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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 \(35 \mathrm{~km}\)

Short Answer

Expert verified
The thickness is approximately 4394 meters.

Step by step solution

01

Calculate Present Mass of Continents

The present mass of the continents is calculated using the formula for mass, which involves the volume and density of the continents:\[M_c = A_c \times h_c \times \rho_c\]Substituting the provided values:\[A_c = 1.9 \times 10^8 \text{ km}^2 = 1.9 \times 10^{14} \text{ m}^2\]\[h_c = 35 \text{ km} = 35,000 \text{ m}\]\[\rho_c = 2700 \text{ kg/m}^3\]So,\[M_c = 1.9 \times 10^{14} \text{ m}^2 \times 35000 \text{ m} \times 2700 \text{ kg/m}^3 = 1.79625 \times 10^{23} \text{ kg}\]
02

Determine Mass of Subducted Sediments

Given that the mass of subducted sediments is half of the present mass of continents, we calculate it as:\[M_s = \frac{1}{2} \times M_c = \frac{1}{2} \times 1.79625 \times 10^{23} \text{ kg} = 8.98125 \times 10^{22} \text{ kg}\]
03

Calculate Volume of Subducted Sediments

The volume of the subducted sediments can be found using its mass and density. Rearrange the formula \( M = \rho \times V \) to find:\[V_s = \frac{M_s}{\rho_s}\]Where \(\rho_s = 2400 \text{ kg/m}^3\). Substitute the known values:\[V_s = \frac{8.98125 \times 10^{22} \text{ kg}}{2400 \text{ kg/m}^3} = 3.7421875 \times 10^{19} \text{ m}^3\]
04

Find Thickness of Subducted Sediments

The thickness of the sediments, \(h_s\), can be calculated using the subducted area and volume. First, find the subducted area:\[A_s = \text{rate of subduction} \times \text{time}\]Convert 3 Gyr to seconds (using 1 Gyr = \(10^9\) years, and 1 year = \(3.1536 \times 10^7\) seconds):\[3 \text{ Gyr} = 3 \times 10^9 \text{ years} = 9.4608 \times 10^{16} \text{ seconds}\]Calculate the area:\[A_s = 0.09 \text{ m}^2/\text{s} \times 9.4608 \times 10^{16} \text{ s} = 8.51472 \times 10^{15} \text{ m}^2\]Now find the thickness:\[h_s = \frac{V_s}{A_s} = \frac{3.7421875 \times 10^{19} \text{ m}^3}{8.51472 \times 10^{15} \text{ m}^2} = 4393.74 \text{ m}\]
05

Conclusion: Thickness of Subducted Sediments

Based on the calculations, the thickness of the sediments that would have been subducted over the last 3 Gyr is approximately 4393.74 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continental Mass
The concept of continental mass refers to the total mass of the continents on Earth. It's essential in understanding geodynamics because it plays a critical role in tectonic activities. To determine the present mass, we use the formula:
  • Mass = Area \( \times \) Thickness \( \times \) Density
This allows us to consider the continental area, average thickness, and density. In the given problem, the present mass is evaluated using an area of \(1.9 \times 10^{14} \text{ m}^2\), a thickness of 35,000 meters, and a density of 2700 kg/m\(^3\). These values reflect current scientific understandings and provide a basis for calculations in geodynamics.
Understanding continental mass is crucial for reconstructing Earth's geologic past and predicting future changes.
Subduction Rate
The subduction rate is a measure of how quickly tectonic plates are sinking into Earth's mantle. It's a vital factor in the recycling of the Earth's crust and influences many geodynamic processes. In this example, the subduction rate is given as \(0.09 \text{ m}^2/\text{s}\).
Subduction zones are regions where one tectonic plate moves under another, which leads to the formation of trenches, volcanic arcs, and earthquakes. This rate is crucial for determining how much material has been subducted over time.
  • Consider how the rate can impact the amount of material recycled into the mantle
  • This helps in calculating the total volume of subducted sediments
Hence, knowing the subduction rate provides insight into past and future tectonic activities that shape the Earth's surface.
Sediment Density
Sediment density is a measure of how much mass is contained within a given volume of sediment. It is vital to geodynamics as it affects how sediments interact with tectonic processes. The density of the sediments in our problem is 2400 kg/m\(^3\).
Density helps in converting the mass of subducted sediments into volume, which is essential for further calculations in understanding subduction processes. By knowing sediment density:
  • We can calculate the volume of the sediments
  • It becomes possible to determine the thickness if the subducted area is known
Understanding sediment density is key to analyzing how sediments contribute to and are transformed by tectonic activities.
Continental Area
The continental area is the total surface area of Earth's continental crust. It plays a crucial role in geodynamic calculations involving mass and volume of continents. In the exercise, the continental area is given as \(1.9 \times 10^8 \text{ km}^2\).
Continental area is fundamental because:
  • It directly influences the calculation of continental mass
  • It provides a measure for how much land is available for geological and tectonic processes
This area is a key variable when understanding the dynamics of Earth's surface and forms a base for calculating the thickness of materials like sediments in tectonic models.
Geologic Time Scale
The geologic time scale is an essential framework for understanding Earth's history and the timing of geological events. It spans billions of years, helping us measure changes over extensive periods. In the given exercise, "Gyr" stands for gigayears, each signifying a billion years.
This long-term view is indispensable when considering processes like subduction that occur over vast timescales:
  • It allows scientists to piece together the history of Earth's tectonic activities
  • By understanding these time scales, it becomes possible to comprehend the volume and thickness of materials like sediments moved over epochs
The geologic time scale essentially bridges our knowledge of Earth's ancient past with its dynamic present and future.

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Most popular questions from this chapter

The measured declination and inclination of the paleomagnetic field in Lower Cretaceous rocks at \(45.5^{\circ} \mathrm{N}\) and \(73^{\circ} \mathrm{W}\) are \(D=154^{\circ}\) and \(I=-58^{\circ}\). Determine the paleomagnetic pole position.

Consider the TTT triple junction illustrated in Figure \(1-38 .\) This triple junction is acceptable because the relative velocity between plates \(\mathrm{C}\) and \(\mathrm{A}, \mathbf{u}_{\mathrm{CA}},\) is parallel to the trench in which plate \(\mathrm{B}\) is being subducted beneath plate \(\mathrm{C}\). The trench between plates \(\mathrm{C}\) and \(\mathrm{B}\) has an azimuth of \(180^{\circ}\) so that \(\mathbf{u}_{\mathrm{CA}}\) has an azimuth of \(0^{\circ}\); assume that \(u_{\mathrm{CA}}=\) \(50 \mathrm{~mm} \mathrm{yr}^{-1}\). Also assume that the azimuth and magnitude of \(\mathbf{u}_{B A}\) are \(315^{\circ}\) and \(60 \mathrm{~mm} \mathrm{yr}^{-1} .\) Determine the azimuth and magnitude of \(\mathbf{u}_{B C}\).

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If the area of the oceanic crust is \(3.2 \times\) \(10^{8} \mathrm{~km}^{2}\) and new seafloor is now being created at the rate of \(2.8 \mathrm{~km}^{2} \mathrm{yr}^{-1}\), what is the mean age of the oceanic crust? Assume that the rate of seafloor creation has been constant in the past.

Show that a triple junction of three transform faults cannot exist.

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