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So if i dig a hole to the other side of the earth

  1. Apr 4, 2008 #1
    here's my question:

    A hole large enough for a human to jump into is dug from one side of the earth to the other side, the hollow tube it forms goes through the center of the Earth. Describe the air pressure distribution of the hollow tube, particularly around the center and halfway between the center and the surface.

    No geothermal activity at center
    No Rotation
    Mass is uniformly distributed
    The mass the hollow tube displaces is trivial

    I'm an electrical engineer, however i work in an office full of aerospace engineers. Anyways, i asked them this question and a debate erupted, i still don't know what the answer is.

    most popular answers in the office were:
    1) uniform pressure equal to surface pressure through the tube
    2) pressure increases as you get closer to the center, and at the exact center is zero, but very high just around that point.
    3) a theory about a conic pressure differentials which is too hard to explain, but i like it the best. (basiclly the pressure varies not only along the length of the tube, but also the radius of the hole.)
  2. jcsd
  3. Apr 4, 2008 #2
    Since dp=-ro*g*dh and ro(density) is proportional to p,
    air pressure in constant g decreases exponentialy with hight. So it increases exponentialy with depth, as long as uniform g is a good aproximation (in fact g increases lineary with distance from the center of the Earth).
    The hight/depth where p changes by a factor e is about 8 km, so presure should increase by exp(10) just 80 km under Earth's surface. Even before that depth pressure would probably be high enough to liquefy air, so we could not use ideal gas equation anymore.
  4. Apr 4, 2008 #3


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    This is incorrect because it makes the false assumption that gravity increases as the depth increases. In fact, gravity is maximum at the surface of the Earth, and decreases with increasing depth, until it is zero at the centre.
  5. Apr 4, 2008 #4
    it says "g increases lineary with distance from the center of the Earth" wich is the opposite from what you say it says. The point that in the first 80 km, g would be nearly constant, and that the pressure would become so high that you couldn't use the ideal gas equation is also valid. (altough air can't liquify above the critical temperature)
  6. Apr 4, 2008 #5
    The pressure should increase. Here is my reasoning:
    The air pressure is basically from the weight of air above you. Since we are keeping the area over which it acts constant, then the pressure is directly proportional to weight of air above you. When do you have more air above you? When you are deeper down!

    You can also look at it this way. Pressure at a point below a specific level can be described by [tex]P=f(h)+P_o[/tex], where P_o is the pressure at the specific level and f(h) is some complicated function that takes into consideration some weird stuff like changing g, density of air and so on. But f(h) does not really matter, but instead the +Po. It means that we will be always adding to that pressure, so any pressure below has to be higher.

    I hope I didn't miss anything in my reasoning.
    Last edited: Apr 4, 2008
  7. Apr 4, 2008 #6


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    No, that's exactly what I said.

    g increases with distance from the centre i.e. the farther from the centre you are, the greater g is. That's the same thing as the closer to the centre the less g is.

    In other words, g is maximum at the surface of the Earth, and zero at the centre.
  8. Apr 4, 2008 #7
    Yes, and this was also said by Lojzek, in a post you declared to be incorrect.
  9. Apr 4, 2008 #8


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    Yes, but air near the centre of the Earth does not weigh as much as it does at the surface of the Earth because gravity is dropping off.

    A cubic metre of air at the surface weighs 1.2kg. A cubic metre of air at the core weighs 0 - which means it contributes 0 to the weight of air over you, which means it contributes zero to the air pressure over you. A cubic metre of air at 2000 miles depth weights about half its surface weight - 0.6kg, which mean it cointributes only that much to the air pressure.

    So the air pressure will increase, but the rate at which it increases will drop off. By the time you are near the centre of the Earth, the increase in air pressure with increased depth will be almost zero.
    Last edited: Apr 4, 2008
  10. Apr 4, 2008 #9
    Yes, this is what I would bet on.

    Option A doesn't work, since if you dig a 50ft hole, there will be more pressure than at sea level.

    Option B doesn't work, because there seems to be a discontinuity from going to "high" pressure to no pressure at all. And would it be at a singular point in the center or what?

    Option C is total BS. Even if it's not uniform throughout the area of the hole, it won't make enough of a difference to be noticeable compared to how much depth effects the pressure. You're not assuming that air is traveling through the hole with it, it just stays there while you move on. If it was moving, then sure, all sorts of fluid dynamics would apply.
  11. Apr 4, 2008 #10


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    No, what was incorrect was that he claimed that the air pressure increased exponentially. It doesn't.
  12. Apr 4, 2008 #11
    what would be the weight of 100 lb human-at center of whole? ( a whole through center of earth)
  13. Apr 4, 2008 #12


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    Zero. A human - and everything else at the centre of the Earth - is weightless.
  14. Apr 4, 2008 #13
    Yup, I agree with that. I made sure not to make any statements about the rate at which the pressure increases, but only that it increases. Also another question: wouldn't the density of air also increase as we go deeper down? If so, this would mean that a cubic meter of air near the center has more mass, and so weighs more. This would help to increase the rate of increase of pressure, but I'm not sure if it would do so by a significant factor.
  15. Apr 4, 2008 #14


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    See attached diagram.

    The air pressure at 50 ft above sea is X.
    The air pressure at sea level is X+Y.
    The air pressure at the bottom of a 50ft hole will NOT be X+Y+Y, it will be ~ X+Y+.99Y.
    The air pressure at the bottom of a 100ft hole will be ~ X+Y+.99Y+.98Y etc.

