10
$\begingroup$

I have an idea that the space elevator can be build in the following conditions:

  1. Not a classic theory. No cable. No Geostatic orbit. Not equator platform position
  2. The possible platform / base position is Earth's south magnetic pole - Antarctica, or north magnetic pole in Arctic.
  3. The lifting force is Earth's magnetic field.
  4. The lift design is a huge ring (more than 100 m) with a magnetic coil.
  5. The wire of the coil is a superconductor for a high current.
  6. The surface of the ring is closed with parabolic solar energy collector with lens in the focus. The lens is the laser beam receiver from the ground platform.
  7. Near the focus are located payload spacecrafts with engines for changing the motion vector after dumping and during the fall.

Let us clarify the question in the following way: 1) “What conditions and parameters of the electromagnet ring or coil and of other components in the assembled structure (elevator) will satisfy the condition that the Earth’s magnetic field will be enough to lift up this structure.” The parameters can be like the following: radius of the ring R (m), number of circuits in a coil N (pcs.), current I (A), weight M (kg) and others required for calculations.

The additional questions: 2) “What height can be reached if the method is possible at all” 3) “Is it possible somehow to focus the Earth’s magnetic field of some circle area in the pole which will help to increase the chances for questions 1 and 2”

Space elevator magnetic ring

Legends:

Item 4 - ring with coil is red color

Item 5 - wire is inside the coil inside the ring (4)

Item 6 - solar panels - energy collector is transparent blue

Item 6 - lens in the focus - laser beam receiver - purple

Item 7 - payload - green cylinders

Updates (20.11.19). I see one more reason that the elevator will not work out. While the electromagnetic field inside the ring will be oriented in the opposite direction to the Earth’s magnetic field creating the lifting force, the electromagnetic field outside the ring will goes the same direction as the Earth magnetic field and will pull the ring down. Only in case if the radius of the ring will be bigger than the radius of the Earth’s magnet we will not face this issue until some height is reached.

The picture of 3dmodel is updated with better rendering (20.11.19)

$\endgroup$
  • 7
    $\begingroup$ There's not much point putting a space elevator at the pole as there is no rotation of the earth at that point so anything you lift will have no orbital velocity and will just fall down again. ok - you can have rockets to accelerate to orbital velocity, but that defeats a lot of the advantages of the elevator. $\endgroup$ – rghome Nov 18 at 8:58
  • 1
    $\begingroup$ 1. The advantage of the space elevator of such a type is that it will help to save up to 90% of fuel, required to overcome the gravity and the shape of the space carrier can be any, even cube, without perfect aerodynamic coefficient. 2. The picture (diagram) with legend is added. Thank you for the advice. 3. I need some stronger reasons that this model of space elevator will not work to sleep again. $\endgroup$ – Alexander Spacelifter Nov 18 at 12:19
  • 5
    $\begingroup$ @AlexanderSpacelifter The hard part isn't getting up; it's going sideways fast enough. Your elevator doesn't help with that at all. The elevator isn't in orbit, it isn't in free fall, it's basically just a very tall tower. That doesn't help much with anything space-wise. $\endgroup$ – Luaan Nov 18 at 14:29
  • 2
    $\begingroup$ @AlexanderSpacelifter regardless of all the other aspects, the assumption that "it will help to save up to 90% of fuel" is totally false; getting cargo to that position would be helpful and might save something like 20% of total fuel, but that wouldn't help with the vast majority of the energy investment required to accelerate it to orbital velocity. $\endgroup$ – Peteris Nov 18 at 17:04
  • 1
    $\begingroup$ The magnetic north pole is in the Artic. The south pole is in Antarctica. Also, the magnetic north pole is over an ice sheet that floats on water. $\endgroup$ – DwB Nov 18 at 20:47
27
$\begingroup$

Setting aside the technical details, the biggest problem with this idea would be Earnshaw's theorem, stating that it is impossible to stably levitate a magnetic dipole in another magnetic dipole's (Earth's) field. In other words, you could theoretically generate enough magnetic force for lift-off with your dipole (the coil), but you would have trouble stabilizing it so that it doesn't topple over and fly back into the Earth.

Edit, after bringing up active stabilization in the comments:
While in theory, yes you could actively stabilize the whole setup. And just for the sake of arguing, you could also stabilize the ring by (mechanically) spinning it around its axis. The torque that would act to topple over the dipole would now cause precession instead. Voila: the ring is stabilized.

However, I still don't think that this whole idea would work in reality.
The magnetic moment of the dipole is $m_{B}=IS$, where $I$ is the current through the loop and $S=\pi R^2$ is the enclosed area by the loop of radius $R$. Since we're talking about a setup on the magnetic axis of the earth, the geomagnetic field can be written as $B_{\rm E}=B_0(1+h/R_{\rm E})^{-3}$, where $B_{0}\approx62\,$µT is the field strength at the surface of the Earth, $R_{\rm E}\approx6400\,$km is the radius of the Earth, and $h$ is the height above ground. The force generated by our dipole lifter would be: $$ F=-\frac{\rm d}{{\rm d}h}\left[m_{B}B_{\rm E}\right]\ =-\pi IR^2B_{0}\frac{\rm d}{{\rm d}h}\left[(1+h/R_{\rm E})^{-3}\right] =\frac{3\pi IR^2B_{0}}{R_{\rm E}}(1+h/R_{\rm E})^{-4}, $$ We can also approximate that $h\ll R_{\rm E}$, so that $F\approx3\pi IR^2B_{0}/R_{\rm E}$.

