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Homework Help: Pressure and density

  1. Jul 11, 2004 #1
    What would be the height of the atmosphere if the air density (a) were uniform and (b) decreased linearly to zero with height? Assume that at sea level the air pressure is 1.0 atm and the air density is 1.3 kg/m^3.

    I solved part a, my question is about part b.

    Every approach I have tried with this problem - and I've tried a lot of approaches - has resulted in a useless or wildly incorrect answer. How can this problem be solved?

    The most logical thing seems to be using the equation for pressure variation with height:

    [tex]p_2 = p_1 + \rho g (y_1 - y_2)[/tex]

    Setting level 1 to be sea level, and level 2 to be any level above it, that means:

    [tex]p_1 = p_s, ~y_1 = 0, ~y_2 = h[/tex]

    Where the subscript s denotes sea level, and h represents the height above sea level. Now for the part where I'm probably going wrong, I need to come up with an expression for density that "linearly decreases to zero with height". That doesn't seem like enough information, since it could lose density at any constant rate and still be "linearly decreasing". The atmosphere could be a centimeter high or it could be a trillion miles tall, and could still have its density versus height graph be a simple line.

    Taking my best guess and trying to generalize, I come up with the following expression:

    [tex]\rho = \rho_s \cdot (1 - \frac{h}{h_{max}})[/tex]

    Rho is the current density, the subscript s denotes the density at sea level, h is the current height, and Hmax is the height at the edge of the atmosphere.


    [tex]p_2 = p_s -gh \rho_s \cdot (1 - \frac{h}{h_{max}})[/tex]

    [tex]p_2 = p_s -g \rho_s \cdot (h - \frac{h^2}{h_{max}})[/tex]

    If the pressure is zero, then we're at the edge of the atmosphere. Then h = Hmax, which means h - h² / Hmax = Hmax - Hmax² / Hmax = Hmax - Hmax = 0, which results in the contradiction 0 = Ps.

    I've tried a lot of other approaches, like trying to view it as a limit and use L'Hospital's rule to evalute it, or assuming [itex]d \rho / dh = -1[/itex] and integrating from there to build up an expression for density, but none of these approaches has worked, so I'll spare you all the pointless transcription of my failed work.

    How can this be done correctly?
  2. jcsd
  3. Jul 11, 2004 #2


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    Staff Emeritus
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    Actually, you should have

    [tex]P(h) = g \int_h^H \rho (h) dh[/tex]


    [tex] \rho (h) = \rho (0) (1- \frac {h} {H}) [/tex]


    [tex] H = h_{max} [/tex]
  4. Jul 11, 2004 #3
    Er. Should I have seen this somewhere before?
  5. Jul 11, 2004 #4

    Doc Al

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    Staff: Mentor

    Gokul43201's integral shows how to find the pressure as function of height when the density varies. You were using [itex]p_2 = p_1 + \rho g (y_1 - y_2)[/itex], which only applies when the density is constant.
  6. Jul 11, 2004 #5
    Thanks, I really appreciate that. I want to know, though, whether the integral he gave is something I would be expected to have just intuitively known, or whether it is something that normally would have been covered in the material preceding this problem?
  7. Jul 11, 2004 #6

    Doc Al

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    Staff: Mentor

    Beats me. But I would say that if you understand the meaning of pressure and the concept of integration, then you should be able to figure it out. After all, it's just a generalization of [itex]p_2 - p_1 = \rho g (y_1 - y_2)[/itex] to get [itex]dp = \rho(y) g dy[/itex].
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