P = Po + Density of the fluid (air) * g (gravity constant) * h (height of the fluid above)

A couple of problems -- the density of air would no longer be virtually constant for that column but we will ignore that for a second. (we can use 1.293 kg/ cubuc meter)

the local gravity constant would no lonegr be 9.8 throughout (but we will ignore to get a ballpark)

the radius of the Earth is about 6.33 *10^6 meters

Po = 101.3 kPA or 1.013 *10^5 PAscals

So ignoring the calculus needed to adjust for the constants no longer being constant

P = 1.013E5 + 1.293*9.8*6.33E6 = 8.03 * 10^7 Pascals which is about 792 atmospheres of pressure

as the air spills into the tube it gets compressed. so there would be more moles of air molecules in the tube than at standard pressure which implies more mass and stronger pressure.

So that would add another complication.

when i looked online i got all types of answers.

would the air remain a gas at that pressure?

I walked into the Physics department, and a bunch of professors were having a heated debate. I asked one who was more on the sidelines what was going on. They were arguing whether a point in the center of the bottom of an opened but full bottle of Coca-Cola was under the same amount of pressure as one more to the side, or would the curvature of the bottle affect it.

I think any high school student who had studied physics before around 1960 would have the answer readily, but these professors probably never studied fluid mechanics.

I read that the study of fluid mechanics in the US physics education program died by accident. In the early '60s Halliday and Resnik's text became a standard at the university level, and when they chose content they basically skipped fluids as, at the time, physics-ready college students had by and large received a good understanding at the high school level. Then the folks who write standards for high school physics looked and saw that the colleges didn't include the topic in their textbooks and dropped it from the high school curriculum. And so with the exception of adding some elementary fluids for pre-med oriented physics classes, instruction in fluid mechanics at the undergrad level was largely left to the engineering curriculum.

would be depth in feet/2.309. The same principal would hold for air.

Since air pressure is caused by gravity, and at the center of the earth, there is no gravity (that I know of) since you are surrounded in all directions by earth, wouldn’t it be approximately zero?

Who will check?

You're right that the local gravity would be negligible at Earth's center, but the pressure comes from the collective weight of the air above it. I discuss this in more detail in my long post.

Air pressure is caused by gravity, but is not proportional to the local gravity.

would be depth in feet/2.309. The same principal would hold for air. But then you have the old gravity issue mucking up the works. It seems to me, that there would be a cancellation of gravitational effects at the precise center of the mass as the pull would be equal all around resulting in weigtlessness of the colum of air or water, therefore zero PSI.

But, hey, I am a plumber, not a scientist.

I remember posing a similar question to a teacher when I was in maybe 6th grade. The question was, "If you drilled a hole completely through the earth and dropped a ball down the hole what would happen. My teacher sternly told me to keep my mind on my classwork. Maybe that is why I became a plumber.

it would accelerate until it hit the center of the earth and then start to slow on as it passed the center until its velocity would be zero at teh othe rside. then it would fall back down and repeat. (ignoring friction and any coriolis force)

or a near stop, at center of the mass. If no other forces were in play, I suspect that it might osolate forever. My point was that my teachers didn't want to be asked questions, especially theoritical questions, that did not have the answers in the text book. I was under the impression that asking a hard question was a challenge to their authority. (which is probalbly why I asked it)

a liquid for the most part isnt (except under extrme pressure)

that compressed air would be liquid and would move into solution in the mantle.

At the core, it would be a solid.

It would be compressed and very, very hot.

Check out the Kola deep borehole. They got a quarter of the way through the crust before pressure and heat kept softening the drill head.

it would be hot thats a given. but would the pressure be able to overcome th eoutward pressure of the heat to form a liquid or solid.

ive gotten several different answers from people that have had a lot more science than me

was including temperature factors, so I assume room temp.

i know pounds are weight instead of mass but the same calculations can be made using pascals which is kg/ms^2 and kg is a unit of mass

is the total weight of the water column on one squrare inch of space. That is measuring weight. A column of water 2.309 feet tall, will produce a (weight) pressure of 1 PSI.

... since it's the force due to gravity, which would be 0 at the center of the Earth. The weight of anything falling into the hole would decrease as it fell.

Pounds is a unit of measure for weight, pounds per square inch is a unit of measure for pressure.

It's just like miles vs miles per hour. The former is a unit for distance, the latter is a unit of measure for speed.

On edit:

Using your example, increase the cross sectional area of the column to 10 sq. in.. The weight (pounds)will increase by a factor of 10. The pressure (psi) will remain the same.

