Talk:2908: Moon Armor Index

2024-03-19T06:35:58Z

172.69.195.118:

Can someone hurry up/w the explanation?[[Special:Contributions/162.158.159.162|162.158.159.162]] 22:43, 18 March 2024 (UTC)
:Did it :) --[[User:1234231587678|1234231587678]] ([[User talk:1234231587678|talk]]) 00:16, 19 March 2024 (UTC)

According to https://sl.bing.net/kR6wrqrekg0 it would be 43.1 meters. [[Special:Contributions/172.70.174.117|172.70.174.117]] 23:17, 18 March 2024 (UTC)

Bing was wrong, it screwed up the units [[Special:Contributions/172.70.38.181|172.70.38.181]] 23:39, 18 March 2024 (UTC)!

Anyone figure out if this takes the recently-discovered moons into account? I'd expect as much but it would make a good addition to the explanation. [[Special:Contributions/172.70.131.155|172.70.131.155]] 01:39, 19 March 2024 (UTC)
:The new moon around Uranus is 8 km in diameter, and the moons around Neptune are 23 km and 14 km in diameter. The inventory of outer moons is believed to be complete down to 2 km for Jupiter, 3 km for Saturn, 8 km for Uranus, and 14 km for Neptune. And the total combined mass of smaller moons (e.g. in Saturn's rings) is also constrained.

:All these moons are round, and thus approximately ball-shaped. The volume of a 3-ball with radius r₀ is 4⁄3 πr₀³. Uranus and Neptune are also approximately ball-shaped with radii of 25,559 km and 15,299 km, respectively. (I don't know exactly how these radii are defined, but I assume optically. Uranus and Neptune don't have solid surfaces.) The volume of a spherical shell is just the difference of the outer and inner spheres, so 4⁄3 π(R³−r³) if the outer radius is R and the inner radius is r. These volumes are equal if the whole moon is converted into a spherical shell. So for Uranus, we have 4⁄3 πr₀³ = 4⁄3 π(R³−r³), where r₀ is the radius of the moon, r is the radius of Uranus, and R−r is the thickness of the shell. Solving gives R−r = ³√(r₀³+r³)−r. Plugging in r₀ = 8 km and r = 25,559 km gives R−r = 0.26 mm. If we laid it on top of the other moons instead of the "surface" of Uranus itself, it would make practically no difference. Doing the same calculation for each newly-discovered moon of Neptune gives thicknesses of 17 mm and 3.9 mm (for a total of 21 mm).

:In other words, they are tiny rounding errors. [[User:EebstertheGreat|EebstertheGreat]] ([[User talk:EebstertheGreat|talk]]) 03:17, 19 March 2024 (UTC)

I like that turning the Moon into a spherical shell coating the Earth is not definitely stated to be impossible with current technology. There's so much hedging going on I feel like I'm trapped in a maze in ''The Shining.'' [[User:EebstertheGreat|EebstertheGreat]] ([[User talk:EebstertheGreat|talk]]) 03:17, 19 March 2024 (UTC)

The formula used seems to give the instantaneous technical distance, but in reality, there would be a rate of change of the surface area of the planet as each layer of thickness x was added. Does anyone know if this is significant with the distances we are talking, or does it just turn out to be a rounding error? [[Special:Contributions/172.68.0.254|172.68.0.254]] 03:34, 19 March 2024 (UTC)
:For most, I suspect it is indeed the roundingest of rounding errors. Obviously, Earth+Moon and Pluto+(Charon+the others) would be the most ''out'', but subtending difference of area at (say) sea-level radius and sea-level plus 43km doesn't sound like much to account for.
:A=4πr², so Adif of A2-A1 would be (4πr2²)-(4πr1²) or 4π(r2²-r1²) ((which looks like you could work it out as a pythogorean calculation, i.e. model a new line-length that would go at a tangent out from r1 until it hits the endpoint of the r2 radius elsewhere ... but that's probably not useful!)).
:Given Earth at a normal 6371km (between equatorial and polar radii, to simplify as a true sphere), Earth+Moon therefore 6371+43 (using figure stated by comic), that gives ...if I've done it right... now an extra 7 million km² on top of the roughly 510 million that it normally has. An increment of 5%, by the time you start spreading your arbitrarily thin final layer (so approximate back to being 2.5% extra by volume, without actually using Eebster's alternate direct shell-volume calculation or doing an integration).
:Pluto (saying 44km of layering, as slightly more than Earth's 'pile', on its far smaller radius) isn't that much more 'off'. It would increase the surface by about 8% (so says my mental arithmatic, at least) so maybe 4% more volume than a "flat surface raised up primstically".
:(Not quite the same as "wrap a string around a tennis ball, add an inch to its length, what is its additional radius? / wrap a string around the Earth, add an inch ..." sort of thing, due to the extra dimensionality involved, but I don't feel like doing the full algebraic differentiations necessary to establish the trend of departure.).
:It certainly initially looks like the '≈'ing of the result holds fairly well under even the two most extreme examples (cases of particularly large moons-by-volume). And, at a certain point, a planet's (single largest) moon cannot be made bigger without drifting into double-planet territory (indeed, Pluto/Charon may be considerd double-dwarfs!), and then, soon after, you're switching their roles around and dismantling the 'planet' (really a moon) to armour the 'moon' (now the planet). So that probably suggests we're at our limit, with twin-binary capping our one-satellite scenarios, until you get into 'busy' N-ary systems with many not-insignificant moons but somehow an identifiable 'main body' planet in the midst of them.
:I don't think "armour the Sun with all the planets (''and'' their moons), dwarf-planets, minor-planets, random detritus, etc" will strain that relationship. Top of my head estimate is that it'd be nowhere near as high as Earth/Pluto examples, if the Oort cloud isn't oddly massive in total. But someone can correct me if I've goofed or overly hand-waved something. [[Special:Contributions/172.69.195.118|172.69.195.118]] 06:35, 19 March 2024 (UTC)

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Talk:2908: Moon Armor Index