Food for thought

# How far did the earth move? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Energy solution

The magnitude scale is based on horizontal displacement, $M_L=\log_{10}\left(\frac{A}{A_0}\right) \text{, where } A_0=\quantity{1}{\mu m} .$

But the destructive power of an earthquake is better indicated by the amount of energy dissipated. It has been suggested that the energy, $E$, is related to the displacement, $A$, by a simple power law, $E=E_0 \left(\frac{A}{A_0}\right)^p$ where $E_0$ and $p$ are constants.

Can you use the following data from the British Geological Survey and the US Geological Survey to verify this suggestion and find values for $E_0$ and $p$?

Earthquake Magnitude, $M_L$ Energy, $\quantity{E}{(GJ)}$
San Francisco, USA, 1906 $7.8$ $6\times10^7$
Chile, 1960 $9.5$ $1\times10^{10}$
Kobe, Japan, 1995 $6.8$ $1\times10^6$
Kent, 2007 $4.3$ $200$
Lincolnshire, 2008 $5.2$ $3000$
The given data shows how $E$ changes with $M_L$, so first we need to use the formulae above to find a relationship between these two. Using logarithms on the suggested power law, then combining with the definition of $M_L$, \begin{align*} \log_{10}E&=\log_{10}\left[ E_0 \left(\frac{A}{A_0}\right)^p \right]\\ &=\log_{10}{E_0}+p\log_{10}{\left(\frac{A}{A_0}\right)}\\ &=\log_{10}{E_0}+p M_L . \end{align*}

If we plot a graph of $\log_{10}E$ against $M_L$, this should give a straight line with gradient $p$ and intercept $\log_{10}{E_0}$.

Indeed the points do lie roughly on a straight line, so we can confirm the suggested relationship. Reading the gradient and intercept from the graph, we find \begin{align*} p&=1.5\\ \log_{10}{E_0}&=-4.2\\ E_0&=10^{-4.2}\approx \quantity{6.3\times10^{-5}}{GJ}\\ &=\quantity{63}{kJ} . \end{align*}

This value of $p$ agrees with Charles Richter’s empirical model that states $E\propto A^{\frac{3}{2}}$. A consequence of this is that an increase of $2$ in magnitude corresponds to an increase of energy by a factor of $1000$.

Estimate the energy dissipation of the 2011 New Zealand earthquake, magnitude $6.1$.

This can either be read from the graph, or calculated using the formula which can now be rewritten as \begin{align*} E&=E_0\times10^{\tfrac{3M_L}{2}}\\ &=63\times10^{\tfrac{3\times6.1}{2}}\\ &\approx\quantity{8.9\times10^{10}}{kJ}\\ &=\quantity{8.9\times10^{4}}{GJ} . \end{align*}

This is a little more than the energy released by the atomic bomb that was dropped on Hiroshima in 1945.