This question was touched upon in this thread but never answered.

What would be the atmospheric pressure at the bottom of a hole to the centre of the Earth, assuming of course that such a thing were possible. Ignore the heat issue - purely the pressure at the bottom of the hole. I guess it needs some kind of fancy integration calculation as the weight of each unit height of air will decrease the closer you get to the centre. So we would effectively have a sum of ever-decreasing weights exerted by each “slice” of air, until the very lastslice exerts no pressure at all on its own (but is of course supporting the weight of all that air above). It’s got to be a pretty hefty pressure, but can any maths wizard work it out?

Extra points for showing working. Points deducted for non-standard units such as “metric buttloads”.

Bonus question: Does the answer in any way depend on the amount of available air? It seems that if you hypothesize a hole that is wide enough (think Krispy Kreme Earth) you are going to rarefy the atmosphere and drop pressure at the surface, no?

You can choose your own elements of reality to ignore, and get any number you want. I chose to ignore one set, and used real mathematics to get this one. It is entirely meaningless, so I won’t bother to explain how I got that figure.

Wouldn’t the pressure depend on the temperature of the air in the whole? Presumably, air would be at the same temperature as the core and mantle surrounding it. Since the air would expand at that temperature, perhaps the pressure is not as high as Triskadecamus’s (assuming that he did not take temperature into consideration).

Cecil did a column once in which, I think, he said the gravity at the exact center of the earth is zero. So wouldn’t the atmospheric pressure at that point be zero?

No. As I said above - the bottommost slice of air will be exerting no pressure, because it is weightless. But the air one inch above the centre will be exerting pressure from its own weight… as well as transmitting the weight of the ~4000 mile column of air above it, each inch of which becomes progressively heavier up towards the Earth’s surface.

You might as well ask, why isn’t the pressure at the centre of the Earth zero, with no hole in it? Obviously there is the huge weight of rock bearing down from all sides, so assuming a uniformly dense sphere, the pressure would obviously be greatest at the centre.

I dunno. If air pressure is simply the weight of a column of air above a given point, and the force of gravity at this particular measuring point (the center of the earth) is zero, then I’m inclined to say no. But physics is not my forte.

That’s where you’re going wrong. Only the very bottom of the column is at a point with no gravity. The rest of the 4,000-mile column of air weighs plenty, and isn’t just going to magically levitate because the very bottom few millimetres of it are effectively weightless.

The fly in the ointment here is that we’re considering a hole that stops at the center of the earth, whereas Cecil was talking about one that went through. Since the atmosphere in the hole would have less mass than the mass of the earth opposite of the hole - well, this is out of my league. Tell ya what - name the place, and I’ll bring the shovels and we can figure this out the hard way.

I was thinking about this while driving to work today and came up with some thoughts but no real answer (not that there IS a real answer, seeing as how we are missing a comprehensive list of assumptions). I still feel like sharing so somebody can at least check and point out where I’m wrong.

I realized that if you assume we’re not going to suddenly produce more air, we have a firm constant that’s not a silly assumption: total mass of the atmosphere. According to this site we can assume that is 5E18 kg. Seems somewhat light, but there is multiple cites so I am not going to argue. For nice round numbers let’s assume that the atmosphere is 100km thick all around, earth is about 12800km in diameter and that makes a nice 13000km radius atmosphere-included. Just for simplicity, let’s define Pi to be 5.

Volume of the athmosphere:
4/3 * 5 * (6500km)^3 - 4/3 * 5 * (6400km)^3 =~= 8E18 m^3
which just happens to be one metric fluid buttload

6400km long cylindrical hole of a certain diameter D would have a volume of
5*(r^2)*6400km = 32000km * r^2

For a diameter of 10km (you didn’t want an elevator shaft, did you?) we get
along the lines of 3E15 m^3 or 375 fluid microbuttloads.

So the increase in volume is negligible unless we’re making a really huge hole (extreme radius of 6400km is all athmosphere and no earth left). Decrease in average density then should also be negligible. The reason I dug out the mass of the athmosphere above is for a sanity check, and 3/8ths kg/m^3 seems like very reasonable average density for the athmosphere (sea level is about 1.2 kg/m^3).

