An electromagnetic space elevator? 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.
 A: 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.
A: 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.
