Air pressure at Earth's center

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I wonder if anyone has any ideas as to how one might figure this out. If there were a tunnel from one side of the earth, through the center of the earth, and on out the other side, what would the air pressure be in the center? I have some ideas, but one catch is that the air pressure and air density would seem to be mutually interdependent..

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Martoonsky pretty much addressed it. The weight of the Air at the Lip of the hole is ~14 psi.  Even if there were no gravity all the way from the surface to the center, the Air filling that hole would

Somewhere near the centre of Jupiter (many factors might put it at other than the exact centre of mass) the pressure would hit a maximum. At that point the pressure gradient would be zero.

I admit this one does have me scratching my head. I wouldn't expect heat to change the pressure once temperature equilibrium is reached but I'm not sure of it. But if you have an imaginary borehole yo

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I think if the Earth were to stop spinning and had all of its mass evenly distributed in a sphere, it would be close to zero at the exact center.

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6 minutes ago, Endy0816 said:

I think if the Earth were to stop spinning and had all of its mass evenly distributed in a sphere, it would be close to zero at the exact center.

???

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59 minutes ago, Janus said:

???

Basing this on there being the same amount of mass on all sides. Not 100% sure here, however, so feel free to correct me if I'm wrong lol

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56 minutes ago, Endy0816 said:

I think if the Earth were to stop spinning and had all of its mass evenly distributed in a sphere, it would be close to zero at the exact center.

Thanks, but I'm pretty sure that's not correct. Air pressure is due to the weight of the air column above you. As you go down into the earth, the air pressure should rise above the atmospheric pressure at the surface. Among the factors involved, you would have to account for the acceleration due to gravity decreasing as you go down; also, assuming the walls of the tunnel allow heat to flow through, you'd have to allow for the temperature increasing as you go down.

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4 hours ago, Endy0816 said:

Basing this on there being the same amount of mass on all sides. Not 100% sure here, however, so feel free to correct me if I'm wrong lol

Martoonsky pretty much addressed it. The weight of the Air at the Lip of the hole is ~14 psi.  Even if there were no gravity all the way from the surface to the center, the Air filling that hole would have to be at a pressure of 14 psi just to support the weight of the air above the Hole.   But since air at every part of the whole except for exactly at the center is going to add its weight to the mix which has to be supported by the air underneath it the air pressure has to go up as you near the center.

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17 hours ago, Endy0816 said:

Basing this on there being the same amount of mass on all sides. Not 100% sure here, however, so feel free to correct me if I'm wrong lol

This would be why the air at the centre would be essentially weightless not contribute to the pressure, but as  per Janus and Martoonsky the pressure is still higher than at the surface.

The pressure gradient goes to zero at that point.

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29 minutes ago, J.C.MacSwell said:

The pressure gradient goes to zero at that point.

Only if there's no planet involved.

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I agree that pressure must go up as you go down towards the centre of the Earth. I also agree that temperature is important and must go up (however gradually) as you go down. Up to about 6000 Celsius at the centre.

I've made a quick check with the following model:

1) the amount of air within the shaft does not deplete the atmosphere significantly

2) Gravitational field is obtained inside the Earth by applying Gauss's theorem

3) Exponential atmosphere is assumed with T as a function of r (distance from centre of Earth) and V(r) the expression for the interior potential

$P\left(r\right)=P_{0}e^{-V\left(r\right)/k_{B}T\left(r\right)}$

The expression for the interior potential is,

$V_{\textrm{int}}\left(r\right)=G\frac{M_{\oplus}}{2R_{\oplus}}\left[\left(\frac{r}{R_{\oplus}}\right)^{2}-3\right]$

And the model doesn't seem to work. It gives a ridiculously high value for the pressure; much, much bigger than 3.8 trillion psi, which is the value for the centre of the Sun! 😲 To me, this means that at least one of hypotheses 1), 2) or 3) is plain wrong. The most suspect to me is 1). Maybe what would happen is that the whole atmosphere would be swallowed up by the shaft density deficit, and it would cram itself into it, thereby depleting the outer atmosphere. Assumption 1) is implicit in that P0 keeps its current value, even though the whole atmosphere would have been sucked into the shaft.

P0 is the 14 psi reference value that @Janus has mentioned.

A more realistic model should involve the distinction between crust and mantle (discontinuous layers). But I don't think that would make that much of a difference. Maybe I made a mistake or the simple substitution in exponential atmosphere relation with the temperature as a function of r simply does not hold.

On 7/23/2020 at 5:03 PM, Martoonsky said:

I have some ideas, but one catch is that the air pressure and air density would seem to be mutually interdependent..

Why do you see the density-pressure dependence as a problem, @Martoonsky? What particular model are you trying to apply?

Edited by joigus
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8 hours ago, dimreepr said:

Only if there's no planet involved.

Why would there be a gradient at the centre with a planet involved?

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7 hours ago, joigus said:

1) the amount of air within the shaft does not deplete the atmosphere significantly

I don't see this as much of a problem theoretically. You could make the shaft very narrow, or add air. If the atmosphere were depleted, it would reduce the value of the pressure at the center.

I haven't yet tried to do a detailed calculation but the above graphic provides the answer. P-sub-zero is the pressure at the center (at radius r = 0). R is the radius at the surface. Rho = rho(r) is the density and g = g(r) is the acceleration due to gravity.

I thought I would divide the integral into sections for the inner and outer cores, mantle and crust. I think I could come up with an average g for each section. I'm not sure what to do with rho yet since it varies with temperature and pressure.

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Yes, I think you're right that depletion is not that important. I was so stumped by the humongously big number that I didn't know what to think.

