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Derivation of exponential density function for air

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data
    If atmosphere can be treated as an isothermal ideal gas of constant mean molecular mass m, show that density drops exponentially with height - ρ= [ρ0]e^-z/h - where h is a constant

    2. Relevant equations
    ρ= [ρ0]exp^-z/h (derivation of ...)

    ρ0=initial density at sea level
    z = height
    h = scale height (found in later question to be 8.5km)

    The previous question was also a derivation -->
    dP/dz = -gρ which I managed. May or not be a starting point to this question

    (P = pressure, z = distance, g= acc due to grav, ρ = density)
    3. The attempt at a solution

    I've read my lecture notes about 100x but can't even begin to see where this derivation can come from. A previous derivation was the equation
    dP/dz = -gρ
    (P = pressure, z = distance, g= acc due to grav, ρ = density)

    Sorry ese's, I posted this in the wrong section to start with.
  2. jcsd
  3. Nov 19, 2007 #2


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    Staff: Mentor

    If one has P(z) = ρ(z) g (h-z), the since h >> z, P(z) ~ ρ(z) g (h) => dP(z) = dρ(z) gh

    and using P(z) = ρg (h-z) then dP(z) = ρg -dz, but this assumes that ρ(z) is more or less constant.

    one wants to end up with

    dρ(z) gh = ρ(z) g -dz, which yields,

    dρ(z)/ρ(z) = -dz/h

    At the moment, I forget the argument for this approach. Hopefully, ones notes addresses the key parts.
  4. Nov 19, 2007 #3
    Cheers, I got the rest myself. Thanks.
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