### escaping a black hole?

Posted:

**Fri Aug 10, 2012 2:11 pm UTC**could you drop a camera attached to a rope inside a black hole and have it take pictures, then pull it back out?

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Posted: **Fri Aug 10, 2012 2:11 pm UTC**

could you drop a camera attached to a rope inside a black hole and have it take pictures, then pull it back out?

Posted: **Fri Aug 10, 2012 2:14 pm UTC**

No.

Posted: **Fri Aug 10, 2012 2:34 pm UTC**

Why not and what would happen if you tried?

Posted: **Fri Aug 10, 2012 2:38 pm UTC**

>-) wrote:Why not and what would happen if you tried?

The camera would be torn to bits by tidal (is that the right way of putting it?) forces before you could "lower" it all the way in. As would the rope, and also your body unless the rope was long enough.

There's also the problem that if the camera and rope somehow survived, you wouldn't have the energy to pull the camera back out.

Posted: **Fri Aug 10, 2012 3:10 pm UTC**

ElWanderer wrote:>-) wrote:Why not and what would happen if you tried?

The camera would be torn to bits by tidal (is that the right way of putting it?) forces before you could "lower" it all the way in. As would the rope, and also your body unless the rope was long enough.

There's also the problem that if the camera and rope somehow survived, you wouldn't have the energy to pull the camera back out.

even if, for the sake of argument, it was an indestructible digital camera and the rope was some sort of indestructible data cable, once the camera reached the event horizon, not only would pulling the camera out require basically unlimited energy, so would sending back data up the data cable.

Posted: **Fri Aug 10, 2012 3:13 pm UTC**

For a blackhole of 2e10 solar masses, the schwarzschild radius would be 5.906e13 meters. If you lowered it in just 6e10m inside the event horizon, the tidal forces pulling on an object weighing 0.1 kg and 5 cm long would be only about 3.8 newtons. The tidal force on the rope would be about assuming a rope weight of 1gm/cm would be about 0.763 n/m, and the breaking strength of some steels is about 1e6 newtons (per cm^2), so the cable should be able to stay in one piece. I could hook the other end of the rope onto a spacecraft, and then fire thrusters to pull the camera out of black hole?

Since the camera is pretty far from the blackhole, it should only take about 77+ newtons to pull it out? assuming 100g mass. Why would it require unlimited energy? And if it did, would you be able to at least keep the camera from falling in more?

Since the camera is pretty far from the blackhole, it should only take about 77+ newtons to pull it out? assuming 100g mass. Why would it require unlimited energy? And if it did, would you be able to at least keep the camera from falling in more?

Posted: **Fri Aug 10, 2012 3:22 pm UTC**

Once the camera has passed the event horizon, there is no way to pull it back out, period. For any object that passes the event horizon, collision with the singularity is inevitable. One way you can think about it is that, at the event horizon, space is rushing into the black hole at c, so nothing can possibly move fast enough to get out. For a sufficiently massive black hole, it will not be torn apart by tides as you note. However, that holds only for a object that is freely falling into the black hole. Holding the camera stationary, as in your scenario, requires an infinite amount of force to resist the inward flow of space. The rope will break.

Posted: **Fri Aug 10, 2012 5:35 pm UTC**

The rope might not break.

But the alternative is that you get pulled in to the event horizon by it.

better to just let go.

But the alternative is that you get pulled in to the event horizon by it.

better to just let go.

Posted: **Fri Aug 10, 2012 7:23 pm UTC**

>-) wrote:For a blackhole of 2e10 solar masses, the schwarzschild radius would be 5.906e13 meters. If you lowered it in just 6e10m inside the event horizon, the tidal forces pulling on an object weighing 0.1 kg and 5 cm long would be only about 3.8 newtons. The tidal force on the rope would be about assuming a rope weight of 1gm/cm would be about 0.763 n/m, and the breaking strength of some steels is about 1e6 newtons (per cm^2), so the cable should be able to stay in one piece. I could hook the other end of the rope onto a spacecraft, and then fire thrusters to pull the camera out of black hole?

Since the camera is pretty far from the blackhole, it should only take about 77+ newtons to pull it out? assuming 100g mass. Why would it require unlimited energy? And if it did, would you be able to at least keep the camera from falling in more?

What's your rope made of that it only weighs 1g/m in a region exposed to (beyond) 311 gravities?

