What are the physical upper bounds on the magnitude of an earthquake?

Question

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.)

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Notes

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 $\boxed{M \approx 25}$.

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)

or

3.6E23 Nm

which is about Mw9.7.