C) How High would they be if 10km of material were eroded away? Don\'t do A, jus
ID: 286016 • Letter: C
Question
C) How High would they be if 10km of material were eroded away?
Don't do A, just worry about B and C
2. A mountain range 4 km high is in Airy isostatic equilibrium. (10 points) (a) During a period of erosion, a 2 km thickness of material is removed from the mountains. When the new isostatic equilibrium is achieved, how high are the mountains? (b) How much material must be eroded to bring the mountains eventually down to the sea level? Use the crust and mantle densities of 2800 kg/m3 and 3300 km/m3, respectively.Explanation / Answer
Ans:
(A). Height of mountain range -4 Km
For erosion of 2 km of material to attain isostatic equilibrium
the depth of compensation before and after erosion:
c g hc + m g hm = c g h’c + m g h’m = c g ( hc - he ) + m g h’m ------------------------ eq. 1
(here c – crustal density, m- Mantle density, hc – crustal height, h’c – Crustal height after erosion,
hm – Mantle thickness, h’m – Mantle thickness at depth of compensation, he – eroded height )
The equation can be expressed as
h = h’m - hm ------------------------ eq. 2
m (h’m - hm) = c he
h = (c/ m) he ------------------------ eq. 3
Using the value of crustal density - 2800 and mantle density of 3300 and he 2 Km
h = (2800/3300) * 2 km
h = 0.85 * 2 Km
h = 1.70 Km
(Hence for the erosion of 2 Km the loss in height will be only.30 Km)
The new height = 4 – 0.3 Km
= 3.7 Km
(B) To bring back the mountain to sea level the required change in height = 4 Km
Change in height of mountain for erosion of 1 Km
Using eq 3 –
=1 – ((2800/3300) * 2) Km
= 0.15 Km
Erosion needed for 4 Km change in height
= 4/.15
=26.66 Km
(C). The elevation of this mountain mass above sea level.
Given- pc=2.8*103 kg/m3, pm= 3,35*103 kg/m3, eE ~10km
r=[pc/(pm-pc)]e
all values substitute in above equation then...
r=[2.8*103/(3.35*103-2.8*103)](~10000)
r=(2.81/0.55)(~10000)=~51,090m=~51.09km
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