    The air pressure will increase as you go down, but each 50 feet will contribute less and less weight, until near the core, any additional 50 feet will contribute virtually zero weight to the column (which means the air pressure will stop rising - it will be flat.)

    Attached Files:

    Last edited: Apr 4, 2008
  16. Apr 5, 2008 #15
    Some nice incompressible air models!
  17. Apr 5, 2008 #16
    Looks like I must explain my solution: the compressibility of air has MUCH greater effect than the change of gravitational acceleration.

    As long as ideal gas equation can be used, density is propotional to pressure:


    We put it into dp=-ro*g*dh equation:


    Solution for constant g is exponential function:


    If you want to consider g=g(h)=g(0)*(1-h)/R, then you can define a new variable:

    z=Integral(g(h')*dh') from 0 to h

    z(h)=g*(h-h^2/2) (for h<0)

    Then the differential equation is:


    And the solution is:


    But there is no need for that, since the exponential function increases so fast (p(0)/(ro*g)=8km), that we lose linear dependence between ro and p (the comment about critical temperature is true, so air would not liquify).
    Last edited: Apr 5, 2008
  18. Apr 5, 2008 #17


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    Alas, I cannot follow the math. I don't suppose you have an explanation in mere English?
  19. Apr 5, 2008 #18
    @Lojzek: So what is your conclusion? When you posted your math, I was in the middle of doing my own derivation that took care of changing density and and changing gravity, and my solution showed that that rate of change of pressure decreases as height increases. It does not show if density predominates over gravity or vice versa, but I would assume that it is the latter. Because if we assume that increasing density would increase rate of pressure and decreasing gravity would decrease rate of pressure change, then an decrease rate of pressure change indicates that gravity dominates over density. That is opposite to what you claim.

    I will post my solution soon, it will just take some time to type up.
  20. Apr 5, 2008 #19
    it is quite extrodinary that a human is weightless at center of earth-and the effect of bringing a mass of equal mass of the earth(could be more compact)-over the human(standing on top the earth)could render the same human weightless--. The fact is gravity is so weak it takes the total earth to cause the effect of the 100 lb on the human. The 100 lb condition is reversed by bringing equal total mass of earth just above 100 lbs humans head--the humans weight would be zero
    Last edited: Apr 5, 2008
  21. Apr 5, 2008 #20
    Here is my solution. I skipped steps from time to time, because the forums didn't let me insert new lines in my formulas, but you can see the complete thing here.

    The pressure at different heights will depend on two variables: the gravitational field and the density of air. So the first step is to derive both in terms of the depth, h. For gravitational field, we can use the Shell Theorem to claim that the mass of the earth at a specific radius does not have any effect on the gravitational field below it. In other words, gravity only comes from the mass at the radius R-h, where R is the total radius of the Earth, or less. Since we are assuming constant density of Earth, the effective mass is given by:

    [tex]M_{earth} & = & \rho_{earth} \frac{4}{3} \pi (R-h)^3[/tex]

    Then we can just use Newton's famous equation to find the have of g as a function of height:

    [tex]g & = &\frac{G \rho_{earth} \frac{4}{3} \pi (R-h)^3}{(R-h)^2}[/tex]
    [tex]g & = &\frac{4 G \rho_{earth} \pi (R-h)}{3}[/tex]

    As for the changing density, I will make two simple assumptions: the air will stay a gas under any pressure and behave like most gases where [tex]\rho \propto P[/tex]. I am unsure how likely the former assumption is, but I will pretend it is true to simplify the calculations greatly. So the proportionality relation gives us the following equation where A is some constant. We can then divide it by the same equation where the density and pressure are known, such as at the surface. Thus we have:

    [tex]\frac{\rho_{air}}{\rho_0} & = & \frac{AP}{AP_0}[/tex]
    [tex]\rho_{air} & = & \frac{\rho_0 P}{P_0}[/tex]

    Now is the time to calculate the change the total change in pressure as we go down our imaginary hole. We will do this by considering the changes in pressure due to very small disks of air of infinitely small thickness dh and cross-sectional area A. We can do this, because at such a small thickness means that the density and gravitational field are uniform through out. Then if we integrate all those disks from depth 0 to depth H and integrate the pressures from surface pressure to pressure at height H, then we get the absolute pressure at height H! Just look below:

    [tex]AdP & = & g\rho V_{air}[/tex]
    [tex]\int^P_{P_0} \frac{dP}{P} & = & \frac{4 G \rho_{earth} \pi \rho_0}{3P_0} \int^H_0 (R-h) dh[/tex]
    [tex]P & = & P_0 e^{-\frac{4 G \rho_{earth} \pi \rho_0 [(R-H)^2-R^2]}{6P_0}}[/tex]

    Now if we set each of the constants to 1, we end up with the graph below.. I also included the derivative to help observe the rate of change in pressure as height increases.


    As you can see, the rate of pressure increases slowly at the beginning, which indicates that the increased density of air predominates over the decreased gravitational field. However as we go even deeper, we notice that the rate of pressure change slowly decreases until we reach the center of the Earth. Another thing to note is that this graph is symmetric around h=R, meaning that as we pass the center the pressure will slowly decrease until we reach the surface again.
    Last edited: Apr 6, 2008
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