# How would atmospheric pressure change in a giant hole through the earth?

A giant hole exists in the earth surrounded by a super material that can withstand the extreme temperatures and pressures from the crust to crust through the center core of the earth. It is perfectly straight. The material that keeps the hole open maintains an even cave temperature throughout.

How would atmospheric pressure change from the surface to the core? If the hole joined the North and South pole, wouldn't that be different than if it was opposite ends of the Equator, given centripetal forces of the earths rotation?

First Edit:

I'm not holding answers to the hard-science tag, but I would appreciate seeing the math or explanations.

This question is only about atmospheric pressure. We can imagine this hole is perfectly maintained (no rain gets in) and keeps temperature steady (unless having temperature that matches the earths layers makes the question easier to answer).

Second Edit:

My interest in this question is to imagine a person traversing such a hole. When would it be lethal? Looks like it would be around 3-4 bars or about 1000km down the hole? The best "real world" data we have on this is from the Galileo spacecraft's probe:

• What's the diameter of the hole? Because the volume relative to that of the atmosphere matters.
– user86462
Commented Nov 2, 2022 at 17:49
• @Ansible: The other objection someone will make is that the hole will fill up (most likely with water). I assume this is excluded?
– user86462
Commented Nov 2, 2022 at 18:21
• @JBH - sounds a bit too restrictive. While I'd prefer to see some attempt at a scientific explanations, I'll remove the tag. Commented Nov 2, 2022 at 18:27
• @SeanOConnor, that's fine, lets say it's covered up and on top of a pointed island. No water incoming at least from surface rain anyway. Commented Nov 2, 2022 at 18:28
• Answer requires calculus I don't remember -- because force of gravity drops off as more of the Earth's mass is above (vs. below) a given location -- to microgravity at the center. Still, even if you only drill a couple hundred kilometers (not enough for gravity falloff to matter) you'll get much more than 1 bar per 100 km, because the air you're adding is denser than what's above sea level by the same distance. Commented Nov 2, 2022 at 18:45

Equator to equator or pole to pole will be roughly the same as the centrifugal force difference is only about 0.3% that of gravity.

My calculus is not up to the maths, but it would be fairly easy to come up with an approximate number using a spreadsheet with a line for every 100km down to calculate g. I haven't done the maths but the pressure will be very high indeed.

https://www.researchgate.net/figure/The-temperature-pressure-phase-diagram-for-nitrogen_fig1_315888614

Assuming the air is just composed of nitrogen (probably a reasonable approximation) anyone falling into the hole would experience ever increasing pressure as they descended towards the centre with no liquid or solid surface being encountered, just a slow imperceptible transition to supercritical fluid at very great depth.

Any seriously large radius hole of this type could also swallow the entire atmosphere.

• The temperature of the inner core is thought to be 5,200℃. What do atmospheric gasses do when the temp gets that hot (and hotter...)? I could imagine a band of solidified atmospheric gases giving way to gas again and eventually plasma as the temp rises. Is there a chart showing exotic plasma states for high P/high T? However, +1 for your final comment, a quick calculation shows that radius is 407 meters.
– JBH
Commented Nov 2, 2022 at 21:24
• @JBH Yes I was aware of that, but the question is a little ambiguous about temperature. It says it is only interested in pressure so its not clear to what extent the surrounding core is insulated by the tube, I took it as completely insulated and assumed the energy of any infall would have been allowed to reach equilibrium or be ignored in some way. Too many uncertainties and variables. There are high pressure phase diagrams for nitrogen but only behind a pay wall as far as I can see link.springer.com/article/10.1007/s10765-018-2376-1 Commented Nov 2, 2022 at 22:17
• The critical point of ordinary diatomic nitrogen is at −147 °C and 3.4 MPa. There is no such thing as liquid nitrogen at a temperature above −147 °C, at any pressure; I have no idea where the 25 °C came from. Commented Nov 2, 2022 at 22:42
• @Slarty That's a good point. I jumped to the conclusion that the magic tube could hold its shape at all times, but otherwise allowed the planet to be the planet. If it is insulated to temperature, it's believable that it's insulated to pressure, too. In that circumstance, other than swallowing up a considerable fraction of the Earth's atmosphere, the only issue is the weight of the atmosphere along the gravity gradient.
– JBH
Commented Nov 3, 2022 at 0:02
• That last point is really interesting... There would be some feedback from the oceans of course as they would evaporate and refresh the sunken atmosphere. Commented Nov 3, 2022 at 1:55

This is a bit more nuanced than it seems. On the surface of the Earth, we approximate a linear relationship between height and pressure, but even that can be seen to be less than accurate even in the troposphere. (BTW, the width of the hole doesn't make an appreciable difference unless the hole holds enough air to significantly lower the extent of the atmosphere...the increase in weight is cancelled by the increase in area.)

