# 3.4: Isostasy

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.

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

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

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

3. If oceanic plate ia old, 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

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

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 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:

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

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