1
$\begingroup$

If a horseshoe magnet was attached to a satellite orbiting the Earth, would the interaction of the Earth's magnetic field and the horseshoe magnet's magnetic field result in the magnet being propelled in the direction of travel and adding momentum to the satellite? If so, this magnet should help the satellite to maintain its orbit over its lifetime.

To illustrate how this propelling force would be created, please reference the drawing below:

enter image description here

The horseshoe magnet shown in the drawing (and the satellite) is moving in a West-to-East orbit around the Earth. This propelling force is created via the magnetic repulsion of the two magnetic fields.

Also, this horseshoe magnet could be mounted to the end of a ten foot plastic rod with the other end of the rod attached to the satellite so the magnet's magnetic field would not interfere with the satellite's electronics.

EDIT

I have thought about the issue of the magnet and satellite being rotated by the Earth's magnetic field instead of the magnet adding force/momentum in a West-to-East direction.

I believe a solution to this issue would be to place a magnet behind the satellite and another one in front of it, with each one being attached to the end of a long plastic rod so its magnetic field would not interfere with the satellite's electronics.

I have illustrated this concept in the new drawing below:

enter image description here

Would this be a viable solution to the rotation issue and would there be a propelling force which would add momentum to the satellite? These two magnets could be either permanent magnets or electromagnets.

$\endgroup$
  • 2
    $\begingroup$ Using these amazing new magnet physics, you can also get infinite energy. The possibilities are endless! $\endgroup$ – SE - stop firing the good guys Mar 6 at 14:26
  • 1
    $\begingroup$ For one thing, consider what forces would be acting prior to deployment of the magnet. $\endgroup$ – Carl Witthoft Mar 6 at 14:32
  • 1
    $\begingroup$ The arrangement in your modified question is not really any different. You still have dipoles parallel to Earth's and a net repulsive force. It doesn't matter if they are ahead of or behind the spacecraft. $\endgroup$ – uhoh Mar 8 at 9:28
  • $\begingroup$ @uhoh, I see what you are saying. Perhaps then the only viable solution would be to rotate each magnet 90 degrees so the open end of each horseshoe magnet would be pointed towards the surface of the planet. In this new arrangement, both magnets should experience a magnetic repulsion force that should push the satellite away from the Earth, which should help maintain and/or prolong its current orbit. $\endgroup$ – user255577 Mar 8 at 13:22
  • 1
    $\begingroup$ @user255577 my answer explains that while you can get a one time, very very tiny bump in altitude, this will never work to maintain and/or prolong the orbit. Atmospheric drag keeps sapping away the angular momentum so the orbit keeps dropping lower and lower until it burns up. You need a thrust force to push the satellite forward in its orbit in order to maintain altitude, and magnets on the spacecraft just can't do that. $\endgroup$ – uhoh Mar 8 at 13:45
3
$\begingroup$

Dipole-dipole repulsion

The drawing depicts the right idea; if you hold two parallel magnetic dipoles close together there is a strong repulsive force. If you let go, they will fly apart.

In order to imagine what would happen if this was in Earth orbit, let's make the horseshoe magnet an electromagnet. It's the same shape but instead of being made from permanent magnet material plus iron, it's only iron and we wrap a solar powered coil around it so we can turn it on once in its circular orbit.

When that happens there will be a radial force pushing the spacecraft outwards. This doesn't change the velocity because it's perpendicular to it, so now the spacecraft is going too fast for this circular orbit, and it will start rising in altitude until apoapsis, and then will return back to it's original state. You've raised the circular orbit into a slightly elliptical orbit with the same periapsis.

If you ramped up the field slowly, you'd raise the altitude slightly.

It's a little bit like being in a heliocentric orbit and pointing a giant solar sail directly at the Sun. Not the same because the Sun's intensity falls off as $1/r^2$ and the repulsion between two dipoles falls off as $1/r^4$.

Doesn't fight drag

This repulsion has no further effect long term. Once you've slightly raised the orbit you have a new balance between attractive and repulsive forces.

This doesn't stop the drag force from slowly sapping the spacecraft's angular momentum, because drag acts in the direction of motion whereas our repulsion is perpendicular to it. The spacecraft will continue to lose altitude and eventually burn up in the atmosphere. Depending on the kind of magnet used, it might be so large and able to absorb heat that chunks survive reentry and fall on someone's house.

