Conditions at the Centre of the Earth

[John100]

If one could drill a hole to the centre of the earth, presumably the gravitational force on the body would decrease as one passed down this theoretical hole until one could float freely right at the centre where the gravitational force is equal and opposite in all directions with a net effect of zero. But what would be the air pressure there?

 

[DaveC426913]

If one could drill a hole to the centre of the earth, presumably the gravitational force on the body would decrease as one passed down this theoretical hole until one could float freely right at the centre where the gravitational force is equal and opposite in all directions with a net effect of zero. But what would be the air pressure there?

The pressure would be very, very high. While net gravitational force is indeed zero, there is still 3900 miles of rock pushing down from above, even if not all that rock is pushing down with the full weight it would have if it were near the surface.

To ultra-simplify, divide into large sections and use fancifully round numbers.
The top 1300 miles of rock pushes down with a trillion tonnes of weight.
The middle 1300 miles of rock pushes down with a half trillion tonnes of weight.
The bottom 1300 miles of rock pushes down with an eighth trillion tonnes of weight.
So, the entire column pushes down with - not 3 trillion tonnes, but a mere 1 5/8 trillion tonnes.

 

[Oldfart]

He's asking about air pressure in the hole, not rock pressure,

OF

 

[Dale]

He's asking about air pressure in the hole, not rock pressure

I guess then that depends on the construction of the walls. If you are assuming walls of unobtanium that can handle any pressure then you would do something like Dave's calculation but with compressible air. Otherwise, for realistic materials Dave's rough estimate seems reasonable.

 

[John100]

Yes, it is really about the air pressure at the centre. Does it go to zero along with the net gravitational force?

 

[Dale]

No, most definitely not.

 

[DaveC426913]

OK, let's assume magically strong walls. Now we simply have a hole that's 3800 miles deep.

The air pressue at the bottom will be equal to the weight of the entire 3800 mile tall column of air above it.

Again, let's divide it into large chunks.
We know that the 100 mile tall column of air from space down to Earth's surface weighs 14.7lbs for each square inch, right?
To make it simple let's say your tunnle is only one square inch in area. That means the weight at Earth's surface is 14.7 pounds, pressing down on anything in the tunnel.

Now the next 100 miles, from 0 down to -100: that column of air weighs juuuuuust slightly less than 14.7lbs. Let's call it 14.6. So, at 100 miles depth, you have a total weight of 29.3lbs (about 1.95 atmospheres) pushing down on you.

Now the next 100 miles, from -100 down to -200; that column of air might weigh 14.5lbs. That's a total weight of 43.8lbs. or about 2.9atm (I am totally faking these numbers; they are merely for illustration purposes).

How much do you think 36 more columns of air are going to weigh, even if each is slightly less heavy than the last?

Well, it's not going to come in at 38atm - the sum of 38 identical 100mile-tall columns of air, since they're dropping off to zero. I take a wild stab and say it's going to come in at half way - 19atm, or 280lbs. per sq. inch.

 

[Lsos]

He's asking about air pressure in the hole, not rock pressure,

OF

Rocks air water feathers, Dave's analysis applies all the same. Just replace the word "rocks" with whatever you want.

 

[DaveC426913]

Rocks air water feathers, Dave's analysis applies all the same. Just replace the word "rocks" with whatever you want.

No, he's making a valid point. Gases are highly compressible; solids can, in principle, hold up to pressure. It is (again, in principle) possible to construct a chamber at the centre of the Earth that, withstanding millions of tonnes of pressure outside, can maintain an internal air pressure of a mere 1 atm. (True, in practice, the rock would compress the air bubble until its pressure was the same as the rock around it, but...) the distinction is important to the OP. The air pressure he wants to know about is due to the weight of the air above it, not due to the weight of the Earth above it.

 

[Lsos]

No, he's making a valid point. Gases are highly compressible; solids can, in principle, hold up to pressure. It is (again, in principle) possible to construct a chamber at the centre of the Earth that, withstanding millions of tonnes of pressure outside, can maintain an internal air pressure of a mere 1 atm. (True, in practice, the rock would compress the air bubble until its pressure was the same as the rock around it, but...) the distinction is important to the OP. The air pressure he wants to know about is due to the weight of the air above it, not due to the weight of the Earth above it.

