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Guests of this large space station enjoy Earth-like gravity (80-90% minimum).

How? Well the space station is so massive it has its own gravity. The disk shaped "core" is made with ultradense materials with an overall of 20-30 g/cm3 similar to Iridium or Osmium (the densest elements).

I thought that working with material with such density could provide decent gravity with relatively little amount of material so I tried crunching some numbers...

Turns out I suck at math. How massive/large does a disk with these density have to be to provide similar gravitational pull close to its surface? (Ignore edge effects and the core does not have to keep its own atmosphere, even if you can just provide some formulas it would be great since right now I don't have much time for this)

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • $\begingroup$ Just some questions, do the astronauts evolve directly on the surface of the disc ? Are they far away ? do they walk on the side of the disk ? Furthermore a real disc needs a height because we are in a 3D world, what height (thickness) do you want for your disc ? $\endgroup$ – EngelOfChipolata Aug 31 '16 at 20:08
  • $\begingroup$ @EngelOfChipolata Evolve? The astronauts are just astronauts, they came and go on spaceships, maybe some have very long shifts but mostly don't complete a biological cycle on the station. They do however "stand" on the two surfaces of the disk. The thickness can be whichever is enough to generate enough pull. $\endgroup$ – SilverCookies Aug 31 '16 at 20:16
  • $\begingroup$ Sorry, I was unclear, by "evolve" I meant "moving, dancing, living, sleeping" not the Darwin kind of things :S. So to be clear your astronauts are "walking" "on" and "under" the disc right ? And you have to be conscious that your disc needs to be a cylinder (even with a really little height) to exists, even a sheet of paper has a thickness. So I can set the thickness to 1m, only the radius of the cylinder will be calculated. Are you okay ? (I will probably just post the result with not many explanation but tomorrow I will ad more and you will set the values that you want) $\endgroup$ – EngelOfChipolata Aug 31 '16 at 20:22
  • $\begingroup$ @EngelOfChipolata ahahaha yes the do everything "onto" the disk (sorry I'm a biologist). I don't care about the thickness if it doesn't influence the surface gravity though I guess at least for structural reason it's best if it is at least a few meters. $\endgroup$ – SilverCookies Aug 31 '16 at 20:27
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    $\begingroup$ Yes it influence the gravity a lot, because what matters is the mass of your cylinder, the more height the less it needs to be large. $\endgroup$ – EngelOfChipolata Aug 31 '16 at 20:30
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A disk won't really work in the sense that you want it to. Supposing that your material was massive enough to create an appreciable gravitic field, a body would be attracted to the centre of the disk. The centre of mass needs to be directly below your astronauts; they need to walk on the edge of the disk, not the flat dorsal or ventral surfaces.

Additionally, the increase in density is only a four-fold increase, while the decrease in volume will be exponential by going from a sphere to a disk. Logically, your prospective disk will have to be an order of magnitude larger in radius than the earth in order to achieve earth-like mass (and therefore gravitational attraction).

You need some fantastic substance to make this idea viable, either to take the associated interstellar punishment and associated integrity issues that would be dealt out to a disk sized like an enormous net spread out across a solar orbit, or to have some fantastically hyperdense material that will give you the gravity you want. In the latter case, this fantastic substance is entirely under your control, and you can handwave its required properties away.

EDIT: This doesn't even address the problem of how you would prevent your disk from collapsing in on itself. There's a reason celestial bodies are shaped roughly like spheres.

