superconductor levitating in earth's magnetic field? 
Possible Duplicate:
Can superconducting magnets fly (or repel the earth's core)? 

I've seen superconductors levitating on magnets. But is it possible for superconductors to levitate on Earth from Earth's magnetic field?
 A: The lift generated by magnetic field B on a superconductor of area S is:
\begin{equation}
F = \frac{B^2S}{2\mu_0}
\end{equation}
disregarding lateral forces and assuming superconducting cylinder (or similar shape) with area S at the top and bottom and height h, we need three forces to remain in the equilibrium: magnetic pressure on top, bottom and gravity force:
\begin{equation}
F_{b} - F_{t} = F_{g}
\end{equation}
denoting density of the superconductor as ρ, Earth' gravity as g and magnetic field at the top and bottom of the object as Bt and Bb, we have
\begin{equation}
\frac{1}{2\mu_0}(B_{b}^2-B_{t}^2)=\rho gh
\end{equation}
assuming the vertical rate of change of magnetic field is nearly constant and denoting the average magnetic field as B, we have
\begin{equation}
-B\frac{dB}{dz}=\mu_{0}\rho g
\end{equation}
Compare with diamagnetic levitation (superconductor's magnetic susceptibility is -1).
Now, Earth magnetic field is between 25 to 65 μT. For the derivative I have found this survey from British Columbia with upper point on the scale being 2.161 nT/m. Assuming this to be the maximum for vertical derivative we get the required density of 1.1394e-08 kg/m3. For comparison air density at the sea level at 15C is around 1.275 kg/m3, so required density is 8 orders of magnitude smaller.
Even assuming a very high vertical derivative where B goes from its maximum 65 μT to 0 on 1 m of height results in density required of 0.00034272 kg/m3.
