Abekta

The Encyclopédie of CASSA

User Tools

Site Tools


courses:ast401:4

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision
Previous revision
courses:ast401:4 [2025/07/21 01:41] asadcourses:ast401:4 [2025/07/21 02:22] (current) – removed asad
Line 1: Line 1:
-====== 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 ===== 
- 
-===== - 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 ===== 
  
courses/ast401/4.1753083676.txt.gz · Last modified: by asad

Donate Powered by PHP Valid HTML5 Valid CSS Driven by DokuWiki