Different but related:

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @ uhoh, as stated in my latest reply to your comment up above, I believe the only way a horseshoe magnet(s) could help a satellite stay in orbit would be that the open end of the horseshoe magnet(s) would need to be pointed directly at the surface of the planet. $\endgroup$ – user255577 Mar 8 at 13:31
3
$\begingroup$

No. What can be done with a magnetic field in Earth's orbit is to torque, or rotate the satellite. But for propulsion of some kind, you would actually need to have some kind of a varying magnetic field, and a tether of some kind. This website talks a bit about how such a thing could work.

| improve this answer | |
$\endgroup$
  • $\begingroup$ thanks for that link. That’s a very interesting SpaceNews article and I’m going to do some research on it to see what the results were because that article was dated July 12, 2010. $\endgroup$ – user255577 Mar 6 at 16:16
  • $\begingroup$ A varying magnetic field is talking about sharp gradients right? The variations around Earths magnetosphere are too macro to have micro effects on a satellite, correct? $\endgroup$ – Magic Octopus Urn Mar 9 at 15:00
3
$\begingroup$

Let me put this in a different frame. First recall that magnetic fields also store magnetic energy:

$$ E_{mag} = \int_V H. B\, dV $$

Now, from Lagrangian mechanics, note that:

$$ L = U-V = E_{mag}+E_{grav}-\frac{mV^2}{2}-\frac{\omega^TI\omega}{2} $$

And let's forget some dissipative effects and the angular terms for now.

Recalling Euler-Lagrange's equation:

$$ \frac{\partial L}{\partial r}-\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{r}} \right)=0 $$

And understanding that the energy terms do no depend upon the time-derivative of the generalized coordinate, we have:

$$ \frac{\partial E_{mag}}{\partial r}+\frac{\partial E_{grav}}{\partial r}=m\dot{V} $$

No news there, the gradient of the gravity potential energy is just the gravitational force everyone here knows well about. The new term, which intuitively we think could provide some orbit control is the derivative of the magnetic field:

$$ \frac{\partial E_{mag}}{\partial r}=\frac{\partial}{\partial r} \left[ \int_V H. B\, dV\right] $$

Now, this terms represent the instantaneous rate change of magnetic stored energy per unit of length in the position of the spacecraft. So, this disregards changes in attitude of the spacecraft (we know well the effects of magnetic torques on satellites, and dropped any attitude related coordinate before). Thus, we can assume that the magnetic induction field created by the satellite ($B$) is independent of the spacecraft position. So:

$$ \frac{\partial E_{mag}}{\partial r} = \int_V \left[\frac{\partial H}{\partial r} . B\, dV\right] $$

And the Earth's magnetic field on its orbit $H$ is not constant over space. Right? Well, I'd like to find a better reference to this value, but picking to this paper ate 324km (which is very low for an orbit) You get at most 0.16nT/km gradient vertical in a magnetic anomaly. Thus, converting to SI base units, we're talking about $1.6 \times 10^{-11}$ T/m. Dividing by the magnetic permeability of vacuum we get 0.042 A/m². So to get the same the average thrust used by GOCE (which I'm assuming around 2mN), we'd need 2mN = 0.042 A/m² * B, resulting in B ~ 47T. Sounds like much to you?

Well, incidentally I had the hunch but not a reference to tell you if this value was big or not. My google search however tells me that the record holder for world's strongest magnet as of 2019 produces "only" a magnetic flux of 45.5 T.

Remember when people tell you not to bring metal objects close to an MRI machine? Well according to this reference "Across the MR industry, most scanners are 1.5T or 3.0T, however there are varying strengths below 1.5T and more recently, up to 7.0T.". They generate force enough to mechanically break components on a spacecraft.

At one hand, my reference for magnetic gradient is not so great, but you can search data for other orbital altitudes and redo the calculations yourself. Maybe one needs less thrust for higher orbit, but then the base magnetic field and its variations are also smaller.

But wait, there's more! (Trust me, I'm an Engineer). Strong magnetic fields especially if varying over time create electromagnetic interference which may mess up every wire onboard the spacecraft. Also, see this question for extended discussion. I'd even wager that a field this strong could mechanically damage the satellite by breaking its structure, I haven't done the math here, but I'm convinced by this video.

So, in summary: In orbit, and over the size of a satellite, and over the variations of the Earth's magnetic fields, magnetic forces (and not only torques) do exist all the time, but they are really small, to the point were any usable effect would require an insane amount of energy to generate an absurdly strong magnetic field which would probably destroy the satellite.

| improve this answer | |
$\endgroup$
  • $\begingroup$ @ Mefitico, I appreciate that you have pointed out that the Earth's magnetic field is simply too weak at an orbital height about the Earth to have any beneficial effect on the horseshoe magnet(s) and therefore there is really no advantage in using one. $\endgroup$ – user255577 Mar 9 at 12:29
  • 1
    $\begingroup$ @user255577 : Please notice a slight subtleness: It's not the magnitude of the field itself that matters for the purpose of my post, but rather the spatial variations of it. The intensity itself is more relevant on the reasoning of uhoh's post. $\endgroup$ – Mefitico Mar 9 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.