Geology Reference
In-Depth Information
where
g
is the acceleration due to gravity,
ρ
l
is the density of the lithosphere and
ρ
w
is the density of water. We must consider a left-hand column that extends to the
same depth, so the pressure under the left-hand column is
P
l
=
gρ
m
(
d
−
h
+
h
I
)
,
(6.8)
where
ρ
m
is the density of the mantle. If we equate these two pressures and rearrange
we get
h
I
(
ρ
m
−
ρ
w
)
=
(
ρ
l
−
ρ
m
)(
d
−
h
)
.
(6.9)
This expression makes sense if you notice that the left-hand side is (minus) the
mass deficit in the upper part of the right-hand column, relative to the left-hand
column, and the right-hand side is the mass excess in the lower part, and these must
sum to zero.
We have one more step to complete. The lithosphere is denser than the underlying
mantle because of thermal contraction. Therefore we can use Eq. 5.3 to relate
them:
ρ
l
−
ρ
m
ρ
m
h
d
.
=
αT
=
Substituting in Eq. (6.9) gives
h
I
(
ρ
m
−
ρ
w
)
=
hρ
m
(1
−
h/d
)
and because
h/d
is much less than 1 it can be neglected, so
h
I
(
ρ
m
−
ρ
w
)
=
hρ
m
and we get our expression for the subsidence with isostatic balance:
h
ρ
m
ρ
m
−
.
h
I
=
(6.10)
ρ
w
Finally, to get the subsidence as a function of the age of the sea floor, we can
substitute for
h
from Eq. (6.6b) and for
d
from Eq. (6.5), so we end up with
αT
4
κt
ρ
m
ρ
m
h
I
=
π
.
(6.11)
−
ρ
w
Again, the subsidence is proportional to the square root of the age of the sea floor,
but we have taken the trouble to account for isostasy, which changes the answer by
the factor
ρ
m
/
(
ρ
m
−
ρ
w
). Using
ρ
m
=
1000 kg/m
3
, this factor
is 1.43. The isostatic subsidence is then 3.5 km. This is quite a good approximation
to the observed subsidence of 100 Ma old sea floor, which is a little more than 3 km
(Figure 2.5).
3300 kg/m
3
and
ρ
w
=
