gravity as a constant?

Just curious.

You're confusing gravitational force with the Gravitational Constant.

F is the gravitational force that is felt between two bodies. G is the gravitational constant.

 

[Indi SUmmers]

Some Other Curiosities:

* The Spin Problem
http://discovermagazine.com/2008/whole-universe/09-a-lenticular-galaxy-reveals-spinning-black-holes
http://discovermagazine.com/2002/jul/cover
https://en.wikipedia.org/wiki/Rotating_black_hole
http://www.space.com/24936-supermassive-black-hole-spin-quasar.html

So, the way I understand what's being phrased is that we calculate escape velocity from a black hole as being c - is this because of the observable light at the event horizon?

My question is this: if something is spinning, can't it just be ricocheting photons off of it and onto the event horizon if the speed of that ricocheting force is greater than c?

Kind of like trying to throw an object into a fan - if something is light enough and the fan is spinning fast enough, won't it just kick the object back out at you? If you kick an object out at zero-G in the vacuum of space, it'll just stay suspended, won't it? What if black holes are doing this with photons?
If they are doing this, then how can we assume that the observable light around the discs of black holes is strictly from photons escaping the gravitational pull within the singularity?ANSWERED HERE:

We KNOW the "escape velocity" of a black hole. It is the speed of light. Since nothing can travel at the speed of light, nothing with mass can escape from a black hole (so it isn't really an "escape" velocity) and even light can only maintain a position exactly at the event horizon because locally it is traveling outward at c and globally, it is being held in place by the gravity of the black hole.

* Spaghettification
http://science.howstuffworks.com/science-vs-myth/what-if/what-if-fell-into-black-hole2.htm
https://en.wikipedia.org/wiki/Spaghettification

It was my understanding that once you pass the Event Horizon, you're traveling faster than c. And due to extreme tidal forces (i.e. spaghettification), you're accelerating toward the singularity due to the extreme pull.

With that being said, are we calculating this as though the force of gravity is remaining constant inside of a black hole or are we accounting for the acceleration of an object as it nears the singularity?

Am I making any sense? Are we calculating still using the right equation?

Disclaimer
c, >c, >c2 and <c3 are hypothetical numbers and not actual variables for this equation. They're just meant to illustrate a hypothetical acceleration of gravity and a necessary velocity to obtain by light to escape it.

Double Disclaimer
Trust me, I don't know *@#! about science (which may be readily apparent to the learned mind).

 

[phinds]

Some Other Curiosities:

* The Spin Problem
http://discovermagazine.com/2008/whole-universe/09-a-lenticular-galaxy-reveals-spinning-black-holes
http://discovermagazine.com/2002/jul/cover
https://en.wikipedia.org/wiki/Rotating_black_hole
http://www.space.com/24936-supermassive-black-hole-spin-quasar.html

So, the way I understand what's being phrased is that we calculate escape velocity from a black hole as being c - is this because of the observable light at the event horizon?

No it has nothing to do with light at the event horizon, it is an effect of the gravitational force at the event horizon

My question is this: if something is spinning, can't it just be ricocheting photons off of it and onto the event horizon if the speed of that ricocheting force is greater than c?

Forces don't have speed so I have no idea what you are talking about.

 

[Indi SUmmers]

http://discovermagazine.com/2008/whole-universe/09-a-lenticular-galaxy-reveals-spinning-black-holes
http://discovermagazine.com/2002/jul/cover
https://en.wikipedia.org/wiki/Rotating_black_hole
http://www.space.com/24936-supermassive-black-hole-spin-quasar.html

So, the way I understand what's being phrased is that we calculate escape velocity from a black hole as being c - is this because of the observable light at the event horizon?

No it has nothing to do with light at the event horizon, it is an effect of the gravitational force at the event horizon.

I guess what I mean is...if a little shiny object that emitted light were traveling past the event horizon, it would no longer be observable as it passed over because the gravity would suck in all the light, lumosity, radiation, etc. But if it escaped and traveled back outside of the event horizon, wouldn't it become visible again? Is this how they're calculating Gravitation Force inside of a Black Hole? Is this how they're saying that the escape velocity necessary to get out of a Black Hole is 'c' - because to me, it seems like it would be multiples of c.

How are we getting the answer 'c?'

Can someone help me make sense?

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* The Spin Problem
http://discovermagazine.com/2008/whole-universe/09-a-lenticular-galaxy-reveals-spinning-black-holes
http://discovermagazine.com/2002/jul/cover
https://en.wikipedia.org/wiki/Rotating_black_hole
http://www.space.com/24936-supermassive-black-hole-spin-quasar.html

So, the way I understand what's being phrased is that we calculate escape velocity from a black hole as being c - is this because of the observable light at the event horizon?

