What would happen if a Richter magnitude 13 earthquake occurred? I read (at http://www.tulane.edu/~sanelson/Natural_Disasters/impacts.htm; about halfway down the page) that if a comet were to ever hit Earth, it would create a Richter magnitude 13 earthquake.
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3This question would be improved if it had a link to the source stating a comet could produce a 13 magnitude earthquake. – Fred Mar 17 '16 at 23:12
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2See also this answer to Are Richter-magnitude 10 earthquakes possible? – gerrit Mar 18 '16 at 11:26
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The last entry in https://en.wikipedia.org/wiki/Richter_magnitude_scale#Energy_release_equivalents shows the energy release of a Richter 13 magnitude and compares it to an event that occurred 65 million years ago. – Mar 19 '16 at 14:27
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2Not sure, but I feel like if a comet hit earth, the resulting earthquake probably wouldn't be the worst of our worries... – naught101 Mar 22 '16 at 00:13
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1Well to start it will probably set off every other faultline on the planet under even the most modest pressure. Not to mention volcanoes, land slides, a few other marginally stable systems. – John Dec 20 '18 at 04:16
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when it did when i was young it was very scary – gn22 Nov 15 '21 at 19:46
1 Answers
It can be problematic to use an earthquake magnitude scale for such purposes. To explain why, you have to look at the different scales how they were produced. The magnitude scales are normally used related to earthquakes on earth, resulting from elastic rebound effects from fault activity. The Richter scale (local magnitude, ML) is, easily said, measured by the amplitude of a seismograph in < 100 km distance the centre. It gets inaccurate at a magnitude greater than 6.5. From a magnitude greater than ca. 8, even the teleseismic calculated surface-wave magnitude scale (MS) gets inaccurate. For greater magnitudes, the magnitude has to be modelled in a completely different way: Here, the moment magnitude scale (MW) is used. It is defined in relation to the seismic moment:
$M_{W} = \frac{2}{3} log (M_{0}) - 10.7 $
The seismic moment M0 is defined as:
$M_{0} = D *A * m$
where $D$ is mean displacement at the total fault, $A$ is the fault and $m$ the rigidity (shear strength) of the faulted rock. It can only be modelled by visible surface offset/displacement.
So, let's come back to the question: The magnitude of an earthquake is related to the length of the fault. The problem of a magnitude 13 is, that it is not possible according to this concept due to the earth's physical limitations. Keep in mind, that with one magnitude higher, a quake has about 32 times more energy. Of course, you can compare the energy for instance to that of an impact event - which is also often done. However, it is not easy to presume the intensity.
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1The moment magnitude scale is defined in: Hanks, T. C., and H. Kanamori (1979), A moment magnitude scale, J. Geophys. Res., 84(B5), 2348–2350, doi:10.1029/JB084iB05p02348. A graph showing the relationship between different magnitude scales can be seen in Fig. 2 here or here – carnendil Mar 22 '16 at 07:21
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1This seems like you could compare the energy released during an earthquake with the binding energy that the crust has around earth. Doing this would fortify your argument. – AtmosphericPrisonEscape Dec 19 '18 at 21:51