The simplest kind of space elevator would be a large mass somewhat beyond geosynchronous orbit tethered to earths surface such that the tether is always under tension from centrifugal force overpowering gravity, but I feel like the equator would be a problematic place to put it because of satellites and space debris. Is there any reason a space elevator can’t be put at high latitudes while still extending away from the axis of rotation?
-
2$\begingroup$ If you consider the other extreme orbit, a polar orbit, geosynchrony seems impossible intuitively. Is there some threshold where geosynchrony is no longer possible? $\endgroup$– AtogJun 21, 2022 at 22:20
-
3$\begingroup$ Obligatory XKCD: what-if.xkcd.com/126 & what-if.xkcd.com/157 $\endgroup$– TrishJun 22, 2022 at 6:47
-
$\begingroup$ Only if you want maximum efficiency. What did your own research suggest? $\endgroup$– Robbie GoodwinJun 22, 2022 at 23:43
10 Answers
The center of mass of your space elevator is going to need to be in the geostationary orbit, no matter where you put the base anchor. As a result, while you can put the foot elsewhere other than the equator, at least at that center-of-mass point, it will be over the equator.
Well, technically, a bit higher than that; the whole point of a Space Elevator is using the tension of having the center of mass above the synchronous orbit to pull up the rest of the structure below it. Either way, though, it's going to have to pass through the equator... because all orbits pass through the equator. And if your space elevator doesn't have a point on it that traces out an ordinary Keplerian orbit where you can put your docking station, warehouses, and so on at (which is typically that geostationary orbit), you've decided not to use the primary utility of building one in the the first place.
And if you're building a space elevator, you're committing to doing something about satellites and debris in orbit around your planet, regardless of where the foot is, so unless there are political, geographical, or logistic reasons preventing you from putting the base on the equator, might as well put it there so you don't have to make it longer than you need to.
-
2$\begingroup$ Is this actually fully true? If I have a cylinder with a string tied near the top and a weight on the end and spin the cylinder along it's axis, the weight will be "orbiting" in line with the anchor point. If I make the cylinder progressively more curved, it's not going to shift the weight's orbit closer to the center of the curvature. The curve won't affect the orbit at all. In reality, it would actually start orbiting slightly further from the curve as the lower part of the string is bent or rubbed upward by the curve. $\endgroup$ Jun 22, 2022 at 16:46
-
2$\begingroup$ Once you add gravity, I would expect it to orbit slightly toward the equator simply because of the extra gravity vector pulling at an angle to the cable, but the gravity would still be working against the inertial vector, it wouldn't completely cancel it out, I would think. $\endgroup$ Jun 22, 2022 at 16:47
-
1$\begingroup$ @user34314 I feel the issue is a bit more than "slightly". Whatever the geometry, the bodies must revolve around their barycenter. Furthermore, if the geometry creates three distinct moments of inertia, one must consider the Dzhanibekov effect: en.wikipedia.org/wiki/Tennis_racket_theorem $\endgroup$– sdenhamJun 23, 2022 at 13:02
-
1$\begingroup$ If we are talking about just one tether, then if the surface anchor point is anywhere other than on the line between the satellite's and the earth's center of mass, wouldn't the tension of the tether have a lateral component pulling the satellite out of its orbit? I feel the only stable single-tether configuration must have the anchor directly under the satellite. $\endgroup$– sdenhamJun 23, 2022 at 13:17
-
$\begingroup$ I guess it would ultimately depend on how much of the force keeping the satellite in "orbit" is simply gravity vs centripetal. If the majority component is a normal gravitational orbit, with only enough centrifugal force to pull up the cable, then the orbit will have to be around earth center. But if the centrifugal force is basically holding the cable out on its own and controlling the orbiter positioning (which is what I'm imagining) then only the axis would matter, mostly. I don't know in reality which forces are primary. $\endgroup$ Jun 24, 2022 at 21:01
The easy answer is "no, it's just a lot simpler and cheaper"
The Earth is spinning and you're taking advantage of that spin to create your elevator. On the equator, the forces along the shaft of the elevator (ignoring things like climate) are perpendicular to the surface of the Earth. When you're lifting, you can take advantage of the maximum centripital force to minimize energy costs. When descending, you can take advantage of maximum gravity force.
