Understanding Isostasy and Crustal Thickness

Isostasy is a fundamental concept in geology and geophysics, describing the 'floating' equilibrium of the Earth's lithosphere (crust and upper mantle) on the denser, more fluid asthenosphere below. This principle, particularly as described by Airy's hypothesis, suggests that thicker areas of the crust float higher, while thinner areas float lower, much like icebergs in water. The depth to which the crust extends downwards into the mantle (the 'root') depends on its thickness and density relative to the underlying mantle.

Key factors influencing this equilibrium include:

• Crustal Thickness (T): Thicker crust exerts more pressure but also has more buoyant support.
• Crustal Density (ρ_c): Continental crust is generally less dense than oceanic crust.
• Mantle Density (ρ_m): The density of the material the crust is floating within.

The problem asks us to consider a scenario where the crustal thickness is doubled, with no erosion occurring, and determine the percentage by which the Moho boundary (the base of the crust) is lowered due to isostatic adjustment.

Calculating Moho Lowering with Doubled Crust Thickness

Let's denote the initial crustal thickness as T₁ and the final thickness after doubling as T₂. According to the problem statement, T₂ = 2T₁.

We can use a simplified model of Airy isostasy based on pressure balance at the Moho. Assuming uniform densities for the crust (ρ_c) and mantle (ρ_m), and considering the thickness T above a certain reference level and the root depth D below it, a simplified equilibrium condition can be represented as:

ρ_c × T = ρ_m × D

This implies that the depth of the Moho boundary (D) is proportional to the crustal thickness (T):

D = T × (ρ_c / ρ_m)

Let D₁ be the initial Moho depth and D₂ be the final Moho depth.

Initial State: D₁ = T₁ × (ρ_c / ρ_m)

Final State (Crust thickness doubled): T₂ = 2T₁ D₂ = T₂ × (ρ_c / ρ_m) = (2T₁) × (ρ_c / ρ_m) = 2 × D₁

The amount by which the Moho boundary is lowered is the change in depth:

ΔD = D₂ - D₁ = 2D₁ - D₁ = D₁

So, the Moho boundary is lowered by an amount equal to the original Moho depth (D₁).

Determining the Percentage Lowering

The question asks "how much percentage of crust would be used to lower the Moho boundary". This can be interpreted as finding the ratio of the Moho boundary lowering (ΔD) to the *original* crustal thickness (T₁), expressed as a percentage.

Percentage = (ΔD / T₁) × 100%

Substituting ΔD = D₁ and D₁ = T₁ × (ρ_c / ρ_m):

Percentage = ( T₁ × (ρ_c / ρ_m) / T₁ ) × 100%

Percentage = (ρ_c / ρ_m) × 100%

Using typical densities for continental crust (ρ_c ≈ 2700 kg/m³) and mantle (ρ_m ≈ 3300 kg/m³):

Percentage ≈ (2700 / 3300) × 100% ≈ 0.818 × 100% ≈ 81.8%

This value is approximately 80%. Therefore, doubling the crustal thickness causes the Moho boundary to lower by an amount equivalent to about 80% of the original crustal thickness, due to isostatic adjustment.