Posted: **Fri Aug 10, 2012 7:37 pm UTC**

Robert'); DROP TABLE *; wrote:What's your rope made of that it only weighs 1g/m in a region exposed to (beyond) 311 gravities?

300 grams is 300 grams no matter what the local gravity happens to be. Grams are a measure of mass, not weight.

It'll take more than 900 newtons of force to shift it in 311 gravities, though, compared to about 3 newtons at 1 g.

Posted: **Fri Aug 10, 2012 7:52 pm UTC**

Whoops. I got distracted by his describing the weight as 1g/m. The massive decrease in the ability for any physically real rope to actually stay together was what I was trying to focus on anyway.

Posted: **Fri Aug 10, 2012 7:55 pm UTC**

No electrons can flow up the cable past the event horizon even if you somehow had an indestructible cable fixed in place magically.

Posted: **Fri Aug 10, 2012 8:24 pm UTC**

Okay, but assume the rope and the camera are both indestructible and it's Bruce Schneier pulling on the rope.

What would the camera show after having been inside the event horizon?

What would the camera show after having been inside the event horizon?

Posted: **Fri Aug 10, 2012 8:35 pm UTC**

You really couldn't get the camera out without having hypotheticals equivalent to "If an unstoppable object hits an immovable (and indestructable) wall, what happens?".

A reasonable conclusion is that if you magic-ed the camera out though, the pictures would be completely black since no light would reach the lens.

An equally reasonable conclusion is that you'd see unicorns riding around on a cthulu made of rainbows. When you're in the realm of magic unrealistic stuff, anythings possible.

A reasonable conclusion is that if you magic-ed the camera out though, the pictures would be completely black since no light would reach the lens.

An equally reasonable conclusion is that you'd see unicorns riding around on a cthulu made of rainbows. When you're in the realm of magic unrealistic stuff, anythings possible.

Posted: **Fri Aug 10, 2012 8:46 pm UTC**

cphite wrote:Okay, but assume the rope and the camera are both indestructible and it's Saladin's Mom pulling on the rope.

What would the camera show after having been inside the event horizon?

Still nothing because all the photons would be flying straight towards the singularity. The only way to get an exposure would be to turn the camera around to face outside the black hole, and you'd see what was happening outside the event horizon. Or what Dopefish said.

Posted: **Fri Aug 10, 2012 9:07 pm UTC**

JBJ wrote:cphite wrote:Okay, but assume the rope and the camera are both indestructible and it's Saladin's Mom pulling on the rope.

What would the camera show after having been inside the event horizon?

Still nothing because all the photons would be flying straight towards the singularity. The only way to get an exposure would be to turn the camera around to face outside the black hole, and you'd see what was happening outside the event horizon. Or what Dopefish said.

This is incorrect.

An infalling observer notices nothing weird at the horizon. If nothing else, because all observers agree on the value for c, an infalling observer must see light coming from below them, otherwise you'd either have to have a discontinuity where the light suddenly stopped and discontinuities are never good in physical situations or you must have it moving at less than c. Neither are possible.

If you could magic the camera out (or if you just fell in with it), you'd see nothing particularly out of the ordinary.

Posted: **Fri Aug 10, 2012 10:11 pm UTC**

"No light coming from below" is merely the point at which the black hole takes up more than half of the sky.

Even outside the black hole ( http://i.dailymail.co.uk/i/pix/2010/12/ ... 34x493.jpg ), there is part of the sky where no light comes from - it comes from the side instead, squished by gravitational lensing. We call this part of the sky "the black hole"

Even outside the black hole ( http://i.dailymail.co.uk/i/pix/2010/12/ ... 34x493.jpg ), there is part of the sky where no light comes from - it comes from the side instead, squished by gravitational lensing. We call this part of the sky "the black hole"

Posted: **Sat Aug 11, 2012 12:05 am UTC**

Except the point is that it never does if you're in freefall. If instead you slowly lower yourself into it, using your rockets to thrust in the other direction as hard as you can or whatever, then the sky slowly concentrates into a bright disc above your head. However, that is due to relativistic beaming, not spacetime curvature.Charlie! wrote:"No light coming from below" is merely the point at which the black hole takes up more than half of the sky.

Posted: **Sat Aug 11, 2012 12:49 am UTC**

So, if you have a small black hole, the tidal forces are more than enough to tear the rope apart.

