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out of the two columns and only the mantle density counts). Thus the force per
unit area, or pressure P l , at the base of the left-hand column is m D , the weight
exerted on the base of area 1 m 2 . Suppose the depression is filled with water, of
density ρ w . Then the right-hand column, which is in the middle of the depression,
has water to a depth d and rock for the rest of its depth, which is D
d . Thus
its mass is ρ w d
d ) and its weight is g times that mass. Now we can
subtract the pressure exerted by the right-hand column from the pressure exerted
by the left-hand column to obtain the pressure deficit, P , under the depression:
+
ρ m ( D
P
=
P l
P r =
m D
g [ ρ w d
+
ρ m ( D
d )]
=
g ( ρ m
ρ w ) d.
We can think of this as g times the mass deficit created by replacing rock with
water in the depression. Now pressure is one form of stress, so we can write
τ d =
g ( ρ m
ρ w ) d.
(4.4)
This is, to a rough approximation, the stress that drives the flow in the mantle that
must occur as the floor of the depression rises to the level it had before the Ice Age.
That flow is depicted by the arrows in Figure 4.2(b).
This driving stress or pressure acts over the area of the depression, which is about
π R 2 , assuming the depression is roughly circular with a radius of R . If pressure is
force per unit area, then pressure times area yields a total force, the driving force
π R 2 g ( ρ m
F d
=
ρ w ) d.
(4.5)
The driving force is resisted by a force arising from viscous stresses in the mantle.
According to Eq. (4.3), viscous stress is proportional to strain rate, and according to
Eq. (4.1) strain rate is proportional to velocity gradient. If we can identify a typical
velocity gradient within the mantle, then we can estimate the viscous resisting
stress. If the rate of uplift of the floor of the depression is v , then it is plausible
that the upward velocity of the mantle immediately under the depression is close
to v . It is also plausible that as you go away from the depression the uplift velocity
declines, so that far from the depression the velocity is close to zero. How far do
you have to go for the velocity to drop substantially? That will depend on the width
of the depression. The Fennoscandian depression has a radius of about 1000 km,
so if you go 1000 km away the velocity should have decreased significantly. On
the other hand, the North American depression has a radius of about 2500 km, so
its effect in the mantle should extend further, vertically and horizontally, and you
would have to go 2500 km for the velocity to drop significantly. This suggests that
a representative velocity gradient within the mantle flow is v/R , in the sense that
the velocity changes by a significant fraction of v over a distance R . It may be that
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