Wild idea for new Space Elevator we (claimed) can build now.
September 12, 2019 10:17 AM Subscribe
From MIT Tech Review, a new Space Elevator called "Spaceline". The difference is this elevator has its anchor point not on the earth but on the moon. And it just dangles "close" to earth.
This makes it run out through the Lagrange point where the Earth and Moon's gravity cancels and where the authors claim would be a great new construction point for all things space. The authors (Zephyr Penoyre from the University of Cambridge in the UK and Emily Sandford at Columbia University in New York) say a demonstration project could be built for "billions".
Orig. Paper:
The Spaceline: A Practical Space Elevator Alternative Achievable With Current Technology
This makes it run out through the Lagrange point where the Earth and Moon's gravity cancels and where the authors claim would be a great new construction point for all things space. The authors (Zephyr Penoyre from the University of Cambridge in the UK and Emily Sandford at Columbia University in New York) say a demonstration project could be built for "billions".
Orig. Paper:
The Spaceline: A Practical Space Elevator Alternative Achievable With Current Technology
xkcd offshoot "What If?" touched on this.
posted by hanov3r at 10:32 AM on September 12, 2019 [14 favorites]
posted by hanov3r at 10:32 AM on September 12, 2019 [14 favorites]
This is cool! The design is really profoundly different from the usual space elevator setup. As the article notes; this cable would be held in pace by Earth's gravity. Not centrifugal force per usual, which puts a lot more strain on the cable since it's moving so much faster. It's also quite different in what it accomplishes. It lets you get from geostationary orbit to the Moon; a space elevator lets you get from Earth to a geostationary orbit.
I'm confused about something. How does it work that one end is fixed on the moon, which most definitely is not geostationary, but the other end is hovering in one spot in a geostationary orbit? I feel like I'm missing something fundamental to even be asking the question. Most of the paper talks about "the altitude of a geostationary orbit", so maybe that's the wiggle-room I need to make sense of it, but then other parts specifically say "geostationary orbit".
posted by Nelson at 10:35 AM on September 12, 2019 [2 favorites]
I'm confused about something. How does it work that one end is fixed on the moon, which most definitely is not geostationary, but the other end is hovering in one spot in a geostationary orbit? I feel like I'm missing something fundamental to even be asking the question. Most of the paper talks about "the altitude of a geostationary orbit", so maybe that's the wiggle-room I need to make sense of it, but then other parts specifically say "geostationary orbit".
posted by Nelson at 10:35 AM on September 12, 2019 [2 favorites]
Rumors spreadin' 'round
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
posted by Greg_Ace at 10:38 AM on September 12, 2019 [15 favorites]
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
posted by Greg_Ace at 10:38 AM on September 12, 2019 [15 favorites]
I don't want to look at the moon's damned, lit-up tail, on autumn nights.
And from the moon the vessels come and go, The Sphinx, The Michelangelo...
posted by Oyéah at 10:38 AM on September 12, 2019 [2 favorites]
And from the moon the vessels come and go, The Sphinx, The Michelangelo...
posted by Oyéah at 10:38 AM on September 12, 2019 [2 favorites]
Nelson: The only way I can think of is the fixed point on the moon => to *a* point in geosynchronous orbit height (but really the point at that height is moving around the earth to stay at that height above ground).
posted by aleph at 10:45 AM on September 12, 2019 [1 favorite]
posted by aleph at 10:45 AM on September 12, 2019 [1 favorite]
I'm confused about something. How does it work that one end is fixed on the moon, which most definitely is not geostationary, but the other end is hovering in one spot in a geostationary orbit?
I suspect that the answer is that the end at geostationary orbit altitude is not actually in a geostationary orbit. So to get on to the tether from geostationary orbit, you would need to accelerate to catch it as it swings by. Perhaps they would use a smaller rotating tether.
posted by surlyben at 10:46 AM on September 12, 2019 [1 favorite]
I suspect that the answer is that the end at geostationary orbit altitude is not actually in a geostationary orbit. So to get on to the tether from geostationary orbit, you would need to accelerate to catch it as it swings by. Perhaps they would use a smaller rotating tether.
