Pressure at the center of the Earth?

[ninjayeti]

My dad threw this one at me a while ago, and I haven't quite been able to wrap my head around the calculus needed to figure it out.

Say you have a small hollow sphere at the gravitational center of the earth. Suppose then that tunnels are bored from the surface of the Earth to the sphere. The tunnels and the sphere fill with air from the atmosphere. What would the measured air pressure be inside of the sphere?
(I think for the sake of simplicity, we'll ignore temperature -- assume 25C)



My thoughts:
There are current approximations for pressures at the center of the earth -- somewhere around the 360 GPa range. Since air 'weighs' less than molten iron, the answer would have to be less than this.
Air pressure is defined by the force of the 'column of air' pushing down on a particular location. At the center of the earth, you would certainly have a substantially-sized column pushing down, but you have to take into account the fact that the acceleration due to gravity and the density of the air in the column both change as you go down. Now, in the center of the sphere itself, there is no acceleration due to gravity, but that certainly doesn't mean there is no air pressure.

I took one fluids class years ago, but I just can't relate everything to make an answer come out. Thoughts?

 

[imported_Hamzter]

Interesting thought, I suspect the pressure would rise dramatically as you drop, there'll be a point where the air could be approximated to be incompressible (if we considered it compressible at 360GPa then its density would be in the region of 4 million kg/m3 according to ideal gas laws!!). It may well be a liquid from the pressure at the centre, I can't find much on the density of air at high pressures though, anyone know about what density the air would be? I suspect your pressure may be in the same range as with rock, more than a couple of 100km down and you're probably approaching the density of rock, I'd guess that the actually answer is no more than an order of magnitude off. So that means no trips through the centre of the earth unless you want to come out as tomato juice at the other end

I think even if you could do the calculus (which I also had a quick stab at but trying to calculate gravity at depth made my head hurt!), I think it'd be useless because of the compressibility!

 

[Cogman]

What I think is interesting, is if the sphere in the center of the earth is perfect and the tunnels insignificant, the gravity in the center of the earth would be zero (physics teacher pointed this out to us students). I would assume that the air pressure in the center of the earth would be fairly low.

 

[Ruptga]

Yes, gravity at any point inside a hollow sphere, regardless of the thickness of its walls, is zero. But, there's lots of air outside this sphere in the tunnels, and gravity is still pulling it down, so there would still be air pressure, there just wouldn't be gravity inside the sphere.

I have *no* idea how to do the calculations, but I'd be surprised if air wouldn't be (at least) liquid at whatever pressure it turns out to be.

edit: At 96GPa Oxygen becomes metallic. But there's no guarantee that the air pressure would get this high without doing the math, and like I said I don't know how to do that.

http://en.wikipedia.org/wiki/Tetraoxygen

 

[RichPLS]

I would suspect that the air pressure would be at least double one atmosphere, since one atmosphere would be on each hole, meeting at the center, seems double pressure at that point, at least the point at which pressure is greatest, beit liquid or what not.

 

[highwire]

The shell theory makes it simple to deduce that gravity will decrease linearly to zero at the center of the earth, assuming the the earth is of uniform density. (r^3/r^2)

Air density and pressure will increase proportionally, given that the temp is to remain constant at 25°C.

As the density increases with depth, the pressure and density will increase at a greater rate per each additional unit of depth.

Programming assignment:

Divide the air tunnel into 1Km increments from the surface to the center of the earth.

Air density doubles from 6 km altitude to the surface. Using this relation for each Km of air column depth, the air density and pressure will increase by a factor of [edit]1 +.122 times the local gravity fraction of surface gravity. Halfway down the pressure is increasing by 6.1% per Km. But the high density makes the absolute pressure increase much greater per Km than at the surface

Starting at the surface with a pressure of 1 bar, add the incremental pressure at each step until you reach the center of the earth ( 6370 km )

The winner will receive a free trip to the AT podium for a round of applause.

My initial estimate is a VERY large number. Someone mentioned a metallic phase. I think so.

 

[fire400]

if you have an ATI card, download the 9800pro demo from ati.com

"CAVES"

install and run it

"ooh... lala"

the smaller holes in the earth are unstable. blows up constantly and recreates itself almost infinetly.

 

[highwire]

Did a few lines of basic. Interesting.

The density of air at a mere 60 Km would be greater then 1.2g/cc, denser than water!! pressure is~14,700psi ~1000 bar or 100MPa. At these pressures and densities, air would be departing radically from gas law assumptions.
If gas laws held, at 90 Km air would be about twice as dense as uranium at a pressure of 32.7Kbar or 3.27Gpa.

90 Km down and 6280 Km to go!!

 

[silverpig]

Air pressure goes exponentially. To do a very quick rough calculation look up the air pressure at the top of everest and use that as a data point (8848 m, pressure). Then take STP at sea level (0, 101.325 kPa), and fit an exponential. Then solve for the point (-6.38 * 10^6, pressure?)

That probably won't be right because of the change in gravity, and because the air will probably liquefy/solidify by then.

 

[highwire]

Originally posted by: silverpig
Air pressure goes exponentially. To do a very quick rough calculation look up the air pressure at the top of everest and use that as a data point (8848 m, pressure). Then take STP at sea level (0, 101.325 kPa), and fit an exponential. Then solve for the point (-6.38 * 10^6, pressure?)

