Talk:2750: Flatten the Planets

I have to wonder, would you slide down to the sun, or be flung outwards? SDSpivey (talk) 19:39, 15 March 2023 (UTC)

The discs are centered on the orbit of the parent planet, and presumably rotating at the same frequency as the parent planet's orbit. That means the inner edge of each disc is going slower than you'd need to orbit the Sun at that distance, and the outer edge faster. If you moved inward from the original planet's orbit, the Sun's gravity would pull you in, but when you crossed the boundary to the next disc, you'd get flung back outward.162.158.62.61 19:58, 15 March 2023 (UTC)

No Each planet fills out the space within their orbit into the next planet. Easy to see as the outer edge of Neptune's orbit is the same as with the planet flattened. There is a distance from Mercury to the Sun indicated. Maybe because it would melt if it got any closer? --Kynde (talk) 20:03, 15 March 2023 (UTC)

Take another look at the top and bottom images, you'll notice that the Neptune disc is significantly larger than Neptune's orbit (especially on the left hand side of the image). I suspect that, as the other response suggested Mercury and Neptune takes the inner edge of the disc as the average between Neptune and Uranus's orbital radii, and then the outer radius the same distance on the other side of Neptune's orbit. Similar for Mercury.

First, they're rings not discs, but I'm skeptical of the math. And it looks to me like the ring's edges are halfway between the orbits, with Neptune extended outwards the same distance as halfway to Uranus's orbit. 172.69.22.4 20:08, 15 March 2023 (UTC)

This makes about as much sense as other Flat Earth theories. 172.70.200.137 20:00, 15 March 2023 (UTC)

But this would actually be a flat Earth. Albeit with a rather larger surface area ;-) --Kynde (talk) 20:03, 15 March 2023 (UTC)

And..the Earth-ring is not a disc and it's also in the same plane as the sun. Meaning If you were to stand on the surface of this ring earth , there would be a perpetual sunrise / sunset... And similar for everything else in the plane of the ecliptic. Iggynelix (talk) 12:36, 16 March 2023 (UTC)

But what does the plot of surface gravity vs distance from the Sun look like? Gravity of an infinite plane and all that?--Brossa (talk) 00:01, 16 March 2023 (UTC)

The explanation currently says that it would require "several solar system's worth" of matter, but isn't there enough matter in the actual solar system? --Purah126 (talk) 00:49, 16 March 2023 (UTC)

That was said in reference to the Alderson Disk, which requires 1000km or so of thickness. Clearly more than the proposal here that gives a minute thickness (relatively) from the actual planetary mass in the solar system. Even if you reduced its extent (smaller outer, bigger hole for the Sun) it wouldn't thicken up enough. The prior (non-xkcd) version would require a mass of material rivaling, if not exceeding, that of the Sun itself. 172.70.162.222 02:07, 16 March 2023 (UTC)

One of the reasons NASA rejected this could've been the use of inches. 172.71.102.13 02:26, 16 March 2023 (UTC)

Except for Mars. I can only imagine that use of the metric system for the Mars ring is a reference to the Mars Climate Orbiter fiasco, which certainly would not endear Randall, or his proposal, to a NASA granting agency program officer. 172.70.214.150 02:45, 16 March 2023 (UTC)

I assume the use of microns there is simply because 5/512 is a really awkward fraction. 172.71.223.25 05:48, 16 March 2023 (UTC)

Awkward? Its vulgar! 172.70.162.56 08:05, 16 March 2023 (UTC)

Oh right, the Mars Climate Orbiter reference makes sense! I was wondering why Randall would mix imperial and metric units like that. No sane physicist would do that, especially not Randall. 172.71.250.88 12:52, 16 March 2023 (UTC)

And Randall rubs more salt into the wound by using "micron", when the formal/correct SI unit name is "micrometer".172.70.206.187 17:13, 16 March 2023 (UTC)

If the planets of the solar system were to become disks centered on the respective planet's current orbit, how do we deal with the different orbital eccentricities? For example, per That Other Wiki, Venus has an orbital eccentricity of 0.006772, Earth has 0.0167086, and Mars has 0.0934. Not to mention Neptune's 0.008678 and Pluto's 0.2488; Pluto's orbit actually crosses Neptune's. Surely that would cause issues with the disks? 172.71.98.5 08:33, 16 March 2023 (UTC)

Pluto isn't involved, so at least that difficulty doesn't have to be dealt with. Maybe Pluto and other dwarf planets could be used to supplement the asteroid ball bearings.172.71.242.63 10:55, 16 March 2023 (UTC)

It appears that there is enough material in the asteroid belt to do this, since a ring of asteroid ball bearings with a 1 trillion kilometer diameter where each ball bearing was a cube 1 meter by 1 meter (clearly more than enough!) would be less than 10 trillion cubic meters. Since the total mass of the asteroid belt is 10^21 kg, and the average density is around 2 g/cm^3, = 2000 kg/m^3, then the amount of matter required is 2,000*10 trillion = 2 quadrillion which is much less than 10^21. (Not sure if this is actually correct) --Purah126 (talk) 12:17, 16 March 2023 (UTC)