EDIT: @TLW points out that for a sufficiently large disk, the gravitational field changes so that gravity can no longer be approximated by a point. Very cool, but nevertheless huge, and prone to all the problems of a system-sized megastructure.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ Very roughly, at 1m thick, your iridium disk would need to be about 4.64 million kilometers in radius. By comparison, Earth is just 6,371 Km in radius. That disk will be about two-thirds the size of the sun; in other words, massive. The gravitational force of anything else in the vicinity, like a star that it orbits or a moon that orbits it, will tear it apart, never mind the constant bombardment from comets and other astronomical bodies. $\endgroup$ – Lord Dust Aug 31 '16 at 20:55
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    $\begingroup$ " a body would be attracted to the centre of the disk" This is, perhaps surprisingly, false. The near-field of a disk is such that you'd be attracted towards the surface. It's only in the far-field that you're attracted towards the center of the disk. Relevant: tp4.rub.de/~jk/science/gravity/chapt_alderson.html $\endgroup$ – TLW Aug 31 '16 at 23:03
  • $\begingroup$ Interesting. Never heard of it, and I thought I was familiar with all the classic megastructures. Very cool. Nevertheless, it reinforces the point, in order to achieve the correct gravity, the disk would have to be on a solar system scale, and by the math presented at your link, about 100 km thick. Not feasible for a "space station", and still requires a fantastic material to accomplish. $\endgroup$ – Lord Dust Aug 31 '16 at 23:53
  • $\begingroup$ If you had a method of stabilizing neutronium you could build it on a sane scale. Mind you, said stabilization likely doesn't exist. $\endgroup$ – TLW Sep 1 '16 at 1:54
  • $\begingroup$ Would making it ultradense in that fashion mitigate the requirement for the disk to be built to solar system scale? If I parsed that article correctly, the edge effects are proportional to the thickness of the disk, not the radius, meaning that shrinking the radius will mean that a higher proportion of the disk will be subject to edge effects. Perhaps I'm not understanding it correctly. $\endgroup$ – Lord Dust Sep 1 '16 at 2:24
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I believe the space station would induce nausea due to the changing distribution of the amount of gravitational force in different directions as the astronauts walk over the surface. In the center of the disk, for example, I imagine the astronaut would not feel pulled downward as much as they would feel pulled to the edges of the disk. Any effort to fix this likely results in a space station that is spheroid.

To simplify the math, let's set aside the disk shape idea and just think about a spheroid space station made entirely out of material at the upper end of the density range you provided, 30 g/cm3.

The earth has a mass of $5.972 * 10^{24} kg$. $80\%$ of this is $4.778 * 10^{24} kg$, or $4.778 * 10^{27} g$ because there are $10^3 g / kg$. To get this much mass with a $30 g / cm^3$ material, you need:

$$ \frac{4.778 * 10^{27} g}{30 g / cm^3} = \frac{4.778 * 10^{26} g}{3 g / cm^3} = \frac{1.593 * 10^{26} g}{1 g / cm^3} = 1.593 * 10^{26} cm^3$$

In a cubic meter, there are $100^3$ cubic centimeters, and in a cubic kilometer, there are $(10^3)^3$ meters or $10^9$ meters. We end up with $1.593 * 10^{11} km^3$. The earth is approximately 1 trillion $km^3$, so our space station, if spheroid, would be about $16\%$ the size of the earth.

Smashing that sphere into a disk does things to the gravitational field that I am not qualified to answer; I can only imagine it would be very disorienting to walk across.

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    $\begingroup$ "In the center of the disk, for example, I imagine the astronaut would not feel pulled downward as much as they would feel pulled to the edges of the disk" - nope, that's not how gravity works. $\endgroup$ – user2357112 Aug 31 '16 at 23:28
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    $\begingroup$ By the way, if your space station is orbiting the Earth, having a structure that has the same gravitational force than the body you are orbiting around is may not a good idea. Because remember that if you are attracted by a body (the Earth), this body is also attracted by you. This would probably result in a change in the Earth orbit around the Sun and a huge amount of energy would be needed to your station to reach the orbital speed. $\endgroup$ – EngelOfChipolata Sep 1 '16 at 6:15
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    $\begingroup$ Sorry, but everything in this answer is wrong. As mentioned, you wouldn't feel weird on a disk (unless black-hole sized). Gravity will always pulls you toward one point. Also, as pointed in another answer, your math is wrong ; gravity decreases with the squared distance, so the same mass with half the radius will increase surface gravity four fold. For the same surface gravity with a density 5 time higher, you need 20% of earth radius (1275km), which is 0.8% of its volume and 4% of its mass. $\endgroup$ – ElderBug Sep 1 '16 at 8:44
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    $\begingroup$ @ElderBug The inverse square law for gravity applies to point masses, and is only an accurate approximation outside the bounding sphere of the object in question. Standing near the centre of the flat surface of a cylinder, the gravity is approximately inversely proportional to your height above the surface, for a sufficiently large radius cylinder, and sufficiently close to the surface. (Just pointing this out - I still disagree with the answer, and wouldn't expect any nausea.) $\endgroup$ – trichoplax Sep 1 '16 at 10:32
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    $\begingroup$ @TLW that's a nice way of picturing it. I mentioned the centre since that's where it doesn't need to be an approximation - down is down to any order there. $\endgroup$ – trichoplax Sep 1 '16 at 20:04
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Gravitation on the surface scales linearly with radius and density. The earth has a radius of 6,400 km and a density ~ 5g/cm3. Irridium and osmium have densities about 20g/cm3, thats 4 times earths density.

