SUPERMAN SHIFTS TECTONIC PLATE
PREVENTS DEVASTATING EARTHQUAKE
This feat is often brought up in discussions, but it's rarely ever been analyzed in depth. I think we should try to figure out the kinetic energy of such a feat
First and foremost, this calculation is inspired by an older one done by Comicvine user cvdebater666, and originally credit belongs to him.
Ok, so according to the newspaper headline, Superman shifted a tectonic plate, thus preventing a devastating Earthquake. Of course this is a valid feat, as it is something that happened within the DCEU narrative. Dismissing its legitimacy would be akin to dismissing War Machine lifting a tank, or Cthulhu exploding stars, simply because they are not shown on-screen.
Of course, the headline doesn't give much information, but through simple probability and logic we can make some reasonable assumptions...
Deducing the Earthquake[]
Firstly, we need to figure out the duration in which he needed to shift the tectonic plate and thus the duration of said earthquake. And the duration we can extrapolate from the magnitude of said quake, and the magnitude we can establish through logic.
From the headline we can assume he only prevented one earthquake from happening. It doesn't state "Superman prevents yet another earthquake". As such, he likely would have only prevented a quake that would've caused major casualties, giving us an indication of the scale of said earthquake. Furthermore, it's doubtful that Superman would specifically go out of his way to stop 1 out of the about 150 earthquakes that happen every year with a magnitude of 6.5 on the Richter scale.
Likewise, we have to assume that if he went out of his way to solve this particular earthquake it would have been one with significant casualties. He's not called "Earthquake-Man" and he can keep up a regular job at the Daily Planet. It would've likely been in the range of 8.0 to 8.4 since they cause a thousand or more casualties, and roughly 10% more than one happen each year (most not in inhabited regions). So let's keep it average and say it was an 8.2.
That can give us a time-frame in which he did it, assuming he arrived there the about the moment it started. This is because we have no way to reliably predict where earthquakes are going to happen in a way that would allow Superman to fly there before it started happening. Also, reliance on foreshocks is dubious when it comes to earthquakes with a magnitude of 8.0 or higher.
The longer the earthquake the longer the time frame in which Superman is able to push one tectonic plate away. The duration of how long Superman has to prevent the quake is actually quite short as an earthquake is determined by the duration of its fault rupture. The larger the earthquake, the larger the fault length, and the longer the duration of the fault rupture.
The rupture propagates at roughly 3 km/s along the fault line. A magnitude 8.2 earthquake will have a typical fault rupture length of 200 kilometers. That means the duration of the fault rupture is 200 / 3 = 66.67 seconds. These larger earthquakes have an average fault displacement of about 15 meters.
That short duration of rupture can cause an earthquake that feels lengthy. Superman feasibly had to respond within the first few seconds of the total of 66.67 and push it back in a short time to effectively "prevent" the earthquake as it is happening. The shifting of the rupture is already happening; there is no preventing it via alleviating just the stress point by just a few inches beforehand anymore.
Also, don't complain about me using a Magnitude 8.2 earthquake. Smaller earthquakes would actually require Superman to react faster and move the plate faster in relative figures, which is then compensated by the smaller amount of fault displacement. That would result in a far higher KE at the end of the calc.
The Average Tectonic Plate[]
We need the weight of the average tectonic plate.
509,968,660 (Earth's surface area) / 15 = 33,997,910.67 km² per plate.
33,997,910.67 * 7 km (average thickness oceanic crust) = 237,985,374.69 km³ (volume oceanic crust)
3.0 gm/cm³ (density oceanic crust) = 3,000,000,000,000 * 237,985,374.69 = 713,956,000,000,000,000,000 kg for the oceanic crust portion.
(100 - 7) * 33,997,910.67 = 3,297,797,334.99 km³ for the upper mantle's portion.
3.4 gm/cm³ (density upper mantle's portion) = 3,400,000,000,000 * 3,297,797,334.99 = 11,212,000,000,000,000,000,000 kg for the upper mantle's portion.
713,956,000,000,000,000,000 + 11,212,000,000,000,000,000,000 = 11,925,956,000,000,000,000,000 kg = average weight if they were all oceanic plates, which is not the case. So now for a continental plate.
33,997,910.67 * 37.5 km (average thickness continental crust) = 1,274,921,650.13 km³ (volume continental crust)
2.7 gm/cm³ (density continental crust) = 2,700,000,000,000 * 1,274,921,650.13 = 3,442,280,000,000,000,000,000 kg for the continental crust portion.
(200 - 37.5) * 33,997,910.67 = 5,524,660,483.88 km³ for the upper mantle's portion.
3.4 gm/cm³ (density upper mantle's portion) = 3,400,000,000,000 * 5,524,660,483.88 = 18,783,845,600,000,000,000,000 kg for the upper mantle's portion.
18,783,845,600,000,000,000,000 + 3,442,280,000,000,000,000,000 = 22,226,125,600,000,000,000,000 kg = average weight if they were all continental plates.
Therefore (11,925,956,000,000,000,000,000 + 22,226,125,600,000,000,000,000) / 2 = 17,076,040,800,000,000,000,000 kg is the average weight of one tectonic plate.
Finally, Shifting the Plate[]
S, how much kinetic energy would the shifting have taken?
Velocity (If he takes 10% of the available time to safely diminish most of the kinetic energy) = 15 / (66.67 * 0.1) = 2.25 m/s
Mass = 17,076,040,800,000,000,000,000 kg
Kinetic Energy = 4.3223728274999995e+22 Joules Or just barely Country level.