For a large black hole where the event horizon is relatively flat, space near the event horizon is moving at nearly the speed of light into the black hole. If you have a rope that is "danging under the event horizon", and you start pulling on it, you'd have to pull it faster than the speed of light to avoid pulling it out of the hole (well, practically, you'd snap the rope, because you'd be pulling it faster and harder than the electro-chemical bonds between the rope atoms resist it). If you pull it slower than that, by the time you'd pull the thing on the end of the rope "to" you, you yourself would be in the black hole.

In a sense, as you approach a black hole, "towards" the black hole becomes more and more synonymous with "the future". The closer you are to a black hole, the more absolute that connection is. When you cross the event horizon, this connection is absolute -- near the edge, it is nearly absolute, so you have to be moving at a ridiculously fast speed "away" to pull away from it. And you have to keep accelerating, because the black hole keeps on pulling you towards it, bending your future history towards the event horizon. With something dangling "through" the event horizon, the amount of force you need to apply to get it out is no longer finite -- either you snap the rope, or you fall in behind the rope as you pull on it ridiculously hard.

I am not a physicist -- far from it. So my earlier answer "No" is actually the accurate one, what I describe above is going to be less accurate. "No" still stands: there is no way to get the object you lowered through the black hole out, not with a rope, not with a rocket, not in your pocket, not a hope.

For a large black hole where the event horizon is relatively flat, space near the event horizon is moving at nearly the speed of light into the black hole. If you have a rope that is "danging under the event horizon", and you start pulling on it, you'd have to pull it faster than the speed of light to avoid pulling it out of the hole (well, practically, you'd snap the rope, because you'd be pulling it faster and harder than the electro-chemical bonds between the rope atoms resist it). If you pull it slower than that, by the time you'd pull the thing on the end of the rope "to" you, you yourself would be in the black hole.

In a sense, as you approach a black hole, "towards" the black hole becomes more and more synonymous with "the future". The closer you are to a black hole, the more absolute that connection is. When you cross the event horizon, this connection is absolute -- near the edge, it is nearly absolute, so you have to be moving at a ridiculously fast speed "away" to pull away from it. And you have to keep accelerating, because the black hole keeps on pulling you towards it, bending your future history towards the event horizon. With something dangling "through" the event horizon, the amount of force you need to apply to get it out is no longer finite -- either you snap the rope, or you fall in behind the rope as you pull on it ridiculously hard.

I am not a physicist -- far from it. So my earlier answer "No" is actually the accurate one, what I describe above is going to be less accurate. "No" still stands: there is no way to get the object you lowered through the black hole out, not with a rope, not with a rocket, not in your pocket, not a hope.

Posted: **Sat Aug 11, 2012 3:06 pm UTC**

could you drop a camera attached to a rope inside a black hole and have it take pictures, then pull it back out?

The difficult thing to get your head around when it comes to black holes is the crazy way they distort space-time.

We think of the universe as having three dimensions of space and one of time - and with space and time being quite distinct entities. Well, at the gravities we routinely come across, they are.

As you get to and beyond a black hole horizon, though, the dimensions in a sense 'rotate around' mathematically, so one of the space dimensions becomes time-like, and the time dimension becomes space-like.

You wouldn't actually notice anything special on passing an event horizon, assuming the black hole was huge enough that tidal forces were insignificant there; Everything would seem to be pretty normal. But because of the way the dimensions have 'rotated around', to travel in the 'direction' 'away from' the black hole, you'd actually have to travel a 'negative amount' in a 'time-like' dimension (and we know of no way to travel back in time of course). That's why when you do the maths, as you get closer and closer to the event horizon, the energies required to move directly away from it approach infinity - because once you reach the event horizon, there is no 'space-like' direction away from it to travel along.

[Disclaimer: IANAP. Consult an expert before relying on any information in this posting!]

Posted: **Sun Aug 12, 2012 7:04 am UTC**

Wait a second. I thought that an outside observer would never actually see the camera cross the singularity, but rather it would seem to slow down as it approaches the event horizon (or something to do with the camera's information being "smeared" around it? Sorry, I have no idea...) So wouldn't that make it a bit difficult to infer that the camera would be within the event horizon at any given time in your reference frame? I guess that leads to the quick FAQey question, "Is there a way for an object to ever be inside a black hole's event horizon from the viewpoint of an outside observer?"

