Transcript No Slide Title

More about Isostacy
tom.h.wilson
tom. [email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Recent Sedimentation Record - North Sea
16000
The general trend of
age versus depth
14000
Age (years)
12000
10000
8000
6000
4000
2000
0
10
510
1010
1510
2010
Depth (cm)
Problem 2.11
i) sedimentation rate from 10510 to 1490 years ago was 0.0429cm/yr
or 23.29 yr/cm. T his value is computed directly from the endpoints
defining sedimentation during that period of time
ii) T he 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.
Concentration (C)
200
150
100
50
0
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)
Proble m 2.12
Given that C = CoF(d-1) whe re Co is the initial conce ntration of the e lement
in the liquid be fore crystaliz ation, F is the fraction of the liquid re maining, and
d is the distribution coe fficient. In the prese nt e xample we solve for C whe n
F = 0.5, and d = 6.5
o
5.5
C = C F(d-1) = 200*(05)
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
cl  md  c L  m D
cl  md  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 represents the density of the
mantle,  = m - c (where c is the
density of the crust), and h represents
crustal thickening (r + e).
And -

c
r  
  m  c

e

or
 c
r 
 

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/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  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  e  b) m
After rearrangement
3.3c  bs  (3.3  e  b) m
3.3( c  m )  b( s  m )
or
3.3( c   m )
b
( s  m )
&
b  1.5km
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 undeformed 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 halfkilometer 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
Depth to Top of Sedim ent
Deposits
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/cm3.)
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.
Mountain Elevation &
Mountain Root (km)
Isostatic Response to Erosion
25
20
Root Extent (km)
15
10
Mountain Elevation(km)
5
0
0
5
10
15
20
Amount Eroded (km)
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
Complete your reading of
chapter 3
Think over problem 3.11