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Why, at lower altitudes, does the atm change more rapidly?

  1. Feb 25, 2012 #1
    The lower the altitude, the more rapidly the atmospheric pressure changes with respect to altitude.

    http://www.microwaves101.com/encyclopedia/images/powerhandling/Pressure2.jpg [Broken]

    Why does this happen? I know that the pressure itself is supposed to decrease since the weight of the air column increases as altitude drops, but shouldn't this be a linear relation?

    Does it have anything to do with the air density and the convection of air?

    Thanks!

    BiP
     
    Last edited by a moderator: May 5, 2017
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  3. Feb 25, 2012 #2

    Borek

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  4. Feb 25, 2012 #3
    That's the isothermal case but the troposphere is rather adiabatic:

    [itex]p = p_0 \cdot \left( {1 - \frac{{\kappa - 1}}{\kappa } \cdot \frac{{M \cdot g}}{{R \cdot T_0 }} \cdot h} \right)^{\frac{\kappa }{{\kappa - 1}}}[/itex]
     
    Last edited: Feb 25, 2012
  5. Feb 25, 2012 #4

    Borek

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    It shows why the dependence is not linear without delving into too many confusing details. Much better from the pedagogical point of view.
     
  6. Feb 25, 2012 #5

    DaveC426913

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    A rather simplistic but intuitive way to see why it should not be linear is this:

    Examine the graph in your OP and imagine it as a straight line. It would intersect the X-axis (altitude).

    It would mean that the edge of space -the line between atmosphere and vacuum - would be sharp. You could be at 120,000 feet and be in atmo, and then at 121,000 feet and be in hard vacuum.

    Does that make sense?
     
    Last edited: Feb 25, 2012
  7. Feb 25, 2012 #6
    Why not?
     
  8. Feb 25, 2012 #7

    DaveC426913

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    Why not what? Why does it not make sense?

    I think most people intuitively know that there is no hard boundary at very the edge of the atmo - that it gets tenuous the farther out you go.

    I'm pointing out that, knowing this, one can immediately deduce that the graph must be curved.
     
  9. Feb 25, 2012 #8
    Thank you everyone for your replies. I understand it now.

    BiP
     
  10. Feb 25, 2012 #9
    The adiabatic graph also intersects the X-axis although it is curved and even for a linear pressure gradient there would be no hard boundary. Therefore I can not see the logic in your argumentation. The explanation in Borek's link is much more better. The calculation is made for an isothermal case but the same principle works for a non-isothermal atmospheres too.
     
  11. Feb 25, 2012 #10

    DaveC426913

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    If it intersects the axis then that means at one point there is pressure, and at a point an arbitrarily-small distance away the pressure is zero - no air. That's a boundary.

    Most us can intuitively grasp that this is not the way the atmo tapers off. Once the OP realizes that he already knew the pressure gradient was curved, he would realize that it's probably curved all the way to the ground.

    It's the difference between being told something, and self-discovery.

    It's not a competition. Multiple responses are not penalized. :rolleyes:
     
  12. Feb 26, 2012 #11
    Solving differential equation with ideal gas law and P=ρgh you would see that it follows an exponential pattern which coincides with your graph.

    I think an intuitive explanation would be, with decreasing altitude, more weight of the air above is exerted to the air below.
     
  13. Feb 27, 2012 #12
    Its not a hard boundary because for arbitrarily-small distances the corresponding pressure differences would be arbitrarily-small to.

    Trying to understand physics intuitively is sometimes not a good idea. An adiabatic atmosphere would taper off this way and it is more realistic than the isothermal case described by the exponential curve without boundary. Explaining why the real atmosphere has no upper limit is not as easy as you seem to believe. The pressure gradient depends on the temperature profile and that's a quite complex topic in the upper atmosphere.
     
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