## Huge planet

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The-Rabid-Monkey
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### Huge planet

Now, my friend sent me this link, http://tinyurl.com/kmkhf8. I'm pretty sure that if a planet is 6.38*10^16 m across in radius and has a mass of 5.9742*10^44 kg it would like, crunch in on itself or something. Wouldn't it? Or is it possible that this planet could exist without crunching in upon itself?

Also, I just noticed a planet that size would be a couple of magnitudes heavier than the milk way. It'd need an equally insanely huge star to circle methinks.
Last edited by The-Rabid-Monkey on Wed Jun 24, 2009 10:04 am UTC, edited 1 time in total.
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quintopia
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### Re: Huge planet

Aren't those just about the same as the parameters for earth times 10^13?

Yes, I should think that the pressure from that much mass something that size should very rapidly increase the density at the core until a black hole forms. The "planet" wouldn't remain one very long.

Zamfir
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### Re: Huge planet

It could be a hollow structure.

BlackSails
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### Re: Huge planet

You can stand on the surface of a supermassive black hole - the gravity would be negligible.

Zamfir
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### Re: Huge planet

Waht's the surface of a black hole?

gmalivuk
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### Re: Huge planet

You can stand on a shell just outside the event horizon, then.
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Kow
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### Re: Huge planet

BlackSails wrote:You can stand on the surface of a supermassive black hole - the gravity would be negligible.

How is this?

Technical Ben
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### Re: Huge planet

[Joins in the volly of questions and answers]
You can orbit* around a supermassive black holes event horizon quite easily [/volley]

*or "float" as a really slow orbit just looks like your stationary...
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EricH
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### Re: Huge planet

Technical Ben wrote:You can orbit* around a supermassive black holes event horizon quite easily [/volley]

*or "float" as a really slow orbit just looks like your stationary...

Really slow orbit? Near the event horizon? I must be missing something--at the event horizon, escape velocity is c, so the orbital velocity you need to stay just above the event horizon would be just under....

What's easy about orbiting that close to a supermassive black hole is that its tidal forces won't rip you apart, not that it won't take lots of velocity.
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gmalivuk
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### Re: Huge planet

Kow wrote:
BlackSails wrote:You can stand on the surface of a supermassive black hole - the gravity would be negligible.

How is this?

Basically, because gravitational force drops with the square of the radius, whereas escape velocity drops linearly, and the radius of a black hole increases linearly with mass.

In other words, if you have black holes A and B, and B is ten times more massive than A, then we have the following:
- The radius of the event horizon B is 10 times that of A.
- The escape velocity at the event horizon is c by definition. (Well, sorta...)
- The force of gravity at a great distance from B is 10 times what it is that distance from A, but since the event horizon is 10 times farther out, that's at a distance where the gravity from A would be 1/100 what it is at A's event horizon. So the gravity at B's event horizon is 1/10 what it is at A's.

So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

EricH wrote:at the event horizon, escape velocity is c, so the orbital velocity you need to stay just above the event horizon would be just under....

Actually, since near the event horizon Newtonian equations don't do you much good, the orbital velocity just above the horizon would be quite a bit greater than the speed of light. It's the speed of light 50% farther out, and drops from there. But inside that (and still outside the horizon), you absolutely need to have your rocket engines on if you want to avoid falling in.

They wouldn't need to be pushing that hard, though, because they're only countering the force of gravity, which isn't that great if the black hole's big enough. What gets you is the much shallower gradient, which means that even though they don't have to push as hard, they have to push a hell of a lot longer before you have any chance of escaping.
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The-Rabid-Monkey
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### Re: Huge planet

So, a planet that big would crunch into a gigantic black hole then? Fun. Cheers for your help and interesting discussion on insanely big black holes.
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Charlie!
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### Re: Huge planet

gmalivuk wrote:
EricH wrote:at the event horizon, escape velocity is c, so the orbital velocity you need to stay just above the event horizon would be just under....

Actually, since near the event horizon Newtonian equations don't do you much good, the orbital velocity just above the horizon would be quite a bit greater than the speed of light. It's the speed of light 50% farther out, and drops from there. But inside that (and still outside the horizon), you absolutely need to have your rocket engines on if you want to avoid falling in.

They wouldn't need to be pushing that hard, though, because they're only countering the force of gravity, which isn't that great if the black hole's big enough. What gets you is the much shallower gradient, which means that even though they don't have to push as hard, they have to push a hell of a lot longer before you have any chance of escaping.

