"Propeller-head" polar space elevator?

Question

"Why isn't it possible to build a space elevator at the north pole?" made me think...

Like the funny hat with a propeller on a kid's head - replace the head with Earth and propeller blades with space elevators.

Build a stubby tower / hill / mountain on one of Earth's poles. Add a huge motor, that spins in the same axis as Earth on top. Extend space elevator (or two, in opposite directions, for balance) from the motor's axis, going initially parallel to Earth surface, then out into space as Earth curves away.

Such an elevator could be much shorter than the traditional equatorial space elevator propelled by Earth spin alone - since the rotary speed would be much higher. The huge problem of construction vanishes: lifting the material up to GEO to unroll from there, as segments could be added at the axis instead. There would be the problem of the end moving at hypersonic speeds until there's enough of it extended to slow the spin enough that the part within the atmosphere doesn't exceed speed of sound. And of course the engineering problems of the mighty motor, and continuous energy expenditure to fight atmospheric drag. Never mind logistically awful location.

Would it be possible though, or am I missing something? Say, precession, or maybe Earth's gravity bending it to the ground, or Earth magnetic field, or something like this?

Concur with @gerrit, I am trying to visualize this. The picture I get has less promise than a standard space elevator. — James Jenkins, Mar 15 '17 at 11:24
@gerrit: done. The main promise is the whole thing shouldn't need to exceed Earth radius in length, as opposed to ~35000km of GEO. And should be possible to be fully built on Earth, zero space launches required for construction. — SF., Mar 15 '17 at 12:07
What kind of orbit would a payload dropped off the end of this thing be in? — Organic Marble, Mar 15 '17 at 12:39
@OrganicMarble: That strongly depends on RPM. If this curves to equatorial latitudes, it can be just an equatorial LEO. If it's moving way in excess of that, kept mostly straight, the orbit will be inclined and elliptical - it may also be suborbital, if released "early". — SF., Mar 15 '17 at 12:42
@gerrit: Precession problem: The pivoting elevator would retain plane of rotation regardless of Earth rotation. Placed on equator, spinning parallel to horizon at noon, it would want to spin in vertical plane at sunset. And with this kind of momentum, forcing the plane of rotation to turn would need way more energy than I believe we'd be ever capable to expend. — SF., Mar 15 '17 at 12:45
Would the energy costs to spin this megastructure up leave any advantage over conventional launch? — Organic Marble, Mar 15 '17 at 12:48
@OrganicMarble: I don't know how much would be lost to atmospheric drag and magnetic field, which would be the continuous expenditures, but on top of that, after the initial spin-up, the only expense would be electric energy to propel the motor to recuperate momentum lost to released satellite. — SF., Mar 15 '17 at 12:57
The plane of an orbit must go through the center of the earth. The north pole may be on that plane too as longs as the center of the earth is also on that plane. — Uwe, Mar 15 '17 at 15:59
@Uwe: If you mean the plane of rotation of this elevator, it couldn't - Earth in the way. But in this case not force of gravity (towards the center of Earth) is the main source of centripetal acceleration but the force of tension of the elevator (towards its the mount point / axis). This would compound with force of gravity, tilting it towards equator, but the sum of the two forces would keep it spinning offset relative to normal orbit. And a mass released from the end would enter an inclined orbit. — SF., Mar 15 '17 at 16:18
@Krumia: It seems my idea would require either extremely tall mountain, or very high angular velocities - and as result, quite extreme durability of the tether, possibly well in excess of what standard one needs. Conclusion: idea viable, but inferior to alternatives. — SF., Mar 16 '17 at 13:40
The idea might benefit from attaching "airplanes" on the lower end of the cable. These can be electrically powered from the ground, and provide both lift and torque to overcome drag. This is more efficient than a motor at ground level. — MSalters, Mar 20 '17 at 10:08
@MSalters: Currently it seems that bending is the problem - not like Michael described it though. The rope will be drooping in the initial segment, and would need either excessively high tower or lift - and that is a serious problem as we need it to exit atmosphere ASAP to reduce drag. — SF., Mar 20 '17 at 10:37

score 3 · Answer 1 · answered Mar 15 '17 at 12:29

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Because of bending

You assume that because your carousel is spinning that you do not need to lift it.

But you do.

The "orbit" that this carousel is spinning in is not a real orbit, because it is not centered around the Earth's center of gravity. So the "arms" are subjected to gravity, pulling them down towards the Earth's center of gravity.

