Which planet could we build a space elevator with current technology?

In My Humble Opinion
21

In principle, you could use anything as a counterweight, if it’s far enough out, in much the same way that you can counterbalance anyone on a teeter-totter. But you’ll get the best results from something absolutely huge, like a captive asteroid, much, much larger than the ISS.

And even if you’re completely confident in your equipment, there’s still good reason for getting into low-Earth orbit first, and then boosting from there. There are different properties which make one rocket engine “better” than another. One rocket engine might have higher thrust than another, while another might have a higher exhaust velocity.

High exhaust velocity is good, because it has a very strong impact on the efficiency of the engine: The faster your exhaust, the less propellant you need to achieve a given change in velocity (also called “delta-V”). But in order to take off from a planet, you absolutely need high thrust: Certainly greater than the weight of your vehicle, and ideally much greater, or you’re just wasting a bunch of energy on fighting gravity.

And in practice, there’s often a tradeoff between these two, for rocket engines. At one end are things like ion engines, with extremely high exhaust speeds and extremely low thrust, but even short of that extreme, with normal chemical-burning rockets, there’s still usually a tradeoff. So you use one (or more) set of engines, with high thrust, to get you into orbit, and then once you’re in orbit you can use a different set of engines, with high exhaust velocity, to get you from orbit to wherever you’re going.

The short answer is that low orbit takes half the energy of escaping, and energy is proportional to the square of the velocity. But that just pushes the question back one step, to why low orbit takes half the energy of escape… I think a proper explanation of that would really require calculus, but maybe someone can come up with a way around that.

22

Following on to what @Chronos almost said …

What matters is not size but weight (mass actually). The ISS has large dimensions but was built as light as possible to minimize the cost to get it into orbit. It’s roughly 250 x 350 feet and weighs a bit shy of a million lbs. So about as big & heavy as four empty 747s. It’s big, it’s floppy, its flimsy. It’s not very heavy. None of those things make for good counterweights. It’s only advantage is that the mass is already up at 250 miles above the surface. Which gives us a head start in re-orbiting it up to 30,000 or 40,000 miles up where it could serve as a counterweight. But maybe not much of a head start.

Said another way … 500 tons of counterweight is a tiny drop in the necessary bucket unless you want to put it waaaay out on the end of a very long tether. For best economy of launch you’d like a lot more weight installed much closer in.

An interesting question is whether it’d be cheaper to loft ordinary concrete up from the surface, versus slice off a hunk of the Moon & spiral/ellipse that down, or bring in an asteroid from the Belt. Right now the latter two are impossible, so the economic question is easy: Door #1 FTW. But just thinking from a delta V perspective I wonder …

23

That, and the ISS isn’t very robust, either (because in orbit, it doesn’t need to be). A counterweight would effectively be hanging from the point where the tether attaches to it (“hanging” upwards rather than downwards, but the effect is the same). Try to hang the ISS from a single point, and it’d collapse.

Actually, that’d probably be true of an asteroid or Moon rocks or whatever, too. Our counterweight would probably need some of the tether material wrapped around it to hold it. Which is easier if the counterweight is a compact, dense object like a sphere of rock.

Bringing concrete up from Earth certainly isn’t the cheapest option. When what you want is just dumb rock mass, getting it from the Moon or an asteroid will probably be cheaper even with the R&D costs. And even if the R&D costs turn out to be prohibitive for some unexpected reason and you have to send mass up from Earth, there’s no reason for that mass to be concrete: We’d just send up more of the cable. The longer the cable is, the less counterweight mass you need, and there are also benefits to the cable being significantly longer than geosynch height, anyway: Make it long enough, and you can launch deep space missions just by taking them out to the end of the cable and letting go.

The real game-changer would be if we could make cable material from raw materials found in space. This isn’t completely implausible, on the face of it: The cable would most likely be carbon in some form, and a lot of asteroids are rich in carbon. But we don’t know yet what sort of industrial infrastructure will be needed to turn raw carbon into cable, or how easy it will be to launch that infrastructure and run it in space (heck, it might even turn out that it’s easier to make cable in zero-g and hard vacuum).