True, but OP did specify a hole that was drilled from the surface to the center, and not a bathysphere shielded from outside pressure.

It just seemed to me that the only part of your response that was actually considered was the word "rock"...and the rest was quickly dismissed and/ or not understood simply because of that one word. I believe if your response was actually understood, the OPs question would be clearly answered.

 

[willem2]

No, he's making a valid point. Gases are highly compressible; solids can, in principle, hold up to pressure. It is (again, in principle) possible to construct a chamber at the centre of the Earth that, withstanding millions of tonnes of pressure outside, can maintain an internal air pressure of a mere 1 atm. (True, in practice, the rock would compress the air bubble until its pressure was the same as the rock around it, but...) the distinction is important to the OP. The air pressure he wants to know about is due to the weight of the air above it, not due to the weight of the Earth above it.

You're forgetting the fact that the air in the hole would become more dense. If air was an ideal gas, the air pressure would continue to go up exponentially underground with
about 10% per kilometer.
At some point the air will become much less compressible, but you get a density with the
same order of magnitude as the rocks.

 

[DaveC426913]

You're forgetting the fact that the air in the hole would become more dense.

?

No I'm not. Read post 7, where I walk through the whole process of showing the effects of denser air.

At some point the air will become much less compressible, but you get a density with the same order of magnitude as the rocks.

No you do not. We have established that the rock walls are rigid. The only thing making the air more dense is more air on top of it. The question simply boils down to: how much does a column* of air one inch by one inch by 3900 miles weigh?

*and note that it is a column, not merely a volume

 

[DaveC426913]

True, but OP did specify a hole that was drilled from the surface to the center, and not a bathysphere shielded from outside pressure.

It's an area of easy assumptions. It was important to clarify that - in normal circumstances - the rock would deform, and apply pressure to the hole cum air bubble, compressing it until it reached ambient pressure. We must explicitly declare that the rock walls are magically rigid in order for the OP's experiment to proceed.

 

[willem2]

?

No I'm not. Read post 7, where I walk through the whole process of showing the effects of denser air.

No, your colums of air have the same mass per meter. If the air pressure goes up, the density goes up as well and you need to take the extra mass into account.

Another problem, is that because the atmosphere is ~100 miles high, you say that a 100 mile column of air will produce 1 atmosphere of pressure, but you need only ~10 km of air at surface density to produe 1 atmosphere of pressure.

 

[DaveC426913]

No, your colums of air have the same mass per meter. If the air pressure goes up, the density goes up as well and you need to take the extra mass into account.

Yes. I did not add in that level of detail to my numbers. I did not actually specify what the mass of each column of air was. My numbers are a simplification of the problem.

You're right. That should not be left implicit as I did.

Another problem, is that because the atmosphere is ~100 miles high, you say that a 100 mile column of air will produce 1 atmosphere of pressure, but you need only ~10 km of air at surface density to produe 1 atmosphere of pressure.

My statement is right. As I pointed out in the footnote of post 12, a column of air is distinct from a volume of air. A column of air 100 miles high will have the same mass as a volume of air that is much smaller.

But you are again correct. My statement is not much use - the density or mass of column of air from 100 to 0 does not tell us a lot about the mass or density of the columns of air below it.

 

[DrZoidberg]

Theoretically the pressure at the center would be astronomically high.
The deeper you go the higher the pressure gets. Higher pressure means higher air density. So a certain volume of air will be much heavier further down. That leads to an exponential growth in pressure. Theoretically the pressure at the core would be about 1.26 * 10^161 times atmospheric pressure if air was an ideal gas. But since air will liquify at much smaller pressures the actual pressure at the core would be much smaller then that.
I got that number by letting my computer run a simple algortihm. Starting at the surface I take the surface air pressure, add to that the mass of the top most cubic meter of air multiplied by the gravitational constant at the surface to get the pressure at a depth of 1 meter. Then calculate the mass of the second cubic meter of air using the new pressure, calculate the new gravitational constant at that depth and multiply that to get the force excerted by that second cubic meter and add it to the pressure. And so on until I reach the center.

