AgeOfTheEarthUsingHeatConduction 2025-02-20 04:54:12 2025-02-24 01:22:51 Topic % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Houstoun in [1] in 1929 estimates the age of the Earth using heat conduction and Earth temperature. Let's see how close the estimate comes to current day estimate of $4.54 \pm 0.05 10^9$ years [2]. \\ The general formula for linear flow in a semi-infinite solid is \begin{equation} v = \frac{c}{2\sqrt{\pi \kappa t}} \left[ \int_{0}^{\infty} f(\xi) e^{-\frac{(\xi - x)^2}{4 \kappa t}}\,d \xi - \int_{0}^{\infty} f(\xi) e^{-\frac{(\xi + x)^2}{4 \kappa t}} \,d \xi \right] \end{equation} For this problem let $f(\xi)$ be constant and equal to $c$, $ f(\xi) = c $. \\ Then \begin{equation} v = \frac{c}{2\sqrt{\pi \kappa t}} \left[ \int_{0}^{\infty} e^{-\frac{(\xi - x)^2}{4 \kappa t}}\,d \xi - \int_{0}^{\infty} e^{-\frac{(\xi + x)^2}{4 \kappa t}} \,d \xi \right] \end{equation} sub in $\beta = \xi - x$ \\ \begin{equation} v = \frac{c}{2\sqrt{\pi \kappa t}} \left[ \int_{-x}^{\infty} e^{-\frac{\beta^2}{4 \kappa t}}\,d \beta - \int_{+x}^{\infty} e^{-\frac{\beta^2}{4 \kappa t}} \,d \beta \right] = \frac{c}{\sqrt{\pi \kappa t}} \int_{0}^{\infty} e^{-\frac{\beta^2}{4 \kappa t}} \,d \beta, \end{equation} since $ e^{-\frac{\beta^2}{4 \kappa t}}$ is an even function of $\beta$. Write $\alpha^2=\beta^2/(4\kappa t)$; then the result takes the simpler form \begin{equation} v = \frac{2c}{\sqrt{\pi}} \int_{0}^{\frac{x}{2\sqrt{\kappa t}}} e^{-\alpha^2} \,d \alpha. \end{equation} Tables of values of this integral have been drawn up, and hence $v$ can be determined as a function of the upper limit. \\ If we descend into the earth we find that after we pass the points where the diurnal and annual variation cease to be appreciable, the temperature begins to increase. The rate of increase varies from place to place, but may be taken roughly as $1^{\circ} F$ for every 50 feet of descent for depths up to about 1 mile. This increase of temperature is easily explained on the assumption that the center of the earth is at a high temperature and that heat is flowing outwards. \\ If we assume that the earth was originally at a uniform temperature $c$ and that its surface has been always at a constant temperature zero, we can use the above result to find how long it has taken to cool. We neglect the convexity of the earth's surface. \\ We find from (3) that $$ \frac{\partial v}{\partial x} = \frac{2c}{\sqrt{\pi}} e^{-\frac{x^2}{4 \kappa t}} \frac{1}{2\sqrt{\kappa t}} = \frac{c}{\sqrt{\pi \kappa t}} e^{-\frac{x^2}{4 \kappa t}} $$ Kelvin found by the method indicated in paragraph 81 (need to reference new entry) that $\kappa$ for the material of the earth's surface has the value 400, the units of length and time being the foot and year. Assume that the earth was initially at the temperature of molten rock,\textit{ i.e.}\ about $7000^{\circ} F$. Insert the value for the gradient at the surface, namely, $\frac{dv}{dx} = 1^{\circ} F$ for every 50 feet. Then, writing $ x = 0$ in the exponential, we obtain $$ \frac{\sqrt{t}}{50} = \frac{7000}{\sqrt{\pi 400}}.$$ \textit{ i.e.}\ $$ t = \frac{7000^2 \times 50^2}{400 \pi} = 10^8 \,\, years. $$ If we write $x = 100 $ miles, $$ e^{-\frac{x^2}{4 \kappa t}} = e^{-\frac{(100 \times 5280)^2}{1600 \times 10^8}} = e^{-\frac{3}{2}}. $$ At that depth the gradient is only $\frac{1}{4 \cdot 5}$ of its surface value after $10^8$ years. We see, therefore, from the first result, that according to our assumptions $10^8$ years have elapsed since the earth was at a temperature of $7000^{\circ} F$, and we see from the second result that it is permissible to neglect the convexity of the earth's surface. The assumption throughout all the temperature change of constant conductivity, specific heat and density is, of course, open to questions. Also, the earth may have taken much longer to cool, owing to the liberation of heat due to the radioactive disintegration of some of its material. So $10^8$ vs $4.54 10^9$ is not such a bad estimate. This article is a derivative work of the public domain in [1]. \begin{thebibliography}{9} [1] Houstoun, R.A., "An Introduction to Mathematical Physics" Longmans, Green and CO., London, 1929. \\ [2] Wikipedia contributors, "Age of Earth," Wikipedia, The Free Encyclopedia (accessed February 20, 2025).\\ \end{thebibliography}