posted by surlyben at 10:46 AM on September 12, 2019 [1 favorite]
Rumors spreadin' 'round
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
And the math is right
Most ev'ry night
But now I might be mistaken
Hmm hmmm hmmm
Have mercy
posted by nubs at 10:47 AM on September 12, 2019 [11 favorites]
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
And the math is right
Most ev'ry night
But now I might be mistaken
Hmm hmmm hmmm
Have mercy
posted by nubs at 10:47 AM on September 12, 2019 [11 favorites]
How does it work that one end is fixed on the moon, which most definitely is not geostationary, but the other end is hovering in one spot in a geostationary orbit?
hanov3r's XKCD link covers this quite graphically - the Earth-end would be moving quite rapidly relative to the earth's surface (the exact speed would vary depending on the altitude above the surface). The Moon's speed over the earth's surface is around 450m/s, so how you'd attach your payloads to the bottom of the Spaceline would be... interesting.
This is a space elevator, yes, but not in the traditional sense - it won't help get payload off the surface of the earth any easier or more cheaply. More likely it would be used to shift payload between earth orbit and the moon. Although the orbital dynamics of "catching" the earth end (I'm sure that'll be way less than orbital speed) to attach your payload might be... interesting.
posted by Nice Guy Mike at 10:48 AM on September 12, 2019 [1 favorite]
hanov3r's XKCD link covers this quite graphically - the Earth-end would be moving quite rapidly relative to the earth's surface (the exact speed would vary depending on the altitude above the surface). The Moon's speed over the earth's surface is around 450m/s, so how you'd attach your payloads to the bottom of the Spaceline would be... interesting.
This is a space elevator, yes, but not in the traditional sense - it won't help get payload off the surface of the earth any easier or more cheaply. More likely it would be used to shift payload between earth orbit and the moon. Although the orbital dynamics of "catching" the earth end (I'm sure that'll be way less than orbital speed) to attach your payload might be... interesting.
posted by Nice Guy Mike at 10:48 AM on September 12, 2019 [1 favorite]
How does it work that one end is fixed on the moon, which most definitely is not geostationary, but the other end is hovering in one spot in a geostationary orbit?
It's not geostationary, it orbits with the Moon.
posted by signal at 10:49 AM on September 12, 2019 [1 favorite]
It's not geostationary, it orbits with the Moon.
posted by signal at 10:49 AM on September 12, 2019 [1 favorite]
Because what could possibly go wrong with us tugging the moon out of it's normal orbit.
posted by allkindsoftime at 10:49 AM on September 12, 2019 [9 favorites]
posted by allkindsoftime at 10:49 AM on September 12, 2019 [9 favorites]
Yeah. I see this as more mining the moon for easy materials to the Lagrange point for installations/construction there. With a cable coming down towards Earth as far as the tech permits. (Of *course* all of this is a fantasy, you do yours and I'll do mine.)
posted by aleph at 10:54 AM on September 12, 2019
posted by aleph at 10:54 AM on September 12, 2019
It's not geostationary, it orbits with the Moon.
Yeah, I'm skimming the paper and they're calculating energy needs to meet up with the termination point moving at the moon's orbital speed in low earth orbit.
Man, this is cool. Also from the paper:
"One crucial point to remember about the spaceline is that it is traversable without any energy cost. A solar-poweredclimber, gripping the line with two wheels (in the simplest design) can freely move up and down the thread withoutany cost in fuel."
posted by lumpenprole at 10:57 AM on September 12, 2019
Yeah, I'm skimming the paper and they're calculating energy needs to meet up with the termination point moving at the moon's orbital speed in low earth orbit.
Man, this is cool. Also from the paper:
"One crucial point to remember about the spaceline is that it is traversable without any energy cost. A solar-poweredclimber, gripping the line with two wheels (in the simplest design) can freely move up and down the thread withoutany cost in fuel."
posted by lumpenprole at 10:57 AM on September 12, 2019
Rumors spreadin' 'round
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
And the math is right
Most ev'ry night
But now I might be mistaken
Hmm hmmm hmmm
Have mercy
But how, how, how, how?
Huh? How, how, how??
posted by Greg_Ace at 11:00 AM on September 12, 2019 [11 favorites]
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
And the math is right
Most ev'ry night
But now I might be mistaken
Hmm hmmm hmmm
Have mercy
But how, how, how, how?
Huh? How, how, how??
posted by Greg_Ace at 11:00 AM on September 12, 2019 [11 favorites]
I suspect that the answer is that the end at geostationary orbit altitude is not actually in a geostationary orbit. So to get on to the tether from geostationary orbit, you would need to accelerate to catch it as it swings by. Perhaps they would use a smaller rotating tether.