 

[Dale]

I agree with DrZoidberg. The change in pressure for a differential element of air is given by:
$$dP=-\rho(h) \, g(h) \, dh$$

Using the ideal gas law for the density we get
$$\rho(h) = \frac{m P(h)}{R T}$$

and assuming uniform density for Earth we get
$$g(h)=g\frac{h}{r_e}$$

Solving the differential equation and substituting in the values for the Earth and for air at 20º C and using the boundary condition that the pressure at the surface of the Earth is 1 atmosphere I get
P(0) = 2.4 10^161 atm

Of course, Earth isn't a uniform density, and the behavior of air would be far from ideal, and the temperature would probably not be constant either, but in any case it would be a lot of pressure.

 

[Grep]

Of course, Earth isn't a uniform density, and the behavior of air would be far from ideal, and the temperature would probably not be constant either, but in any case it would be a lot of pressure.

Considering what we're talking about, and the fact that the center is pressurized magma that would well up as soon as you drill to it, creating a new and impressive volcano, I don't think we need to quibble on that. It's one of those "spherical cow of uniform density" kind of questions to start with, really.

 

[Dale]

Yeah, good point. As soon as you are considering using a shaft wall of unobtainum you may as well use a uniform Earth and a perfect ideal gas too.

 

[Subductionzon]

The pressure is even worse when you deal with a realistic Earth rather than a uniform Earth. The force due to gravity actually goes up as you go down, until you hit the outer core mantle boundary. At that point the acceleration due to gravity is roughly 10.8 m/sec^2. Then it will drop off rapidly as you approach the center of the Earth.

 

[DaveC426913]

Solving the differential equation and substituting in the values for the Earth and for air at 20º C and using the boundary condition that the pressure at the surface of the Earth is 1 atmosphere I get
P(0) = 2.4 10^161 atm

Did you do a sanity check on that number?

The pressure of the rock is only on the order of 10^6atm. Your number is 151 orders of magnitude larger.

 

[Dale]

Did you do a sanity check on that number?

Of course not. That is why I carefully specified all of the bad assumptions. I know the number is insane.

The pressure of the rock is only on the order of 10^6atm. Your number is 151 orders of magnitude larger.

Yes, the big problem is the assumption of ideal gas behavior. It implies that the gas is infinitely compressible, whereas the rock was assumed incompressible. Unfortunately, I don't know how to model a real fluid at those pressures.

 

[DaveC426913]

Of course not. That is why I carefully specified all of the bad assumptions. I know the number is insane.

I'm sayin' that's not a bad assumption - I'm sayin' that's an error in your calcs. I'm sayin' a decimal point or three has gone splowee.

 

[Dale]

No, the calculations are good. It is just the assumptions that are wrong. Zoidberg and I wouldn't have gotten so close with bad calculations.

 

[D H]

I get P(0) = 2.4 10^161 atm

Of course, Earth isn't a uniform density, and the behavior of air would be far from ideal, and the temperature would probably not be constant either, but in any case it would be a lot of pressure.

You might want to double-check your assumptions.

A better assumption is that the magical unobtanium walls that keep the walls of the tunnel from closing in on itself completely isolate the tunnel from the Earth: No energy transfer whatsoever. That suggests an adiabatic atmosphere assumption. For an ideal gas, this means

$$\frac{T}{T_0} = \left(\frac{p}{p_0}\right)^{(\gamma-1)/\gamma}$$

Where γ is the heat capacity ratio c_p/c_v, which for dry air is about 1.4. T₀ and p₀ denote the temperature and pressure at some known point such as the top of the tunnel. Combining the above with the ideal gas law $p=\rho RT/\mu$ (here μ is the molar mass, 28.964 grams/mole for dry air) and the hydrostatic equilibrium equation $dp/dr=-\rho\,g(r)$ yields

$$\frac{d(p/p_0)}{dr} &= -\,\frac{\mu g(r)}{RT_0}\left(\frac{p}{p_0}\right)^{1/\gamma}$$

Integration yields

$$\left(\frac{p}{p_0}\right)^{(\gamma-1)/\gamma} = -\,\frac{\gamma-1}{\gamma}\frac{\mu}{RT_0}\int g(r)dr + C$$

Assuming a uniform density (a bad assumption) leads to

$$\left(\frac{p}{p_0}\right)^{(\gamma-1)/\gamma} = 1+\frac{\gamma-1}{\gamma}\frac{\mu r_e g_0}{2RT_0} \left(1-\left(\frac{r}{r_e}\right)^2\right)$$

At the center of the Earth this means a pressure of about 13.5 million atmospheres.