Now, then lets say that you build yourself a ring with radius $R=500\,$m, and drive a current of $I=10\,$MA through it. That would generate a lifting force of about $230\,$N (newton), or about $23\,$kg ($50\,$lbs for the imperial entanglementalists).

But how much would such a device weigh?
We, of course, use superconductors in order to not require extremely thick copper or silver conductors. But the problems with superconductors is that you can't just drive however large current you want through them. At some critical current density, the supercondictivity will break down due to the high magnetic fields generated by them. Lets say we build that loop of niobium-nin, a superconducting material with a reportedly record-breaking current density. It's highest current density is $j_{\rm max}\approx2600\,$A/mm$^{2}$.

We wanted to push $I=10\,$MA through it, so that would require a cross-sectional area of $I/j_{\rm max}\approx4000\,{\rm mm}^{2}=0.004\,{\rm m}^{2}$. If we multiply that by the length of conductor needed, $L=2\pi R\approx3000\,$m, we get a volume of $12\,{\rm m}^3$. I don't have the (mass) density of niobium-tin, but needless to say $12\,{\rm m}^3$ of it will weigh in the order of at least $20{-}100\,$tons, and that's not counting all the cooling apparatus you would need in addition to the pure weight of the superconductors. (And I haven't even gotten to address the mechanical strength required for you to spin the whole ring at hundreds or thousands of rpm, in order to stabilize it.)

Tl;dr
You won't be able to generate enough force to even remotely lift your apparatus. I hope this can make you sleep well tonight.

$\endgroup$
  • $\begingroup$ I thought about the troubles in stabilizing. In theory, it can be solved with reactive stabilizer engines on the perimeter, like shunting engines of the space crafts. $\endgroup$ – Alexander Spacelifter Nov 18 at 12:23
  • 2
    $\begingroup$ @AlexanderSpacelifter I hope the additional material in the edit will satisfy you. $\endgroup$ – Andréas Sundström Nov 18 at 14:04
  • $\begingroup$ Dear Andréas Sundström, Thank you very much for your complete answer. I needed it a lot, really)) It is hard to realize that such a huge magnet as planet Earth has such a weak magnetic field in its pole. Thank you once again. :))) $\endgroup$ – Alexander Spacelifter Nov 18 at 14:59
  • 3
    $\begingroup$ So the lifting force goes as $\sim R^2$ and the weight as $\sim R$. Are you saying the ring should be bigger? ;) $\endgroup$ – jkej Nov 18 at 18:15
  • 4
    $\begingroup$ @AlexanderSpacelifter Well, how much magnetic force do you feel from earth in your daily life? I mean, you are using all kinds of stuff made from magnetic iron. You are using all kinds of magnets. Did you ever feel the influence of earth's magnetic field? Yes, the field is huge, and it contains a surprising large amount of energy. But we absolutely need to suspend a magnet on a string or some other bearing that provides for virtually frictionless rotation to even be able to observe earth's field. (Ok, today we can use hall sensors as well, but you get my drift...) $\endgroup$ – cmaster Nov 18 at 23:26
2
$\begingroup$

Here are some problems that I see :

Scaling issue mentioned by earlier answers, primarily limited by the current density allowed in the superconductor.

The superconductor needs to be high-temperature. Yes, even in the Arctic, it's too warm for known superconductors in 2019. The superconductor needs to operate at low pressure (several have been found at very high pressures, but this is in the upper atmosphere and you cannot afford to bring giant compressors with you). The superconductor needs to be resistant to or protected from atmospheric oxygen, cosmic rays, the Van Allen radiation belts, and other standard problems of rocketry.

Weakness of the Earth's magnetic field. This could be worked around by building a HUGE high-powered ring on the Earth, aligned with the Earth's normal magnetic field, and a smaller reversed ring to fly into space.

Weakness of most ferrous metals and other conductors to induction currents from the rings. The spacecraft is crushed and torn apart.

Finally : Orbit is not a matter of altitude, but a matter of speed. So your spacecraft must take significant rockets along with it. Lofting to a mere 35,000 km , the standard geosync altitude, would require a LARGE amount of delta-V that has to be applied promptly. Trying to go higher would require less rocket delta-V, but your magnetic lifter works worse and worse the higher you go.

$\endgroup$
  • 1
    $\begingroup$ The 3km/s to go from stationary-at-35,000km to orbiting-at-35,000km is still a huge improvement over the 13-14 km/s to go from the surface to orbiting-at-35,000km. The losses due to the thrust not being impulsive are manageable. $\endgroup$ – hobbs Nov 19 at 2:36
1
$\begingroup$

You won't be able to go very far. I am assuming the force that you are assuming will push you is the Earth's magnetic field. The strength of the field is around 0.6 Gauss at the poles and this is not nearly enough to produce force to even lift a single person off the ground. Moreover, even if the field was very huge and the resistance of the superconductor super low, remember that the force generated is directly proportional to the change in magnetic flux. The initial change when you just put the ring is very high but as it goes up, the magnetic field will decrease and the current will start to decrease as the change in flux is in opposite direction. Moreover, suppose you keep the elevator fixed for some time. As Earth's magnetic field is static, the time derivative will just go to 0 producing in no net force after that.

$\endgroup$
  • $\begingroup$ Guys, thank you for your comments and answers. I added a picture (3d model) and some legend. It is not passive ring, it is electric coil with its own magnetic field with the same polarity - N at the bottom edge. $\endgroup$ – Alexander Spacelifter Nov 18 at 10:37

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