# Air pressure at the centre of the Earth. (In a hole.)

No, that’s right. And, though the air is denser at the surface, that converts to 1.03 kg/cm^2 or 10.3 tn/m^2. Yeah, that’s mass/area rather than force/area, but where most of us live it’s about the same.

fwiw, 4pi(6400 km)^2 gives you a surface area of 5.1E+08 km^2 or ~5E+18 cm^2. multiply that by the surface pressure and we get a total atmospheric mass of ~5E+18 kg.

Estimates put earth core pressures at 300-400 GPa, which is about 3-4 million atmospheres or about 154 orders of magnitude greater than zut’s calculation.

I’m too stupid and lazy to go through the calculations but if pressure were to increase linearly (i.e., compression is exactly offset by the decrease in gravitational acceleration) the bottom of the hole would be ~64 atm. This has got to be the lower limit. 3-4 million atm has to be the upper limit. Given those limits, Triskadecamus’ 750 atm seems reasonable.

You can’t assume linearity of compression. Nitrogen has a critical point around 35 bar. There’s a lot of things odd about supercritical fluids, including it’s compressability. The other gases mixed in with air will have an effect, but I’d doubt that the column of air down to the center of the earth will act like a gas, period.

Absolutely you are right. I was just attempting to set up first-order limits for the problem…

I am going to pretend that it acts as an ideal gas all the way down. It may not be realistic, but it’s simple. Also, I know only very basic calculus so I can’t work out the numbers. I’m just going to describe what I’m pretty sure would happen. Then I’m going to go take a shower and maybe think of something more I want to say.

Under a constant gravity, air pressure increases at an exponential rate. You go down a certain distance, it doubles. You go down that same distance again, it doubles again. The complication is that gravity is decreasing as you go lower. So let’s think about what weight is. It’s mass times gravitational acceleration. Density for a gas is proportional to pressure, therefore we can assume that the density raises as we go lower. Density is increasing and gravity is decreasing. Back to exponents. So what’s the exponent? It’s proportional to the depth. What about the base of the exponent? For a given gas or mixture of gasses, it’s proportional to gravity. Obviously the more gravity you have, the less distance you have to go to double the pressure. So in other words, the base decreases as the exponent increases. So I have deduced that the weight of a volume of gas at a particular depth will be (c1G)^(c2d), where c1 and c2 are constants that I won’t even pretend I know how to calculate. According to this formula, the gas at the bottom of the hole will be weightless because of the lack of gravity, which is exactly true.

However, my carefully deduced formula only calculates the weight of a specific volume of air at a specific point, not the weight of the air on top of it. Calculating that would go something like 9.8^1+9.7^2+9.6^3… etc, and even that’s just a rough estimate, leaving out 9.75^1.5 and all that middle-ground. Luckily, we have calculus to solve this for us. Unluckily, I’ve only read one book on the matter. I’m going to take a shower.

No, not entirely randomly. I chose to ignore everything but the standard variation of pressure vs altitude, and pretend that no other factors existed, and solved for an altitude of four thousand miles below sea level.

My answer is exactly as valid as any other, including multiplying random numbers. Ignoring all the factors is identical, as far as validity of computation as choosing several to ignore, and pretending that they don’t matter. The fact that air would no longer have the characteristics of a gas at the pressure indicated is simply another factor to ignore, such as the fact that the hole cannot exist. Temperature is either ignored, or some mechanism must be imagined to change the real facts to match the imaginary situation. No matter what facts you choose to include, you choose to ignore others.

Imagine what you will, and then design your mathematics to give you the answer you desire. Or, ignore mathematics, and make it magic. The problem is not one of physics.

Tris

1.064*10[sup]4[/sup] psi assuming standard temperature of 0C all the way down.

This agrees approximately with Triskadecamus his answer being 753 atmospheres while mine is 723.

I arrived at my answer by integrating the weight of air from the surface down added to the weight of the atmosphere at the surface.