As Subductionzon noted, the assumption of a uniform density is not a very good one. A better assumption is that gravitational acceleration remains constant down to r/r_e=1/2 and then falls linearly to zero. This two-layer approach means that that 13.5 million atmosphere pressure will be achieved halfway down. The pressure at the center of the Earth increases to 55 million atmospheres with this more realistic assumption.

So, still ridiculously high, but not quite as high as 10¹⁶¹ atmospheres.

 

[DaveC426913]

No, the calculations are good. It is just the assumptions that are wrong. Zoidberg and I wouldn't have gotten so close with bad calculations.

So you're stickin' with the idea that a volume of gas the size of Earth would have a core pressure unimaginably higher in magnitude than, not just Jupiter, but the Sun, and even a neutron star - the densest physical object in the universe.

OK, this is not an error in calculations; it is blatant misuse of a formula. It's a matter of punching numbers and not thinking.

You young whippersnappers and your calculators...

 

[Dale]

it is blatant misuse of a formula

Didn't I say that twice already.

 

[D H]

Yes, the big problem is the assumption of ideal gas behavior. It implies that the gas is infinitely compressible, whereas the rock was assumed incompressible. Unfortunately, I don't know how to model a real fluid at those pressures.

The ideal gas assumption does lead to an overly large number: 13.5 million atmospheres assuming uniform density or 55 million atmospheres with a more realistic density model. That overestimate pales to the 10¹⁶¹ atmospheres that you obtained by assuming a constant temperature.

 

[Dale]

I thought about taking temperature into consideration, but I would have said that it was at the same temperature the magma at each height. I guess that you could have drilled in vacuum and then let the air rush in for an adiabatic compression as you describe, but I was thinking of an equilibrium condition. I used a constant temperature simply because I didn't have an easy temperature profile for magma.

 

[D H]

Why magma? You have some magical substance that keeps the tunnels walls from closing in on themselves. Certainly it would make a lot more sense to extend this magic to all forms of energy than to assume that this magical substance exchanges thermal energy but no other energy.

 

[Dale]

Yes, that does make sense, but it isn't what I was thinking. Since it is entirely a work of fiction I think either is fine, but obviously the higher the temperature the lower the density and the more reasonable the numbers.

 

[John100]

Guys, thank you all for so much input and erudition. Frankly, I thought there would be a simple and straight forward answer. It's good to know that scientific debate is alive and well. I find it interesting that things, including air, are all trying to get to the centre of the Earth, but when they get there, if they could, through our theoretical tunnel, they would find that the force wanting them to be there has vanished. But the air would still express a sort of memory of that force through its pressure.
Now, how about a tunnel to the centre of a Black Hole? Will there be differing answers for that, or is it actually a simpler problem?

 

[Dale]

Now, how about a tunnel to the centre of a Black Hole? Will there be differing answers for that, or is it actually a simpler problem?

Yes, the pressure is infinite and even unobtanium can't hold up. Of course, even this relies on a bad assumption, namely that quantum gravitational effects are negligible. We will know more about this when we have a working theory of quantum gravity.

 

[DaveC426913]

...things, including air, are all trying to get to the centre of the Earth, but when they get there, if they could, through our theoretical tunnel, they would find that the force wanting them to be there has vanished...

Let's clear this up. The force wanting them to be there is specifically the weight of everything above them pressing down on them. That is a very real force.

Think of it this way:

You are lying on the ground, with a 10 tonne rock on your chest. Almost. The 10 tonne rock is actually sitting on a block of wood, which is in turn, sitting on your chest. The wood only weighs 5 pounds.

The weight that is right on your chest (the wood) is contributing virtually nothing (a paltry 5 pounds) to the 20,005lb. force that is crushing the life out of you. The lightness of the wood does not eliminate the heaviness of what's on top of it.

Now back to the column of air. Just because the last few hundred miles of air weighs only 5 pounds, does not mean you are not feeling the 10 tonne weight of the 3600 miles of air on top of it.

 

[Oldfart]

Let's clear this up. The force wanting them to be there is specifically the weight of everything above them pressing down on them. That is a very real force.

That seems rather misleading...

Fish swimming near the bottom of the deepest part of the ocean are not pressed down against the bottom, and a small bird would meet no resistance from the weight above it as it flew upward from the center of the earth.

OF