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More about Isostacy
tom.h.wilson
tom. [email protected]
Department of Geology and Geography West Virginia University Morgantown, WV
16000 14000 Recent Sedimentation Record - North Sea The general trend of age versus depth 12000 10000 8000 6000 4000 2000 0 10 510 1010 Depth (cm) 1510 2010 Problem 2.11
i) sedimentation rate from 10510 to 1490 years ago was 0.0429cm/yr or 23.29 yr/cm. This value is computed directly from the endpoints defining sedimentation during that period of time ii) The trend of age versus depth was approximated by the line shown above in the figure. Its slope is ~ 2.24 yrs/cm which translates into a sedimentation rate of 0.447 cm/yr.
iii) Assuming that sedimentation continues at the rate of 0.0429 cm/yr, it will take an additional 459.9 years to deposit the remaining 19.75 cm of sediment. Thus the age at the surface or the time when sedimention ceased is 1490 - 459.9 years or approximately 1030 years ago.
200 150 100 50 0 0.0
0.1
0.2
0.3
0.4
0.5
0.6
Liquid Fraction (F) 0.7
0.8
0.9
1.0
Probl em 2.12
Given that C = CoF(d-1) where Co i s the ini ti al concentrati on of the element i n the l i quid before crystali zati on, F i s the fracti on of the l iqui d remaining, and d is the distri buti on coeffi ci ent. In the present exampl e we sol ve for C when F = 0.5, and d = 6.5
C = C o F(d-1) = 200*(05) 5.5
which yields C = 4.4194
The gravity anomaly map shown here indicates that the mountainous region is associated with an extensive negative gravity anomaly (deep blue colors). This large regional scale gravity anomaly is believed to be associated with thickening of the crust beneath the area. The low density crustal root compensates for the mass of extensive mountain ranges that cover this region. Isostatic equilibrium is achieved through thickening of the low-density mountain root.
Solving isostatic equilibrium problems
c l
m d
c L
m D
c l
m d
c
e
l
r
m D
On Tuesday, from the foregoing starting point, we derived a couple basic relationships governing the isostatic equilibrium processes. These included:
e
m h
Where
m mantle,
represents the density of the =
m -
c (where
c is the density of the crust), and h represents crustal thickening (r + e).
And -
r
m
c
c
e r
or
c
e
from which we must also have
e
c
r
In Class Problem:
A 500m deep depression on the earth's surface fills with sandstone of density 2.2 gm/cm 3 . Assume that the empty basin is in isostatic equilibrium and that normal crustal thickness in surrounding areas is 20km. Calculate the thickness of sediment that must be deposited in the basin to completely fill it. (Use crustal and mantle densities of 2.8 and 3.3 gm/cm 3 , respectively.) Hint: Compute the initial thickness of the crust beneath the empty basin and assume that the crustal thickness beneath the basin does not change.
20
c
b
s
l
c
r
m
Recall that on Tuesday we showed that l=16.7km - hence
3.3
c
b
s
r
m
We also showed that r = 20-e-l-b .. thus
3.3
c
b
s
(3.3
b
)
m
After rearrangement
3.3
c
b
s
(3.3
b
)
m
3.3(
c
m
)
b
(
s
m
)
or
b
3.3( (
s
c
m
m
) )
&
b
1.5
km
Recall that since l = 16.7km and lt does not change as the basin is filled, we now have the depth to the base of the crust in the rifted region (b + l = 18.2km), after isostatic equilibrium has been re-established. The base of the crust now rests 1.8km above the base of the continental crust in the surrounding un deformed area. Recall, that when the basin was empty (0.5km deep) the crust extended down to 17.2km and r (the antiroot) was 2.8. It took 1.5km of sediment to fill our half kilometer deep basin!
As sediment is deposited, the basin floor gradually drops to maintain isostatic equilibrium.
Does this really happen?
Conodant alteration indices from this area of the Appalachians indicate that rocks currently exposed at the surface were once buried beneath 3km of sediment.
Let’s examine the dynamics of this process using EXCEL. Pick up the EXCEL file Isostacy1.xls from my shared directory.
Basin Filling Process
0.6
0.5
0.4
0.3
0.2
0.1
0 -0.1
-0.2
0 0.5
1 1.5
Thickness of Deposited Sedim ents
2
Take Home Problem: A mountain range 4km high is in isostatic equilibrium. (a) During a period of erosion, a 2 km thickness of material is removed from the mountain. When the new isostatic equilibrium is achieved, how high are the mountains? (b) How high would they be if 10 km of material were eroded away? (c) How much material must be eroded to bring the mountains down to sea level? (Use crustal and mantle densities of 2.8 and 3.3 gm/cm 3 .) There are actually 4 parts to this problem - we must first determine the starting equilibrium conditions before doing solving for (a).
The preceding questions emphasize the dynamic aspects of the problem. A more complete representation of the balance between root and mountain is shown below. Also refer to the EXCEL file on my shared directory.
Isostatic Response to Erosion
25 20 15 10 5 0 0
Mountain Elevation(km)
5
Root Extent (km)
10 15
Amount Eroded (km)
20 25 30
A few more comments on Isostacy
A B C The product of density and thickness must remain constant in the Pratt model.
At A 2.9 x 40 = 116 At B
C x 42 = 116 At C
C x 50 = 116
C =2.76
C =2.32