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After seeing this question, I wondered what weapon of choice (and, logically, the grand scale of energy amount) is needed to shatter the moon. Let's take our moon, the Moon, the natural satellite of Earth. As a homage to RWBY, let's say the moon is shattered similarly to theirs, i.e. about 1/4 of the visible disc is in ruins.

(I guess their image is a bit unrealistic in the torn away parts, as they would be probably propelled outside from the orbit and either burn down in the Earth atmosphere or leave the Earth-Moon system. But the point is here in "biting away" a similar part of the moon as in that visual. And the question is the specific "how to".)

Roche limit seems not to be an answer. I have seen an esitmation in the teratonnes of TNT equivalent, which is a lot. Here I am asking rather for Sci-Fi'esque (but meaningful from todays's physics point of view) weapons and tools capable of such an energy output. Asteroids made of depleted uranium traveling at some fraction of c and similar ideas are welcome.

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  • $\begingroup$ Pretty sure this question has been answered. You need enough energy to overcome gravitational binding (IE, shattering doesn't work as the shattered pieces will just recollect due to gravity...you need to 'blow' it up). en.wikipedia.org/wiki/Gravitational_binding_energy I'll try to find the previous thread. $\endgroup$ – Twelfth Nov 23 '17 at 19:49
  • $\begingroup$ Then answer is to break the gravitational binding energy of the moon. Otherwise it will just clump up again due to it's own gravity. The result will be a ring around the earth. The picture you posted is unrealistic because all the piece of the moon will all fall to their center of gravity unless there is continuous force keeping them apart. $\endgroup$ – A. C. A. C. Nov 23 '17 at 19:49
  • $\begingroup$ worldbuilding.stackexchange.com/questions/8951/… flagging as duplicate. Err...wait the duplicate I linked here was closed as off topic? worldbuilding.stackexchange.com/questions/4679/… maybe? $\endgroup$ – Twelfth Nov 23 '17 at 19:50
  • $\begingroup$ Ok, I recalled the value wrong, my estimate is off by multiple orders of magnitude. I know that the picture from that battle fantasy is unrealistic, I wanted a similar degree of destruction. But feel free to close the question as a duplicate and thanks for all the links! $\endgroup$ – Oleg Lobachev Nov 23 '17 at 19:57
  • $\begingroup$ @A.C.A.C. Rings turn into planets, planets don't turn into rings. There's no logical reason why the moon debris would be launched away from the moon's gravity well pro/retrograde to its orbit, but not in any other direction. The only way to do so would be to go to the moon, cut it up, and systematically launch the chunks retrograde (basically Hansel & Gretel with moon rocks instead of bread), and even that won't create a smooth ring of debris unless you somehow slow down the debris once it leaves the moon's gravity well (so that it follows the moon, instead of taking a wildly different orbit) $\endgroup$ – Flater Nov 24 '17 at 9:07
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To achieve that effect? Very. The venerable sd.net Asteroid Destruction Calculator puts the cratering energy at approximately 240 teratons: http://www.stardestroyer.net./Resources/Calculators/AsteroidDestruction.html

However, this answer: https://astronomy.stackexchange.com/questions/2092/what-is-the-minimum-mass-required-so-that-objects-become-spherical-due-to-its-ow Strongly suggests that such a crater wouldn't be stable, even if the debris were conveniently removed. The walls of the crater would collapse, causing the Moon to reform into a slightly smaller sphere.

In addition to your 240 teraton nuke/RKKV/whatever, you're also going to need to infill the crater to make it appear shattered. Most things we think of as transparent aren't very transparent when they're hundreds of kilometres thick, so it also has to be really high-quality transparent infill. The effect you're after isn't possible as a result of warfare, but only an elaborate mega-sculpture.

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    $\begingroup$ How about just leaving it spherical, and painting the bite. Even 1 km pixels would look pretty realistic from earth. $\endgroup$ – Sherwood Botsford Nov 23 '17 at 22:59

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