That probably won't be right because of the change in gravity, and because the air will probably liquefy/solidify by then.

Yes, just about what I did. Using the ~1/2 atmosphere 500 Mb chart average - 5800m, fudging it up to 6Km for adiabatic temp changes, I got a pressure ratio of 1.122 per Km. isothermal. Thats is, 2^(1/6)=1.122

I double checked, taking the incremental 1 meter pressure ratio at ISA, then raised it to the 1000th power for the 1Km ratio. I got a pressure ratio of 1.125 using air density =1221g/m^3. For this prob at 25°C though, 1184 g/m^3 is close to the best value to use. That gives a 1 Km pressure ratio of 1.12141. STP would be off by ~9% for this problem.

So, simplified by neglecting gravity and phase changes, the pressure at n Km depth expressed in atmospheres will be:
pressure = 1.12141^n

The simple exponential relation will only hold for the gas state and for the 1st few 100 Km's due to diminishing gravity. Also, you can throw the gas laws out the window at depths of more than 60 Km or so. ( ~1000bar ) at which point air will have the approximate density of water.

 

[beansbaxter]

Originally posted by: ninjayeti
My dad threw this one at me a while ago, and I haven't quite been able to wrap my head around the calculus needed to figure it out.

Say you have a small hollow sphere at the gravitational center of the earth. Suppose then that tunnels are bored from the surface of the Earth to the sphere. The tunnels and the sphere fill with air from the atmosphere. What would the measured air pressure be inside of the sphere?
(I think for the sake of simplicity, we'll ignore temperature -- assume 25C)

Please note, the radius of the Earth varies because it is NOT perfect circle. Therefore, the pressure will vary, as there will be a bigger radius at the equator.

Using the radius of the earth as 6,356 km the radius thus is 6.356 x 10^6m.

Pressure can be determined by the equation

P=F/A
Where P=Pressure F=Force and A=Area

But, F=ma
where m=mass and a=acceleration

We find pressure in square meters, thus the area is 1 square meter.
Now, we find the mass of the air in the 1 square meter colum (that extends to the ends of the atmosphere). This is where it gets complicated. To find the mass, we must calculate the molar mass. Air contains roughly 78% nitrogen, 21.12% oxygen, 0.93% argon, 0.04% carbon dioxide, in addition to about 3% water vapor.

So, 1 mole of air= .78 moles nitrogen+ .2112 moles oxygen etc.

Molar Mass air=(.78x14g/mol)+ (.2112x16g/mol)+(.093x 39.9g/mol)+(.04x12g/molx16g/mol x2)+(.03x1g/molx2x16g/mol)= 34.33g/mol


The radius of the atmosphere is ~120km

120km + 6,356km= 6, 478km or 6,478,000 meters.
If volume equals height x depth x width then

volume =6,478,000mx 1m x 1m= 6.478x10^6 cubic meters
So, we find the moles in that amount of space. However, I'm bored with this so I'm not doing it (but Ill tell you how to finish)

once you get the moles, you use the molecular weight to find the mass of the air. That would be number of moles times the molecular weight. lets call this xkg

Then, you use F=ma to get the force. F=(xkg)*(9.98m/s^2) this will be in Newtons. Lets call it yNewtons

With this you use P=F/A
P=yNewtons/1 square meter. this will give you an answer in pascals. divide by 1000 to get kilopascals. if you want atmospheres, 1 atm=~101kPa (im to lazy to look up the exact value).


Keep in mind, to get the mass of that colum of air, the higher up you go 1 mole of air will occupy more space. So, you cannont use "the idea gas has a volume of 1 mole per 22.1L, because this is at 1atm and obviously that changes.

 

[Pulsar]

You cannot use the acceleration of gravity as 9.8m/s^2.

You will have to integrate over a gravity function. You also cannot use the newtonian definition of gravitational attraction because it is only valid above the earth's surface. As you move inward, you now have mass OUTSIDE you, which also has a gravitational effect.

Google it. It gets incredibly complicated.

 

[highwire]

Originally posted by: Pulsar
You cannot use the acceleration of gravity as 9.8m/s^2.

You will have to integrate over a gravity function. You also cannot use the newtonian definition of gravitational attraction because it is only valid above the earth's surface. As you move inward, you now have mass OUTSIDE you, which also has a gravitational effect.

Google it. It gets incredibly complicated.

Thanks to Newton and his shell theory, it NOT "incredibly complicated". Gravity will vary LINEARLY within a sphere of uniform density from the 1G surface to the 0G center.

The pressure at n Km depth expressed in atmospheres will be:
pressure = 1.12141^n

Reduce the exponent's base accordingly as you proceed incrementally downward to factor in gravity. Simple.

By the way, if one insists on maintaining that air will remain a gas to the center the the earth, you will calculate densities that will induce a black hole about a third the way down. (1)

So, moles, schmoles. Beans, your natural talent does not appear to be engineering. With pages flying, your attempt stalled out half way though a physics textbook going in the wrong direction.

(1)Re: Schwarzschild Radii
http://www.upscale.utoronto.ca...kHoles/BlackHoles.html
The earth compressed to a 1/3 inch sphere (about 1cc) is a density of 2.5X10^56 g/cc