Hm.

Centripetal acceleration if you move at c is c^2/r. Just barely around a black hole of radius m*2G/c^2 (making centripetal acceleration c^4/(2Gm)), the gravitational acceleration will then be c^4/(4*G*m). If you will note, this means that the centripetal acceleration going very fast near the horizon is twice that of the gravitational acceleration, and decreases as the mass of the black hole increases. Note: the simple stuff is all totally legit in the frame of the orbiting body. I think
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Kow
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### Re: Huge planet

gmalivuk wrote:So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

I still don't see how this would work. Low gravity just above the event horizon, but high enough that light can't escape just inside? What creates such a steep gradient?

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### Re: Huge planet

The-Rabid-Monkey wrote:Now, my friend sent me this link, http://tinyurl.com/kmkhf8. I'm pretty sure that if a planet is 6.38*10^16 m across in radius and has a mass of 5.9742*10^44 kg it would like, crunch in on itself or something. Wouldn't it? Or is it possible that this planet could exist without crunching in upon itself?

*crunches a few numbers*
The Schwartzchild radius of that much mass is 8.87e17 m. It would already be a blackhole...
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### Re: Huge planet

Kow wrote:
gmalivuk wrote:So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

I still don't see how this would work. Low gravity just above the event horizon, but high enough that light can't escape just inside? What creates such a steep gradient?

The problem is a shallow gradient (in the force of gravity), and what creates it is a huge amount of mass. Escape velocity is classically computed from the amount of energy it would take to remove something out to an infinite distance away. Force drops to 1/4 as you double your distance from the center of something, but for a really big something, like the sun, gravity is 1g at something like 400000km, which means you have to go another 400000km to have it drop to a quarter of that. On Earth, by comparison, you only have to go a bit more than 6000km for the force to drop to 1/4g.

Energy is force*distance, and so having the same force pulling you back for a much greater distance means you need to start out going much faster.
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Goemon
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### Re: Huge planet

gmalivuk wrote:So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

That's only a Newtonian approximation... Thanks to relativistic corrections, the acceleration required to hover just outside the EH is infinite regardless of the radius of a black hole.
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Charlie!
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### Re: Huge planet

Kow wrote:
gmalivuk wrote:So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

I still don't see how this would work. Low gravity just above the event horizon, but high enough that light can't escape just inside? What creates such a steep gradient?

The trick is that there's no set acceleration at the event horizon - what determines the radius isn't acceleration, but rather the escape velocity needed (which basically comes from how deep you are in the gravitational potential). So with a big enough mass, you get a huge potential well with more gently sloping "sides" (At a certain depth) than a smaller well whose sides are steep at the same depth.

Goemon wrote:
gmalivuk wrote:So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

That's only a Newtonian approximation... Thanks to relativistic corrections, the acceleration required to hover just outside the EH is infinite regardless of the radius of a black hole.

I'm curious, what relativistic corrections? The main problem I have wrapping my head around that claim is that, to the person expending the energy, there's never a clear boundary. The event horizon is only apparent to someone else. So by checking your energy expenditure you would be able to get information about your absolute location in a potential well, not just relative location.
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massivefoot
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### Re: Huge planet

Alright, there seems to be some (understandable) confusion here. The black hole region of a spacetime is defined to be the region from which no light signal can be sent to an observer at infinity, and the future event horizon is defined to be the boundary of this region. The reference to observers at inifinity make this sound very much like something with only global significance, and indeed it is - an observer passing the event horizon doesn't detect anything special occuring locally.

But you're interested in whether you can stay outside this horizon, particularly in the case of a Schwarzschild black hole, and whether this takes an infinite force, which all depends on where you're standing. Suppose that I stand at infinity (having taken the trouble of walking there), and that I have with me my A-level Mechanics KitTM, which of course include an unit point mass and a light, infinite, inextendible sting. I tie the mass to the string and lower it towards the black hole, and measure the force that I need to hold it stationary at various distances. As it approaches the event horizon, I would find that the force approached 1/4M, which is called the surface gravity (actually there's probably a few constants like G missing there - mathematicians have a bad habit of setting such things to 1.)

However, if I were nearer the black hole (and somehow remaining motionless myself) I would find that this force was larger. And as I position myself closer to the horizon, I would find that it would become infinite! Therefore, if we were actually to sit at infinity and attempt this experiment, our string would snap somewhere near the event horizon.

gmalivuk
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### Re: Huge planet

Right, I wasn't thinking about the other dilation effects that happen to the observer who is doing the accelerating near the horizon. When that's taken into account, what looks like finite force from infinity gets a mite larger...
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### Re: Huge planet

The accretion disks around supermassive black holes can emit so much radiation they are visible across the entire universe. So these are not benign regions of space.

