= Local isostatic response to an added surface layer
{title2=$w=(\rho_b-\rho_f)H/(\rho_m-\rho_f)$}
Let a deposited layer have thickness $H$ and <density> $\rho_b$, replacing a fluid of <density> $\rho_f$, while the original <crust> translates downward by $w$ above <mantle> of <density> $\rho_m$. Assume all deposited material stays in the same fluid environment, unchanged pre-existing layer thicknesses, an external material source, and complete local <isostatic adjustment> without bending resistance.
At a fixed <compensation depth>, added column <mass> is $(\rho_b-\rho_f)H$ per unit area, whereas depression replaces a column contrast $(\rho_m-\rho_f)w$. Equal <hydrostatic pressure> therefore requires
$$
(\rho_b-\rho_f)gH=(\rho_m-\rho_f)gw.
$$
It follows that
$$
w=\frac{\rho_b-\rho_f}{\rho_m-\rho_f}H,\qquad
\Delta z=H-w=\frac{\rho_m-\rho_b}{\rho_m-\rho_f}H.
$$
The pre-existing ground subsides by $w$, while the top of the new layer rises by $\Delta z$ if $\rho_b<\rho_m$. For a subaerial deposit, air <density> is negligible. For a fully submerged deposit, use <seawater> <density>. A deposit that crosses the fluid surface requires separate treatment of its submerged and emerged portions. Specifying only volume $V$ leaves the answer dependent on loaded area through $H=V/A$.
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