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The Basic Isostatic Calculation
Let’s start with the basic ”iceberg” calculation. Suppose we want to know how high a block of crust
would ”float” in a viscous fluid mantle. (This is not a very realistic depiction of the Earth’s crust
and mantle. It’s a little bit more relevant to a hot planet like Venus, where block-like ”plateau
highlands” rise from plains thought to be underlain by ductile mantle that comes closer to the
planet’s surface than on Earth).
The first thing to do is choose a ”depth of compensation”. Lithostatic stress is the same everywhere
at this depth, so it needs to be in the fluid layer, and deeper than hc .
∆h
ρc
ρm
hc
hm
h'm
depth of isostatic compensation
(lithostatic stress is the same everywhere at this depth)
Figure 1: Set-up for the basic problem.
To solve the problem we equate the lithostatic stress in a column of mantle rock at left with the
combined lithostatic stress due to the crust and mantle, at right. This gives:
ρm g hm = ρc g hc + ρm g h0m
(1)
The only unknown is the thickness of mantle h0m that equalizes the lithostatic stress at our chosen
depth of compensation. We can re-arrange equation 1 to get an expression for h0m :
h0m = hm −
ρc
hc
ρm
(2)
Noting that ∆h = h0m + hc − hm , you can substitute and re-arrange to get:
∆h = hc (1 −
1
ρc
)
ρm
(3)
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The Sedimentary Basin Calculation
In this calculation we asked how much isostatic subsidence occurs if we add a 1 km thick sedimentary
section on to the surface of the crust. If the crust subsides by more than 1 km, this would be a
viable method of making a sedimentary basin. If the crust subsides by less than 1 km, then the net
result is actually a hill, not a depression, and we wouldn’t expect it to remain a center of deposition.
ρsed = 2500
kg /m 3
∆h
hsed = 1 km
ρc = 2850
kg /m 3
ρc
hc = 40 km
hc
ρm = 3300
kg /m 3
hm
h'm
depth of isostatic compensation
Figure 2: Left: the initial state. Right: the crust has subsided under the weight of the 1 km
sedimentary layer. We want to know the amount of subsidence ∆h.
Again, we set up the problem by equating the lithostatic stress at some chosen depth of compensation before and after the sediment load is added. We solve it by finding an expression for ∆h =
hm − h0m :
ρc g hc + ρm g hm = ρsed g hsed + ρc g hc + ρm g h0m
(4)
As before, g cancels out everywhere, and this time, the crustal load cancels out on both sides,
leaving:
ρm (hm − h0m ) = ρsed hsed
(5)
which simplifies to:
ρsed
hsed
(6)
ρm
Substituting the numerical values given in the diagram, we find that the crustal block only subsides
760 m for every 1 km of sediment load piled up on top. In other words, we’ve made a 240 m high
mountain, not a sedimentary basin.
hm − h0m = ∆h =
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The Response of the Crust to Erosion
The final calculation we did in class considered what would happen to a block of crustal rock
thinned by eroding 1 km from the surface. We want to know what the change in the altitude of the
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crustal surface would be once isostatic equilibrium is reached. This will also tell us how the height
of a marker fixed at some initial height on the crustal block (e.g. a shoreline deposit, marking sea
level) will change in response to the crustal thinning.
he = 1 km eroded from the
surface of the crust
∆h
ρc
ρc = 2850
kg /m 3
h'c = 39 km
hc = 40 km
ρm = 3300
kg /m 3
∆h
hm
h'm
depth of isostatic compensation
Figure 3: Left: the initial state. Right: after thinning, the crustal block ”floats” at a new position.
We want to know if the surface is higher or lower than before, and how far a marker fixed to the
crust moves up or down (∆h).
As always, we set up the problem by equating the lithostatic stress at the depth of compensation
before and after the erosion takes place:
ρc g hc + ρm g hm = ρc g h0c + ρm g h0m = ρc g (hc − he ) + ρm g h0m
(7)
Rearranging terms, we get an expression for ∆h = h0m − hm :
ρm (h0m − hm ) = ρc he
(8)
ρc
he = 0.86 he
ρm
(9)
Which shows that:
∆h =
This is a surprising result! Although erosion may remove a kilometer from the crust, isostatic
rebound will restore the surface elevation by 860 m. In other words, assuming isostatic equilibrium
is regained, a kilometer of erosion will only lower the surface elevation of the crust by 140 m. A
shoreline deposit, or other marker fixed to the crust before the erosion took place, will end up 860
m above its original altitude.
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