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--- courses:ast401:4 [2025/07/21 01:39] – [5. Using dimensional analysis for lithosphere] asad
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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.
-===== - Modeling planetary interiors =====
-===== - Examining planetary data =====
-===== - Ploting the earth's density profile =====
-===== - Deriving the equation for temperature =====
-===== - 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 =====