You could try to work it out with the barometric density formula:

Both expressions, for pressure and density, are basically the same,

$\rho\left(r\right)=\rho_{0}e^{-V\left(r\right)/k_{B}T\left(r\right)}$

But they are isothermal expressions. The leap of faith I've tried is assuming that you can substitute T by T(r). I'm not sure that would hold. You could always take a look at the wiki derivation and try to work it out for little slices of width dr and constant temperature for each slice. My sub_zero is value at the surface. Maybe that's the ticket.

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6 hours ago, Martoonsky said:

I don't see this as much of a problem theoretically. You could make the shaft very narrow, or add air. If the atmosphere were depleted, it would reduce the value of the pressure at the center.

I haven't yet tried to do a detailed calculation but the above graphic provides the answer. P-sub-zero is the pressure at the center (at radius r = 0). R is the radius at the surface. Rho = rho(r) is the density and g = g(r) is the acceleration due to gravity.

I thought I would divide the integral into sections for the inner and outer cores, mantle and crust. I think I could come up with an average g for each section. I'm not sure what to do with rho yet since it varies with temperature and pressure.

Perhaps you would like to consider the implications of this formula?

I agree that it is best to place the origin at the centre, as you have done but consdier the implications for your integral.

You have correctly identified that both density and the gravitational acceleration are functions of the radius.

At the origin, gR = 0.

Thus the integral must be zero at r = 0

At r = R, the integral must equal the surface pressure, by definition

P0  =  PR  + PR  = 2 PR

I think this difficulty arises because you have tried to include a constant of integration in a definite integral.

Edited by studiot
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9 hours ago, J.C.MacSwell said:

Why would there be a gradient at the centre with a planet involved?

There is on Jupiter.

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At Jupiter's core, you would feel as much as 650 million pounds of pressure pressing down on every square inch of your body. That would be like having approximately 160,000 cars stacked up in every direction all over your body!

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32 minutes ago, dimreepr said:

There is on Jupiter.

Somewhere near the centre of Jupiter (many factors might put it at other than the exact centre of mass) the pressure would hit a maximum. At that point the pressure gradient would be zero.

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6 minutes ago, J.C.MacSwell said:

At that point the pressure gradient would be zero.

Perhaps we're talking past each other, but how is that pertinent?

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21 minutes ago, J.C.MacSwell said:

Somewhere near the centre of Jupiter (many factors might put it at other than the exact centre of mass) the pressure would hit a maximum. At that point the pressure gradient would be zero.

I see no way in which this could be wrong! +1

Edit: Sorry; I can only see one: The system is not in equilibrium.

Edited by joigus
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4 minutes ago, dimreepr said:

Perhaps we're talking past each other, but how is that pertinent?

It was pertinent (in my mind at least) to Endy's post. He seemed to intuitively recognize that at the centre something with regard to pressure went to zero, and for the reasons he stated, which made that point unique. (Jupiter OTOH could have any number of points of zero gradient at local maximums or local minimums at any given time)

It wasn't the pressure itself that went to zero though. It was the gradient.

3 minutes ago, joigus said:

Edit: Sorry; I can only see one: The system is not in equilibrium.

It has to hit a maximum somewhere down there...

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6 minutes ago, J.C.MacSwell said:

It was pertinent (in my mind at least) to Endy's post. He seemed to intuitively recognize that at the centre something with regard to pressure went to zero, and for the reasons he stated, which made that point unique. (Jupiter OTOH could have any number of points of zero gradient at local maximums or local minimums at any given time)

It wasn't the pressure itself that went to zero though. It was the gradient.

It has to hit a maximum somewhere down there...

I agree. "Out of equilibrium" I didn't mean as a serious objection. It was meant for completeness.

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3 hours ago, studiot said:

P0  =  PR  + PR  = 2 PR

Sorry, but this is not correct. If rho and g were constants, then the integral would evaluate to P-sub-R, but they are not. Check out the graph below of two possible curves for rho x g vs r. They have the same values at r=0 and r=R, but obviously their integrals are not the same.

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24 minutes ago, Martoonsky said:

Sorry, but this is not correct. If rho and g were constants, then the integral would evaluate to P-sub-R, but they are not. Check out the graph below of two possible curves for rho x g vs r. They have the same values at r=0 and r=R, but obviously their integrals are not the same.

Are you seriously suggesting that the density (blue or red  ???) is zero at the centre?

I see that I have misinterpreted your graph.

Sorry.

So red and blue are different (proposed) values of the product of density and gravity (both of which are functions of r).

Fine so you still have both = 0 at the centre so my original statement hold good when substituted into your equation, but P0 now = PR

If that maths is incorrect, please provide the correct maths.

Edited by studiot
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4 minutes ago, studiot said:

Are you seriously suggesting that the density (blue or red  ???) is zero at the centre?

Red is V; blue is rho times g. g is zero at the centre.

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13 minutes ago, joigus said:

Red is V; blue is rho times g. g is zero at the centre.

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3 minutes ago, studiot said:

$V_{\textrm{int}}\left(r\right)=G\frac{M_{\oplus}}{2R_{\oplus}}\left[\left(\frac{r}{R_{\oplus}}\right)^{2}-3\right]$

$V_{\textrm{ext}}\left(r\right)=-G\frac{M_{\oplus}}{r}$

Edit: It's the line going down with distance linearly.

Edit 2: Sorry, that's g. The line going down linearly with distance is the gradient, that is, g.

Edited by joigus
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2 minutes ago, joigus said:

Vint(r)=GM2R[(rR)23]

Vext(r)=GMr

Edit: It's the line going down with distance linearly.

This is not what Martoonsky says.

36 minutes ago, Martoonsky said:

two possible curves for rho x g vs r.

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