Question

If we assume that the current rate of subduction, \(0.09 \mathrm{~m}^{2} \mathrm{~s}^{-1}\), has been applicable in the past, what thickness of sediments would have to have been subducted in the last 3 Gyr if the mass of subducted sediments is equal to one-half the present mass of the continents? Assume the density of the continents \(\rho_{c}\) is \(2700 \mathrm{~kg} \mathrm{~m}^{-3},\) the density of the sediments \(\rho_{s}\) is \(2400 \mathrm{~kg} \mathrm{~m}^{-3},\) the continental area \(A_{c}\) is \(1.9 \times 10^{8} \mathrm{~km}^{2},\) and the mean continental thickness \(h_{c}\) is \(35 \mathrm{~km}\). A MATLAB code for the solution of this problem is provided in Appendix \(D\). An introduction to the use of MATLAB is given in Section 11.2 .

Short answer

About 19.75 km of sediments have been subducted.

Step-by-step solution

Calculate Mass of Current Continents

To find the mass of the continents, use the formula \( M_c = \rho_c \times A_c \times h_c \). The given values are:\( \rho_c = 2700 \ \text{kg/m}^3 \),\( A_c = 1.9 \times 10^8 \ \text{km}^2 = 1.9 \times 10^{14} \ \text{m}^2 \) (converted from km^2 to m^2),\( h_c = 35 \ \text{km} = 35000 \ \text{m} \) (converted from km to m).Thus, the mass is:\[ M_c = 2700 \times 1.9 \times 10^{14} \times 35000 \]Calculating this gives:\[ M_c = 1.7955 \times 10^{22} \ \text{kg} \]

Determine Mass of Subducted Sediments

The mass of subducted sediments is given as half of the present mass of the continents, i.e., \( M_s = \frac{1}{2} M_c \).Using our earlier result:\[ M_s = \frac{1}{2} \times 1.7955 \times 10^{22} \ = 8.9775 \times 10^{21} \ \text{kg} \]

Convert Subduction Rate from Area to Volume

The subduction rate given is in \( \text{m}^2/\text{s} \). To convert this to a volume rate, multiply by time (in seconds):The total subduction time is 3 billion years, which is:\[ 3 \times 10^9 \times 365.25 \times 24 \times 60 \times 60 \ = 9.46728 \times 10^{16} \ \text{s} \]

Calculate Sediment Volume Subducted

Use the rate of subduction and time to find volume subducted:\[ V = 0.09 \ \text{m}^2/\text{s} \times 9.46728 \times 10^{16} \ \text{s} = 8.520552 \times 10^{15} \ \text{m}^3 \]

Calculate Thickness of Subducted Sediments

Relate volume and density to find the thickness. The volume is linked to mass and density as \( V = \frac{M_s}{\rho_s} \).Using density \( \rho_s = 2400 \ \text{kg/m}^3 \):\[ h = \frac{M_s}{A_c \cdot \rho_s} = \frac{8.9775 \times 10^{21}}{1.9 \times 10^{14} \times 2400} \]Calculating this gives:\[ h \approx 1.975 \times 10^1 \ \text{m} \]Therefore, the thickness of sediments subducted is approximately 19.75 km.

Key concepts

Subduction

Subduction is one of the pivotal processes in geodynamics and involves the sinking of one tectonic plate beneath another.
This process primarily occurs at convergent plate boundaries, contributing to the recycling of the Earth's crust.

• As one plate dives beneath another, it takes with it sediments and rocks that have accumulated over millions of years.


• This phenomenon is responsible for some of the Earth's most dramatic seismic activities, including earthquakes and the formation of mountain ranges.

Understanding the rate of subduction is crucial, as seen in the exercise where a rate of 0.09 ext{ m}^2/ ext{s} provides insight into the mass and volume of sediments that could have been subducted over geological time scales.
Such studies help us comprehend not only the dynamics of the Earth's tectonics but also the history and evolution of continents as they experience a cycle of renewal and destruction.

Continental Mass

Continental mass is an essential element in understanding the composition and evolution of Earth's crust.
Continents consist of massive blocks of land primarily composed of granitic rocks, characterized by a lower density compared to oceanic basalts.

• In the context of the exercise, the mass of the continents is calculated using their density, area, and thickness.


• For instance, a mean density of 2700 kg/m^3 and a calculated mass of 1.7955 imes 10^{22} kg highlight the enormity of these landmasses.

The mass of these continents not only informs us about their current state but serves as a reference for understanding geological processes such as erosion and sediment deposition.
By determining the mass of subducted sediments as half of the present continental mass, we can infer the extent of material contributions to subduction zones, which plays a significant role in the tectonic reshaping of the Earth.

Sediment Thickness

Sediment thickness plays a crucial role in geodynamic studies, providing insights into past environmental conditions and tectonic activities.
In the exercise, calculating the thickness of sediments subducted over 3 billion years helps assess the geological activity at subduction zones.

• The thickness is derived by considering both the mass and density of the sediments, as well as the area they cover.


• In this case, a calculated thickness of approximately 19.75 km signifies substantial contributions to the Earth's tectonic processes.

By understanding sediment thickness, we gain knowledge about the Earth's geological history, helping to reconstruct past environments and infer the forces shaping the Earth's crust.
It is crucial not only for academic research but also for practical applications, such as resource exploration and assessing seismic risks.

MATLAB

MATLAB is a potent tool utilized in geodynamics for modeling and solving complex mathematical problems.
Its ability to handle arrays and matrices with ease makes it especially useful in simulating geological processes like those discussed in the exercise.

• The MATLAB code provided in the appendix can calculate sediment thickness, using input parameters like subduction rate and sediment mass.


• Its computational power is harnessed to perform large-scale numerical calculations, bringing efficiency and accuracy to geophysical studies.

Understanding how to use MATLAB enables students and professionals in geodynamics to engage with data driven approaches and analyze the Earth's structural characteristics more effectively.
As highlighted in Section 11.2, beginners can learn the basics of MATLAB, illustrating its significance as a versatile and essential tool in scientific computation.