# 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 D_{c}, P_{1}=P_{2} pressure, where D_{c} is the depth of compensation and P=\(\rho\)gh.

\[P=\sum_{i=1}^{n}\rho_ih_ig\;with\;n\;layers\]

Isostasy⇒topography supported by viscous stresses⇒pressures at D_{c} balanced in "static" state, ie slow state

Isostatic balance goal⇒to use known \(\rho_i\) and h_{i}'s to find an unknown \(\rho_i\) or h_{i}.

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 P_{1}=P_{2}

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

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:

\[\rho_{cc}h_{cc}+\rho_{cL}h_{cL}=\rho_{air}h_{air}+\rho_{w}h_{w}+\rho_{oc}h_{oc}+\rho_{oL}h_{oL}\]

3. If oceanic plate ia old, d_{cm}=d_{om}⇒D_{c} 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⇒h_{air}. So we will solve for h_{air}.

D_{c} at the bottom of the continental crust

\[\rho_{cc}h_{cc}=\rho_{air}h_{air}+\rho_{w}h_{w}+\rho_{oc}h_{oc}+\rho_{oL}h_{oL}\]

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.

\[h_{cc}=h_{air}+h_{w}+h_{oc}+h_{m}\]

and

\[h_{m}=h_{cc}-h_{air}-h_{w}-h_{oc}\]

Now substitute and solve for h_{air}.

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

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
- P
_{1=}P_{2} - Determine D
_{c} - Use total thickness to get rid of extra unknowns
- Solve for unknown

For example, given H_{a}, we solve for H_{cc}. What thickness of oceanic crust is needed?

Example Time Changes

**FIGURE time changes**

T_{1} h_{a}=4000 m and after erosion T_{2} h_{c}=h_{c1}-2000 m. -2000 m is the change in height. So what stays the same? h_{w} and h_{oc}. h_{a} and h_{m} change. Now writing an equation:

\[h_{a1}=h_{cc}(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})+h_w(\frac{\rho_m-\rho_{N}}{\rho_m-\rho_a})-h_{oc}(\frac{\rho_m-\rho_{oc}}{\rho_m-\rho_a})\]

\(h_w(\frac{\rho_m-\rho_{N}}{\rho_m-\rho_a})\) is A_{1} 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.