Given what we know about the physical mechanisms underlying earthquakes, what do the theoretical upper bounds on the magnitude of an earthquake look like? What physical phenomena impose those upper bounds?

Assume that we're talking about the "modern" Earth, i.e. from the past 4 billion years, after the Earth was fully formed. Also assume that only natural phenomena are involved - no nuclear weapons or Deep Children or anything of the sort.

If it makes a difference, I would like to know what the answer looks like in two cases:

  • No impact events are involved - the earthquake is generated by purely terrestrial phenomena
  • Arbitrarily large impact events are involved, provided that the Earth still looks "roughly the same" (defined however makes sense to you) after the impact

We have a trivial upper bound of moment magnitude ~25 for the former case, since the energy released by a magnitude-25 earthquake is roughly equal to the total mass-energy of the Earth1 - but I'm sure you folks know of tighter bounds than that. :)

(cf. the question Are Richter-magnitude 10 earthquakes possible? - but most of the answers there are from a statistical perspective, not a physical perspective.)


1 As we know, the total mass-energy content of a system is given by

$$E = \sqrt{(mc^2)^2 + (pc)^2}$$

Since the Earth's motion (both translational and rotational) is non-relativistic, we discard the momentum term as negligible and simply write $E = m_\text{Earth}c^2$. Using $m_\text{Earth} = 6 \times 10^{24}\ \mathrm{kg}$, this gives us $E = 5 \times 10^{41}\ \mathrm{J}$. Converting this to a moment magnitude using

$$M = \frac{2}{3} \log_{10} \left(\frac{E}{1\ \mathrm{N} \cdot \mathrm{m}}\right) - 2.9$$

we get $\boxed{M \approx 25}$.

  • $\begingroup$ Just to be clear, are you talking about Earth is it is currently, or Earth as it might theoretically exist? For example, are you attempting to understand the upper bounds of what a fault might be, or as they are currently? Also, please cite a source for the "total mass-energy of the Earth" claim. Thanks! $\endgroup$
    – blunders
    Commented May 14, 2014 at 17:04
  • $\begingroup$ @blunders Either way would work (as it is now, or as it could plausibly be at some point). I'll include the calculation for the mass-energy of the Earth in my post if you'd like, though I mostly mentioned that as a mild joke. $\endgroup$
    – senshin
    Commented May 14, 2014 at 20:43
  • $\begingroup$ Single event earthquakes ? I'll try to find the paper, but I remember reading that large earthquakes, Alaska region, are offset by many many smaller earthquakes, and that's because of the nature of both fault size and fault clustering that large amounts of stress are not executed in one event. $\endgroup$
    – Neo
    Commented May 14, 2014 at 21:34
  • $\begingroup$ The world building stack exchange considered What would happen to the Moon if it was lowered onto the Earth? One answer included a magnitude 17 earthquake. $\endgroup$
    – Jasper
    Commented Jan 26, 2019 at 3:13

1 Answer 1


Earthquakes involve slip on faults, most of which occur on plate boundaries.

Now the width of a fault is usually limited by the downdip limit of the brittle ductile transition. E.g., in case of vertical strike slip faults this limit is between 10-20 km. In subduction zones this limit can be larger as the fault/interface is inclined.

The theoretical length of the fault is limited by the length of the plate boundary. It is impossible to have a fault with a length greater than the circumference of earth.

The final variable is slip. Average slip, based on empirical relationships, doesn't exceed more than 5-15m (rocks can accumulate only so much strain). Though locally it is possible to have slip as high as 50m (e.g., in the case of the Tohoku earthquake.

So for a plate boundary fault which exactly splits the earths brittle upper crust into two halves (in a single earthquake) with an average slip of 15 m, assuming a modulus of rigidity of 30GPa (mean for upper crust) we have:

Moment = Fault_area x rigidity x slip

which for our hypothetical scenario is equal to

= ((2*pi*6371000)*20000)*(30e9)*(15)


3.6E23 Nm

which is about Mw9.7.

  • 1
    $\begingroup$ Do you have references for some of this? $\endgroup$
    – user889
    Commented Dec 7, 2014 at 19:26

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