# Could a satellite levitate above the magenetic fields of Earth?

Like the toy the base of the levitation could the Earth be the base and the space station the levitron?

https://earthscience.stackexchange.com/questions/13854/where-on-earth-is-the-magnetic-field-intensity-stronger

https://worldbuilding.stackexchange.com/questions/108896/could-a-city-be-built-out-of-balloons

https://engineering.stackexchange.com/questions/21284/can-this-version-of-the-levitron-work

Is there a place where the Earth's gravity well is present but the electromagnetic field strong enough to allow a object to stay suspended in Earth's magnetic field like a levitron?

• the magnetic field of earth is too weak for this to work,it will work over a magnetar but there it is an other set of problems. Commented Mar 22, 2018 at 7:39
• You might like this. It's not really related to the physics in the question, but it's cool youtu.be/0tJfqMYHaQw?t=950
– uhoh
Commented Dec 11, 2018 at 4:38

The basic calculation we need is set out here. The force that would need to counteract gravity is given by equation 17 $f = \nabla(\mu.B)$ where $\mu$ is the dipole moment of the magnet and $B$ the Earth's magnetic field. Now a good neodymium permanent magnet has a moment equal to its volume times about 875 $kA/m$ (as discussed here) and a density of 7400 $kg/m^3$ so a specific magnetic moment of very roughly 100 $A/kg m$. Since the Earth's magnetic field is roughly a $10^{-4} T$ dipole at the surface, and the strength of a dipole drops as $1/r^3$ the gradient will drop as $1/r^4$ and so be something like $B/r$, so it will be very roughly $10^{-13} T/m$. So the force on each kilogram of magnet is something like $10^{-11} N$, about one part in a million million of the gravitational force.

So lets consider instead a superconducting magnet. It seems the critical current density of niobium film is about $10^{11} A/m^2$ while the magnetic dipole moment of a current loop is $2IS$ where $S$ is the area and $I$ the current, so a loop of niobium wire of radius $R$ and cross-sectional area $a$ could carry a current of about $10^{11} a$ amps and a would have a dipole moment of about $10^{12} a R^2$. Its density is about 8500 $kg/m^3$ so its mass is about $10^4 a R$. The magnetic force on it, from the above calculation, will be about $0.1 a R^2$ so we find that $R$ needs to be about $10^5 g$ or about a million meters. So the loop is nearly as large as the Earth. The loop would also need to be strong enough to hold the "pressure" of the magnetic field within it. I can't find numbers for that, but I'm pretty sure you'd need something a lot stronger than niobium.

A secondary problem, as already pointed out in another answer, is that Earth's magnetic field does not provide a stable minimum where the magnet could sit, so it might experience lift, but would need active stabilisation.

• I think you've chosen an extremely heavy and spatially tiny dipole for your example, and so this is not very conclusive. How about a superconducting loop, or even two in a Helmholtz configuration. You'll see that the force scales with size and mass much more favorably than a compact and dense rare Earth magnet. On the other hand, the lack of a minimum and therefore any stability is already conclusive.
– uhoh
Commented Mar 22, 2018 at 12:21
• Edited to explore that case. Commented Mar 22, 2018 at 12:53
• Wow, looks really great!
– uhoh
Commented Mar 22, 2018 at 13:05

Probably not.

There are several differences between the laboratory example and the situation on Earth.

The magnet shown in the example is not a dipole. It is a large, ring shaped magnet with the height of the levitated object above it roughly equal to the diameter of the ring. It is not anything like the shape of the Earth's field or the field of a dipole. It has a convenient local minimum in the transverse plane.

Also, the large drawing of the Earth's field shown in your question (https://i.sstatic.net/zsz9D.jpg) is wrong and completely unrealistic. It does not show the correct shape of a dipole field or the field of the Earth.

I'll add a plot and a quantitative analysis in a few minutes er, hours...

• absolutely not?
– Muze
Commented Mar 22, 2018 at 4:22
• hang on... I'll see if I can prove it conclusively.
– uhoh
Commented Mar 22, 2018 at 6:30
• take a look here and see the intensity part en.wikipedia.org/wiki/Earth%27s_magnetic_field earths magnetic field is simply way too weak. Commented Mar 22, 2018 at 7:49
• @trondhansen without stating mass, size, and dipole moment of the device, there's no way to draw that conclusion. Anyway, the Earth's field is the wrong shape. It lacks the transverse minimum that the torus has. This is why the example uses a torus and not a dipole.
– uhoh
Commented Mar 22, 2018 at 10:29