Ahh yes, the classic cubic-bearing. Just what we need in this planetary ring system we've created. Since Randall elects to eschew spheres for the planets, let's go all in and refuse them for the bearings as well. Bravo. ;-) Iggynelix (talk) 12:40-12:47, 16 March 2023 (UTC)

I just want to say that this line of logic really tickled my funnybone. Well done! ...I've got no other valid contribution at this time, just that. 172.70.90.101 13:11, 16 March 2023 (UTC)

That was to further overestimate the material needed, since a cube is more mass than a sphere. --Purah126 (talk) 15:09, 16 March 2023 (UTC)

Did anyone notice that this came out just after pi-day? Iggynelix (talk) 12:40, 16 March 2023 (UTC)

That's really crusty of Randall. It does explain the rolling pin. He probably also knows, and chooses (for cause) not to disclose, that pronunciation of the Greek letter as "pie" doesn't conform to modern language usage. 162.158.90.232 17:01, 16 March 2023 (UTC)

for the same reasons outlined in the link, beta and phi are also pronounced differently, and I'm pretty sure zeta, eta, theta, xi and chi are too 172.71.26.105 22:03, 16 March 2023 (UTC)Bumpf

Although all these terms were deliberately taken from ancient Greek, because that's when they were first proposed! I believe Pi was devised around 250 BC by the Greek mathematician Archimedes, and is sometimes called Archimedes' Constant. There is no reason to use more modern pronunciation, except to make jokes about "Pee Day", I suppose. Why are we talking of hypotheticals of Randall's knowledge and secret humor, anyways? Do we need such an elaborate justification to quote and link to this dude, "Amoebe"? Please wait til next time, next year!Cuvtixo (talk) 22:40, 16 March 2023 (UTC)

Mathematicians have recognized that the ratio of a circle's circumference to its diameter is a constant since around 2000 BCE, but it was only given the symbol π in 1647 CE (and even then as the *p*erimeter of the circle; π wasn't used for the ratio until 1706). So if you want the pronunciation as it was when the usage was first proposed, you're looking for how English people would have said it in the 17th and 18th centuries, not how the ancient Greeks would have said it.172.70.114.88 17:55, 20 March 2023 (UTC)

I don't know if there's a non spoilery-way to mention that there are similar ideas explored in the novel Death's End by Liu Cixin. Nedlum (talk) 13:22, 16 March 2023 (UTC)

What would even happen here? Would the rings collapse into planets again? Where will the atmospheres go? Are the rings a uniform material like rock or many small pebbles? What happens at the borders? Would i be squished or will all life still be intact? If i a squished, do i have to put up with my worst enemy next to me? Will it be like the flat skins from //All Tomorrows//? Will i die? I expect to see this in «What If 2» coming out 13th october. 172.68.134.38 14:31, 16 March 2023 (UTC)

Presumably for the gas giants to hold a disc shape, the gas would have to be encased in hollow discs made out of the solid cores. How thick would the walls be? What if we used thinner cavities to store the inner planets' atmospheres as well? And how much would the core material decompress as a result of not being a core? Promethean (talk) 18:18, 16 March 2023 (UTC)

isn't it hypothesized that they're "solid cores" are only solid because of the immense pressures? Like isn't much of Jupiter's inner core (or outer-inner core) metallic helium or something? My reading about this is outdated, but it's mentioned in the explanation that it requires "tensile strength beyond what is likely physically possible for any known form of matter." Actually I'm a little annoyed that this statement doesn't get "citation needed", rather than "planets of our solar system would not be suitable for this endeavor", because obviously if the first statement is true, the second needs no citation if the first is true, because no planet is made of unknown forms of matter. Correct?" Cuvtixo (talk) 22:19, 16 March 2023 (UTC)

Is there an element of Pluto discourse here? IIRC, one of the complaints about the "clearing your orbit" was that different metrics would 'punish' (for those who view the reclassification as a punishment) objects orbiting at a greater distance. I find the distribution of thicknesses really intriguing, and I wonder how the various dwarf planets at different parts of the system would stack up, thickness-wise. [newbyBoredAtWork]

I've replaced 'roller pin' with 'rolling pin' because a quick online search indicates that a roller pin facilitates relative motion, linear or rotary, between objects whereas a rolling pin is used to make a mass of material thinner, especially in culinary contexts.

I did an estmate of max 210m for the bearing sizes which is clearly enough even if I was off by a factor of a thousand, the proceedings are here if someone wants to put it into the main article or check for errors. Xkcdjerry (talk) 12:48, 18 March 2023 (UTC)