24

One of the assumptions underlying using a counterweight at all is that tether material is stupid expensive. So that as betwen a few 10,000s of miles of tether and a few 10,000s of tons of rock, moving the rock is cheaper than making the tether.

If that cost equation favors tether over rock, then as you say, the best move is to skip the rock entirely and build a very long tether as an interplanetary mass launcher / slingshot.

25

But part of the reason why tether material would be expensive is that it has to be launched into space, and a lot of it needs to be launched with old, expensive rockets.

26

Assuming there’s no counterweight, but just a really long tether going out. How long will that outward tether be?

If there’s a station† at the end of the tether (which will shorten the tether but not by all that much), it’ll experience an outwards force which, to the station personnel, will feel like gravity. How strong will that force be?

†Not the ISS or anything like it. Something more robust, meant to handle gravity.

27

I’ve considered it in the past, but now it seems I’ve got adequate push for Kerbal :slight_smile:

I…see. Makes sense! It boils down to the capacity of a rocket to facilitate mass flow through its nozzles. In theory, if we could scale the ion drive to spit out a good chunk of mass in short time, that’d be the ideal rocket because it suits all situations. But since rocket science already has been ~100 years old, I’m not betting on a breakthrough.

My goodness, that’s right! I completely forgot that nuance.

Wow, I can imagine it pretty clearly now: the way to go is 1st build a space base on some body (Moon, Mars, asteroid) using old-fashioned rockets, then manufacture and transport the cable (and a big rock) from there, and in the final stage just reap benefits from the SE. The whole process like that would delay the SE dozens of years and probably out of our lifetime, but it also increase the plausibility by like an order of magnitude. It’s almost like I can see the bittersweet future.
The Moon is obviously the prime candidate for such space industrial base, being close and having similar composition to Earth. I came across this article before opening this thread, but its relevance only dawns now. Dangling a tether from our big ol’ satellite toward us to facilitate the Earth’s SE building is tempting, but how would we deal with the wobbling of the Moon? At such distance, it’d turn to a nasty whip.

Quick calc in my head: an all-tether SE will have roughly the same lengths extending both from GEO, with the outer part longer. Since the inner part is 36k km, I’ll assume the outer part is 45k km. At that distance, an orbit would be around twice the length of GEO, so ~2 days. The outward ‘feeling’ is the difference between that free-fall orbit (2 days) and the actual SE rotation rate (exactly 1 day), thus ~1 day. Since we can note that the centripetal force(s) is roughly 2 times normal gravity at that point of space, we can use just that gravitation value for the ‘outward-feeling’ pressure. Which is around (6371+36k+45k) / 6371 = 13.5 times weaker than on Earth. In other words, a 82kg person would feel a 6kg ‘pressure’ on such a station.
Well, inb4 the experts here correct my calc. It’d be great to learn where I’m wrong :slight_smile:

28

OK, I was contemplating setting up the differential equation to find the needed length of cable for an all-cable counterweight, and then I realized that it’d depend on how much payload weight you want the cable to be able to carry, and then I realized that I’m neither getting paid nor course credit for this, and I don’t do differential equations without pay nor course credit, but…

Of course, for a rough estimate, this is probably good enough.

That said, your calculations from there are a bit off. At twice geosynchronous distance, the centrifugal force on the elevator will be twice as great as at geosynch (F_c = m\omega^2r), while the gravitational force will be one quarter of what it is at geosynch (F_G = G\frac{Mm}{r^2}).
…Except it looks like your errors in the inverse-square law mostly cancel each other out. Using equal lengths on both sides, I find that weight at the counterweight station would be about 7% of what it is on the surface of the Earth.

29

I’d pay you, but I’m all out of QuatlooCoin†

A rough estimate is all I really was interested in, so good enough.

My intuition is very wrong in this situation, since I figured it would be roughly 1g.

Thanks for the work.

†Unsurprisingly, someone has created a Quatloo cryptocurrency. Most recent quote is $0.00493 … don’t think I’m going to invest.