As gmalivuk rightly pointed out, the surface gravity of a black hole is inversely proportional to its mass, and thus very small for a supermassive black hole. But the distances involved are correspondingly big. So matter falling into a supermassive black hole is still accelerated to relativistic speeds - a small force over a huge distance still adds up.

So you might be able to stand on the boundary of a supermassive black hole without feeling significant gravity. You still need a stupendously powerful engine to do so (see massivefoot's excellent post). And you'll be bathed by huge amounts of radiation. Which is not healthy either
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Kow
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### Re: Huge planet

gmalivuk wrote:
Kow wrote:
gmalivuk wrote:So a big enough black hole can have an arbitrarily small force of gravity at the event horizon.

I still don't see how this would work. Low gravity just above the event horizon, but high enough that light can't escape just inside? What creates such a steep gradient?

The problem is a shallow gradient (in the force of gravity), and what creates it is a huge amount of mass. Escape velocity is classically computed from the amount of energy it would take to remove something out to an infinite distance away. Force drops to 1/4 as you double your distance from the center of something, but for a really big something, like the sun, gravity is 1g at something like 400000km, which means you have to go another 400000km to have it drop to a quarter of that. On Earth, by comparison, you only have to go a bit more than 6000km for the force to drop to 1/4g.

Energy is force*distance, and so having the same force pulling you back for a much greater distance means you need to start out going much faster.

I understand you on the escape velocity part, it's the event horizon part that gets me. If you were to enter from stationary (relative to the black hole) just above the event horizon, and go just inside (yes, I know there's a problem with time dilation doing this, etc) you'd suddenly feel a gravitational force "sucking" you in? Or is the event horizon defined by the location where light can't escape past when emitted from the center? This would pose a problem though, 'cause you'd be able to go inside the event horizon from the outside and still return, since gravity isn't pulling so hard?

I think I'm looking at this wrong.

massivefoot
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### Re: Huge planet

Kow wrote:I understand you on the escape velocity part, it's the event horizon part that gets me. If you were to enter from stationary (relative to the black hole) just above the event horizon, and go just inside (yes, I know there's a problem with time dilation doing this, etc) you'd suddenly feel a gravitational force "sucking" you in? Or is the event horizon defined by the location where light can't escape past when emitted from the center? This would pose a problem though, 'cause you'd be able to go inside the event horizon from the outside and still return, since gravity isn't pulling so hard?

I think I'm looking at this wrong.

You are looking at it wrong, but it's a very understandable mistake that you're making. It's one of such magnitude and subtlety that the realisation that we're "looking at it wrong" was a significant step in formulating general relativity.

Suppose you're in a laboratory with no windows or other obvious means of interacting with the outside world, and let's suppose you do some experiments that only involve stuff inside this lab. Then the result of those experiments will be the same regardless of whether the laboratory is sat at rest on the Earth's surface, or being accelerated at a rate of [imath]9.81 m s^{-2}[/imath] somewhere in deep space. There's no way you can unambiguously determine the graviational field strength within your laboratory.

However, you can measure how much it varies across the laboratory - the tidal force. And this is exactly what happens near the event horizon - there's no sense in which you can measure the gravitational "suck" locally. If the black hole is nice and big, then the tidal force is low, and you don't see anything special happen as you cross the horizon.

But if nothing wierd and wonderful happens, why can't you switch on a torch as you pass the horizon and send a light ray outwards? The answer to this is basically to do with coordinate systems. You can shine a light ray in a direction that is "outwards" relative to your "freely falling", but this isn't outward in the coordinate system of an observer at infinity. If you know anything about lightcones, it might help to know that, if you drew a spacetime diagram, the lightcones near the event horizon would be "tipped over" so that no portion of the future lightcone was pointing outwards.

Kow
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### Re: Huge planet

massivefoot wrote:You are looking at it wrong, but it's a very understandable mistake that you're making. It's one of such magnitude and subtlety that the realisation that we're "looking at it wrong" was a significant step in formulating general relativity.

