Gravity Anomalies: Calculate Isostatic Root Thickness

This solution outlines the steps to determine the thickness of the isostatic root using given gravity anomaly information and density parameters.

Given Information:

• Elevated land mass thickness: $h = 1.0 \text{ km}$
• Density of elevated land mass: $\rho_l = 2.7\text{ gm/cc}$
• Relationship between anomalies: $FAA = 0.5 \times BA$
• Density contrast at crust-mantle boundary: $\Delta\rho = 0.3\text{ gm/cc}$

Deriving Bouguer Anomaly (BA)

The Bouguer anomaly ($BA$) corrects for the gravitational effect of the overlying mass (topography), unlike the Free Air anomaly ($FAA$). The relationship is approximated as:

$ BA \approx FAA + 2\pi G \rho_l h $

Using the given condition $FAA = 0.5 \times BA$:

$ BA \approx 0.5 \times BA + 2\pi G \rho_l h $

Rearranging the equation to solve for $BA$:

$ 0.5 \times BA \approx 2\pi G \rho_l h $

$ BA \approx 4\pi G \rho_l h $

Relating BA to Root Thickness (R)

In isostasy, the Bouguer anomaly ($BA$) reflects the gravitational effect of deeper density structures, such as the compensating root. This effect is approximated by:

$ BA \approx 2\pi G \Delta\rho R $

where $\Delta\rho$ is the density contrast at the crust-mantle boundary and $R$ is the root thickness.

Calculating Root Thickness (R)

Equating the two expressions derived for $BA$:

$ 4\pi G \rho_l h = 2\pi G \Delta\rho R $

Simplify the equation:

$ 2 \rho_l h = \Delta\rho R $

Solve for $R$:

$ R = \frac{2 \rho_l h}{\Delta\rho} $

Final Calculation:

Substitute the given values:

• $\rho_l = 2.7\text{ gm/cc}$
• $h = 1.0\text{ km}$
• $\Delta\rho = 0.3\text{ gm/cc}$

$ R = \frac{2 \times (2.7\text{ gm/cc}) \times (1.0\text{ km})}{0.3\text{ gm/cc}} $

$ R = \frac{5.4}{0.3} \text{ km} $

$ R = 18.0 \text{ km} $

The thickness of the root is 18.0 km.