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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 =====
-When a planet forms, it gains a large amount of gravitational potential energy as material collapses inward. A portion of this energy is converted into heat, raising the temperature of the forming body. In this section, we estimate the temperature $T$ a planet would reach if all this energy were used to heat the planet’s interior.
-==== Step 1: Total Heat Released by Accretion ====
-We assume the planet formed through the gradual accretion of smaller bodies called planetesimals. As each piece of matter falls onto the growing planet, its gravitational potential energy is converted into thermal energy.
-Let:
-  * $R_p$ be the final radius of the planet
-  * $\rho$ be its uniform density
-  * $M$ be its total mass
-The mass of a sphere is:
-$$
-M = \frac{4}{3} \pi R_p^3 \rho
-\tag{1}
-$$
-To calculate the total gravitational energy released as the planet forms from radius 0 to $R_p$, we integrate the incremental gravitational energy for a shell of thickness $dr$:
-$$
-E = \int_0^{R_p} \frac{G (4 \pi r^2 \rho)(\frac{4}{3} \pi r^3 \rho)}{r} dr
-$$
-This simplifies to:
-$$
-E = \int_0^{R_p} \frac{G 16 \pi^2 \rho^2 r^4}{3} dr
-$$
-Evaluating the integral:
-$$
-E = \frac{16 \pi^2 G \rho^2}{3} \int_0^{R_p} r^4 dr = \frac{16 \pi^2 G \rho^2}{3} \cdot \frac{R_p^5}{5}
-$$
-So the total energy is:
-$$
-E = \frac{16 \pi^2 G \rho^2 R_p^5}{15}
-\tag{2}
-$$
-==== Step 2: Relating Heat to Temperature ====
-Now we relate the gravitational energy $E$ to the internal energy needed to raise the temperature of the planet. If $C_p$ is the specific heat capacity (assumed constant) and $T$ is the resulting temperature, then:
-$$
-E = C_p M T
-\tag{3}
-$$
-Substitute Equation (1) and (2) into (3):
-$$
-\frac{16 \pi^2 G \rho^2 R_p^5}{15} = C_p \cdot \left( \frac{4}{3} \pi R_p^3 \rho \right) \cdot T
-$$
-Canceling common terms and solving for $T$:
-$$
-T = \frac{4 \pi G \rho R_p^2}{5 C_p}
-\tag{4}
-$$
-Thus, the estimated temperature of a newly formed planet is:
-$$
-T \simeq \frac{4 \pi G \rho R_p^2}{5 C_p}
-$$
-==== Step 3: Interpretation ====
-This formula gives a rough estimate of the planet’s temperature assuming all gravitational energy is retained as heat. It shows that larger and denser planets tend to be hotter upon formation. However, in reality, some of this heat may be lost through radiation during accretion, especially if the planet forms over an extended period.
-===== - 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 =====