Übung zu Globale Geophysik II 02 Answers
Übung zu Globale Geophysik II: Thursday, 14:00 – 16:00, Theresienstr. 41, Room C419
Lecturer: Heather McCreadie
Companion class to Globale Geophysik II: Wednesday, 9:00 – 11:00, Theresienstr. 41, Room C419
Lecturer: Rocco Malservisi
All details are taken directly from Chapter 5, Fowler, C. M. R. (1990), The Solid Earth.
Cambridge University Press, reprinted 1992.
Answers:
Q1a) What is Pratt’s hypothesis?
[10]
1. The depth of the base of the upper layer is a constant.
[5]
2. Isostatic equilibrium is achieved by allowing this upper layer to be composed of
columns of constant density.
[5]
Taking the base of the upper layer as the compensation depth and equating the masses
above this level in each column of unit cross-sectional area gives:
ρu D = ( h1 + D ) ρ1
= ( h2 + D ) ρ1
(1.1)
= ρwd + ρd ( D − d )
[10]
Thus in this model, compensation is achieved by mountains consisting of and being
underlain by material of low density,
⎛ D ⎞
ρ1 = ρu ⎜
⎟
⎝ h1 + D ⎠
(1.2)
[5]
And the oceans being underlain by material of higher density,
ρ D − ρwd
ρd = u
D−d
(1.3)
[5]
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Übung zu Globale Geophysik II 02 Answers
Q1b) What is Airy’s hypothesis?
[10]
1. The rigid upper layer and the substratum are assumed to have constant density.
[5]
2. Isostatic compensation is achieved by mountains having deep roots.
[5]
Taking an arbitrary compensation depth that is that is deeper than the deepest mountain root
in the substratum and equating the masses above that depth in each vertical column of unit
cross-sectional area, one obtains
t ρ u + r1 ρ s = ( h 1 +t + r1 ) ρ u
= ( h 2 +t + r2 ) ρ u + ( r1 − r2 ) ρ s
(1.4)
= d ρ w + ( t − d − r3 ) ρ u + ( r1 + r3 ) ρ s
[10]
A mountain of height h1 would therefore have a root r1 given by rearranging equation (1.4)
hρ
(1.5)
r1 = 1 1
ρ s − ρu
[5]
Similarly a feature at depth d beneath the sea level would have an anti root r3 given by
r3 =
d ( ρu − ρ w )
ρ s − ρu
(1.6)
[5]
These equations apply to any system of layered bodies.
Q1c) What is Archimedes principle of hydrostatic equilibrium and what does it have to do
with isostasy?
Archimedes principle of hydrostatic equilibrium states: a floating body displaces its own
weight of water. A mountain chain can therefore be compared with an iceberg or cork
floating in water.
[5]
Q1d) Describe the term compensation depth.
The depth below which all pressures are hydrostatic is termed the compensation depth.
[5]
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Übung zu Globale Geophysik II 02 Answers
Q2a) Verify the isostatic equilibrium calculation for continents and ocean basins, using
densities of sea water, crust and mantle of 1.03, 2.9 and 3.3 (x103 kgm-3),
respectively, an ocean-basin depth of 5km and an oceanic crustal thickness of 6.6km.
Hint: show that the thickness of the continental crust is 35km.
Using Airy’s hypothesis, let the thickness of the continental crust be t . Because
there is no root r1 = 0 therefore equation (1.4) becomes
t ρ u = d ρ w + r3 ρu + ( t − d − r3 ) ρ s ⇒ t =
=
d ρ w + r3 ρu − ( d + r3 ) ρ s
ρu − ρs
5 × 103 • 1.03 ×103 + 6.6 ×103 • 2.9 ×103 − ( 5 ×103 + 6.6 ×103 ) 3.3 ×103
2.9 ×103 − 3.3 ×103
⎡ 5 • 1.03 + 6.6 • 2.9 − ( 5 + 6.6 ) 3.3 ⎤
3
3
=⎢
⎥ ×10 = 35 ×10 m
2.9
3.3
−
⎣
⎦
[10]
Where
[5]
d
ρ u = density of crust = 2900 kgm -3
ρ s = density of mantle = 3300 kgm -3
ρ w = density of sea water = 1030 kgm-3
r3 = ocean crust depth = 6600 m
t
r
d = ocean basin depth = 5000 m
⇒ t = 35 km
Q2b) Using the Pratt and Airy hypotheses find the density beneath the mountain and the
depth of the root for a mountain range 5km high in isostatic equilibrium?
Pratt’s hypothesis equation (1.2) gives:
⎛ D ⎞
⎛ 35 ⎞
−3
density beneath mountain ρ1 = ρu ⎜
⎟ = 2900 ⎜
⎟ = 2500kgm
⎝ 5 + 35 ⎠
⎝ h1 + D ⎠
[5]
Airy’s hypothesis equation (1.5):
root beneath mountain r1 =
h1 ρ1
5 × 2900
=
= 36km
ρ s − ρu 3300 − 2900
total thickness of crust in mountain is 71km
[5]
[5]
Total marks 120
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