This means that you have to deal with huge bending forces. And bending is a much harder problem to solve than just compression / tension forces.

answered Mar 15 '17 at 12:29

MichaelK

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Since it's not a rigid structure, it's stretching due to centrifugal force; sure it will bend towards equator due to gravity - but sufficient speed (and initial altitude) should keep it from hitting the surface. Plus if it's constructed as a lifting surface, it can fly until its ends exit the atmosphere. – SF. Mar 15 '17 at 12:32
@SF. Needs more detailed calculations before we decide to build it ;-) – gerrit Mar 15 '17 at 12:36
@SF Well that does not help your case here. Centrifugal forces only adds a horizontal force... it does not add any vertical lifting force. And you still have the problem that you are not in any kind of equatorial plane but offset to those. – MichaelK Mar 15 '17 at 12:37
@MichaelKarnerfors Shouldn't centrifugal forces add a force along the line of his "rods", thus having both a horizontal and a vertical component? – gerrit Mar 15 '17 at 12:39
@gerrit No, the centrifugal force go straight out. The combined forces of the centrifugal force and gravity will add together and the rod will align itself with the resulting force vector. – MichaelK Mar 15 '17 at 12:42
@MichaelKarnerfors But straight out in the drawing is neither horizontal nor vertical. Therefore I don't understand why it would only add a horizontal force. – gerrit Mar 15 '17 at 12:42
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@MichaelKarnerfors: and yet as you spin a rock on a long string above your head, it doesn't hit the ground. The centrifugal force pulls it away horizontally, preventing the drop. Also, old helicopter-style lifting could be quite helpful in the initial segment, before enough centrifugal force kicks in. (add lifting surface, imagine the structure is a helicopter rotor). – SF. Mar 15 '17 at 12:48
@SF No, because of bending. Your rods are not infinitely rigid, nor infinitely strong. If they were, there would be no problem building an equatorial space elevator to begin with. – MichaelK Mar 15 '17 at 12:50
@gerrit: centrifugal force will be in horizontal plane parallel to the ground at the pole, not "local horizontal" wherever. Gravity is perpendicular to that. Yes, gravity will bend it towards the ground, but with sufficient centrifugal force it will be pulled out away from the axis stronger, not allowing the pull to make it fall. – SF. Mar 15 '17 at 12:51
@SF And the spinning-rock-on-a-string example does not help your case because it is not the centrifugal force that provides the lift... it is your hand that does that. Your hand lifts no more, nor any less just because you start spinning it. The downward force is still the same. – MichaelK Mar 15 '17 at 12:52
@MichaelKarnerfors: "Tower/hill/mountain". – SF. Mar 15 '17 at 12:54
@SF And then there is the issue of conservation of angular momentum. As we know from other space ventures, if you release a weight on a tether, that very effectively stops the rotation. So: no... no no no no. Your concept is cute, but of it was physically possible, then building a space elevator on the equator would not be an issue... it would be even easier. You are only making it more complex and adding a whole lot more force that the structures must put up with. – MichaelK Mar 15 '17 at 13:00
@MichaelKarnerfors: If we release a small part of the weight, the tether loses a little bit of the momentum. Obviously the payloads would be a small fraction of the mass of the tether! - and then we can recuperate it simply by putting some extra torque into the motor, spending cheap, ubiquitous electricity. – SF. Mar 15 '17 at 13:04
Let us continue this discussion in chat. – SF. Mar 15 '17 at 13:04
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Don't forget that you also have to rendezvous with this thing, which would be difficult if it isn't in a regular orbit. – Steve Mar 15 '17 at 13:09
@Steve: Why? Landings would be performed through normal reentry procedure; it would only be for launches. – SF. Mar 15 '17 at 16:09
@SF. I suppose science fiction is playing a part in my comment... If someone was in orbit and needed to get to the top of the elevator, they would have to re-enter and take the elevator back to the top? – Steve Mar 15 '17 at 18:25
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@Steve: Or perform a very tightly timed docking maneuver; there are orbits tangent to the trajectory of the top, with matching speed at the contact point, but that would be very much like grabbing the skyhook except without aid of air to help maneuvering. Definitely not like current Soyuz-ISS docking, more like SpaceX hover-slam. – SF. Mar 15 '17 at 20:56

"Propeller-head" polar space elevator?

1 Answers1

Because of bending