30

At that price, like a space tether, it’s got nowhere to go but up. :wink:

31

For what it’s worth, that was my intuition, too. I imagined a “Topsy Turvy Station” at the counterweight, where everyone walks about upside-down (that is, with their heads pointing towards the Earth). But however you get around in 7% g, it probably wouldn’t much resemble “walking”.

32

It’s easy enough to type F=GMm/r^2.

It can be hard to grok what that really means as we get well away from our current r of ~4000mi and the corresponding F we’re all so used to considering a constant.

33

That’s the gravitational force towards the Earth. But out at the far end of the tether, the net force will be the other direction.

34

Wait a minute while I count up how many hands I have… (two, so I guess this is still just “the other hand”): The assumption of approximately equal lengths of tether above and below geosynch was based on the assumption of equal weight at both ends. But it’s clearly not equal weight on both ends, so that assumption is bad.

The actual needed length must be somewhere in between the equal-lengths value and the equal-weights value. It’ll still be less than the equal-weights value, because we need to add up the weight of the cable along the cable’s whole length.

This would still require some calculus to get right, but with the right simplifying assumptions, it could be a mere integral, not a differential equation. An integral, I just might do for free, sometime later today.

35

Yes. Agree completely. Gravity is just half the big story; centripetal force is the other half. Which is true all along the tether, with local gravity decreasing as you climb and the local centripetal force increasing as you climb.

I was speaking to what I understood was @dtilque’s initial assumption in his ballpark math that the acceleration of Earth gravity was 9.8ms^-3 everywhere along the tether’s length. Which you pointed out accurately by the relevant formulas was not correct. Which he immediately acknowledged.

My point was simply that when trying to ballbark a solution to a complicated problem in a more exotic realm, it’s easy for any of us to inadvertently apply simplifications that work fine for terrestrial living at the bottom of an atmosphere and a gravity well. But that badly invalidate our ballpark work.

Sorry if I was unclear. I now see my comment made a lot more sense within the context in my head than within the context visible in the thread. Oops.

36

Unrelated to any of the above, one of the questions I’ve always had about space elevators is the electrodynamic and magnetodynamic consequences.

If the tether is (monomolecular?) carbon fiber, it will be electrically conductive, albeit with probably high resistivity. Although given a big enough tether cross section in the fatter parts, the resistance per meter of length might be pretty darn low. For sure the antenna length, at multiple tens of thousands of miles, is plenty big.

The tether will be substantially stationary relative to the static component of the Earth’s intrinsic fields. But the Earth’s fields are not 100% static on their own. More significantly the Earth fields are constantly and erratically being buffeted by the solar wind, etc. As the Earth rotates, dragging the tether with it, the tether’s axis relative to that solar wind will be constantly changing.

I’m not sure what kind of charges and forces might build up on the tether, nor how they might safely be dissipated. But I expect that too will be an obstacle to design & deployment.

37

I’ve long wondered how the ascent cars will be powered on a space elevator. It’s 36,000 km long, so you’re not going to be able to power it only from the two ends. Resistance losses over thousands of kilometers would make it impossible unless someone invents a room-temp superconductor. Adding a large solar array every 50-100 km would add a lot of weight, perhaps too much weight. The cars could carry their own solar panels, but that means either they stop at night or carry a big heavy battery.

But your post gave me the idea that they could harvest energy from the tether itself. Combine that with solar panels on the cars and maybe they don’t need any other source of power.

38

Carbon nanofiber is highly conductive, but it’s not that conductive. Over a range of tens of thousands of kilometers, resistive losses will kill you.

The usual proposal I’ve seen is microwave power beamed up from a powerplant on the surface. But you can also get away with a lot by connecting the cars together with another tether. Remember, in the rotating reference frame, the cars are only going uphill until they reach geosynch height. Past that, they’re going downhill. So what you can do is build one or two extra cables, each with an elevator car attached every hundred km or so. These cables will be balanced in place, only loosely connected to the stationary cable, just enough to stabilize them. Cable A will start off with its lower end at the surface, and Cable B will start off with its lower end 100 km above the surface. Fill up the bottom car on A with whatever your cargo is, and lift cable A up 100 km (which takes very little energy, since the cable as a whole is very close to its balance point). Meanwhile, you’re lowering cable B to the surface, which likewise takes very little energy. Then you transfer cargoes from A to B (including a new cargo from Earth loading onto the bottom of B), and then lower A again and raise B again. Net result is almost no energy expenditure: The cargoes are effectively being siphoned over the hill.

