Editing Openai/6924d898-a68c-8009-beac-944d19ca0daf (section)

===== Toneri: =====
* Slices the Moon in half and causes both halves to move apart by a visibly large distance. Amino Apps<ref>{{cite web|title=Amino Apps|url=https://aminoapps.com/c/scanning_calcs/page/item/toneri-cuts-the-hollow-moon-in-half/Q7QP_1wHYIlVPqjENEW02gG3Xv8EW7qVRP|publisher=Amino Apps|access-date=2025-11-26}}</ref>

We’ll heavily simplify:
# Assume he gives each half an outward speed vvv.
# Mass of a half-moon: Mhalf≈12Mmoon≈3.675×1022 kgM_{\text{half}} \approx \frac{1}{2}M_{\text{moon}} \approx 3.675\times 10^{22}\ \text{kg}Mhalf≈21Mmoon≈3.675×1022 kg.

Let’s say (for a lower-bound):
* The two halves separate by about one lunar diameter (Δx≈3.5×106 m\Delta x \approx 3.5\times 10^6\ \text{m}Δx≈3.5×106 m)
* over a time Δt≈4 hours=1.44×104 s\Delta t \approx 4\ \text{hours} = 1.44\times 10^4\ \text{s}Δt≈4 hours=1.44×104 s

Average relative speed of separation:

vrel=ΔxΔt≈3.5×1061.44×104 m/sv_{\text{rel}} = \frac{\Delta x}{\Delta t} \approx \frac{3.5\times 10^6}{1.44\times 10^4} \ \text{m/s}vrel=ΔtΔx≈1.44×1043.5×106 m/s
vrel≈243 m/sv_{\text{rel}} \approx 243\ \text{m/s}vrel≈243 m/s
If both halves move away from the center, each has speed roughly:

v≈vrel2≈120 m/sv \approx \frac{v_{\text{rel}}}{2} \approx 120\ \text{m/s}v≈2vrel≈120 m/s
Now compute kinetic energy for both halves:

Etotal=2×12Mhalfv2=Mhalfv2E_{\text{total}} = 2 \times \frac{1}{2} M_{\text{half}} v^2 = M_{\text{half}} v^2Etotal=2×21Mhalfv2=Mhalfv2
Plug in:

Etotal≈3.675×1022 kg×(120 m/s)2E_{\text{total}} \approx 3.675\times 10^{22}\ \text{kg} \times (120\ \text{m/s})^2Etotal≈3.675×1022 kg×(120 m/s)2
(120)2=14400(120)^2 = 14400(120)2=14400
Etotal≈3.675×1022×1.44×104 JE_{\text{total}} \approx 3.675\times 10^{22} \times 1.44\times 10^{4}\ \text{J}Etotal≈3.675×1022×1.44×104 J
Multiply coefficients:

3.675×1.44≈5.2923.675 \times 1.44 \approx 5.2923.675×1.44≈5.292
Add exponents:

1022×104=102610^{22} \times 10^{4} = 10^{26}1022×104=1026
So:

Etotal≈5.3×1026 JE_{\text{total}} \approx 5.3\times 10^{26}\ \text{J}Etotal≈5.3×1026 J
That’s a lower bound, because:
* The moon in the movie is at least partially solid (not massless shells).
* We ignored cutting the moon (fracture/vaporization energy) and only did bulk kinetic.
* If the time is actually shorter than 4 hours (likely), vvv — and thus EEE — get much bigger (E∝v2)(E\propto v^2)(E∝v2).

So Toneri’s feat reasonably sits in:

EToneri∼1027 J (Moon-level) or higherE_{\text{Toneri}} \sim 10^{27}\ \text{J}\ \text{(Moon-level)}\ \text{or higher}EToneri∼1027 J (Moon-level) or higher