Posted: **Sun Aug 12, 2012 1:23 pm UTC**

I thought that an outside observer would never actually see the camera cross the singularity, but rather it would seem to slow down as it approaches the event horizon

So, what is happening here is that as you get closer and closer to the event horizon, there is still light that is escaping, but it is taking longer and longer to escape. So it seems to an outside observer that the object slows down, until as it reaches the event horizon it takes an infinite amount of time for the light to escape, so the object appears stopped just before the black hole (because there is always a small amount of light still escaping).

Well, when I say that I am considering space to be continuous. Quantum mechanics tell us that space is not continuous, so it would seem that you would see stuff fall into the black hole eventually (when is impossible to say). But, people don't really understand how space works in the regions of black holes, because the gravitational effects become really important, relativity becomes really important, and quantum mechanics become really important. Unfortunatly these two theories are fundamentally incompatible, and it is one of the big big areas in physics to try and reconcile them (think string theory).

s there a way for an object to ever be inside a black hole's event horizon from the viewpoint of an outside observer?

Maybe? But not according to relativity, to the best of my knowledge.

Posted: **Sun Aug 12, 2012 4:12 pm UTC**

Diemo wrote:So, what is happening here is that as you get closer and closer to the event horizon, there is still light that is escaping, but it is taking longer and longer to escape.

This is misleading - light always travels at c. It really does work as if time flows differently at different places.

Posted: **Sun Aug 12, 2012 4:14 pm UTC**

You can't hover "just outside the event horizon" to dip a camera on a fishing rod down into the interior of a black hole. Firstly, If you're stationary outside of a black hole, the Event Horizon is ALWAYS infinitely far below you, due to the way the black hole stretches space. No matter how many steps you take, you can't cross the EH at a walking pace. Your fishing line would have to be infinitely long to dangle the camera below the horizon.

The second problem is that the acceleration due to gravity experienced by any object attempting to hover above the horizon approaches infinity as you approach the Swarzschild radius. You'd need infinitely powerful engines to remain stationary.

If you turn off your engines and fall inward, the local gravity accelerates you downward. Your speed increases. Thanks to SR length contraction, the distance between you and the EH shrinks. At the EH, your acceleration becomes infinite, your velocity reaches c, and the length contraction due to SR becomes infinite, cancelling out the infinite length expansion in the spacetime surrounding the black hole. In you go. But at that point, there's no way to stop unless your engines can generate infinite force to oppose the gravitational acceleration.

Actually, yes: an object slows to a stop (as seen by an outside observer) at the EH, so long as the space around the black hole is otherwise empty. If more mass falls inward later, the approach of the increased mass causes the event horizon to expand, engulfing the first object. That's how a black hole forms in the first place when a star collapses.

The second problem is that the acceleration due to gravity experienced by any object attempting to hover above the horizon approaches infinity as you approach the Swarzschild radius. You'd need infinitely powerful engines to remain stationary.

If you turn off your engines and fall inward, the local gravity accelerates you downward. Your speed increases. Thanks to SR length contraction, the distance between you and the EH shrinks. At the EH, your acceleration becomes infinite, your velocity reaches c, and the length contraction due to SR becomes infinite, cancelling out the infinite length expansion in the spacetime surrounding the black hole. In you go. But at that point, there's no way to stop unless your engines can generate infinite force to oppose the gravitational acceleration.

cyanyoshi wrote:"Is there a way for an object to ever be inside a black hole's event horizon from the viewpoint of an outside observer?"

Actually, yes: an object slows to a stop (as seen by an outside observer) at the EH, so long as the space around the black hole is otherwise empty. If more mass falls inward later, the approach of the increased mass causes the event horizon to expand, engulfing the first object. That's how a black hole forms in the first place when a star collapses.

Posted: **Thu Aug 16, 2012 1:39 pm UTC**

I've always been a bit curious about what happens when 2 event horizons intersect.

if you have black holes who's schwarzschild radiuses graze each other are they doomed to combine or is it possible for the 2 to seperate again?(lets say they're both supermassive black holes, both are moving at close to c in opposite directions and the intersection is tiny)

would the event horizons of the 2 black holes retreat from each other at the location where the forces from both cancel out?

if they did combine I imagine there would be a vast amount of potential energy being lost in such a situation, would his manifest as an increase in the mass of the black holes or in some other manner?

if you have black holes who's schwarzschild radiuses graze each other are they doomed to combine or is it possible for the 2 to seperate again?(lets say they're both supermassive black holes, both are moving at close to c in opposite directions and the intersection is tiny)

would the event horizons of the 2 black holes retreat from each other at the location where the forces from both cancel out?

if they did combine I imagine there would be a vast amount of potential energy being lost in such a situation, would his manifest as an increase in the mass of the black holes or in some other manner?

