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What is the richter scale of a super-nova?

If one could measure compare it to standard earthquakes measured in logarithmic richter scale, what would be the value for a super-nova?

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  • $\begingroup$ The Richter scale has various shortcomings, so it isn't actually used much these days. Instead, other logarithmic scales, like the moment magnitude scale, are used, although the news media tend to use the old Richter name when reporting those numbers. A unit used for supernova energies is the foe, $10^{44}$ joules. An average supernova can release a foe just in radiant energy. Also see en.wikipedia.org/wiki/Orders_of_magnitude_(energy) $\endgroup$
    – PM 2Ring
    Commented Feb 27, 2019 at 7:07

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Wikipedia (and the comment from PM 2Ring) tell me that it's reasonable to assume an energy release from a supernova of about $10^{44}$ joules. The energy released by an earthquake is sometimes given as $10^{4.4+1.5M}$ where $M$ is the Richter scale measurement. This gives $44=4.4+1.5M$ or $M=26.5$.

You might want to modify this a bit depending on exactly how you want to define the "energy release" associated with an earthquake; you can see some discussion about this here.

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    $\begingroup$ That 1 foe is just for the radiant energy, the total energy released by a supernova can be quite a bit more than 1 foe, as I described here. $\endgroup$
    – PM 2Ring
    Commented Feb 27, 2019 at 8:14

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