Chapter 4: Problem 15
Derive an expression for the temperature at the center of a planet of radius \(a\) with uniform density \(\rho\) and internal heat generation \(H .\) Heat transfer in the planet is by conduction only in the lithosphere, which extends from \(r=b\) to \(r=a\). For \(0 \leq r \leq b\) heat transfer is by convection, which maintains the temperature gradient \(d T / d r\) constant at the adiabatic value \(-\Gamma\). The surface temperature is \(T_{0}\). To solve for \(T(r),\) you need to assume that \(T\) and the heat flux are continuous at \(r=b\).
Short Answer
Step by step solution
Understand the Problem
Apply Heat Generation Equation to the Lithosphere
Solve the Integrated Equation
Integrate Again and Apply Boundary Conditions
Find Constants Using Continuity at Convection Boundary
Formulate Expression for Temperature at the Center
Final Step: Verify and Confirm Derivation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Conduction
The process can be modeled by a differential equation that describes how temperature varies with distance inside a spherical object, like a planet. This equation accounts for the heat flow between different layers of the lithosphere. It is given by:
- \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \)
In simpler terms, think of conduction as the planet trying to equalize temperatures within its layers, but the rate and efficiency of heat flow vary with depth due to factors like thickness and material properties.
Internal Heat Generation
In our problem, the internal heat generation rate is denoted by \(H\). This value significantly affects how the temperature changes through the lithosphere. The higher the internal heat generation, the warmer the planet will be, and it must be conducted or convected towards the surface.
The heat generation affects the planet's temperature profile by contributing an additional source term in the heat conduction equation:
- \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + \frac{H}{k} = 0 \)
Thermal Conductivity
In our spherical planet model, thermal conductivity (denoted as \(k\)) plays a crucial role in determining how efficiently the heat generated internally can be conducted to the planet's surface. A high thermal conductivity means the planet can easily conduct heat, resulting in a more uniform temperature distribution. Conversely, low thermal conductivity means that the planet might have significant temperature differences between its core and surface.
Knowing the thermal conductivity helps solve the heat conduction equation, as it appears in the expression for internal heat generation:
- \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + \frac{H}{k} = 0 \)
Adiabatic Gradient
The adiabatic gradient is significant because it tells us how temperature naturally changes without external influences, driven by pressure differences as you move through layers of different depths in a planet. At the boundary layer where convection gives way to conduction, the adiabatic gradient provides a condition for continuity.
In terms of mathematics, the adiabatic gradient \(-\Gamma\) is used when setting boundary conditions at the depth \(r=b\), where conduction in the lithosphere abuts convection in the core:
- \( \frac{dT}{dr} = -\Gamma \)