At the poles you have a little less efficiency, but still some advantages. The elevator is still perpendicular to the surface of the planet, but you no longer have the advantage of centripital force when you lift the car. We'll ignore the twist to the story due to the rotation of the planet — it's not that big a problem.
But if you put that elevator anywhere else, benefits decrease and problems increase pretty quickly.
let's anchor that sucker on the Tropic of Capricorn
Let's assume that Australia gets its knickers in a twist and decides to throw the proverbial bird at the whole world and build its own elevator — smack dab on the Tropic of Capricorn. Now what happens?
The elevator won't be perpendicular to the surface of the planet because the planet's surface is curved and the force wants to push the elevator straight-out from the axis of spin. That makes it, what, lean more-or-less 45°? This puts uneven stresses on the elevator, exacerbates climate effects, and basically throws half of the benefit of gravity out the window. (You still get centripital force, though!)
But what does all that translate to?
Cost!
In the fight between physics and economics, economics wins more often than you might think
When the Obama Administration pulled the trigger and disallowed manufacture of 60, 75, and 100 watt incandescent bulbs, it caused people to howl! Was it the right choice? From the perspective of forcing people off of the addiction to energy-hogging incandescent light bulbs and forcing them to use energy-efficient LEDs, yes. But from the perspective of forcing people to stop using cheap 50¢ light bulbs in favor of, at that time, still very expensive (\$20-\$50) LED bulbs, no! People on fixed incomes were unimpressed with the argument that LEDs would last longer. They couldn't live without light waiting to save up to buy one.
Cost is a big deal, and the cost of building and maintaining a space elevator at the equator is as cheap as it gets. It doesn't hurt that in many ways the climate effects are also minimized. But build that darn thing anywhere else, and you must be prepared to pay for it.
Edit
A number of commenters have suggested that centripital force can be ignored. Well... kinda. There's a tether between the top of the elevator and the base. It's not enough to simply get the top of the elevator moving in a valid orbit — you need to keep it there with the drag of that tether due to everything from bad weather conditions to centripital force. Yes, you could use thrusters to do that... at a cost.
Every dollar spent to keep the elevator operating away from the equator is a dollar that didn't need to be spent.
-
3$\begingroup$ Wait, what? On a pole? What would support the weight of a space elevator at a pole? I mean, the elevator will need to be on orbit, and just "lightly" (meaning many orders of magnitude less than the weight of the elevator) anchored so it doesn't float away. And you don't have an orbit, which would go anywhere near a pole, yet allow the elevator to be anchored. $\endgroup$– hydeJun 22, 2022 at 11:44
-
1$\begingroup$ Any off-equator elevator COULD be held up with enough thrusters, but it would certainly beg the question of "why?!" $\endgroup$ Jun 22, 2022 at 15:57
-
2$\begingroup$ The problem with a space elevator above the Tropic of Capricorn is that it wouldn't stay above the base station. The elevator would stand perpendicular to its own orbital axis, which would not be the same as the earth's axis of rotation. It's simply not possible to have a passive orbit that remains exclusively above the southern hemisphere, the notion of an elevator orbiting above some point on the Tropic at a 45 degree angle makes no sense. Centripetal force isn't much of a factor either, it would only save about 1% of the fuel cost of ascending the elevator. $\endgroup$ Jun 22, 2022 at 16:07
-
1$\begingroup$ @abestrange Thrusters? You mean... continuously pump hydrogen and oxygen up the elevator shaft, burn it in rocket engines along the way to keep the elevator at any desired position, and gravity be damned? :-D Now there's a sci-fi concept to write a story around! $\endgroup$– hydeJun 23, 2022 at 7:32
-
1$\begingroup$ @abestrange I don't argue this point with respect to orbital station location, only with respect to tethering point location. Off-equator tethering would be less efficient, but as we can see, less than ideal efficiency does not stop projects. $\endgroup$ Jun 23, 2022 at 17:24
As noted in other answers, while the center of mass of the tether (the core of the space elevator) must be geostationary (hence, above the equator), the anchor point at the surface need not be -- but there's a limitation on how far from the equator it can be: the excess strength of the tether material.
