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)
- 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)
- 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)
My mistake, Randall's math is correct, sorry. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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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!
- 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)
172.70.162.56 08:05, 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 plants, let's go all in and refuse them for the bearings as well. Bravo.