I’m not sure I agree with that. A great deal of physics involves ignoring factors that complicate the problem you are analysing so that you can arrive at a solution that, while not exact, is a useful approximation to reality. Once you have a first order-approximation, you can then refine your mathematical model in order to introduce factors which you had left out, to see if they have a significant effect.

As an example, consider the result that g = 9.8 m/s^2, used happily and extensively by large numbers of people. It’s not true because a) it varies from place to place b) is not constant over altitude c) ignores air resistance entirely. It’s still a useful approximation to reality, though, and I believe this question is being asked in the same spirit. Examining a problem can also reveal what factors might be relevant to the situation. To dismiss it as “not a problem of physics” defeats the very aim of physics, which is to enquire about the universe around us.

Looks like I opened up something of a can of worms here. Nice to see the finest minds on teh internets kept busy

Oh by the way BJMoose (and does that name conjure an odd mental image or what?) the snippet of Cecil’s column you quoted:

is true, but not really relevant to the pressure question, because the air acts as one continuous column pushing into the centre. In fact, even if the centre of the Earth were not molten, I would say it would be impossible to have a “cave” at the centre because the immense pressure of rock from all sides would quickly squish any putative hole out of existence.

You mean like this formula for variation of pressure with altitude? If so, check your math. Or a linearized version of it? If so, nothing wrong with that, but that’s effectively ignoring compressibility, which is good to state up front.

Pretty much what Devorin said. Intentionally ignoring or including specific factors is done all the time. Even when ignoring a particular factor leads to wildly “wrong” answers (as it does in this case!), it still bounds the problem, and lends insight into the effects of various parameters. Comparing the validity of an intentionally methodical procedure to that of a completely random answer is ignorant in the extreme.

The calculation diverges from reality because real gases are not infinitely compressible – at a certain point, it stops getting denser in proportion to pressure (all else being equal), and then the exponential increase in pressure with depth breaks down.

If you’re talking about planetary-scale gas columns, why not look at Jupiter and Saturn for inspiration? It’s pretty clear that at some depth you’ll get an ocean surface. I think it would be nitrogen first, since it will have the highest partial pressure, but it’s been over 10 years since high school chemistry, so feel free to beat me with blunt sticks.

At its finest, the SDMB is all about canned worms!

Yeah. I’ve heard about my username before. (I’ll skip the usual jokes involving Moosehead Beer.) I’ll just say that, when I enlisted, I figured Bullwinkle was already taken (it was) and settled on this form of the Blessed Name, not noticing its potential for levity.

What pressure at the center? Doesn’t the equal mass pulling at the center from all vectors cancel each other out? (Besides, around here we usually don’t let practical realities mess up our hypotheticals.)

Now to muddy up things further with notions from the old moosegut. It seems to me that maximum gravitation force is to be found at the surface, where you have the most mass below you. As you descend, you have less mass pulling at you from below and more mass pulling at you from above, giving a net gravity of less than 1G. When you get to the center, you reach 0G. I suspect then that air pressure would decrease as you drop below the surface.
Sounds like this question is ripe for submission to Cecil (tho’ one could hardly blame him if he ducked and covered).

You’re mixing up pressure and gravitational acceleration. Not the same thing. The air pressure is a function of the weight of the air above you. That weight is a function of the gravitational acceleration where the air is, not where you are.

Try this analogy on: Take a 1000 pound block of steel. Set it on a scale. What’s it weigh? 1000 pounds, right?

OK, now. Dig a hole ten feet deep. Put the scale in the bottom of the hole. Now (and here’s the logical jump) construct a massless column ten feet tall. Put the column on top of the scale. Put the weight on top of the column. What does the scale read? 1000 pounds, right?

OK. Dig a hole one mile deep. Put the scale in the bottom of the hole. Construct a massless column one mile tall. Put the column on top of the scale. Put the weight on top of the column. What does the scale read? Remember the weight is still at the surface, so it’s experiencing the same gravitational acceleration as it did without the hole. So it transmits the same force to thecolumn, and the column will transmit the same force to the scale. So the scale will read… 1000 pounds. Right?

