Geology Reference
In-Depth Information
you have to go a distance 2
R
for the velocity to drop by, say, 80%, but let's not
worry about factors of 2 for the moment.
If we have a representative velocity gradient of
v/R
, then from Eq. (4.1) a
representative strain rate is
s
=
v/
2
R
. Then from Eq. (4.3) a representative viscous
stress is
τ
r
=
2
μs
=
2
μv/
2
R
=
μv/R.
(4.6)
This is the stress within the mantle that resists the driving stress due to the pressure
deficit, as given by Eq. (4.4). It also acts over an area that is comparable to the area
of the depression, so we can say that the resisting force is roughly
π
R
2
μv/R.
F
r
=
(4.7)
The driving and resisting forces must balance. This follows from Newton's
second law of motion, which says that force
acceleration. If acceleration
is zero, then (net) force must be zero. Strictly speaking, the uplift velocity is
changing, according to Figure 4.2(a), so the acceleration is not exactly zero. Yet
the velocity is very small, and it changes only over thousands of years, so the
acceleration is tiny. Let's put numbers to those statements. From Figure 4.2(a),
about 100 m of uplift has occurred over about 10 000 years, so the average uplift
rate is about 1 cm/yr. A year is about 3.12
=
mass
×
10
7
×
seconds (go on, work it out for
10
-10
m/s. This velocity changes over about
3000 years, which is about 10
11
s, so the acceleration is about 3
yourself), so the velocity is about 3
×
10
−
21
m/s
2
,a
very small quantity. It is characteristic of mantle flow that accelerations are totally
negligible. It follows that everywhere in the mantle forces and stresses balance, to
a very good approximation.
So the driving force should equal the resisting force, or
×
F
d
−
F
r
=
0
,
so
π
R
2
g
(
ρ
m
−
ρ
w
)
d
−
π
R
2
μv/R
=
0
.
We can rearrange this to give
μ
=
g
(
ρ
m
−
ρ
w
)
dR/v.
(4.8)
=
1000 kg/m
3
Using the values already mentioned as well as
ρ
w
and taking
d
10
21
Pa s. A more rigorous solution of Eq. (4.8) is
given in Appendix A (Section A.2), and it yields
μ
=
to be 100 m yields
μ
=
8
×
10
21
Pa s. The original
analysis by Haskell [38] in 1937 for the Fennoscandian region yielded 10
21
Pa s.
A recent analysis by Mitrovica [37] in 1996 confirmed Haskell's value, though it
3.3
×