Assuming a cooled dead earth (so let’s say our athmosphere is uniformly 300K) and a solid structure supporting the hole, the only thing creating this pressure is the weight of the gas. Density in the column will be a function of depth – however it depends both on gravity and mass of the air column above, and once you are digging a deep hole gravity will decrease, and mass of the air column increase. It’s going to be a nasty differential equation I’m afraid, unless I’m missing something.

Gravity inside Earth sphere can be approximated by taking a plane perpendicular to the radial line through your location, and pretending that you are dealing with two point masses m1 and m2 which are equal to the mass to either side of that plane. The location of the point masses would be the uniform center of mass of each chunk respectively. Err, this is actually just a guess, so somebody correct me.
Theoretically one could also try to fudge the barometric formula
to replace g with g(z) but I am not solving that either.

Being at the point of no gravity doesn’t have any bearing on pressure. The air on the exact center wil reflect the applied pressure of the entire gavity pulled column. Think of it as a solid pillar if you must. The weight of all the air above the center is applied to that zero G point.

The nice thing about choosing the boundaries of a thought experiment is that the resulting answers can lead to greater understanding of effects in the real world. Care to describe the elements you ignored, or did you just multiply two numbers together randomly?

The problem here is that air is compressible, so, near the Earth’s surface, each slice of air actually would weigh more than the slice above it, because the air density gets larger the farther you go down.

If I assume the Earth is of uniform density , then gravitational acceleration decreases linearly toward the center of the Earth. If the Earth is 6360km radius and surface gravity is 9.81m/s[sup]2[/sup], you can figure out acceleration as a function of depth.

Now, the air pressure at the surface is 100 kPa, and density is 1.23 kg/m[sup]3[/sup]. I’m not a great mathemetician, but I can construct an approximate spreadsheet.

A mass of air, one meter by one meter by one kilometer high, at 1.23kg/m[sup]3[/sup], masses 1230kg. At 9.81m/s[sup]2[/sup], that would weigh 12000N, so the increase in pressure is 12kPa (or 12%). That’s not exact, because I’m not accounting for the continuous increase in pressure and density, but it shouldn’t be far off. And it doesn’t seem like an unreasonable number.

If I continue to calculate downwards, kilometer by kilometer (ignoring things like excess heat and so forth), then the pressure doubles to two atmospheres about six kilometers down.

Then it doubles again at twelve kilometers. And doubles again at eighteen.

You get the picture. At the center of the Earth (6360km down) I get a pressure of 2.2E+160 atmospheres. Yeah, 160 zeroes, assuming I did my math right. That clearly doesn’t pass the laugh test; I imagine the air will liquify long before (like a million billion trillion quadrillion quintillion googol times before) that. Does that help?

Meh. I alway thought air pressure (14.7 psi at sea level) had to do with the weight of the column of air being measured, and that weight is a function of gravity. Perhaps I err.

I think the laugh test is not that it would liquify, but that to get such densities you’d need all of the atmosphere to be at the lowest levels. Digging a big hole should not really suck all of the atmosphere into it, even if air is compressible.

Right, that is entirely correct. 1 atm pressure is created from the weight of the air column. Just because there is a hole underneath does not significantly reduce gravity at sea level over the hole. So the air pressure over the hole would still be 14.7 PSI. As you go down, it can only increase, because until you reach the center the only force the air is experiencing is downward pull of gravity towards the center. Yes, that pull slowly reaches zero as you reach the center, but that doesn’t make the air pressure there zero. That entire column of air is being pushed down by gravity into that center!

Imagine a blimp with a rope dangling below. The longer rope you hang below the blimp, the more the rope will weigh, the more force the blimp will experience. Even if the tip of the rope touches the ground so that the force of gravity is balanced by the normal force at the end of the rope, doesn’t mean that the rope just became weightless in respect to the blimp. However, if you put that tip on a scale it will barely weigh anything. I guess this is a bizarre analogy but you can assume that the only forces the air experiences is gravity pulling it down and the pressure preventing futher compression. That air over the hole that’s at 1 atmosphere can only have denser air below it or it would no longer be at 1 atmosphere. Just because gravity is not pulling the air at the center down doesn’t mean that it suddenly stops pulling the entire column down. At which point do you imagine that would occur?

Gravity at the center would be zero, but pressure would not because