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Does a 9 magnitude earthquake have a wave height of 1,000 meters?
Does this imply the earth raised upward 1,000 meters?
10 pow 9 = 1,000,000,000 microns or 1,000 meters
10 pow 9 = 1,000,000,000 microns or 1,000 meters
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Not quite.
Loosely described, the Richter assigns a numeric value to an earthquake based on the amplitude of vibrations recorded on a seismograph at a specific distance from the epicenter.
The formula is something like Magnitude = log (amplitude) + correction factor (1), where the correction factor accounts for the distance of the seismograph from the epicenter.
So, in an real situation, an amplitude of 23 mm (or 23000 microns) could result in a Richter magnitude of 5.
http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html
To get a Richter magnitude of 9, keeping the correction factor the same as in the example, requires that A be on the order of 1,000,000 mm (1,000,000,000 µm), as you suggest (pretty much off the charts for the setup given).
The problem is that this is the amplitude as recorded on the seismograph paper *not the amplitude in terms of actual earth displacement.* To translate the recorded amplitude into the *real* amplitude is a bit more complex.
For a short answer, "The magnification of Richter's seismometer was about 2,800, so one millimeter on the paper record corresponds to about 0.36 microns of actual ground motion" (2). In this framework, a magnitude 9 would correspond to 1,000,000 mm, which translates into about 360 mm of actual ground motion.
Loosely described, the Richter assigns a numeric value to an earthquake based on the amplitude of vibrations recorded on a seismograph at a specific distance from the epicenter.
The formula is something like Magnitude = log (amplitude) + correction factor (1), where the correction factor accounts for the distance of the seismograph from the epicenter.
So, in an real situation, an amplitude of 23 mm (or 23000 microns) could result in a Richter magnitude of 5.
http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html
To get a Richter magnitude of 9, keeping the correction factor the same as in the example, requires that A be on the order of 1,000,000 mm (1,000,000,000 µm), as you suggest (pretty much off the charts for the setup given).
The problem is that this is the amplitude as recorded on the seismograph paper *not the amplitude in terms of actual earth displacement.* To translate the recorded amplitude into the *real* amplitude is a bit more complex.
For a short answer, "The magnification of Richter's seismometer was about 2,800, so one millimeter on the paper record corresponds to about 0.36 microns of actual ground motion" (2). In this framework, a magnitude 9 would correspond to 1,000,000 mm, which translates into about 360 mm of actual ground motion.
source(s):
http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html
http://tremor.nmt.edu/faq/how.html
http://www.scientificamerican.com/article.cfm?id=how-was-the-richter-scale
http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html
http://tremor.nmt.edu/faq/how.html
http://www.scientificamerican.com/article.cfm?id=how-was-the-richter-scale
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Your a genius
voted helpful: bunnyphuphu, adrianarcher
No I wouldnt think so. Im not even sure how that would be possible. Either all the crust would have to do it or some of it would do it and there would be crust breaking, and you know the media would be taking pictures of that. But that is just my opinion. Im not sur if my mind is in the same area yours is right now. But the "earth raising" is somewhat vague too. I am taking it to mean literally raising which doesnt seem really possible
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