Now dig a hole to the center of the Earth. Put the scale in the bottom of the hole. Construct a massless column of a height equal to the radius of the Earth. Put the column on top of the scale. Put the weight on top of the column. What does the scale read? The weight is still at the surface of the Earth, so it’s experiencing the same gravitational acceleration as it did without the hole, and the force is transferred through the column to the scale. So how could the scale read anything but 1000 pounds?

And that’s why there would be air pressure in the same hole. The weight of the air would still bear down on the entire column of air.

True, but the thing we’re trying to wrap our head around is the 1000lbs weight on the surface weighs 999lbs(or whatever) if put at the bottom of the hole on tope of the scale. I’m no math wiz, but I imagine that you would solve the problem first as if gravity was constant, and then multiplied that result by a fraction representive of the decrease in gravity on the way down.

Given that gravity would not decrease linearly on the way down, I doubt that’s a good idea. The first thing you have to do is make sure you have the right equation for the job, taking into account the different masses involved as you go deeper: there’s less mass below than there was before, and there’s more above. Once you have that, you integrate (with luck, the final equation will be integrable) from the core to the surface, and there’s your answer.

It’s definately possible that it isn’t linear, but even if the decrease is reflected as a bell curve, my suspicion is that could be reflected as a fractional amount.

I dunno, I’m guessing it would be more simple to just use gravity as the mass is the result of gravity and it’s variation on the way down. Fluid dynamics is a tricky subject though.

Gravity (more properly, gravitational attraction) is inversely proportional to the square of the distance; how on earth could you get a linear relationship from that? Any inverse relationship is not going to be computed well with a simple fractional factor.

Inside a planet of uniform density, the gravitational acceleration varies linearly. That’s because the mass of the shell “above” you cancels out. Then as the distance between you and the center of the planet decreases, the effective mass that’s attracting you also decreases. Mass is a cubic function of distance, which, when divided by the inverse square of the distance, results in a linear function.

A much more in-depth explanation is included in this very, very cool page on gravity and pressure inside planets. In fact, it’s worth a look just to see what the actual gravitational acceleration is in real, non-uniform-density planets.

The hole cannot exist.

An atmosphere extended four thousand miles is not an atmosphere of gas, nor can it have the same chemical make up as Earth’s atmosphere. It cannot exist on a planet of the same size as Earth. It cannot, in fact, exist at all.

You find the fact that I applied a linear function to determine a value for a condition that cannot exist at all to be less valid than using a different function to determine the value for a condition that cannot exist. Pardon me while I chuckle at your understanding of the meaning of the word valid.

So, we imagine that one part of physics doesn’t apply. Then we have to decide that some aspects of chemistry are also not operating. And we have to assume a magical level of physical strength and . . . it goes on and on. Why not just imagine the air to be balmy, and cool, and a constant pressure the entire way to the center of the Earth? Are you claiming that that is not valid, but some number you like is valid?

Pardon me while I continue to chuckle, and begin to guffaw at your understanding of the term valid.

It’s magic. It’s imaginary. It isn’t physics.

Mathematics doesn’t apply, except on an imaginary basis, and you can imagine what you want, and so can everyone else. But that doesn’t make one answer more valid than another. Like I said, pick two numbers, then either multiply, or add, or divide them. Then, use the result, or some other number you like more. That has precisely the same level of validity as any other number.

You don’t like my first number, Okay, I will make up another. Seven.

## Tris

“As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.” ~ Albert Einstein ~

OK, so a hole to the centre of the Earth is out. (Darn.)
But, slightly more within the realms of the real world, anyone care to make an informed estimate as to how deep a hole you’d have to dig before the air at the bottom of it liquefied under the pressure?

(Might have to assume constant temperature here, I suppose.)

By the way, in case you’re wondering what made me think of this question… I have an altimeter watch which also displays the local atmospheric pressure (and uses it to either display sea-level pressure, if you are at a fixed altitude, or altitude if you fix the SLP). The SLP in the UK is pretty high at the moment: 1042mb, or about 30.8inHg. When I go on the Tube, below sea level, the pressure reading obviously goes up even more. That got me wondering how deep below the ground you could actually go before air pressure got too much…