Isostatic Equilibrium Calculation for Water Thickness

This problem involves calculating the thickness of the water column required for isostatic equilibrium between continental and oceanic crusts. We will use the principle of isostasy, which states that the pressure exerted by the crustal column at a certain depth (compensation depth) must be equal for both continental and oceanic regions.

Applying Isostasy Principle

We assume Airy's model of isostasy, where the columns have different depths but exert equal pressure at a specific compensation depth ($D_C$). The pressure at this depth under the continental crust must equal the pressure under the oceanic crust (including the water column).

Let:

• $T_{CC}$ = Thickness of Continental Crust = $30 \text{ km}$
• $\rho_{CC}$ = Density of Continental Crust = $2.5 \text{ gm/cc}$
• $T_{OC}$ = Thickness of Oceanic Crust = $10 \text{ km}$
• $\rho_{OC}$ = Density of Oceanic Crust = $3.0 \text{ gm/cc}$
• $\rho_{W}$ = Density of Water = $1.0 \text{ gm/cc}$ (assumed standard)
• $T_{W}$ = Thickness of Water Column (to be calculated)
• $\rho_{M}$ = Density of Mantle = $3.5 \text{ gm/cc}$
• $g$ = acceleration due to gravity

The pressure condition at the compensation depth ($D_C$) is:

Pressure (Continental) = Pressure (Oceanic + Water)

$ \rho_{CC} \cdot g \cdot T_{CC} + \rho_{M} \cdot g \cdot (D_C - T_{CC}) = \rho_{W} \cdot g \cdot T_{W} + \rho_{OC} \cdot g \cdot T_{OC} + \rho_{M} \cdot g \cdot (D_C - T_{OC} - T_{W}) $

We can cancel $g$ and $D_C$ terms since they appear on both sides related to mantle pressure:

$ \rho_{CC} \cdot T_{CC} + \rho_{M} \cdot D_C - \rho_{M} \cdot T_{CC} = \rho_{W} \cdot T_{W} + \rho_{OC} \cdot T_{OC} + \rho_{M} \cdot D_C - \rho_{M} \cdot T_{OC} - \rho_{M} \cdot T_{W} $

Simplifying by removing $\rho_{M} \cdot D_C$ from both sides and rearranging:

$ (\rho_{CC} - \rho_{M}) \cdot T_{CC} = (\rho_{OC} - \rho_{M}) \cdot T_{OC} + (\rho_{W} - \rho_{M}) \cdot T_{W} $

Substituting Values and Calculating

Now, substitute the given values into the simplified equation:

$ (2.5 \text{ gm/cc} - 3.5 \text{ gm/cc}) \cdot 30 \text{ km} = (3.0 \text{ gm/cc} - 3.5 \text{ gm/cc}) \cdot 10 \text{ km} + (1.0 \text{ gm/cc} - 3.5 \text{ gm/cc}) \cdot T_{W} $

$ (-1.0 \text{ gm/cc}) \cdot 30 \text{ km} = (-0.5 \text{ gm/cc}) \cdot 10 \text{ km} + (-2.5 \text{ gm/cc}) \cdot T_{W} $

$ -30 \text{ km} \cdot \text{gm/cc} = -5 \text{ km} \cdot \text{gm/cc} - 2.5 \cdot T_{W} \text{ km} \cdot \text{gm/cc} $

Rearrange to solve for $T_{W}$:

$ 2.5 \cdot T_{W} \text{ km} \cdot \text{gm/cc} = -5 \text{ km} \cdot \text{gm/cc} + 30 \text{ km} \cdot \text{gm/cc} $

$ 2.5 \cdot T_{W} = 25 \text{ km} $

$ T_{W} = \frac{25 \text{ km}}{2.5} $

$ T_{W} = 10 \text{ km} $

Therefore, the thickness of the water column in the oceans is $10 \text{ km}$.