My question is this: if something is spinning, can't it just be ricocheting photons off of it and onto the event horizon if the speed of that ricocheting force is greater than c?

Kind of like trying to throw an object into a fan - if something is light enough and the fan is spinning fast enough, won't it just kick the object back out at you? If you kick an object out at zero-G in the vacuum of space, it'll just stay suspended, won't it? What if black holes are doing this with photons?
If they are doing this, then how can we assume that the observable light around the discs of black holes is strictly from photons escaping the gravitational pull within the singularity?

ANSWERED HERE:

We KNOW the "escape velocity" of a black hole. It is the speed of light. Since nothing can travel at the speed of light, nothing with mass can escape from a black hole (so it isn't really an "escape" velocity) and even light can only maintain a position exactly at the event horizon because locally it is traveling outward at c and globally, it is being held in place by the gravity of the black hole.

So with that being said, I guess my real question becomes:

What velocity is needed to not only escape a Black Hole, but to travel away from the event horizon? It seems that this would correlate directly with the mass of the Black Hole (which I believe you addressed in your FORCE equation and not your CONSTANT one as I was confused about)? But are we calculating or are able to measure the pull of the Black Hole?

How hard does a black hole "suck?" ;-)
Do they all 'suck' the same? ;-)
Does the equation you posted...

You're confusing gravitational force with the Gravitational Constant.

F is the gravitational force that is felt between two bodies. G is the gravitational constant.

Because if it doesn't account for the rate of "suck" in a black hole (as I understand all black holes "suck" differently?), it seems like we're calculating this incorrectly?

Am I missing this in the math?

 

[AgentSmith]

Because if it doesn't account for the rate of "suck" in a black hole (as I understand all black holes "suck" differently?), it seems like we're calculating this incorrectly?

Am I missing this in the math?

The suck of black holes is not a physical observable.

 

[davenn]

ANSWERED HERE:

We KNOW the "escape velocity" of a black hole. It is the speed of light. Since nothing can travel at the speed of light, nothing with mass can escape from a black hole (so it isn't really an "escape" velocity) and even light can only maintain a position exactly at the event horizon because locally it is traveling outward at c and globally, it is being held in place by the gravity of the black hole.

So with that being said, I guess my real question becomes:

What velocity is needed to not only escape a Black Hole, but to travel away from the event horizon? It seems that this would correlate directly with the mass of the Black Hole (which I believe you addressed in your FORCE equation and not your CONSTANT one as I was confused about)?

so what part of the answer you quoted ( and the other responses) do you not understand ?

NOTHING traveling at c or less escapes

The mass of a black hole is irrelevant ... there are small black holes, there are large blacks holes, there are huge ones at the cores of many/most galaxies
note the common thing in all of them ... They are ALL black. what does that tell you ?Dave

 

[stedwards]

What about definitions? What is a black hole?

 

[rootone]

What about definitions? What is a black hole?

An object that has so much gravity that light, (or anything else), cannot escape from it once it has crossed the event horizon.
The event horizon is the 'point of no return'.
An object with sufficient force applied to it can (in principle anyway) escape from the black hole's gravity while it is still outside of the EH.

 

[stedwards]

An object that has so much gravity that light, (or anything else), cannot escape from it once it has crossed the event horizon.
The event horizon is the 'point of no return'.
An object with sufficient force applied to it can (in principle anyway) escape from the black hole's gravity while it is still outside of the EH.

You know, I don't recall who I was directing this question to. I think one of the best definitions is "A black hole is a region from which nothing can escape; not even light," making the OP question moot.

Apparently this shouldn't preclude virtual particles enabling space-like interaction (effectively v>c). The Coulombic forces of electromagnetism and gravity should be in effect, so that the charge and mass within a black hole is measurable outside the event horizon.

Give this, the weak and the strong force should not be precluded and the black hole "no hair" argument should be revises to include the virtual weak and strong forces.

 

[davenn]

I think one of the best definitions is "A black hole is a region from which nothing can escape; not even light," making the OP question moot.

and that is what everyone has repeatedly been telling him for 3 pages now, but it still doesn't seem to be sinking in !

 

[stedwards]

and that is what everyone has repeatedly been telling him for 3 pages now, but it still doesn't seem to be sinking in !

Nah. I did a little checking. The original poster is long gone. Thereafter, it's been just us chickens trying to impress each other without knowledge of black hole physics.

What do you think of strong or weak charges as variable quanities that should be included in the description of a black hole?