The problem is that air is compressible. As a result, the higher the pressure becomes, the denser the air is...so the increment in pressure is proportional to pressure. As such, the relationship becomes more exponential. There's math to account for that, but instead of fleshing it out here, I'll just use this handy calculator.

The average radius of earth is (on average) 6378100 meters. Entering that as a negative altitude gives us about 228 billion atmospheres.

Now, this tool probably doesn't take into account the fact that the gravitational pull of the earth decreases linearly as we approach the center. This reduces both the direct and incremental contributions of the densest regions of air, reducing that number considerably. For instance, if we shave 10% off the (negative) altitude, it shows us less than 132 billion atmospheres, a reduction of about 43% based on the weight of the highly compressed air at the bottom. However, the pull of gravity on that highly compressed air is, on average, only 1/20 of that affecting the surface air. In turn, due to the assumption of constant gravity, the total amount of air in the tunnel is less, further reducing the resulting pressure.

If I get the chance to work out that math, I'll update this answer. Until then, I'll do some math handwavium and propose that the correct answer is on the order of magnitude of what the calculator offers in the area of halfway down the hole (-3189050m), where its overestimation of the gravitational effect seems like it should be about right for the rest of the correction. For this depth, the calculator offers 6,186,467,404.49 atm.

Of course, it will probably go solid well before that point...there may be the possibility that the amount of solid air packed into the hole could severely deplete the atmosphere.

# Some Approximations

I invoke the holy mathematical rite of symmetry to reduce this problem down to 1 dimension: depth.

As you may recall, the standard equation for pressure-at-depth in a fluid is $$P=\rho g h$$ Some people will look at this and say "hey, the pressure must be 0 at the core of the earth because the force of gravity equals 0 in the middle of a sphere." I had that ungodly reaction at first, but the light of reason penetrated my mind.

While it is true that gravity does equal 0 in the middle of a sphere (and likewise 0 inside a hollow shell, for you devotees of the hollow-earth), it is not the case that there is absolutely no pressure in the exact middle of each planet. This is an affront to common sense: what about the many atmospheres worth of air on either side of you, each side equally attracted by gravity to the other half?

If we consider only a hemisphere of the earth at once, we eliminate these infernal conundrums and maintain the applicability of the familiar equation.

# The "Back-of-Napkin" Solution

Let's substitute g for the portion of the universal gravitational law it represents, $$\frac{Gm}{r²}$$.

We can also substitute $$\rho$$ for mass and volume.

Using the above substitutions, we get... $$P= \frac{G m² h}{Vr²}$$

Which is actually supported by others who did more complicated mathematics! This gives us 347(10^9) Pa at the center of the earth for rocks. Figuring your gas would be easy- just insert the nominal density of your air. This introduces some complications, though...

# Some Homework

You can take the above equation and account for things, like the effect of heat increasing or decreasing air density, but that is really beyond what I can do for a internet post.

Does the atmosphere in the hole do something weird which invalidates the above equation? Entirely possible that it becomes a solid (or some sort of fancy "liquid") as you go down. Such gems of knowledge such as the density of nitrogen and other atmospheric components at extreme temperatures and pressures is sadly unavailable to me. Otherwise, one could figure out the pressure of each gas/liquid/solid as you go down, and use that to determine overall pressure with the approach I outline here.

• The fact that you got something similar as people in your link should ring some alarms, as they were calculating a pressure at the center of Earth, not a pressure of air in a hole going trough the center of Earth. Source of their pressure is solid (mostly metals), while source of your pressure is gas. And because of that the problem is way harder to calculate, as density of gas change drastically by pressure, unlike solids where is more or less constant. The proper solution to this problem would take a differential equation. Commented Nov 3, 2022 at 12:38
• @Negdo which is exactly why I gave a "back of napkin" solution. For those unused to that phrase, it indicates a somewhat reasonable first stab at a solution, not a full solution. Also, confirming my equation is similar to what others achieve (when using values for rock) means it's not entirely off base. Commented Nov 3, 2022 at 15:45
• Its not a back of napkin solution if you calculate the pressure of rocks instead of pressure of gas. You are off by orders of magnitude. In physics, that is not a reasonable first stab at a solution. Commented Nov 4, 2022 at 6:17
• @Negdo I do see how one can misinterpret that, so I clarified it. Commented Nov 4, 2022 at 16:15