Suppose you're in a laboratory with no windows or other obvious means of interacting with the outside world, and let's suppose you do some experiments that only involve stuff inside this lab. Then the result of those experiments will be the same regardless of whether the laboratory is sat at rest on the Earth's surface, or being accelerated at a rate of [imath]9.81 m s^{-2}[/imath] somewhere in deep space. There's no way you can unambiguously determine the graviational field strength within your laboratory.

However, you can measure how much it varies across the laboratory - the tidal force. And this is exactly what happens near the event horizon - there's no sense in which you can measure the gravitational "suck" locally. If the black hole is nice and big, then the tidal force is low, and you don't see anything special happen as you cross the horizon.

But if nothing wierd and wonderful happens, why can't you switch on a torch as you pass the horizon and send a light ray outwards? The answer to this is basically to do with coordinate systems. You can shine a light ray in a direction that is "outwards" relative to your "freely falling", but this isn't outward in the coordinate system of an observer at infinity. If you know anything about lightcones, it might help to know that, if you drew a spacetime diagram, the lightcones near the event horizon would be "tipped over" so that no portion of the future lightcone was pointing outwards.

I think the part I don't get is HOW, at the event horizon, the pull becomes so strong when outside the event horizon it's a relatively low amount of gravity.

Charlie!
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### Re: Huge planet

gmalivuk wrote:Right, I wasn't thinking about the other dilation effects that happen to the observer who is doing the accelerating near the horizon. When that's taken into account, what looks like finite force from infinity gets a mite larger...

Muh? Still confused. What other dilation effects?

Or are you talking, like massivefoot seemed to be, about lowering a string down to the event horizon, and how that force will increase as you get closer to where you see the event horizon (since that means getting closer to the center at a faster rate)? Because that would confuse me and my interpreting event horizon to mean "distance about a schwartzchild radius from the black hole where a very outside observer sees the event horizon."
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### Re: Huge planet

Kow wrote:I think the part I don't get is HOW, at the event horizon, the pull becomes so strong when outside the event horizon it's a relatively low amount of gravity.

There's no sudden change in pull. The pull just below the event horizon is about the same (marginally stronger) then the pull just outside it.

The problem is the warping of space. The closer you get to a black hole, the more warped space becomes. The event horizon is the point where space is so warped that there are no longer any routes going out. You can certainly start moving out, but you will find that the road is so curved you aren't actually going anywhere but in.

A flawed, but insightful, analogy is a threadmill. Imagine you're walking on a giant threadmill, trying to reach the other side to get off. The faster threadmill goes, the more energy you have to expend to get off. Trying to escape from a black hole is like running on such a threadmill. But the closer you get to the event horizon, the faster the threadmill runs. So the faster you need to run outwards to escape. At the horizon itself the threadmill goes with the lightspeed, and no matter how fast you run you'll never be fast enough.
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gmalivuk
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### Re: Huge planet

Okay, so now I'm confused myself.

Was massivefoot right, that the apparent force is different depending on where you're looking at it from? Or was I right all along, that a small force on your part just outside a huge event horizon can keep you stationary w.r.t. the rest of the universe, but once you're just inside, it's no longer possible for any force to do that?

Because I'm now inclined to believe I was wrong before. If I was right, then there's a distance outside the event horizon from which continuous application of some relatively small (subjective) thrust, albeit for perhaps a very long time, can get me out to wherever I want in the outside universe. But if I let the thrust dip for a fraction of a second and fall a bit closer to the hole, no amount of thrust will ever get me anywhere but the singularity. And that doesn't make a whole lot of sense to me...
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### Re: Huge planet

gmalivuk wrote:Was massivefoot right, that the apparent force is different depending on where you're looking at it from? Or was I right all along, that a small force on your part just outside a huge event horizon can keep you stationary w.r.t. the rest of the universe, but once you're just inside, it's no longer possible for any force to do that?
massivefoot has it right. If you try to hover in place some distance above the horizon, the acceleration required to maintain your position, as you measure it, becomes infinite as your desired location gets closer to the horizon.

Kow
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### Re: Huge planet

Diadem wrote:The problem is the warping of space. The closer you get to a black hole, the more warped space becomes. The event horizon is the point where space is so warped that there are no longer any routes going out. You can certainly start moving out, but you will find that the road is so curved you aren't actually going anywhere but in.

This is the answer I was looking for. Thanks. I was thinking in terms of simple gravitational force and not bent spacetime.

