Suppose I could dig a hole to the center of the earth. How would I calculate the air pressure at an arbitrary depth in the hole?

How to compare this to water pressure experienced by divers?

Suppose I could dig a hole to the center of the earth. How would I calculate the air pressure at an arbitrary depth in the hole?

How to compare this to water pressure experienced by divers?

Water pressure for divers is mostly at the surface of the Earth. To put it in perspective, the deepest part of the ocean is a bit over 6 1/2 miles, but the Earth’s diameter is close to 8,000 miles. Water pressure for divers is basically linear, increasing by one atmosphere for every 33 feet or so of depth.

A hole to the center of the Earth isn’t possible. Think of the Earth as a pan of hot liquid soup that has been sitting on the stove for so long that a crust of dried soup has formed across its surface. Sure, you can poke holes in the crusty bit, but once you get down to the liquid bit then it’s not going to work very well.

But for the “assume a spherical cow” type of physics question, if you could create a hole to the center of the Earth, calculating the air pressure actually gets to be a bit tricky. If you are actually at the center of the Earth, the mass of the Earth is pulling you in every direction, so there’s no felt gravity. The air pressure in your hole is still going to have all of that air from the surface pushing down on it, but it’s not a trivial thing to calculate.

And I don’t actually know off the top of my head how to calculate it, so I’ll stop by just saying it’s a much different set of conditions that your divers in water, once you get deep enough that a significant chunk of the Earth is above your head.

You can do it incrementally in a spreadsheet. Assume constant temperature of 68F. Top row is sea level: elevation = 0 feet, density = 0.0765 lb/ft^3, pressure = 14.7 psi.

Next row down: altitude = altitude at previous row -100 feet, pressure = pressure at previous row + (density at previous row * change in elevation), density = sea-level density * current pressure divided by sea level pressure.

Repeat those calculations until you reach the required depth. If you’re doing things right, then you’ll come up with a pressure at the bottom of the **Tautona Mine** (2.4 miles down) of about 22 psi (7.3 psi above sea-level atmospheric).

EDIT: as engineer_comp_geek points out, local gravity will decrease as you get farther below the surface of the earth. You would need to account for this in the calculation of the pressure increase due to each layer of air. By the time you get very close to the center of the earth, even though the air is very dense, the rate of pressure increase with depth will approach zero as you get closer and closer to the center of the earth.

The pressure will rise at a faster and faster rate as you go down. You’ll get some ridiculously high number at the center of the earth, I’m guessing millions or billions of psi. But I don’t have a grasp for what will happen in reality, as the behavior of oxygen and nitrogen will deviate from the ideal gas law eventually, and at some point O2 and N2 will actually solidify. For O2 at room temperature, that’s around 725,000 psi:

For reference, ocean divers experience a pressure increase of 1 atmosphere per 33 feet of depth. The recreational dive limit is about 130 feet. The deepest dive that didn’t use a hard-shelled suit was 1752 feet:

That’s about 780 psi above atmospheric.

It seems to me that this is the answer here if you are actually located at the center of the “spherical cow” earth. There would be zero air pressure. The gravity of the mass of the planet is pulling on the air in all directions outward. Maybe if you made a spherical chamber and calculated the air pressure at the wall(you would have something roughly like 1/2 the planet behind the wall)

No, there will be plenty of air pressure. You’ll be sitting at the bottom of a 4000+ mile-tall column of air, and the part above sea level, by itself, will be providing a pressure of 14.7 psi. And the part below sea level will also provide its own pressure; it’s just that you can’t assume constant gravity all the way down to the center of the earth.

If I have some time this weekend, I’ll lay it out in a spreadsheet and see what we get.

I have not done any math yet but 14.7 is the pressure at sea level due to gravity pulling the air down. As you reach the center of the hole, gravity is no long pulling air into the hole (more accurately is it pulling equally in all directions). There is a point somewhere between the surface and the center where there is a maximum air pressure and then it decreases as you continue down the hole.

Yes, thank you. Assume the earth is a solid sphere of uniform material that you can drill through without the material collapsing back into the hole. This is a question about air pressure, and not geology.

It is true that the air near the center of the earth is not being pulled down very much (or not at all, for the air right at the center). But it must bear the weight of all of the air above it. If the pressure at the earth’s surface is 14.7 psi, then the pressure at the center definitely will not be less than that. In fact, it will be quite a bit more, because of the weight of all the air far from the center of the earth pushing down on it.

TIL that subterranean gravity does not scale linearly with distance from earth’s center in the way that you would expect for a spherical planet of uniform composiition. Because the earth is actually composed of different materials at different depths, it turns out to be quite complicated:

Turns out gravity is approximately constant (~9.81 m/s^2) from the surface down to a distance of about 4500 km from the center, then actually *increases* to 10.7 m/s^2 at the core/mantle boundary (3470 km from center). Below that depth, it’s almost (but not quite) a straight line down to zero gravity at the center.

Another complication - water is essentially incompressible, so has the same density and weight at every depth (almost). So for divers, it;s a simple calculation - every 33 feet, add one more atmosphere (14.7psi).