Every degree the foot is moved north or south of the equator, is a degree of off-plumb attachment in the tether, which increases the tether weight (which is one of the limits -- the tether must support its own weight, that of the counterweight, the elevator structure, and any payload on the elevator, plus weather loads -- likely very small relative to static and dynamic loads from normal operation, but need to be considered) and introduces shear loading. Any mechanical or materials engineer will tell you shear loading is bad, when you're operating near the absolute limits of the strength of your material (and you will be). Shear itself is bad enough for a tensile-only material like the carbon nanotubes currently considered the best candidates for tether material, but it introduces a biaxial load (both shear and tension) which unavoidably weakens the material, which is already pretty weak in shear.
While I've read that IFF we can produce carbon nanotubes in lengths of tens of kilometers, defect free, and bond them together into cable without weakening them, we can pretty readily make a space tether, I've also read that there's not a huge safety factor in the strength. If you choose to handwave a super-material (like Niven's scrith or Sinclair monofilament), you make your own rules, but if you're abiding by real world physics, every 95 km (~ one degree) you move your base anchor away from the equator will have a disproportionate effect on the tether's ability to support space elevator operation.
There's one potential practical exception to this: if you mount a large enough solar sail at the top of the "beanstalk" -- well beyond the neutral point, out at the counterweight -- you might be able to continually adjust its angle to provide a reasonably constant force pushing the top end out of an "orbit" and into a powered path in a circle that doesn't cross the equator. How big your solar sail can be, and how much northward or southward force it can produce would determine how far off the equator your anchor can be and still keep the tether locally vertical.
There are other engineering problems to be solved in doing this -- not least that I don't have the math to calculate what it would do the stresses on the tether or the anchor point -- but it's at least not a violation of orbital mechanics; in fact, it's akin to the situation we already have with a beanstalk where we use a counterweight at greater than orbital speed to keep tension on the tether.
-
$\begingroup$ This is the correct answer. Nothing prevents us to design an off-equator tethered elevator except the tether's own mass. $\endgroup$ Jun 22, 2022 at 18:19
-
$\begingroup$ @Alexander ... and all the other hazards noted in this answer. $\endgroup$– Ian KempJun 23, 2022 at 14:17
-
$\begingroup$ @IanKemp Of course. I wanted to emphasize the thing that hadn't been brought up yet. $\endgroup$ Jun 23, 2022 at 14:18
-
$\begingroup$ I don't see why it has to be geostationary. Afaik, that is only true for actual "free" orbits - without a tether. The tether can provide a lateral force creating a stable non-geostationary orbit can't it? $\endgroup$ Jun 23, 2022 at 16:01
-
1$\begingroup$ @KlaasvanAarsen The side force of an off-equator tether will be much too small to prevent north-south movement -- it may warp the orbital plane, but you'll trade off altitude, which will alter period; the top end will wander in a figure 8 or chaotic figure as seen from the foot (depending how far off it is). However... $\endgroup$ Jun 23, 2022 at 16:03
If strictly ideal, then yes equatorial.
If it is desirable to have the idealized ladder that has a fixed earth connection, it must be geostationary which requires positioning above/along the equator.
If not touching earth, no.
If it is acceptable to have the earth end mostly in one region with no earth contact. Ie earth end sits at an elevation 10 to 40 Km above earth. Then geosynchronous orbit is a reasonable possibility/option. The high elevation makes a big difference to ease of construction thus cost.
Space hooks could be almost wherever.
