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-====== 4. Interiors of Planets ======
-By the end of this chapter, you should be able to:
-  - **Model** planetary interiors using measurements of mass, moment of inertia, $J_2$, and $\Lambda$.
-  - **Examine** available data on the interiors of the solar system planets.
-  - **Plot** Earth's density profile using seismic information and the Williams-Adams algorithm.
-  - **Derive** an equation to estimate a planet’s internal temperature from its radius, density, and heat capacity.
-  - **Use** dimensional analysis to determine the region of the lithosphere from which heat escapes.
-  - **Explain** why Earth remains hot and tectonically active using results from learning goals 4 and 5.
-The following notes have been created using a book written by **Stephen Eales**.
-===== - Modeling planetary interiors =====
-===== - Examining planetary data =====
-===== - Ploting the earth's density profile =====
-===== - Deriving the equation for temperature =====
-===== Deriving the Initial Temperature of a Planet from Gravitational Accretion =====
-When a planet forms, it gains mass by accumulating smaller bodies like planetesimals. These bodies fall into the planet’s gravitational field and release energy, which is converted into heat. If we assume that this heat is retained, we can estimate the initial temperature of the planet.
-==== Step 1: Expression for Planetary Mass ====
-Assume the planet has:
-  * Radius $R_p$
-  * Constant density $\rho$
-  * Mass $M$
-The volume of a sphere is $V = \frac{4}{3} \pi R_p^3$, so the total mass of the planet is:
-$$
-M = \frac{4}{3} \pi R_p^3 \rho
-\tag{1}
-$$
-==== Step 2: Energy Released by Accretion ====
-As mass is added gradually, the energy released during the accretion of a shell of mass $\delta M$ at radius $r$ is due to gravitational potential energy. The gravitational potential energy per unit mass at radius $r$ is:
-$$
-U(r) = -\frac{G M(r)}{r}
-$$
-where $M(r)$ is the mass already assembled within radius $r$. Since the density is constant:
-$$
-M(r) = \frac{4}{3} \pi r^3 \rho
-\tag{2}
-$$
-An infinitesimal shell at radius $r$ with thickness $dr$ has mass:
-$$
-dM = 4 \pi r^2 \rho \, dr
-\tag{3}
-$$
-So the gravitational energy released when this shell falls onto the growing planet is:
-$$
-dE = \frac{G M(r) \, dM}{r}
-= \frac{G \left( \frac{4}{3} \pi r^3 \rho \right) \left( 4 \pi r^2 \rho \, dr \right)}{r}
-\tag{4}
-$$
-Simplifying:
-$$
-dE = \frac{G \cdot \frac{4}{3} \pi r^3 \rho \cdot 4 \pi r^2 \rho}{r} \, dr
-= \frac{16}{3} \pi^2 G \rho^2 r^4 \, dr
-\tag{5}
-$$
-Now integrate this from $r = 0$ to $r = R_p$ to get the total gravitational energy released during the planet's formation:
-$$
-E = \int_0^{R_p} \frac{16}{3} \pi^2 G \rho^2 r^4 \, dr
-\tag{6}
-$$
-This is a straightforward integral:
-$$
-\int_0^{R_p} r^4 \, dr = \frac{R_p^5}{5}
-$$
-Therefore:
-$$
-E = \frac{16}{3} \pi^2 G \rho^2 \cdot \frac{R_p^5}{5}
-= \frac{16}{15} \pi^2 G \rho^2 R_p^5
-\tag{7}
-$$
-This is the total energy released and assumed to be converted entirely into heat.
-==== Step 3: Internal Energy Required to Heat the Planet ====
-The total thermal energy needed to raise the temperature of the planet from 0 K to $T$ is:
-$$
-E = C_p M T
-\tag{8}
-$$
-Where:
-  * $C_p$ is the specific heat capacity of the material (assumed uniform and constant)
-  * $M$ is the total mass from Equation (1)
-Substitute Equation (1) into Equation (8):
-$$
-E = C_p \cdot \left( \frac{4}{3} \pi R_p^3 \rho \right) \cdot T
-= \frac{4}{3} \pi C_p \rho R_p^3 T
-\tag{9}
-$$
-==== Step 4: Equating Energies and Solving for Temperature ====
-Set the gravitational energy (Equation 7) equal to the thermal energy (Equation 9):
-$$
-\frac{16}{15} \pi^2 G \rho^2 R_p^5 = \frac{4}{3} \pi C_p \rho R_p^3 T
-\tag{10}
-$$
-Cancel out common factors:
-  * $\pi$ from both sides
-  * One power of $\rho$
-  * $R_p^3$
-This leaves:
-$$
-\frac{16}{15} \pi G \rho R_p^2 = \frac{4}{3} C_p T
-\tag{11}
-$$
-Multiply both sides by 3:
-$$
-\frac{48}{15} \pi G \rho R_p^2 = 4 C_p T
-\tag{12}
-$$
-Simplify:
-$$
-\frac{16}{5} \pi G \rho R_p^2 = 4 C_p T
-\tag{13}
-$$
-Now solve for $T$:
-$$
-T = \frac{16}{5} \cdot \frac{\pi G \rho R_p^2}{4 C_p}
-= \frac{4 \pi G \rho R_p^2}{5 C_p}
-\tag{14}
-$$
-==== Final Expression for Temperature ====
-$$
-T = \frac{4 \pi G \rho R_p^2}{5 C_p}
-\tag{15}
-$$
-This equation estimates the average internal temperature a planet would reach if all gravitational energy during its formation were converted into heat and retained.
