Question

Using Airy's hypothesis, calculate the thickness of the root beneath a 4km high mountain in isostatic equilibrium with a 40km thick continental crust of density 2800kg/m³ and a mantle of density 3300kg/m³. Express your answer in km. (Round off to one decimal place)

Answer 1

Correct Answer: 22.4

$\text{1. Define Variables and Airy's Hypothesis}$

Airy's hypothesis states that isostatic compensation occurs due to vertical movements of crustal columns of varying thicknesses, all of which have the same density ($\rho_c$), but float on a higher-density, uniform substratum (mantle, $\rho_m$).

The pressure at the compensation depth ($P_c$) must be equal for two columns: the Mountain Column ($M$) and the Normal Column ($N$).

$\text{Variables}$

Height of the Mountain ($h$): $4 \text{ km}$

Normal Crustal Thickness ($T$): $40 \text{ km}$

Crustal Density ($\rho_c$): $2800 \text{ kg/m}^3$

Mantle Density ($\rho_m$): $3300 \text{ kg/m}^3$

Root Thickness ($r$): The unknown thickness of the crustal root extending below the normal crustal thickness. (This is the required value).

$\text{Pressure Equation}$

The pressure exerted by a column of rock is $P = \rho g H$, where $\rho$ is density, $g$ is gravity, and $H$ is thickness. Since $g$ cancels out, we equate the mass per unit area.

$$\text{Pressure}_{\text{Mountain}} = \text{Pressure}_{\text{Normal}}$$

$$P_M = P_N$$

The mountain column pressure is the sum of pressure from the mountain above the normal crustal surface ($h$), the normal crustal thickness ($T$), and the root ($r$):

$$P_M = \rho_c (T + r) + \rho_m (X)$$

Where $X$ is the thickness of the mantle column beneath the mountain.

It is simpler to equate the mass above the compensation depth $C$ (which is $T+r$ deep in the mountain column):

Mountain Column Mass: (Crust thickness $T+r$)

$$M_M = \rho_c (T + r)$$

Normal Column Mass: (Normal Crust thickness $T$) + (Mantle thickness $r$)

$$M_N = \rho_c T + \rho_m r$$

Equating the pressures ($P_M = P_N \implies M_M = M_N$):

$$\rho_c (T + r) = \rho_c T + \rho_m r$$

$$\rho_c T + \rho_c r = \rho_c T + \rho_m r$$

This equation is incorrect because it implies $0 = (\rho_m - \rho_c)r$, which is only true if $r=0$. The correct way to set up the isostatic balance is to consider the mass of the mountain column above the compensation depth versus the mass of the normal column above the same depth.

$\text{Correct Airy Balance}$

The pressure at the compensation depth ($T + r$) must be equal in both columns.

The mountain's topography $h$ is supported by the root $r$. The vertical displacement $h$ above the normal crustal level is compensated by the vertical displacement $r$ below the normal crustal level.

$$P_{\text{Topography}} + P_{\text{Normal Crust above } T} + P_{\text{Root below } T} = P_{\text{Normal Crust}} + P_{\text{Mantle at depth}}$$

It is much simpler to use the geometric relationship derived from the balance of buoyancy forces:

$$\rho_c h = (\rho_m - \rho_c) r$$

This formula directly relates the excess mass column (mountain height $h$, density $\rho_c$) to the deficit mass column (root thickness $r$, density difference $\rho_m - \rho_c$).

$\text{2. Calculate the Root Thickness } r$

Rearrange the equation to solve for $r$:

$$r = h \left(\frac{\rho_c}{\rho_m - \rho_c}\right)$$

$\text{3. Substitute Values}$

$h = 4 \text{ km}$

$\rho_c = 2800 \text{ kg/m}^3$

$\rho_m = 3300 \text{ kg/m}^3$

$$r = 4 \text{ km} \times \left(\frac{2800 \text{ kg/m}^3}{3300 \text{ kg/m}^3 - 2800 \text{ kg/m}^3}\right)$$

$$r = 4 \text{ km} \times \left(\frac{2800}{500}\right)$$

$$r = 4 \text{ km} \times \frac{28}{5}$$

$$r = 4 \text{ km} \times 5.6$$

$$r = 22.4 \text{ km}$$