Gases, OTOH, compress and so become denser with pressure. IIRC from pilot school the air pressure doubles every 18,000 feet, it’s more logarithmic. (I.e. air pressure at 18,000 is one-half atmosphere, (14.7)/2 psi; at 18,000 feet deep it would be (14.7)x2 psi)

So gas around a spherical chicken at the center of the earth will not be zero - it simply won’t be increasing as you get to the center. Gravity inside a solid sphere is the integral of the material around the pint being considered - so the radial vector component of every tiny piece of the sphere.

If let’s say you’re halfway between the surface and the center, gravity is the sum of the radial component of the attraction of each little particle - for the dome at or above the tangent of your position, those pieces are pulling you upward; below that, they are pulling you down. As you reach the center, there is an equal amount pulling you both ways. “up” and “down”.

Maybe a rough rule of thumb would be to consider the outer core and mantle to be rock, density approx 3, and the iron center core to be … wait for it… iron, with a density of 6. So combine 2 calculations, the one for the whole earth and and add an additional density-3 core sphere to the calculation.

Waddya know, some time opened up this afternoon. I already had a standard atmosphere spreadsheet that calculates pressure versus altitude above sea level, assuming a standard atmospheric temperature profile and calculating upward from sea level in a stepwise fashion. So I extended it downward to see what happens below ground. I incorporated the subterranean gravity model I linked to in my previous post, and assumed a constant temperature of 68F all the way down.

The results get weird very quickly. Just 37 miles below sea level, the pressure is already up to about 14,000 psi, and the density of the air is about the same as that of liquid oxygen. I’m guessing it’s a supercritical fluid at this point, and I’m not clear on what this means for compressibility. at some point it’s going to be difficult to achieve any increase in density, but where that is I don’t know. If we assume we can compress it enough to match the density of the earth’s mantle (specific gravity = 4.5), my spreadsheet says that happens at a depth of around 44 miles. If we assume a constant density from there to the center of the earth, I calculate a pressure of around 31.8M psi at the center. This is a bit less than **the estimated actual pressure of about 53M psi**, probably to do with the core being much more dense than the mantle (remember, I assumed mantle density for our air column at depths below 44 miles, i.e. more than 99% of the total column height).

The basic equation to do the calcs is simple :

dP = rho. g . dz (where dP is change in pressure as you go down a mine from earth’s surface to the center)

where

P= pressure at a given depth

z = depth from the surface of the earth

rho = density of air at a given depth (function of temperature assuming ideal gas)

g = acceleration due to gravity at a given depth (changes with depth and is 0 at the center of the earth)

You can use the relationship between g and z to integrate the above :

Density of air changes with temperature. You can assume that the earth is at the same temperature at all depths (bad assumption) and can replace

rho = PM/RT, where P = Pressure at depth z, M = molecular weight of Air, R = Universal Gas Constant and T is Air temperature

NM didn’t read the whole post.

This is not a good assumption. I am not a physicist, but you have to use the equations in this link “6.3 THE CENTRAL TEMPERATURE OF THE SUN”

Basically you need pressure and weight of the column of gas to match up.

Sorry, forgot to attach link and missed the edit window.

https://www.astronomy.ohio-state.edu/weinberg.21/Intro/lec6.html

Isn’t this essentially what happens on Jupiter, Saturn, etc. to some extent? Just a little cooler.

They show four steps:

(1) Divide the sun into an inner half and an outer half. We’ll demand that the pressure in the inner half support the weight of the outer half.

(2) Estimate the weight of the outer half from Newton’s law of gravity.

(3) The gas pressure in the inner half depends on the density of atoms and on the temperature. We estimate the density by dividing the number of atoms in the sun by the volume of the sun.

(4) Finally, we find the temperature that is needed to provide the necessary pressure.

In their step #3, they estimate the density based on how many atoms are available in the sun. That gives them their starting point. Our starting point is that earth’s sea-level atmospheric pressure is known to be 14.7 psi, and then we work our way down from there. We can assign whatever temperature we want to any section of the fluid column we choose. I picked 68F because I installed a cooling system in my borehole to maintain that temperature. Also because it makes the math easier. You certainly could use the expected temperature of the earth at any particular depth if you wanted to, but it’s not necessary.

Agree with that. You are right.

Well that just made my day in a good way. Like the spherical rigid cow.

But even if you assume a constant temperature, the ideal gas law is no longer valid at higher pressures, so the expressions break down.

One way I like to picture the “the pressure at the center can’t be zero” argument is thusly*:

Scenario 1: The kinetic energy situation. Take the usual idealized situation of a hole all the way thru the Earth from one side, thru the center and out the other. Imagine dropping two rocks into the holes from each end. They zip towards the center and go “Bang!” against each other at the center. Note if you drop only one rock and there’s no air pressure (idealized piled up on idealized) then the rock will zip thru the center up to the other whole, turn around and repeat.

Scenario 2: The potential energy situation. Imagine the rocks at the openings connected by a long, long rod (if you want to continue the idealization, let the rod have negligible mass and arbitrary stiffness). The two rocks are pushing against each other via the rod. A pressure gauge inserted into the rod at the center will register the two rocks’ masses pushing the rod in.

With air there’s just a zillion tiny rocks and instead of a rod they are pushing against each other.

* I apologize profusely for the “thusly”.

Air or no air, I think it is widely known that the pressure at the center is so great that the inner core is solid, despite the great heat.