Question

In the last Problem Set, we looked at Mount Kilimanjaro, and you were told to assume it was isostatically compensated. Let's re-examine that assumption to see if it makes sense. For this analysis, you will need to know that Mount Kilimanjaro is about 50 km wide. You will also need to know that the lithosphere in this region has a Young's modulus of 100GPa, a Poisson's ratio of 0.25 , and an elastic thickness of 50 km. (a) [5 points] In class, we discussed the flexural equation: rho c

​
gh=Δrhogw+D dw 4
d 4
w
​
and its solution for periodic loads w= Δrho+ g
Dk 4
​
rho c
​
h
​
where w is the lithospheric deflection, h is the topographic height above the surrounding crust (which you may recall from the last Problem Set is 4895 m),Δrho=(rho m
​
−rho c
​
) is the density contrast at the base of the crust, rho m
​
=3300 kg/m 3
is the density of the mantle, rho c
​
=2700 kg/m 3
is the density of the crust, g is the surface gravity, k=2π/λ is the wavenumber appropriate to the horizontal length scale of the load ( λ ), and D is the flexural rigidity, whose definition you can find in your notes. Assuming we can treat Kilimanjaro as a periodic load, how much lithospheric deflection should we expect under the volcano? (b) [3 points] Let C be the compensation factor, which compares lithospheric deflection to the theoretical upper limit of complete isostasy (i.e. zero rigidity in the lithosphere): C= 1+ Δrhog
Dk 4
​
1
​
When C=0 the load is completely uncompensated and when C=1, the load is fully compensated (isostatic). Show that C is the ratio of the deflection w relative to complete isostatic deflection, w 1
​
. Do this by setting D=0 in equation 2 to obtain w 1
​
. Then divide equation (2) by the expression for w 1
​
. (c) [3 points] What happens to C when λ (the load wavelength) is very large? (d) [3 points] If the mode of support transitions to isostatic at C=0.5, was it reasonable to assume that Mount Kilimanjaro is isostatically compensated? Explain in 1-2 sentences.

Answer 1

The flexural equation relates the lithospheric deflection (w) to various parameters such as the density contrast, gravity, load wavelength, and flexural rigidity.

Step-by-step explanation:

Given the equation:

w = Δρg * λ^4 / (D * c * h),

where:

Δρ = Density contrast at the base of the crust (ρ_m - ρ_c)

g = Surface gravity

λ = Wavelength of the load

D = Flexural rigidity

c = Speed of light (to account for units)

h = Topographic height

Substituting the given values and solving for w:

Δρ = 3300 kg/m^3 - 2700 kg/m^3 = 600 kg/m^3

g = 9.81 m/s^2

λ = 50 km = 50000 m

D = (E * h^3) / (12 * (1 - ν^2)), where E = 100 GPa (given Young's modulus), h = 50 km (elastic thickness), and ν = 0.25 (Poisson's ratio)

c = Speed of light

h = 4895 m

Plug these values into the equation to calculate w.

(b) The compensation factor C represents the ratio of the lithospheric deflection (w) relative to the theoretical upper limit of complete isostatic deflection (w_1) when D = 0. In other words, it measures how much the actual deflection deviates from the fully compensated state.

To show this relationship, set D = 0 in the flexural equation to find w_1. Then divide the original equation (with D ≠ 0) by w_1.

(c) When the load wavelength (λ) is very large, the denominator in the expression for C becomes very large. As a result, C approaches 0, indicating that the load becomes more uncompensated. In other words, as the load becomes large compared to the wavelength, the compensation factor decreases.

(d) If the mode of support transitions to isostatic at C = 0.5, it means that the actual deflection (w) is half of the theoretical upper limit of complete isostatic deflection (w_1). If Mount Kilimanjaro were fully isostatically compensated (C = 1), w would equal w_1. Since C ≠ 1, the assumption of complete isostatic compensation for Mount Kilimanjaro may not be entirely reasonable. The value of C suggests some level of uncompensation or deviation from fully isostatic behavior.