Talk:2908: Moon Armor Index

Can someone hurry up/w the explanation?162.158.159.162 22:43, 18 March 2024 (UTC)

Did it :) --1234231587678 (talk) 00:16, 19 March 2024 (UTC)

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

Bing was wrong, it screwed up the units 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. 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. 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. 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? 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 A_dif 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 prismatically".

(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 considered 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. 172.69.195.118 06:35, 19 March 2024 (UTC)

I'm glad there are at least links to them, but shouldn’t there be at least ONE sentence HERE on explainxkcd saying what the heck the last five ‘worlds’ are? I’d bet that’s what most people needing an explanation come here to find out! and all there are are links. 162.158.186.98 09:59, 19 March 2024 (UTC)