-==== Notes ====
-  * The temperature increases with the square of the radius: larger planets get much hotter.
-  * It also increases with density and decreases with specific heat.
-  * In reality, some energy would be lost via radiation during accretion, so this gives an upper bound.
-===== - Using dimensional analysis for lithosphere =====
-To estimate how deep thermal energy could have conducted through rock since the Solar System formed, we apply a simple dimensional analysis approach. Suppose heat conduction is the dominant process of energy loss from a planet’s interior. We aim to estimate the maximum depth, $L_{\text{max}}$, from which significant energy could have diffused away.
-We assume that $L_{\text{max}}$ depends on a few key physical parameters:
-  * $K_T$ – the thermal conductivity of the rock
-  * $C_p$ – the specific heat capacity
-  * $\rho$ – the density of the material
-  * $\tau$ – the age of the Solar System (roughly $4.5 \times 10^9$ years)
-Let’s assume:
-$$
-L_{\text{max}} \sim f(C_p,\, \rho,\, K_T,\, \tau)
-$$
-We now use dimensional analysis to combine these variables into a length. The units are:
-  * $[K_T] = \mathrm{W\,m^{-1}\,K^{-1}} = \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$
-  * $[C_p] = \mathrm{J\,kg^{-1}\,K^{-1}}$
-  * $[\rho] = \mathrm{kg\,m^{-3}}$
-  * $[\tau] = \mathrm{s}$
-We look for a combination of these that yields units of length ($\mathrm{m}$). The expression:
-$$
-\frac{K_T \tau}{\rho C_p}
-$$
-has units of $\mathrm{m^2}$. Taking the square root gives a quantity with units of length:
-$$
-L_{\text{max}} \sim \sqrt{ \frac{K_T \tau}{\rho C_p} }
-$$
-This gives a simple estimate of how far heat can travel by conduction in time $\tau$. Using typical rock values:
-  * $K_T \sim 3\, \mathrm{W\,m^{-1}\,K^{-1}}$
-  * $C_p \sim 1000\, \mathrm{J\,kg^{-1}\,K^{-1}}$
-  * $\rho \sim 3000\, \mathrm{kg\,m^{-3}}$
-  * $\tau \sim 4.5 \times 10^9\, \mathrm{years} \approx 1.4 \times 10^{17}\, \mathrm{s}$
-We find:
-$$
-L_{\text{max}} \sim \sqrt{ \frac{3 \times 1.4 \times 10^{17}}{3000 \times 1000} }
-\approx \sqrt{1.4 \times 10^{11}} \approx 3.7 \times 10^5\, \mathrm{m} = 370\, \mathrm{km}
-$$
-So, the maximum thermal diffusion depth is about **300–400 km**, suggesting that small bodies like asteroids may have cooled completely by conduction over the Solar System's lifetime, but larger planets could retain significant internal heat.
-===== - Explaining why earth is still hot =====
-Earth formed over 4.5 billion years ago from the gravitational collapse of planetesimals. That process released a large amount of energy, raising Earth’s initial temperature significantly. But surprisingly, even after billions of years, Earth’s interior is still hot and geologically active. Why?
-This can be explained using three key physical ideas:
-==== 1. Initial Heating from Accretion ====
-The temperature of Earth immediately after its formation is estimated by:
-$$
-T = \frac{4\pi G R_p^2 \rho}{5 C_p}
-$$
-Where:
-  * $G$ = gravitational constant
-  * $R_p$ = Earth's radius
-  * $\rho$ = Earth's average density
-  * $C_p$ = specific heat of rock
-Using typical values, this gives $T \approx 4000\,\text{K}$ — hot enough to melt rock. So, Earth started as a molten body.
-==== 2. Limited Heat Loss via Conduction ====
-Heat escapes from Earth's interior primarily by conduction through the lithosphere. The depth to which heat can conduct over time $\tau$ is given by:
-$$
-L_{\text{max}} \sim \sqrt{ \frac{K_T \tau}{\rho C_p} }
-$$
-Where:
-  * $K_T$ = thermal conductivity
-  * $\tau$ = time (age of Solar System)
-  * $\rho$ = density
-  * $C_p$ = specific heat
-This evaluates to about 300 km. Since Earth’s radius is over 6000 km, most of its interior lies **beneath** the conductive zone and remains well insulated. This means **cooling is very slow**.
-==== 3. Ongoing Heating from Radioactive Decay ====
-Earth's interior contains radioactive isotopes such as:
-  * Uranium-238 ($^{238}$U) — half-life ≈ 4.5 billion years
-  * Thorium-232 ($^{232}$Th) — half-life ≈ 14 billion years
-  * Potassium-40 ($^{40}$K) — half-life ≈ 1.3 billion years
-These decay slowly, releasing energy continuously. Because their half-lives are so long, these isotopes are **still active today**, providing a **steady internal heat source**.
-==== How This Heat Keeps Earth Alive ====
-The combination of initial heat, slow cooling, and radioactive heating powers:
-  * **Mantle convection**
-  * **Plate tectonics**
-  * **Volcanic activity**
-  * **The magnetic field** (via the core geodynamo)
-Without these heat sources, Earth would be geologically inactive — like the Moon or Mercury.
-==== Summary Table ====
-^ Source of Heat ^ Role in Earth’s Interior ^
-| Gravitational Accretion | Heated the early Earth to molten temperatures |
-| Slow Thermal Conduction | Prevents deep interior from cooling quickly |
-| Radioactive Decay | Continually replenishes internal heat |
-**→ Together, these processes explain why Earth still has a hot interior and remains geologically active.**