To be in geostationary orbit you need to revolve around the Earth once per day. The moon revolves around the Earth every 28 days. So you actually need to go much slower to hit the space elevator.
My rough calculations are that true geosynchronous orbit is about 166,000 mph, while the end of the cable would be traveling at only about 6,000 mph. (Relative to a stationary Earth, I guess you could subtract about 1,000 mph to get the acceleration relative to a rotating Earth.)
I don't know how much of the actual fuel budget goes into altitude vs sidereal velocity, but this seems like it might be a lot easier to get to the cable than to get into a true geostationary orbit.
posted by bjrubble at 11:01 AM on September 12, 2019 [4 favorites]
To be in geostationary orbit you need to revolve around the Earth once per day. The moon revolves around the Earth every 28 days. So you actually need to go much slower to hit the space elevator.
My rough calculations are that true geosynchronous orbit is about 166,000 mph, while the end of the cable would be traveling at only about 6,000 mph. (Relative to a stationary Earth, I guess you could subtract about 1,000 mph to get the acceleration relative to a rotating Earth.)
I don't know how much of the actual fuel budget goes into altitude vs sidereal velocity, but this seems like it might be a lot easier to get to the cable than to get into a true geostationary orbit.
posted by bjrubble at 11:01 AM on September 12, 2019 [4 favorites]
Rumors spreadin' 'round
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
And the math is right
Most ev'ry night
But now I might be mistaken
Hmm hmmm hmmm
Have mercy
But how, how, how, how?
Huh? How, how, how??
Just let me know if you wanna go
To that line out in our orbit
They got a lot of nice payloads
posted by nubs at 11:08 AM on September 12, 2019 [10 favorites]
That MIT town,
'Bout that 'vator at Lagrange
(Y'all know what I'm talkin' 'bout)
And the math is right
Most ev'ry night
But now I might be mistaken
Hmm hmmm hmmm
Have mercy
But how, how, how, how?
Huh? How, how, how??
Just let me know if you wanna go
To that line out in our orbit
They got a lot of nice payloads
posted by nubs at 11:08 AM on September 12, 2019 [10 favorites]
"One crucial point to remember about the spaceline is that it is traversable without any energy cost. A solar-poweredclimber, gripping the line with two wheels (in the simplest design) can freely move up and down the thread withoutany cost in fuel."
Yeah, that's bollocks. Or more accurately, that's two separate claims mashed together either accidentally or deliberately and obfuscating the situation.
It's clear that there is, in fact, an energy cost, as they admit in the following sentence - if there was no energy cost, you wouldn't need a solar-powered climber/descender. Now, it may be true that there's no fuel cost in the sense of chemical fuel... but here's my question: How much time does the structure spend in the shadow of either a) the Earth or b) the Moon? What's the energy requirement and how much sunlight do you have to convert to electricity to power the climber/descender?
I don't know how much of the actual fuel budget goes into altitude vs sidereal velocity, but this seems like it might be a lot easier to get to the cable than to get into a true geostationary orbit
The answer is Not Much vs Most of it, respectively. You're only adding a couple of hundred km to your altitude; you're adding about 6.8km/s of "lateral" speed. The problem is, you're effectively lobbing something ballistically at the Earth-side end of the tether from the surface, with nowhere for said payload to go if you miss - you'd be at earth orbit altitude without the velocity. There's only one place to go if you don't catch the tether end - back down, hitting the atmosphere at a relatively low speed (compared to orbital speeds) but potentially fast enough to cause heating problems.
posted by Nice Guy Mike at 11:09 AM on September 12, 2019 [2 favorites]
Yeah, that's bollocks. Or more accurately, that's two separate claims mashed together either accidentally or deliberately and obfuscating the situation.
It's clear that there is, in fact, an energy cost, as they admit in the following sentence - if there was no energy cost, you wouldn't need a solar-powered climber/descender. Now, it may be true that there's no fuel cost in the sense of chemical fuel... but here's my question: How much time does the structure spend in the shadow of either a) the Earth or b) the Moon? What's the energy requirement and how much sunlight do you have to convert to electricity to power the climber/descender?
