tom.h.wilsontom. wilson@geo.wvu.edu

Department of Geology and GeographyWest Virginia University

Morgantown, WV

More about IsostacyMore about Isostacy

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Age

(ye

ars)

Recent Sedimentation Record - North Sea

10 510 1010 1510 2010

Depth (cm)

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.

The general trend of age versus depth

Problem 2.11

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50

100

150

200

Con

cent

rati

on (

C)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Liquid Fraction (F)

C = CoF(d-1) = 200*(05) 5.5 which yields C = 4.4194

Problem 2.12

Given that C = CoF(d-1) where Co is the initial concentration of the element in the liquid before crystalization, F is the fraction of the liquid remaining, and d is the distribution coefficient. In the present example we solve for C when F = 0.5, and d = 6.5

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.

DLdl mcmc

Drledl mcmc

Solving isostatic equilibrium problems

hem

Where m represents the density of the mantle, = m - c (where c is the density of the crust), and h represents crustal thickening (r + e).

On Tuesday, from the foregoing starting point, we derived a couple basic relationships governing the isostatic equilibrium processes. These included:

ercm

c

And -

cr e

or

from which we must also have

c

e r

In Class Problem: A 500m deep depression on the earth's surface fills with sandstone of density 2.2 gm/cm3. 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/cm3, 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 s c mb l r

3.3 c s mb r

Recall that on Tuesday we showed that l=16.7km - hence

We also showed that r = 20-e-l-b .. thus

3.3 (3.3 )c s mb e b

After rearrangement

3.3 (3.3 )c s mb e b

3.3( ) ( )c m s mb

or

3.3( )

( )c m

s m

b

1.5b 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

-0.2

-0.1

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0 0.5 1 1.5 2

Thickness of Deposited Sediments

De

pth

to

To

p o

f S

ed

ime

nt

De

po

sit

s

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/cm3.)

There are actually 4 parts to this problem - we must first determine the starting equilibrium conditions before doing solving for (a).

Isostatic Response to Erosion

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Amount Eroded (km)

Mo

un

tain

Ele

vati

on

&

Mo

un

tain

Ro

ot

(km

)

Root Extent (km)

Mountain Elevation(km)

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.

A few more comments on Isostacy

At A 2.9 x 40 = 116

The product of density and thickness must remain constant in the Pratt model.

ACB

At B C x 42 = 116 C=2.76C=2.76

At C C x 50 = 116 C=2.32

Complete your reading of chapter 3

Think over problem 3.11