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Density as a function of depth

  1. Sep 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the density of water at a depth z in the ocean is related to the surface density rho_s by
    [tex]
    \rho(z) \approx \rho_s [1 + (\rho_s g/B)z]
    [/tex]
    where B is the bulk modulus of water.

    2. Relevant equations

    B = -V (dP/dV)
    B = rho (dP/d rho)

    3. The attempt at a solution


    I've been trying to get this problem for 4 hours...aaargh.

    I started by relating the change in pressure to the change in depth: pressure increases with depth.
    [tex]
    \frac{dP}{dz} = \rho(z) g
    [/tex]


    so

    [tex]
    dP = \rho(z) g \, dz
    [/tex]


    Then, substituting this expression for dP into the second formula above, I got

    [tex]
    B = \rho^2 g \frac{dz}{d\rho}
    [/tex]


    Then I got
    [tex]
    \frac{d\rho}{dz} = \frac{\rho^2 g}{B}
    [/tex]


    This is a separable differential equation, but I don't think it's the right one. I tried solving it with the initial condition rho(0) = rho_s and got
    [tex]
    \rho(z) = \frac{\rho_s}{1 - \frac{\rho_s g}{B}z}
    [/tex]


    which doesn't make sense because density should not become infinite at a certain depth. What did I do wrong?
     
  2. jcsd
  3. Sep 3, 2015 #2

    Geofleur

    User Avatar
    Science Advisor
    Gold Member

    For the answer you got, take a Taylor expansion about ## z = 0 ##. As for the infinite density at a certain depth, maybe it would be good to think about the limitations of the model. Also, you might want to calculate what that depth is for water.
     
  4. Sep 3, 2015 #3
    Hmm...yes, it does look like the second term in the denominator stays small for depths up to 1000 m. At 11000 m (the Marianas trench is about as deep), the denominator is 1 - 0.049, which to me means the model ought not to be used here. Thanks! That was not obvious to me at all.
     
  5. Sep 4, 2015 #4
    Even at that depth, it's not a bad approximation.

    Chet
     
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