Therefore the radius of the sphere is 1/4 that of earth to produce normal gravity. 1,600km radius ball of osmium might be hard to find. If you want 80% gravity then 1,600*80%=1,300km radius. 4/3*Pi*r^3 tels us that the volume is 9.2 billion cubic km and as 1 billion cubic m is 1 cubic km and 1 cubic m weighs 20 tonnes. The volume is 9.2 billion billion m^3 and the density 20 million g/m^3. That gives a weight of 184 million billion billion gram's.

Iridium costs £10 per gram and osmium £5 per gram. (ballpark) This makes the project cost in the region of £1 billion billion billion or £10^27.

With a world gdp of £50 thousand billion this project could be paid for in 20 thousand billion years(assuming all funds are devoted to it and osmium prices and gdp remain constant) About when the last stars are dying out.

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    $\begingroup$ Jup. And our galaxy is not nearly old enough so that neutron stars have cooled down to temperatures where you could mine them for some really dense material. Of course you could use the guy who tries. ;-) $\endgroup$ – Karl Aug 31 '16 at 22:32
  • $\begingroup$ Not sure the market price holds anyway for quantities that large. Is that 10^27 the FOB price for an iridium planet, or just the price for some iridium in a planet-sized vault somewhere to have my name written on it so I can collect at my own expense? $\endgroup$ – Steve Jessop Sep 1 '16 at 18:23
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Given a disk of radius $R$ and thickness $T$ made out of a material of density $\rho$, the gravitational acceleration a distance $h$ directly above (or below) the center of the dist is given by the integral $$\int_0^T\int_0^R\frac{2\pi\rho G}{\sqrt{x^2+(t+h)^2}}\,dx\,dt.$$ The solution is lengthy, but what really interests us is the limit as $h\to 0$, which is $$a=2\pi\rho G\left(R \log \left(\frac{\sqrt{R^2+T^2}+T}{R}\right)+T \log \left(\sqrt{R^2+T^2}+R\right)-T \log (T)\right)$$ If we assume that the thickness of the disk is very small compared to the radius of the disk, We can make some simplifications which reduce this to $$a=2\pi\rho G\left(R(1-\log 2)+R\log(R/T)\right).$$ Anyway, here are the equations you asked for.

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If you can stabilize neutronium, or something equally dense, this will work.

Big if, however.

Otherwise, it won't work - the density simply isn't high enough to build it on a sane scale.

Build the flooring of your station out of a sheet of neutronium ~58nm in thickness and ~100m in diameter.

A cautionary note: this will mass somewhere on the order of three-quarters of a quadrillion metric tons (~7.5 * 1014 kg) - somewhere in the range of a decent asteroid (several km in radius, depending on the density of said asteroid). You had best ensure that it is safely bound to the station, lest it tear the station apart. Remember, it's 60nm thick.

Also, one had best hope that said stabilization is stable, lest it explode. That much evaporating neutronium would release on the order of 5.6*1028 J of energy. (That's somewhere around the amount of energy required to stop the Moon in its orbit around the Earth, just to give you a comparison.) That's enough to melt a cm of aluminum at 45 light-seconds away (!), if I did my math correctly.