The weight of all the air above is what causes the pressure to increase going down the hole, while the inertial mass doesn't change. Yes, PSI is a measure of pressure, but in this case it's also a direct measure of the weight of the air in a square-inch column, because gravity is the only force causing pressure.

I think that matters because the rate at which the pressure would increase is not constant: it would decrease going down the hole. The question at hand is whether it would reach a high enough pressure to liquify air, and if so, would it also reach a level to force a solid. I don't know the answer, but any analysis needs to consider this effect when estimating the pressure.

I don't want to calculate and answer right now but I think the answer depends crucially on assumptions.

One interesting one is the thermal properties of the "unobtanium" - can we assume the temperature of the gas matches the temperature of Earth's interior at any given depth? Or do we assume it is also a perfect insulator, so that the properties of the air in the tube are driven by the atmosphere alone?

I think it's safe to ignore the rotation of the Earth.

JT45242's calculation is a useful starting point. One issue they acknowledged is that "g" diminishes as you go deeper into the Earth. Exactly how it depends on depth depends, in turn, on how you model the interior of the planet. If you treat Earth as having a uniform mass distribution then local "g" is going to be proportional to the distance from the center of Earth, until you hit the surface. But it's understood that the density decreases with the distance from the center, which complicates the calculation of local "g."

Poking around a bit online I found some lecture notes including a calculation of "g" vs. distance from the center of Earth. (The vertical axis is labeled in meters but is obviously supposed to be km.) "g" actually is fairly constant from the surface to around 4000 km, rises to around 10.7 m/s^2 at 3500 km from the center, and smoothly goes toward zero from there (not perfectly linear, but close enough). A serviceable model would have g constant at about 10 m/s^2 to a radius of about 3200 km, then drop linearly to zero as you go to zero radius. So the formula in JT45242's post is pretty accurate until you get to 3200 km radius - assuming a constant air density.

But... that's not a great assumption, either. Air density is not constant in the thin shell of air above our heads, so it wouldn't be constant to a great depth, either. Other folks have commented on air being compressible, and the high pressures JT45242 calculated would certainly suggest that you can't take properties such as density to be constant.

As an aside, some folks note that "g"=0 at the center and suggest that implies a low pressure. But that's not how pressure works. Consider a sliver of air in that 1 square inch channel maybe an inch from the center of Earth. To be in mechanical equilibrium, that little slab experiences a (tiny) downward force of its own weight plus the pressure exerted by all the air above it, which needs to be balanced by the pressure beneath it. What is true is that when you get near the center, the additional contribution to pressure from the weight of a given "slab" of air gets smaller. But there is still the pressure from all the air above that needs to be considered.

When I'm less tired I'll dust off my thermal physics texts and see what I can figure out. It seems likely that you'd wind up leaving the gas phase (model the air as all nitrogen) at some depth. But I don't think you get a solid. If we take JT45242's answer for pressure you get something on the order of 80 MPa, and the phase diagrams I see for hydrogen require a pressure of at least 1 GPa to get solid nitrogen for any plausible temperature for this problem. Though if you wind up having a phase change, that will dramatically change the density... which in turn might really affect the pressure.

I think this problem is more interesting for the discussion of assumptions than for the answer!

For all atmospheric games (excepting trace anthropogenic gases) "room temperature" (never mind core temperature) is above the critical temperature. Thus a liquid phase cannot exist although the system can be supercritical above the critical pressure. The nonzero pressure is as noted in the post to which I'm responding, but stated differently, is a function of asymmetry. Even though the gravitational acceleration at the center point appears to approach zero, the gas is compressed by the apparent "weight" of the gas above it.

The Earth, I believe, is not massive enough to create sufficient pressure for solidifying common atmospheric games but if I correctly recall the large gas giants are, and can in fact induce multiple solid phases, for instance, metallic hydrogen.

I believe some of these pressures have been experimentally observed using devices like diamond anvils. I seem to recall this was a topic discussed by Raold Hoffman in a lecture I attended many years ago on the topic of extreme pressures.

One may intuitively gain insight to this question of whether the pressure is zero at the center of a gas object by asking if Jupiter is hollow.

... as they condensed and froze on their way down the hole.

The mouth of your hole would become an unpredictable geyser, hellish supersonic booming, for a long time. And should your unobtanium tube fail, things might get even more interesting. Maybe make yourself a volcano!

Of, you could take this as a plain old physics problem. Imagine a great earth-sized ball of gas floating in space. It soon dissipates.