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12.002 Physics and Chemistry of the Earth and Terrestrial Planets
Fall 2008
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Isostasy
1. Compensation for extra mass due to topography.
-Pressure must be equal everywhere at same reference level.
Water
reference
Pressure must be equal
everywhere at same
reference level
Figure by MIT OpenCourseWare.
2. Two Hypotheses: Airy –> constant density; MOHO changes
Pratl –> MOHO stays the same; density varies
Airy
Pratt
ρ1
ρc
MOHO
ρ1 = ρ2
ρ2
MOHO = reference
reference
Figure by MIT OpenCourseWare.
3. Ice Cube
balance
column A
(h+d)ρi
column B
=
dρw
h = d(ρw – ρi)/ ρi = d(1000 -900)/900
h = (1/9)d
Figure by MIT OpenCourseWare.
4. High Altitude Regions – Tibet
ρc = 2800 kg/m3
B
A
ρm = 3300 kg/m3
dt
ρc
dt+dc+dr
Sea level
ρc
dc
dr
ρm
reference level
Figure by MIT OpenCourseWare.
Balance :
Column A
Column B
(dc + dt + dv) ρc
=
dcρc + dvρm
dv = dtρc /(ρm-ρc) = 5000(2800)/500 = 28 km
How much is the crust thicker? 5+28=33km
5. Continental Crust at Sea Level
ρ0=3000 kg/m3
A
B
Ocean
dc
ρc
ρw=1000 kg/m3
ρw
dw
ρo
do
ρm
Figure by MIT OpenCourseWare.
balance: Column A
dcρc
=
Column B
dwρw + doρo + (dc – dw – do) ρm
dc = (dw ρw – ρm + do(ρo – ρm))/ (ρc – ρm)
dw=5 km conduction
d0=8 km conduction
dc =(5000(1000-3300) + 8000(3000-3300))/(2800-3300)
dc = 27.8 km
Total Crustal Thickness beneath Tibet
61 km
6. Crustal Thinning – Sedimentary Basins
ρs = 2300 kg/m3
A
B
di
ρ
ρc
ρs
dc
ρm
ρc
df
l
reference
ρc
ρm
dc ρc
l
Figure by MIT OpenCourseWare.
Balance: Column A
Column B
diρ +dcρc + (l – di – dc) ρm = dfρs +dcρc + (l – df – dc) ρm
df = di(ρ – ρm)/( ρs – ρm)
For water: df = (1000-3300)di/(2300-3300) = 2.3di
For Air:
df = (-3300)di/(2300-3300) = 3.3di
7. This relationship between topography and tectonic plates can be applied
to other terrestrial planets.
High topo –> Thick Crust
–>
Process : Crustal Thickening/Shortening
Low Topo –> Thin Crust
–>
Process: Crustal Thinning/Extension
Figure by MIT OpenCourseWare.