Posted: **Thu Aug 16, 2012 3:50 pm UTC**

HungryHobo wrote:I've always been a bit curious about what happens when 2 event horizons intersect.

See also this thread for some discussion on this.

Posted: **Thu Aug 16, 2012 8:12 pm UTC**

Now, I'm no master on theoretical physics, But I have yet to see anything in this thread that suggests that it's theoretically impossible to lift the camera out

suppose you had a mass `m` inside the event horizon of a black hole, and you applied a force 'F' to it pointing out of the black hole whatever is causing 'F' is undetermined. How big does 'F' have to be in order to take 'm' outside of the event horizon

does it literally take more than`Lim as F -> Infinity (F)` to lift the mass out?

If I were to try to calculate this, I really couldn't go too much deeper than a classical mechanic's calculation, and I'd imagine that F would be finite. Is there something about theoretical physics that makes it different?

suppose you had a mass `m` inside the event horizon of a black hole, and you applied a force 'F' to it pointing out of the black hole whatever is causing 'F' is undetermined. How big does 'F' have to be in order to take 'm' outside of the event horizon

does it literally take more than`Lim as F -> Infinity (F)` to lift the mass out?

If I were to try to calculate this, I really couldn't go too much deeper than a classical mechanic's calculation, and I'd imagine that F would be finite. Is there something about theoretical physics that makes it different?

Posted: **Thu Aug 16, 2012 8:23 pm UTC**

sam_i_am wrote:and you applied a force 'F' to it pointing out of the black hole

This is the impossible part, according to my understanding.

It's not just a question of how much force you'd need - spacetime within the event horizon is so strongly distorted that there IS no direction that points out of the black hole. Once you're inside the event horizon, every direction points towards the singularity, and all paths away from the singularity point into the past.

Posted: **Thu Aug 16, 2012 8:46 pm UTC**

sam_i_am wrote:Now, I'm no master on theoretical physics, But I have yet to see anything in this thread that suggests that it's theoretically impossible to lift the camera out

suppose you had a mass `m` inside the event horizon of a black hole, and you applied a force 'F' to it pointing out of the black hole whatever is causing 'F' is undetermined. How big does 'F' have to be in order to take 'm' outside of the event horizon

does it literally take more than`Lim as F -> Infinity (F)` to lift the mass out?

If I were to try to calculate this, I really couldn't go too much deeper than a classical mechanic's calculation, and I'd imagine that F would be finite. Is there something about theoretical physics that makes it different?

We're talking about close proximity to a black hole. In that situation, classical mechanics is close to useless; you really have to use relativistic equations.

There are two answers to this that take at least some portion of relativity into account. The sort of naive one is that yes, it really does literally take more than infinite force to lift the mass out. The probably more correct one is that asking "what direction should I push in" gets a result of "undefined". Or maybe "backward through time", I believe that's a valid interpretation of the math. You can't pull or push something out of the black hole because the direction "towards the outside of the black hole" literally does not exist as a spatial direction. To get something out of a black hole, you would literally need a time machine.

Posted: **Fri Aug 17, 2012 1:16 am UTC**

Whereas most other posters in this thread have focused on the tidal forces tearing the camera apart, or whether the camera and the rope could survive, I'll explain why it would be impossible to pull the camera back out, and why sending information back up the cable would be impossible.

The original post seems to have a misunderstanding or lack of understanding about a. Einstein's Special Theory of Relativity and/or b. what exactly constitutes a black hole.

Einstein's equation for kinetic energy, from his Special Theory of Relativity, is

E=[imath]((mc^2)/√(1-v^2/c^2))-mc^2[/imath]

where E is the total kinetic energy of the object in joules, m is the object's mass in kg, v is the velocity of the object in meters per second, and c is the speed of light in meters per second.

