1 pixel = 3,485km/115 = 30.3043478km (30304.3478m)
30304.3478m X 39 = 1181869.56m
1181869.56m/2 = 590934.78m
30304.3478m X 37 = 1121260.87m
1121260.87m/2 = 560630.435m
Volume as an ellipsoid.
V = 4/3πabc
= 4/3 X π X 590934.78 X 560630.435 X 560630.435
= 7.78003413e17m^3
According to this link and this one, the inner core would weigh around 12,000kg/m^3.
M = 7.78003413e17 X 12000kg
= 9.33604096e21kg
Adding GPE to that first of all.
E = 560630.435 X 9.807 X 9.33604096e21
= 5.13305118e28 joules
When the chunk of the core is dragged up, magma from each of the different layers of the planet trails behind it. This would give us a minimum timeframe for the magma to remain in such a state before it sinks back down again, right? Nothing's holding it there except inertia. So that we can use I believe.
R = (h/2) + c^2/(8h)
= (123/2) + 992^2/(8 X 123)
= 1061.56504 pixels
The radius of the Earth is 6378.1km.
1061.56504 pixels = 6378.1km
1 pixel = 6378.1km/1061.56504 = 6.00820464km
6.00820464km X 50 = 300.410232km (300410.232m)
So we have the height of the magma trail, how long will it take to fall back down again? Using this calculator, we get a timeframe of 247.60s (that doesn't take into account air resistance, but there's no air in the middle of the planet so I think it'll be fine). Once again we find a timeframe from a distance of 2900km.
T = 2900km/247.60s
= 11712.4394m/s
Divide the above by 340.29 and we get a speed of mach 34.4189938. Finally we get the kinetic energy again, once again using the mass I previously calculated of 9.33604096e21kg...
KE = (0.5)mv^2
= (0.5) X 9.33604096e21 X 11712.4394^2
= 6.40364822e29 joules
To finish up, let's add on the GPE from earlier.
E = 6.40364822e29 + 5.13305118e28
= 6.91695334e29 joules
= 165.31915248565963702 exatons
Final Results
Orochi drags up a chunk of the core (speed) = Mach 34.419
Orochi drags up a chunk of the core (energy) = 165.319 exatons