This will give you a difference in acceleration of something like 20cm/s2 between head and feet, which may be annoying. Though you can mitigate this by making it larger.

In actuality, you'll want to vary the thickness depending on the radius, as well as spin it slightly (which reduces the total material required). Otherwise you'll get weird effects anywhere but the center of the disk. But with the spinning and thickness variation you can get something like 90 % of the disk's radius to be locally flat, which is good enough for most purposes.

Note that you could accomplish much the same thing with an array of small black holes - although they would have to be rather small in order to not have a noticeably "bumpy" field, and the required stabilization would be even more absurd for them than for neutronium (among other things, the Hawking radiation would mean that they would decay almost instantaneously unless stopped)

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  • $\begingroup$ "Hawking radiation" -- wrap it in a mirror. Anything that comes out is chucked straight back in ;-) Yeah, I know, distance of mirror from event horizon, light speed, time to evaporate, can't have a static structure "on" the event horizon, check and mate. $\endgroup$ – Steve Jessop Sep 1 '16 at 18:19
  • $\begingroup$ That's a mighty strong and efficient mirror you've got there... $\endgroup$ – TLW Sep 1 '16 at 18:33
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    $\begingroup$ I just have to make sure the "mirror" is massive enough that all light paths point back down towards the original black hole. Oh hang on, I've just invented a bigger black hole. Never mind, add another mirror. $\endgroup$ – Steve Jessop Sep 1 '16 at 18:41
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Robert L Forward wrote about using ultra dense materials for controlling and manipulating gravity, but when he was speaking of ultra dense materials the subject was the sort of degenerate material from White Dwarf stars or neutronium as you would find in a Neutron star. One example in his book "Future Magic" suggests that to nullify a 1g field you could take a 4 million ton mass and compress it to a sphere 32cm in diameter, or a disc 45cm in diameter and 10 cm thick.

Forward was always a bit playful, so some of his suggestions included supporting a disc of degenerate matter with massive diamond pillars to counteract the gravity of Earth under the disk. This would make for an amazing amusement park ride (although the ticket price would be a bit steep). Small matters like how to keep degenerate matter or neutronium from expanding at close to the speed of light outside of their environments were conveniently overlooked.

At any rate, many other posters provided the mathematical tools, so simply plug in the desired "g" field to find out what unnatural material the gravity disc will have to be made from.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • $\begingroup$ Excuse me, "a 4 million mass" Tons? or what? Robert L Forward also put similar ideas in his INDISTINGUISHABLE FROM MAGIC (1995). Both books are veritable gold mines of speculation for SF writers. $\endgroup$ – a4android Sep 1 '16 at 7:34
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    $\begingroup$ Osmium is merely a dense material. Ultradense really does require densities of white dwarf matter or neutronium. This kind of ultradense matter would allow for a disc or plate of the right size for a space station. Charles Sheffield also had ultradense discs in spaceships to compensate for high-acceleration travel. Conceptually plausible, but currently practical impossibilities. $\endgroup$ – a4android Sep 1 '16 at 7:42
  • $\begingroup$ @a4android: I suppose spheres (or spinning spheroids) should still be easier than discs. With a sphere you "just" have to contain your neutron star so that the people walking about on it (well, near it) stay in the "habitable zone" where gravity is tolerable. For a disc you also have to somehow resist its very strong preference to collapse into a sphere, so you need some sort of structural support. Hand-wavium scaffold. $\endgroup$ – Steve Jessop Sep 1 '16 at 9:54
  • $\begingroup$ @SteveJessop Good idea. The notion of spinning discs flashed into my mind. But the spin rate might be relativistic. Thucydides identified that ultradense matter will rapidly collapse into spheres. Perhaps a component of exotic matter is needed to stabilize the UDM. $\endgroup$ – a4android Sep 1 '16 at 10:22
  • $\begingroup$ @SteveJessop Actually small quantities of neutronium would absurdly violently explode, not collapse. A common misconception. Neutronium is not stable at "low" pressures. $\endgroup$ – TLW Sep 1 '16 at 17:43

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