SME
Geophysics · Problem type
The depth of a mountain’s root
Use this for “A range stands 3 km high on 35 km of normal crust (2.8 g/cm³ over a 3.3 g/cm³ mantle). How deep is its root, and how thick is the crust under the peak?”
Diagram
Root
r
=
h
×
crust
density
mantle − crust density
=
3
×
2.8
3.3 − 2.8
=
16.8
km.
Crust under the peak = T + h + r = 35 + 3 + 16.8 = 54.8 km.
Down to the compensation depth both columns weigh the same: 2.8 × 54.8 = 2.8 × 35 + 3.3 × 16.8.
This is an example. Type your own values to start a new problem.
r (km)
=
h (km)
×
ρc (g/cm³)
÷
(
ρₘ (g/cm³)
−
ρc (g/cm³)
)
T
Normal crust
example · km
H
Crust under the peak
calculated · km
Formulas
r
=
h
×
ρc
ρₘ − ρc
16.8
=
3
×
2.8
3.3 − 2.8
H
=
T
+
h
+
r
54.8 = 35 + 3 + 16.8
Assumptions
•
The crust floats on the mantle like ice on water: every column weighs the same down to the compensation depth.
•
The crust has one density everywhere (Airy’s model), so a high mountain needs a deep root.
•
Widths in the picture are not to scale; heights are.
Step-by-step
WE KNOW
h = 3 km, ρc = 2.8 g/cm³, ρₘ = 3.3 g/cm³, T = 35 km
FIND
root (r), crust under the peak (H)
Step 1 · Find root (r)
Use r = h × ρc ÷ (ρₘ − ρc)
The
root’s
missing
weight
(ρₘ
−
ρc)
×
r
balances
the
mountain’s
ρc
×
h
.
r
=
3
×
2.8
3.3 − 2.8
r
=
3
×
2.8
0.5
r
=
8.4
0.5
r
=
16.8
km
Step 2 · Find crust under the peak (H)
Use H = T + h + r
The normal crust, the mountain above it and the root below.
H
=
35
+
3
+
16.8
H
=
54.8
km
Check
Put the numbers back in. Both sides should match.
16.8
=
3
×
2.8
3.3 − 2.8
✓
54.8 = 35 + 3 + 16.8 ✓
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The anomaly over a buried sphere
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Paleolatitude from magnetic inclination
Topic check
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