If center of mass is not at geostationary level/height. Then it would be more correctly labeled a space hook and a space hooks can be in many more possible orbits. Also a small space hook can be constructed and placed into orbit within five to ten years if any of the major space organizations decided to prioritize it.
Space hooks would only make ground momentarily if they do, majority would have closest approach to earth measured in kilometers. Cost to orbit would be the price of a flight of a high altitude aircraft.
Other satellites, debris is an issue for all.
I am surprised no other answer spoke about that (maybe I read to fast), but in the original designs I read, the best place for the "space port"/anchor on earth (if you choose to have it at "ground" level) is neither on the equator nor any other static place, but a platform in the ocean that can move on the surface of the earth. That way :
- you can move it to avoid debris/satellite in orbit
- more importantly, you can avoid storms. In most designs, the tether is a nanotube fiber, which is, among others, extremely sensible to high temperatures that could be caused by lightnings
For more information about the concrete prospects of these space elevators on earth and why it is supposed to be anchored in the oceans, there is the 2003 Nasa Study (this is an old project) and the last 2021 engineering study from ISEC
-
1$\begingroup$ I'd question whether what amounts to a huge barge with a tower on it can be agile enough to avoid a hurricane (or whatever they're calling them in the Pacific these days). Especially since track forecasting is still more art than science and the storms can be a couple hundred kilometers across. $\endgroup$ Jun 22, 2022 at 14:30
-
1$\begingroup$ You mean, like, some other planet? Tropical storms form almost anywhere within 30 degrees of the equator, on the Atlantic, Pacific, and Indian Oceans. And don't forget, if you change longitude, you have to change the orbit of the top station, which invariably requires changes in altitude... $\endgroup$ Jun 22, 2022 at 15:11
-
1$\begingroup$ Space elevators primarily work due to the tension load at the base. It has to hold the elevator structure and space load in place. If you put this on a boat, it would be yanked off the ocean faster than a kitesurfer in gusty winds. $\endgroup$– JimJun 22, 2022 at 15:15
-
2$\begingroup$ @Jim You put your satellite into a high enough orbit such that there is some tension in the cable, but not too much. You need the tension to be enough to support whatever load you want to transport upwards, but not so much that it's going to snap the cable. Think about how fat you'd need the cable to be to lift, say, a cargo ship out of the ocean: it becomes obvious that the cable would snap before it would support the cargo ship, therefore a cargo ship must be heavy enough to support the earth end of the cable. Work down, and anything larger than a few times the max payload would do $\endgroup$– canton7Jun 22, 2022 at 16:06
-
1$\begingroup$ The primary problem with an ocean platform for the base is the massive logistical nightmare of now having to transship everything out to a very inconveniently located base. Similar issues, to a lesser extent, are involved if you pick a high elevation base along the equator (shaving of several thousand feet of length would be nice, but is it worth the inconvenience of everything going up the beanstalk needing to be transported to a remote mountaintop first?). $\endgroup$ Jun 22, 2022 at 19:41
Almost all of the answers presume that the elevator has to stay in place. (Thank you, @Carm) This isn't the only answer, it's just the best one.
An alternative answer would be to have the bottom of the platform perform a figure-8 over a set area of land. If a geosynchronous satellite isn't directly over the equator, it will make a loop over an area every 24 hours. This could be used to take advantage of multiple launch points, with a "space hook," or landing platform that would grab the flying craft and drag it upward.
The challenges of this design are that you have to perfectly balance the orbit so that the end of the tether drags through the atmosphere. The size of the loop would be largely limited by atmospheric drag on the tether which, over time, would attempt to drag the counterweight back to earth.
