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3.4: Isostasy

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  • In the last section, we looked at uncompensated and compensated topography.

    FIGURE gravity lecture 2 uncompensated and compensated

    Another way to talk about this is through isostasy, which is topography supported by viscous stresses ie pressure. Consider a continent ocean system:

    FIGURE continent ocean

    Mantle viscous flow, stop moving "static". At Dc, P1=P2 pressure, where Dc is the depth of compensation and P=\(\rho\)gh.


    Isostasy⇒topography supported by viscous stresses⇒pressures at Dc balanced in "static" state, ie slow state

    Isostatic balance goal⇒to use known \(\rho_i\) and hi's to find an unknown \(\rho_i\) or hi.

    Example Continent Oceans

    We are going to write this scenario out for the continent ocean example then simplify. We can start at the top or the bottom because P1=P2

    \[\rho_{cc}h_{cc}g+\rho_{cL}h_{cL}g+\rho_{m}h_{m}g=\rho_{air}h_{air}g+\rho_{w}h_{w}g+\rho_{oc}h_{oc}g+\rho_{oL}h_{oL}g+\rho_{m}h_{om}g+\rho_{m}h_{m}g \label{ex1}\]

    1. All g's cancel

    2. \(\rho_{m}h_{m}\) on both sides⇒choose \(D_c\) where there are no more density differences

    The Equation \ref{ex1} now reads:


    3. If oceanic plate ia old, dcm=dom⇒Dc moves up to the bottom of continental crust. 

    Now we need to think about what is known vs unknown: What is the height of continent above sea level⇒hair. So we will solve for hair.

    Dc at the bottom of the continental crust


    where \(\rho_{cc}h_{cc}\) is about 30 km, \(\rho_{air}h_{air}\) is ?, \(\rho_{w}h_{w}\) is about 5 km, \(\rho_{oc}h_{oc}\) is about 75 km and \(\rho_{oL}h_{oL}\) is ?. We can now see that we have 2 unknowns, and thus we will need a second equation. 




    Now substitute and solve for hair.

    \[ \begin{align*} \rho_{cc}h_{cc} &=\rho_{air}h_{air}+\rho_{w}h_{w}+\rho_{oc}h_{oc}+\rho_{m}h_{cc}-\rho_{m}h_{air}-\rho_{m}h_{w}-\rho_{m}h_{oc} \\[4pt] -h_a(\rho_m-\rho_a) &=-\rho_{cc}h_{cc}-h_w(\rho_m-\rho_w)-h_{oc}(\rho_m-\rho_{oc})+\rho_mh_{cc} \\[4pt]h_a(\rho_m-\rho_a)&=h_{cc}(\rho_m-\rho_{cc})-h_w(\rho_m-\rho_w)-h_{oc}(\rho_m-\rho_{oc}) \\[4pt] h_a&=h_{cc}\left(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a}\right)-h_w \left(\frac{\rho_m-\rho_w}{\rho_m-\rho_a}\right)-h_{oc}\left(\frac{\rho_m-\rho_oc}{\rho_m-\rho_a}\right) \end{align*}\]

    where \(h_a\) is the total height above and all other terms are a fraction of each others height weighted by the density differences across the columns. 

    We could also have other unknowns. To find them, we use the same process. 

    • g's cancel
    • P1=P2
    • Determine Dc
    • Use total thickness to get rid of extra unknowns
    • Solve for unknown

    For example, given Ha, we solve for Hcc. What thickness of oceanic crust is needed?

    Example Time Changes

    FIGURE time changes

    T1 ha=4000 m and after erosion T2 hc=hc1-2000 m. -2000 m is the change in height. So what stays the same? hw and hoc. ha and hm change. Now writing an equation:


    \(h_w(\frac{\rho_m-\rho_{N}}{\rho_m-\rho_a})\) is A1 and \(-h_{oc}(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})\) is \(A_2\), and there is no change in these terms. We can rewrite the equation as:

    \[h_{a2}=(h_{cc}-\Delta h)(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})+A_1+A_2\]

    \[h_{a2}=h_{cc}(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})+A_1+A_2-\Delta h(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})\]

    where \(h_{cc}(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})+A_1+A_2\) is \(h_{a1}\).

    \[h_{a2}=h_{a1}-\Delta h(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})\]

    Example Depth of an Ocean Basin

    FIGURE depth of an ocean basin

    \[\rho_ch_c+w(o)\rho_L+h_a\rho_a=\Delta d(t)\rho_w+\rho_ch_c+\rho_Lw(t)\]

    We know that \(h_a=\Delta d(t)+w(t)\)

    \[h_c+h_a=\Delta d(t)+h_c+w(t)\]

    The rest of the example you will finish on your homework.