Question

Assume that the continental crust and lithosphere have been stretched by a factor \(\alpha=2\). Taking \(h_{c c}=35 \mathrm{~km}, y_{L 0}=125 \mathrm{~km}, \rho_{m}=3300\) \(\begin{array}{lllll}\mathrm{kg} \mathrm{m}^{-3}, & \rho_{c c}=2750 & \mathrm{~kg} \mathrm{~m}^{-3}, & \rho_{s}=2550 & \mathrm{~kg} \mathrm{~m}^{-3}\end{array}\) \(\alpha_{v}=3 \times 10^{-5} \mathrm{~K}^{-1}\), and \(T_{1}-T_{0}=1300 \mathrm{~K},\) deter- mine the depth of the sedimentary basin. What is the depth of the sedimentary basin when the thermal lithosphere has thickened to its original thickness?

Short answer

The sedimentary basin depth is approximately 12.83 km.

Step-by-step solution

Calculate Initial Thermal Lithosphere Thickness

To calculate the thermal lithosphere thickness before stretching, use the formula \( y_{L} = \frac{y_{L0}}{\alpha} \). Here, \( y_{L0} = 125\, \text{km} \) and \( \alpha = 2 \). Thus, \( y_L = \frac{125}{2} = 62.5\, \text{km} \).

Determine Initial Crustal Thickness After Stretching

The continental crust thickness after stretching by a factor of \( \alpha \) can be calculated using \( h_c = \frac{h_{cc}}{\alpha} \). Given \( h_{cc} = 35\, \text{km} \), so, \( h_c = \frac{35}{2} = 17.5\, \text{km} \).

Calculate the Subsidence Due to Density Differences

Using the formula for sedimentary basin depth, \( D = (\rho_{m} - \rho_{c c})\,(h_{c c} - h_c) / (\rho_{m} - \rho_{s}) \), substitute the values: \( \rho_{m} = 3300\, \text{kg/m}^{3} \), \( \rho_{cc} = 2750\, \text{kg/m}^{3} \), \( \rho_{s} = 2550\, \text{kg/m}^{3} \), \( h_{c c} = 35\, \text{km} \), \( h_c = 17.5\, \text{km} \). \[ D = \frac{(3300 - 2750) \times (35 - 17.5)}{3300 - 2550} = \frac{550 \times 17.5}{750} = \frac{9625}{750} \approx 12.83\, \text{km} \]

Consider Thermal Lithosphere Expansion

Calculate the change in lithosphere due to thermal expansion using \( \Delta_y = \alpha_v (T_1 - T_0) y_L \), where \( \alpha_v = 3 \times 10^{-5}\, \text{K}^{-1} \), \( T_1 - T_0 = 1300\, \text{K} \), and \( y_L = 62.5\, \text{km} \). \[ \Delta_y = 3 \times 10^{-5} \times 1300 \times 62.5 = 2.4375\, \text{km} \].

Determine Total Lithosphere Thickness

The total thickness after thermal expansion is calculated as: \[ y_{LTotal} = y_L + \Delta_y = 62.5\, \text{km} + 2.4375\, \text{km} = 64.9375\, \text{km} \]

Recalculate Sedimentary Basin Depth

If the thermal lithosphere returns to its original thickness, the sedimentary basin depth will again be \( h_{s0} \), or 0, as no stretching occurred. Otherwise, with stretching, re-calculate as in Step 3. Here, continue assuming the aforementioned expansions.

Key concepts

Continental Crust Stretching

Stretching of the continental crust is a fascinating geodynamic process that plays a key role in forming sedimentary basins. When the crust is stretched by a factor, often denoted as \( \alpha \), it leads to thinning of the crust. In our example, with \( \alpha = 2 \), the original crust thickness \( h_{cc} \) of 35 km is halved, resulting in a new thickness \( h_c \) of 17.5 km.
This process is crucial as it results in extended areas where sediment can accumulate, shaping the topography significantly.

• The amount of stretching determines the amount of thinning.
• Stretched regions are prone to subsidences, leading to basin formation.
• This mechanism is a primary driver for subsidence and subsequent sedimentation.

Understanding the factors involved in crustal stretching helps geologists predict geological formations and natural resource deposits.

Thermal Lithosphere Thickness

The lithosphere, comprising the crust and the upper part of the mantle, changes thickness due to temperature variations. Initially, its thickness can be calculated by adjusting for stretching using the formula \( y_{L} = \frac{y_{L0}}{\alpha} \). In this context, before stretching, with \( y_{L0} = 125 \) km and \( \alpha = 2 \), its thickness becomes 62.5 km.
Thermal lithosphere thickness is important because:

• It influences the mechanical behavior of the crust.
• Thicker lithospheres are typically more rigid, impacting tectonic activity.
• The thickness affects heat flow and, consequently, the surface heat gradient.

In scenarios where the lithosphere thickness returns to its original value, it indicates the absence of thermal or mechanical changes.

Density Differences in Lithosphere

Density variations within the lithosphere mainly result from compositional differences between its layers. The mantle \( \rho_m \) is usually denser than the continental crust \( \rho_{cc} \) or sediments \( \rho_{s} \). This density contrast creates differential buoyancy, which leads to subsidence or uplift. In our example, these values are \( 3300 \), \( 2750 \), and \( 2550 \) kg/m³, respectively.
The differences in density are fundamentally responsible for the formation of sedimentary basins, aiding in:

• Increasing the potential for oil and gas reserves.
• Enhancing the understanding of basin dynamics.
• Providing critical insights into past geological transformations.

Calculating the subsidence depths with these densities allows for predictions of basin evolutions over geological timescales.

Thermal Expansion in Geodynamics

Thermal expansion is a critical concept in geodynamics, describing how materials expand when heated. In the lithosphere, this entails the change in thickness due to temperature changes. Using the thermal expansion coefficient \( \alpha_v = 3 \times 10^{-5} \) K⁻¹, along with a temperature differential \( T_1 - T_0 = 1300 \) K, the change in lithospheric thickness can be calculated. For instance, with a pre-stretch thickness \( y_L = 62.5 \) km, thermal expansion adds an additional 2.4375 km.
This phenomenon impacts:

• The stress distribution within tectonic plates.
• Temperature-induced deformational processes.
• Long-term stability of the crust during temperature fluctuations.

Recognizing thermal expansion's influence is key to understanding past and present geodynamic behavior.