-
$\begingroup$ This made me wonder, provided we have some flexible way to attach the cable so that it has some degree of freedom, wouldn't it be possible to have the attachment point in Papua New Guinea, Australia or Japan and let the station draw an 8 around it? Considering the length of the cable, the angle would be barely noticeable. It would also require some creative way to attach the lift, though 🤔 $\endgroup$– armandJun 23, 2022 at 9:43
-
1$\begingroup$ I said something silly. The angle would be up to 20 degree-ish, so quite noticeable. All the more fun. $\endgroup$– armandJun 23, 2022 at 9:49
-
$\begingroup$ The key piece of information is that the cable doesn't move in these scenarios. It's like the needle on a phonograph, with the world moving underneath it. $\endgroup$ Jun 23, 2022 at 15:30
Not necessarily
While it is necessary that a tethered space elevator must stay on equatorial orbit, there is no necessity to attach this elevator at the equator.
@Zeiss Ikon is quite right to point this out, let's me elaborate a little bit more on this.
Think about Tetherball. The ball is attached to the top of a pole, but its plane of rotational does not contain this top - the ball spins somewhere below it. Similarly, it is theoretically possible to design a space elevator which center of mass sits on geostationary equatorial orbit, while its tethering point is located anywhere on Earth's surface. Moreover, our space elevator can have multiple points of attachment, which can be very useful if multiple nations are sharing the same space platform.
What's complicates things though is tether's own mass. Ideally, we should have massive space platform and nearly massless cable, but realistically this is not possible. All space elevator design require massive cable, if this cable is not vertically upright, things get complicated. Now think about tetherball on a massive iron chain. It would still spin, but its behavior would be different. Overall, the requirement to attach the tether not on equator can be a project killer.
space elevators work best at the equator, and at the poles, and everything else is chaos
Considerations about satellites are valid, but once someone is investing in a space elevator they can invest in satellite cleanup or protection.
What you want to consider is the strain caused by the orbit. In an equatorial elevator the strain works in your favor, the constant angular speed at the end of the elevator pulls the whole structure up. The orbit is circular, and in line with your structure. In a polar elevator you have no orbit and so no strain to keep the elevator in place. This isn’t good, but a strong enough elevator can support itself. Further more, the elevator just holds itself upright with very little other strain. In a non-equatorial and non-polar elevator you have a problem. The end and the center of mass have circular orbits, but not in line with the center of mass of the earth. This means that your elevator will strain towards the equator as the elevator tries to follow a normal orbit and has to apply radial force to maintain the different orbit.
This means that the force you have in the each direction is based on the azimuth of the elevator in relation to the earth. The upwards direction is the cos(azimuth)*cos(azimuth) of the azimuth, and the sideways force is the cos(azimuth)*sin(azimuth). So to maximize upward force, azimuth must be 0. To maximize the sideways force, you sit a 45 degrees of the equator, and have half the normal upwards force in the sideways direction, and only an equal half in the upwards direction also. Near the poles at 90 degrees the upwards is zero, but so is the lateral.
Therefore, build at the equator to get to equatorial orbit and for low cost elevators, build at the poles for polar orbits and high strength elevators, and build anywhere else to flex on Kardashev type 1 civilizations.
If you have multiple base stations, the orbiting part of the elevator can be above the equator, with two base stations, both equidistant from the equator at opposite latitudes. It should be simple trigonometry to figure out how far from the equator the stations need to be to make the tethers miss the most debris-filled latitudes. Basically the tethers make a big "A" shape with a ring of debris passing between them.
It could also be possible to move the orbiter by varying the lengths of the cables. Reeling in the cable at the north base and letting out more cable at the south base will move the orbiter north. Reeling in the cable at the south base and letting out cable at the north base will move the orbiter south. If you have 4 ground stations, you can move the orbiter east and west as well.
An elevator cable anchored away from the equator (let's say north) will form a curve asymptotic to the equatorial plane. It will not be straight, because it is not ‘vertical’ with respect to the forces on it; and its top end will be some small but nonzero distance north of the equator, because of the northward component of tension on it, balanced by the southward component of Earth's gravity (which is zero on the equator). Centrifugal force pushes away from Earth's axis, not its center.
There must be a maximum latitude for the anchor, at which the sag makes the cable tangent to the ground. I lack the skill to find what that latitude is; it depends on the amount of taper in the cable.