# 3.4: Isostasy

Now we're going to take a look at uncompensated and compensated topography. Figure $$\PageIndex{1}$$: Uncompensated and Compensated Topography

Another way to talk about this is through isostasy, which is topography supported by viscous stresses i.e., pressure. Consider a continent ocean system: Figure $$\PageIndex{2}$$: A Continent Ocean System

In the figure above, at Dc, P1=P2 pressure, where Dc is the depth of compensation and $$P=\rho gh$$. Note that when mantle viscous flow stops moving, it is "static".

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

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 hi.

Example Continent Oceans

We are going to write this scenario out for the above continent ocean example and 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
• $$P_1=P_2$$
• Determine $$D_c$$
• 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 $$\PageIndex{3}$$: 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 $$\PageIndex{4}$$: Depth of an Ocean Basin

$d(t)=d_o+\Delta d(t) ?$

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

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.