I don't know how much of the actual fuel budget goes into altitude vs sidereal velocity, but this seems like it might be a lot easier to get to the cable than to get into a true geostationary orbit
The answer is Not Much vs Most of it, respectively. You're only adding a couple of hundred km to your altitude; you're adding about 6.8km/s of "lateral" speed. The problem is, you're effectively lobbing something ballistically at the Earth-side end of the tether from the surface, with nowhere for said payload to go if you miss - you'd be at earth orbit altitude without the velocity. There's only one place to go if you don't catch the tether end - back down, hitting the atmosphere at a relatively low speed (compared to orbital speeds) but potentially fast enough to cause heating problems.
posted by Nice Guy Mike at 11:09 AM on September 12, 2019 [2 favorites]
I was curious what the longest rope ever made was. Apparently a 6-in diameter, 24-mile long natural fiber rope was constructed in 1842 to serve as a barrier across the Strait of Gibraltar, by the British Navy.
However, there doesn't seem to be any technical limitation to how long a rope one can create, using existing technology. One assumes it gets to be something of a pain to manage all the material after a certain point.
The longest undersea communications cable is 24,000 miles long, although I don't think it's continuous—it's really more of a series of point-to-point links, I think. Although that's probably not an unfair comparison to a moon-Lagrange cable, because presumably you wouldn't boost the whole thing up in one piece... you'd need to bring it up in sections and assemble it in orbit. If we can do that and lay it across the ocean floor, I believe that it's probably technologically feasible to do it in orbit.
I'm skeptical of the cost, though.
posted by Kadin2048 at 11:16 AM on September 12, 2019 [2 favorites]
However, there doesn't seem to be any technical limitation to how long a rope one can create, using existing technology. One assumes it gets to be something of a pain to manage all the material after a certain point.
The longest undersea communications cable is 24,000 miles long, although I don't think it's continuous—it's really more of a series of point-to-point links, I think. Although that's probably not an unfair comparison to a moon-Lagrange cable, because presumably you wouldn't boost the whole thing up in one piece... you'd need to bring it up in sections and assemble it in orbit. If we can do that and lay it across the ocean floor, I believe that it's probably technologically feasible to do it in orbit.
I'm skeptical of the cost, though.
posted by Kadin2048 at 11:16 AM on September 12, 2019 [2 favorites]
There's only one place to go if you don't catch the tether end - back down, hitting the atmosphere at a relatively low speed (compared to orbital speeds) but potentially fast enough to cause heating problems.
In a way, that's a safety feature, not a bug, since it prevents any terrifying drifting forever in space kinds of situations, and landing safely is a problem that has already been solved in a number of ways.
posted by surlyben at 11:20 AM on September 12, 2019 [1 favorite]
In a way, that's a safety feature, not a bug, since it prevents any terrifying drifting forever in space kinds of situations, and landing safely is a problem that has already been solved in a number of ways.
posted by surlyben at 11:20 AM on September 12, 2019 [1 favorite]
A Lofstrom loop also does not need any new materials, is considerably smaller, and has the advantage of actually being able to get things into orbit.
posted by joeyh at 11:33 AM on September 12, 2019 [1 favorite]
posted by joeyh at 11:33 AM on September 12, 2019 [1 favorite]
Is it grant-writing season?
posted by rhizome at 11:37 AM on September 12, 2019 [3 favorites]
posted by rhizome at 11:37 AM on September 12, 2019 [3 favorites]
Because what could possibly go wrong with us tugging the moon out of it's normal orbit.
The moon is naturally drifting away from the Earth at a rate of 3.8cm/year. "Tugging" it would at most slow that outward drift by an infinitesimal amount, because there is no realistic amount of force we are going to impart that will palpably alter the mechanics of the Earth-Moon system. The masses involved are so far beyond anything we will be capable of affecting in the next few centuries that it's not worth spending further words on.
posted by Ryvar at 11:38 AM on September 12, 2019 [16 favorites]
The moon is naturally drifting away from the Earth at a rate of 3.8cm/year. "Tugging" it would at most slow that outward drift by an infinitesimal amount, because there is no realistic amount of force we are going to impart that will palpably alter the mechanics of the Earth-Moon system. The masses involved are so far beyond anything we will be capable of affecting in the next few centuries that it's not worth spending further words on.
posted by Ryvar at 11:38 AM on September 12, 2019 [16 favorites]
To the tune of Decatur, or, Round of Applause for Your Stepmother!