Of course there’s no free lunch. In the non-rotating reference frame, it’s clear that energy is coming from somewhere: Specifically, doing this will slow down the rotation of the Earth. If, that is, cargoes are only going one way. If we have space elevators (or other cheap launch systems-- Different systems work on different worlds) elsewhere in the Solar System, then we could pay back the energy borrowed from the Earth’s rotation by sending cargoes the other way (even if it’s just boring dumb-mass cargoes like iron ore from Mars).

39

Oops. I totally couldn’t remember the centrifugal formula and thus relied solely on gravity, but I messed up the r^2 part too. Arrived at the ‘correct’ answer not in any proud way, but at least my consolation is that my intuitions are rarely ‘wrong’. Tks @Chronos ! Now I learned that at twice the distance, an object’s angular speed is halved… uh, right?

And if it ends up ~45k km, then I’ll be golden :wink: To my credit, I did think of a tapering gravity (weight) of an extending cable when making the conjecture.

Huh? Assuming that you meant electrical resistivity, the whole doesn’t make sense to me. How could something be both conductive and resistant?? As I rushed to read a bit more about carbon fiber, other properties are mentioned, but few point to electrical resistance. Could you clarify this stuff for me? Cause without it I can’t process the flowing discussion after the post ::
Though 1 thing I suspect is that the cable will likely not be monomolecular. Compounds are usually stronger than pure materials…

Your description draws my imagination to a ski gondola lift, but I have troubles with it too. A link to some rough sketches will help a lot :slight_smile:

40

Every material has electrical resistance. (Ignoring superconductors.) How much varies by the material. A “conductor” is simple a material whose resistance is low enough that we can build something electrically useful out of it without wasting too much electricity as heat along the way.

Copper has resistance, but not much. Silver has even less resistance, so would be a better conductor to wire the whole world with. But silver’s much higher cost means copper gets used instead despite it’s higher resistance. Long distance transmission lines are made of steel and aluminum. The inherent resistance is higher than copper, but the lower cost and greater strength make that the optimal choice.

The thing in your oven or toaster that glows red and makes heat is a conductor. It completes a circuit and current flows through it. But by design it is a really shitty conductor and “wastes” most of the electricity as heat. In this case heat is the desired product, so that’s good.

With that conceptual background now on board …

“Resistivity” is the concept of the inherent obstacle a bulk material makes to electrical flow. See Electrical resistivity and conductivity - Wikipedia

“Resistance” is the practical obstacle a particular sample of a particular dimension offers to electrical flow. As applied to things like wire, a resistance value depends on cross section, where wider has lower resistance, and length, where longer has higher resistance. Once we know those two factors, we can multiply by the resistivity of the material to know what the actual specific resistance value of our conductor (or non-conductor) is .

My point was that a carbon fiber would have high inherent resistivity, but if the “conductor” was 30 feet in diameter, the resistance per unit length would be low. @Chronos tells us the resistivity of carbon nanotubes is actually rather low. So the electrical conductivity of the bulk tether would be even higher than I feared.

Which is generally not what you want for the tether interacting with Earth’s magnetic field, the solar wind, etc. For that situation, you’d prefer something close to electromagnetically inert, not something that wants to enthusiastically couple to all that electromagnetic energy out there.


And with that you missed something vital.

The very best possible bonds between molecules of any sort are vastly weaker than the bonds that hold molecules together internally. AIUI it’s generally accepted that no material can ever be created with ordinary inter-molecular bonds that’s both strong enough and light enough to work as an Earth-based space tether.

The only way to square that circle is to make the material entirely of intra-molecular bonds. Which means the whole tether, umpteen thousands of miles long must be a single molecule.

Or at least that’s one theory.

As you suggest, molecules need not be made from a single element. But it turns out the strongest most regular bond lattice is pure carbon. Adding anything else makes it weaker, not stronger.