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 | \(5.2\) | \(3000\) |

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