A bunch of mad scientists, they just couldn't wait for
a rocketship to ship stuff to our nearest lunar neighbor.
So dangling down from the moon to the equator,
they built a big rope that they called a space elevator.
Well things went bad, and it started to tilt
it kind of whipped backwards and wrapped all around itself!
Now the moon, with its shiny new crater,
is known as the universe's biggest self-flagellator.
posted by oulipian at 11:38 AM on September 12, 2019 [4 favorites]
A bunch of mad scientists, they just couldn't wait for
a rocketship to ship stuff to our nearest lunar neighbor.
So dangling down from the moon to the equator,
they built a big rope that they called a space elevator.
Well things went bad, and it started to tilt
it kind of whipped backwards and wrapped all around itself!
Now the moon, with its shiny new crater,
is known as the universe's biggest self-flagellator.
posted by oulipian at 11:38 AM on September 12, 2019 [4 favorites]
I am reading the publication for this and I understand none of the math but I will say I am going to use the phrase "This is not idle theorycrafting" as much as possible now.
posted by Young Kullervo at 11:59 AM on September 12, 2019 [4 favorites]
posted by Young Kullervo at 11:59 AM on September 12, 2019 [4 favorites]
there is no realistic amount of force we are going to impart that will palpably alter the mechanics of the Earth-Moon system.
I mean, I guess legally you can say this in a world where The Rock exists, but I can't imagine why you'd want to.
posted by Etrigan at 12:20 PM on September 12, 2019 [3 favorites]
I mean, I guess legally you can say this in a world where The Rock exists, but I can't imagine why you'd want to.
posted by Etrigan at 12:20 PM on September 12, 2019 [3 favorites]
It looks like the cable hangs down to lower than geostationary orbit altitude, but the article doesn't give a number.
XKCD points out that space isn't high, it's fast. The hard part isn't getting up there, it's moving rapidly to the right once you do (and hauling up all the fuel that lets you move rapidly to the right).
This cable would seem to make that part easier, because it's moving at a lot slower than orbital speed. If the cable were at geostationary orbit height, for example, the end would be traveling at about 1/30th orbital velocity. Orbital velocity is 3km/s, so you'd be talking 100 meters/second. This is a much more reasonable number.
You fire something up there, give it a little push to the right and grab the end of the cable.
posted by It's Never Lurgi at 12:23 PM on September 12, 2019 [5 favorites]
XKCD points out that space isn't high, it's fast. The hard part isn't getting up there, it's moving rapidly to the right once you do (and hauling up all the fuel that lets you move rapidly to the right).
This cable would seem to make that part easier, because it's moving at a lot slower than orbital speed. If the cable were at geostationary orbit height, for example, the end would be traveling at about 1/30th orbital velocity. Orbital velocity is 3km/s, so you'd be talking 100 meters/second. This is a much more reasonable number.
You fire something up there, give it a little push to the right and grab the end of the cable.
posted by It's Never Lurgi at 12:23 PM on September 12, 2019 [5 favorites]
The hard part isn't getting up there, it's moving rapidly to the right once you do (and hauling up all the fuel that lets you move rapidly to the right).
This is a classic case of looking at the problem all wrong. What we really need is a rapid acceleration to the left.
posted by nickmark at 12:32 PM on September 12, 2019 [8 favorites]
This is a classic case of looking at the problem all wrong. What we really need is a rapid acceleration to the left.
posted by nickmark at 12:32 PM on September 12, 2019 [8 favorites]
There was an experiment a few years ago that unrolled a quarter mile long cable/wire. I think it had to be aborted due to massively high static electricity voltage build up. I'm a total space elevator fanboi but I expect there are a whole bunch of science and engineering tests and experiments needed before an actual attempt is made.
posted by sammyo at 12:43 PM on September 12, 2019 [4 favorites]
posted by sammyo at 12:43 PM on September 12, 2019 [4 favorites]
I am reading the publication for this and I understand none of the math but I will say I am going to use the phrase "This is not idle theorycrafting" as much as possible now.