If you look at

[imath]√(1-v^2/c^2)[/imath]

you will notice that as [imath]v^2[/imath] approaches [imath]c^2[/imath], the entire expression approaches 0, and when [imath]v^2=c^2[/imath], [imath]√(1-v^2/c^2)=0[/imath]. The result is that, as [imath]v^2[/imath] approaches [imath]c^2[/imath], [imath](mc^2)/√(1-v^2/c^2)[/imath] becomes arbitrarily large.

What this means without the math is that, as an object approaches the speed of light, the kinetic energy of that object becomes arbitrarily large, and an object with mass moving at the speed of light would have infinite kinetic energy. This is how we know it is impossible to accelerate an object to the speed of light.

Now for why this means escaping a black hole (for an object with mass) is impossible. A black hole is, by definition, a region of space in which gravity is so strong that it escape velocity is greater than c. Since this means that, for an object with mass, escaping the black hole would require accelerating it to a velocity greater than c, it would require infinite energy to do so, and would therefore be impossible.

As to why light cannot escape, the explanation is simpler: light always moves at c. Since the escape velocity is greater than c, light will never reach the escape velocity.

The original post seems to have a misunderstanding or lack of understanding about a. Einstein's Special Theory of Relativity and/or b. what exactly constitutes a black hole.

Einstein's equation for kinetic energy, from his Special Theory of Relativity, is

E=[imath]((mc^2)/√(1-v^2/c^2))-mc^2[/imath]

where E is the total kinetic energy of the object in joules, m is the object's mass in kg, v is the velocity of the object in meters per second, and c is the speed of light in meters per second.

If you look at

[imath]√(1-v^2/c^2)[/imath]

you will notice that as [imath]v^2[/imath] approaches [imath]c^2[/imath], the entire expression approaches 0, and when [imath]v^2=c^2[/imath], [imath]√(1-v^2/c^2)=0[/imath]. The result is that, as [imath]v^2[/imath] approaches [imath]c^2[/imath], [imath](mc^2)/√(1-v^2/c^2)[/imath] becomes arbitrarily large.

What this means without the math is that, as an object approaches the speed of light, the kinetic energy of that object becomes arbitrarily large, and an object with mass moving at the speed of light would have infinite kinetic energy. This is how we know it is impossible to accelerate an object to the speed of light.

Now for why this means escaping a black hole (for an object with mass) is impossible. A black hole is, by definition, a region of space in which gravity is so strong that it escape velocity is greater than c. Since this means that, for an object with mass, escaping the black hole would require accelerating it to a velocity greater than c, it would require infinite energy to do so, and would therefore be impossible.

As to why light cannot escape, the explanation is simpler: light always moves at c. Since the escape velocity is greater than c, light will never reach the escape velocity.

Posted: **Fri Aug 17, 2012 1:40 am UTC**

Xaphnir wrote:Now for why this means escaping a black hole (for an object with mass) is impossible. A black hole is, by definition, a region of space in which gravity is so strong that it escape velocity is greater than c. Since this means that, for an object with mass, escaping the black hole would require accelerating it to a velocity greater than c, it would require infinite energy to do so, and would therefore be impossible.

I should point out that one need not necessarily hit escape velocity to escape from an object's gravity; escape velocity is the velocity required to escape on a ballistic trajectory. I could escape the Earth's gravity at 1m/s if I wanted to; I just need to be applying thrust the entire time.

Posted: **Fri Aug 17, 2012 4:20 am UTC**

Considering you need General Relativity for black holes, the original poster may not be the only one with a misunderstanding about the underlying theory.Xaphnir wrote:The original post seems to have a misunderstanding or lack of understanding about a. Einstein's Special Theory of Relativity and/or b. what exactly constitutes a black hole.

Posted: **Fri Aug 17, 2012 2:11 pm UTC**

yurell wrote:Xaphnir wrote:Now for why this means escaping a black hole (for an object with mass) is impossible. A black hole is, by definition, a region of space in which gravity is so strong that it escape velocity is greater than c. Since this means that, for an object with mass, escaping the black hole would require accelerating it to a velocity greater than c, it would require infinite energy to do so, and would therefore be impossible.

I could escape the Earth's gravity at 1m/s if I wanted to;

Really?

Posted: **Fri Aug 17, 2012 2:30 pm UTC**

sam_i_am wrote:yurell wrote:Xaphnir wrote:Now for why this means escaping a black hole (for an object with mass) is impossible. A black hole is, by definition, a region of space in which gravity is so strong that it escape velocity is greater than c. Since this means that, for an object with mass, escaping the black hole would require accelerating it to a velocity greater than c, it would require infinite energy to do so, and would therefore be impossible.

