If you surround the equator with a continuous Niobium Tin superconductor ring, and ran somewhere near but less than the maximum current density through, the magnetic field of the Earth would support the ring at low Earth orbit. Could such a ring substitute for a space elevator and space station?

Obviously there are no launch costs associated to such a thing, you can just turn on the current and let it lift itself into orbit. The Earth's rotation would supply the required lift.

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    $\begingroup$ I like the idea, but I can't help imagining this: "Hi, we'd like to build a huge structure right through the middle of your equatorial country. Don't worry, it's only temporary, it'll lift off into space when we switch it on. Of course, it'll come straight back down when we switch it off, but don't worry, we're not planning to do that..." $\endgroup$ – Nathaniel Sep 1 '12 at 11:37
  • $\begingroup$ My guess is that the required lift due to gravity will be hard to overcome with the small magnetic field from the earth but all issues put aside, it might not be stable due to Earnshaw's theorem. $\endgroup$ – Alexander Sep 1 '12 at 13:08
  • $\begingroup$ @Nathaniel: I don't think this is hard, considering the potential tourism dollars. You can also slowly spin up a whole-Earth flywheel once in space to give you a centrifugal lift, and then it's never coming down. The construction through the oceans is a serious problem, since it will have to be on buoys. $\endgroup$ – Ron Maimon Sep 1 '12 at 16:37
  • $\begingroup$ @Alexander: I specified the material and the current in the question, and it overcomes it's own weight. The biggest obstacle will probably be "not enough He on Earth to cool the magnets". $\endgroup$ – Ron Maimon Sep 1 '12 at 16:38
  • $\begingroup$ I much prefer the launch loop idea :) Anyway, @Alexander Earnshaw's theorem holds for static magnets, you can easily overcome it by continuously adjusting the magnetic field strength and/or orientation (see the levitating globe for example. And "centrifugal lift"; although every part of the ring will be in orbit, I think the whole contraption would be inherently unstable...nothing uncorrectable probably, but catastrophic power failure would make that less attractive...Let me do some calculations. $\endgroup$ – Rody Oldenhuis Sep 2 '12 at 6:53

I don't understand why we can't address this in a very simple Physics 101 way. The direction of the field and current seem to work out alright, I mean, they're in the right direction. I think you could do this and get a force upward. The equation for a wire in a magnetic field with a current is:

$$ F = ILB$$

Here we have the force equal to the current times length times magnetic field. The force from gravity is trivial.

$$ F = \lambda L g $$

$$ ILB = \lambda L g $$

Here $\lambda$ is the linear mass density, then we have the length and gravity. We only need a little manipulation of these equations. If the device is working, these two forces will be equal. I'm going to use a quote from the Wikipedia article on Niobium-tin.

In April 2008 a record non-copper current density was claimed of 2643 A/mm² at 12 T and 4.2 K [1]

This gives us a current density. I will introduce $\phi$ to denote this quantity, and it will be defined as such.

$$\phi = \frac{I}{A}$$

We can also get a compatible representation for the linear mass density trivially by using the actual volumetric density.

$$ \rho = \frac{\lambda}{A}$$

Divide the previous equation by length and area to get:

$$ \phi B = \rho g $$

We have values for all of these.

  • magnetic field is 25,000 to 65,000 nT, so I'll use 50,000 nT
  • density of Niobium is 8.57 g/cm^3, I'll call that good enough

$$ \phi B = (2643 \frac{A}{mm^2}) (50 000 \times 10^{-9} \text{Tesla}) = 132,150 \frac{kg}{m^2 s^2} $$

$$ \rho g = (8.57 \frac{g}{cm^3} ) (9.8 \frac{m}{s^2} ) = 83,986 \frac{kg}{m^2 s^2} $$

These numbers are surprisingly close. If we take the lower value for the magnetic field we will find that it couldn't work. Also, since you need <5K temperature to get this, the cooling would add to the linear mass density. We conclude that current technology isn't good enough to do this. Current density isn't good enough by enough of a margin.

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  • $\begingroup$ Because the formula doesn't work, the superconductor expels the field, there is no significant force. $\endgroup$ – Ron Maimon Sep 16 '12 at 6:32
  • $\begingroup$ I did this calculation before asking the question, and I noticed the near coincidence, that's why I asked it. The reason to be at the equator is that the magnetic field is strongest there, and it is about double the weight of the superconductor. But the idea doesn't work, the superconductor expels the field. I just screwed up, and didn't notice until later. $\endgroup$ – Ron Maimon Sep 16 '12 at 6:45
  • $\begingroup$ @RonMaimon, a type II superconductor will have flux pinning that will produce lots of torque, but you need quite some current going on $\endgroup$ – lurscher Sep 16 '12 at 6:52
  • $\begingroup$ @lurscher: It's not important if it's pinned or not, the force on the current carrying superconductor is IBL as always, as can be seen by conservation laws (for example, consider two wires carrying parallel currents, the conservation laws don't allow the force to change when one becomes superconducting). My comment above is the stupid one, the idea is sound. Whether you expel the field or not, the field stresses give you the same magnetic force regardless. Sorry for the stupidity. $\endgroup$ – Ron Maimon Sep 16 '12 at 7:01
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    $\begingroup$ Thanks for the answer, I was wrong about the expelling, and I am sorry for going back and forth, but I was embarassed about asking something so half-baked. I did the physics 101 calculation, of course, and it is correct, so +1 to you for saying the obvious, because it is true. $\endgroup$ – Ron Maimon Sep 16 '12 at 7:13

As the wire goes up, the current is doing work and so the current drops and needs to be replaced to continue the ascent. So the launch cost is the electrical power that needs to be supplied on the way up. Of course the lift is determined by the current times the field, and that needs to be enough to lift the wire, the cooling apparatus and the payload. Lifting the length of the wire and cooling to go around the world will likely require greater power than a rocket would.

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    $\begingroup$ This is not true--- nothing is going up, my idea is stupid. $\endgroup$ – Ron Maimon Sep 15 '12 at 21:24
  • $\begingroup$ @RonMaimon: I beg to differ. It is not a stupid idea - perhaps unusable given the state of technology, but certainly not stupid. Then again, I may be prejudiced because I'd been entertaining a similar thought (+: $\endgroup$ – Everyone Sep 16 '12 at 3:36
  • $\begingroup$ @Everyone: You are right, my idea that it is stupid was the stupid one. The idea works. The superconductor expels the field, but the current-magnet force is the same as if it didn't, the force just acts on the surface of the wire instead of in the bulk (or around a vortex). The field stresses that transmit the force are the same incoming around the wire, and the details of the interior don't matter except to the detailed force distribution. $\endgroup$ – Ron Maimon Sep 16 '12 at 7:02
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    $\begingroup$ Sorry, ignore the top comment above, the expulsion of magnetic field is not important at all, the lift force is determined by conservation laws away from the wire, and it doesn't care about whether the wire is superconducting or not. The lift is provided by the Earth's magnetic field, and the lift force is provided by the Earth's rotation--- the Earth will slow down infinitesimally that's all. $\endgroup$ – Ron Maimon Sep 16 '12 at 7:31
  • $\begingroup$ @RonMaimon - why do you say that the earth's rotation provides the lift force? I assumed the magnetic field would be the one that mattered - and you would have to do work (by supplying current into a giant inductor) in order to get this thing off the ground. The mass of the wire isn't what will kill you - it's the cryostat around it (even without the mass of the helium)... $\endgroup$ – Floris Sep 5 '14 at 5:55

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