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### Re: Huge planet

[ninja'd, but...] The acceleration "due to gravity" felt by someone hovering outside the event horizon at Schwarzschild coordinate r is (in geometric units)

$a = \frac{m}{r^2 \sqrt{1-\frac{2m}{r}}}$

So if you try to descend slowly on rocket power to a point just immediately outside the EH, the acceleration you feel sitting in your Captain Kirk command chair increases without limit as you approach the Schwarzschild radius r = 2m. You'd get crushed into neutronium or worse, if it was possible for your rockets to provide enough thrust to remain stationary. Which they can't, of course, if you get too close. There is no rocket or means of propulsion conceivable that can hover right at the EH.

For the same reason, it's not possible to build a stationary shell immediately outside an EH, even if the black hole is static. The physics says it's theoretically impossible even for unobtanium to withstand the forces there. So if you've ever wondered why you can't stand just outside the EH, dip your yo-yo below the line and pull it back up, one reason is because the pull of gravity is so strong that the string MUST break. (There's also the fact that when you're hovering, length dilation makes the proper distance to the EH always just a bit farther than you can reach.)

[Edit...]

For someone very far away from the black hole, a yo-yo falling toward the EH on a very long string appears to slow to a stop due to time dilation as it approaches the Schwarzschild radius. So, as seen from very far away, the force needed on the string to keep the yo-yo from descending further is approaching zero. But time doesn't stop from the yo-yo's point of view... and as far as local observers are concerned, it takes infinite force to hover just outside the EH.
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Charlie!
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### Re: Huge planet

Goemon wrote:The acceleration "due to gravity" felt by someone hovering outside the event horizon at Schwarzschild coordinate r is (in geometric units)

$a = \frac{m}{r^2 \sqrt{1-\frac{2m}{r}}}$

Okay, why is that?

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### Re: Huge planet

That's the LAW

If you solve Einstein's Equations for a spherically symmetric, static, mass distribution, the solution you get is the Schwarzschild solution. You can even proof that this is the only possible solution. From this solution follows the acceleration law that Goemon gives.

All spherical masses work like this. Black holes, but also stars and planets. It's just that the 1 - 2m/r part is usually neglicible. For example for earth 2m comes out as a few centimeters. So 2m/r is, at the surface, of the order of 10^-8. So the 1 - 2m/r part can safely be left out. And if we do that you find the traditional Newtonian law: a = m / r^2

(For those uncomfortable with me expressing a mass in centimeters: Some constants such as G and c were left out in the above formula. Set to one. If you want to use everyday units you can always put them back in, but it makes the formulas a lot more cumbersome, so it's rarely done)
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Goemon
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### Re: Huge planet

Charlie! wrote:Okay, why is that?

Well, it shouldn't seem too outrageous, if you think about it. The Newtonian acceleration due to gravity looks like

$a = \frac{F}{m} = \frac{GM}{r^2}$

so we'd expect that far away from the black hole, the relativistic formula should reduce to the Newtonian formula and be proportional to m/r^2. (In geometric units, the base units of length, time, and mass conspire to make G = 1, and c = 1.) In the relativistic version of gravity, observers near the black hole experience time dilation and length contraction (as measured by someone far away) proportional to sqrt(1 - 2m/r). So with all those factors of sqrt(1 - 2m/r) flying around, shouldn't be all that surprising that the relativistic version of the acceleration has one showing up in the denominator...
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Charlie!
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### Re: Huge planet

If you solve Einstein's Equations for a spherically symmetric, static, mass distribution, the solution you get is the Schwarzschild solution. You can even proof that this is the only possible solution. From this solution follows the acceleration law that Goemon gives.

All spherical masses work like this. Black holes, but also stars and planets. It's just that the 1 - 2m/r part is usually neglicible. For example for earth 2m comes out as a few centimeters. So 2m/r is, at the surface, of the order of 10^-8. So the 1 - 2m/r part can safely be left out. And if we do that you find the traditional Newtonian law: a = m / r^2

Oh, okay. Guess I'll have to learn me some general relativity later then
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### Re: Huge planet

Goemon wrote:[ninja'd, but...]So if you try to descend slowly on rocket power to a point just immediately outside the EH, the acceleration you feel sitting in your Captain Kirk command chair increases without limit as you approach the Schwarzschild radius r = 2m. You'd get crushed into neutronium or worse, if it was possible for your rockets to provide enough thrust to remain stationary. Which they can't, of course, if you get too close. There is no rocket or means of propulsion conceivable that can hover right at the EH.

Not only that, but Unruh Radiation will disassociate your constituent particles into extra toasty plasma.

Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away.
-- Antoine de Saint-Exupery