I could escape the Earth's gravity at 1m/s if I wanted to;

Really?

I think yurell is right, however the energy required to maintain 1m/s would vary depending on gravity, in a black hole the energy required to go >0m/s within the event horizon would be approaching infinity..... I think.

Posted: **Fri Aug 17, 2012 2:36 pm UTC**

Yes. Climb a ladder, for example. It takes a lot more energy to do it that way, though, since you spend a lot more time fighting high gravity near the surface than if you start out going really fast.sam_i_am wrote:yurell wrote:

I could escape the Earth's gravity at 1m/s if I wanted to;

Really?

This is why the realization that escape velocity is the speed of light (By Newtonian calculations) is irrelevant. You can't even hover at 0 m/s at the event horizon, which is obviously not the case at the surface of other bodies.

Posted: **Fri Aug 17, 2012 2:37 pm UTC**

AvatarIII wrote:I think yurell is right, however the energy required to maintain 1m/s would vary depending on gravity, in a black hole the energy required to go >0m/s within the event horizon would be approaching infinity..... I think.

I believe so.

to move one meter away from the earth per second you'd have to accelerate away from the earth at 10.78 m/s then maintain at least 9.78 m/s of thrust.

To move one meter away from a black hole at the event horizon you'd need to accelerate away from the black hole at 299792459 m/s (speed of light 299792458 m/s) which I'm fairly sure is impossible.

Posted: **Fri Aug 17, 2012 3:19 pm UTC**

HungryHobo wrote:AvatarIII wrote:I think yurell is right, however the energy required to maintain 1m/s would vary depending on gravity, in a black hole the energy required to go >0m/s within the event horizon would be approaching infinity..... I think.

I believe so.

to move one meter away from the earth per second you'd have to accelerate away from the earth at 10.78 m/s then maintain at least 9.78 m/s of thrust.

To move one meter away from a black hole at the event horizon you'd need to accelerate away from the black hole at 299792459 m/s (speed of light 299792458 m/s) which I'm fairly sure is impossible.

I think it's already been said that classical physics doesn't apply, but I don't think I agree with the calculation you made there,

I believe the force of gravity F = GMm/r

According to wikipedia, escape velocity v = sqrt(2GM/r)

so,

c = sqrt(2GM/r)

rF/m = GM/r

c = sqrt(2rF/m)

(c)

m(c)

(c)

So, if classical mechanics applied, the acceleration to hover would more or less be dependent upon where the event horizon was. but regardless i don't believe that, v

Posted: **Fri Aug 17, 2012 4:11 pm UTC**

And on an unrelated note, why would a black hole be really black?

I would imagine that there are a lot of interactions going on outside the event horizon that would produce a photon. I would also imagine that a lot of atoms would be orbiting the hole in their death spiral that would interact. Photons outside the event horizon could escape and be detected, no? Probably a lot of high energy gamma rays, but also wavelengths in the visible spectrum?

I would imagine that there are a lot of interactions going on outside the event horizon that would produce a photon. I would also imagine that a lot of atoms would be orbiting the hole in their death spiral that would interact. Photons outside the event horizon could escape and be detected, no? Probably a lot of high energy gamma rays, but also wavelengths in the visible spectrum?

Posted: **Fri Aug 17, 2012 5:10 pm UTC**

No, they're definitely not, as simple dimensional analysis will tell you (one is a velocity, the other is an acceleration). Also, the fact that Earth's gravity is 9.8m/ssam_i_am wrote:i don't believe that, v_{e}=g is valid.

Posted: **Fri Aug 17, 2012 8:44 pm UTC**

Arkham wrote:And on an unrelated note, why would a black hole be really black?

I would imagine that there are a lot of interactions going on outside the event horizon that would produce a photon. I would also imagine that a lot of atoms would be orbiting the hole in their death spiral that would interact. Photons outside the event horizon could escape and be detected, no? Probably a lot of high energy gamma rays, but also wavelengths in the visible spectrum?

If I recall correctly, you are correct and such emissions are one of the ways astronomers attempt to detect black holes. The black hole itself only emits Hawking radiation, which gets weaker with higher mass (and with how much mass is usually required to form a black hole in the first place...), but in reality there's always other stuff in the process of falling into the black hole and the falling stuff can radiate quite energetically.