Be warned that this is a good way for e-sports enthusiasts to confuse you for one of their own--theorycrafting is a common practice in the days before major announced game changes, such as the imminent arrival of a new expansion or patch to games like HearthStone or World of Warcraft, among other uses. I'm a little surprised they chose to employ the term in this context, but considering the robust history of flight sims and physics puzzles, perhaps it shouldn't be as surprising to me as it is.
posted by jsnlxndrlv at 12:54 PM on September 12, 2019 [1 favorite]
Be warned that this is a good way for e-sports enthusiasts to confuse you for one of their own--theorycrafting is a common practice in the days before major announced game changes, such as the imminent arrival of a new expansion or patch to games like HearthStone or World of Warcraft, among other uses. I'm a little surprised they chose to employ the term in this context, but considering the robust history of flight sims and physics puzzles, perhaps it shouldn't be as surprising to me as it is.
posted by jsnlxndrlv at 12:54 PM on September 12, 2019 [1 favorite]
"within reach of modern technology" (from the abstract of the original paper)
Same as fusion reactors, in other words. Both are always about twenty years from being achieved.
posted by beagle at 1:13 PM on September 12, 2019 [1 favorite]
Same as fusion reactors, in other words. Both are always about twenty years from being achieved.
posted by beagle at 1:13 PM on September 12, 2019 [1 favorite]
What we really need is a rapid acceleration to the left.
posted by nickmark
And then a step to the right.
posted by Splunge at 1:17 PM on September 12, 2019 [11 favorites]
posted by nickmark
And then a step to the right.
posted by Splunge at 1:17 PM on September 12, 2019 [11 favorites]
“it’s” for possessive “its” three times in the first sentence of the paper! I can’t.
posted by nicwolff at 1:31 PM on September 12, 2019 [2 favorites]
posted by nicwolff at 1:31 PM on September 12, 2019 [2 favorites]
I'm dumb so I am simply not getting this at all. A cable based on the moon, with the other end in "geostationary orbit." But we'd still need rocket fuel to get to the Earth end of the cable, which is still very far out of the atmosphere, so what's the point? What am I missing?
posted by zardoz at 2:05 PM on September 12, 2019
posted by zardoz at 2:05 PM on September 12, 2019
It looks like the cable hangs down to lower than geostationary orbit altitude, but the article doesn't give a number.Take a look at the journal article, Figure 5. The height of the lower end of the cable is a parameter, and the amount of cable you need is *very* sensitive to it as it decreases. At the gray vertical line (currently mass-produceable materials) you get an "effective length" (the ratio of total material volume vs minimum cross-sectional area, proportional to cost vs capacity) of 5 or 10 times the distance from the Earth to the moon ... but at half the geosynchronous radius you'd need 100 times the distance, at twice the Earth's radius you'd need a few thousand times, at the Earth's radius you'd need a million times ...
Also interesting is equation B26 and Figure B2, for delta V to the end of the spaceline. By their numbers, docking with the spaceline is never near as cheap as a trip to LEO, but you do save 1 km/s over escape velocity if you're leaving the Earth-Moon system entirely, and if it's the Moon itself you want to reach then getting the ascent and descent there "for free" is a huge saving.
Except ... the graph here seems totally wrong to me. At alpha=1 (super strong spaceline reaching all the way to Earth) the delta V should be approximately .5 km/s (you have to cancel out Earth's faster rotation), not approximately 17 km/s. It looks to me like they calculated the cost of first getting into LEO and *then* getting to the cable, whereas you could almost always do better by taking a suborbital trajectory to the cable. That's a huge (and surely unintentional, since it makes their proposal seem much less attractive) mistake, unless I'm misunderstanding something. Anybody want to tell me what I'm misunderstanding, before I write the authors and make an ass of myself?
posted by roystgnr at 2:14 PM on September 12, 2019 [1 favorite]
I suppose I could just contact them either way. Even if my correction is in error, they would then know that their current draft text, "Questions and suggestions welcome", should be changed to "Intelligent questions and suggestions welcome" in subsequent drafts.
posted by roystgnr at 2:20 PM on September 12, 2019
posted by roystgnr at 2:20 PM on September 12, 2019
zardoz:
The basic currency of space flight is delta-v, which is a really funny way to say 'acceleration'. If you want to change orbits, you need to accelerate. If you want to land on the moon without turning into a pancake, you need to accelerate. If you want to get off the moon after your visit? More acceleration.
This acceleration is usually provided by rocket fuel, and therefore needs to be hauled at great expense into space. Meaning the weight of the craft is heavier, meaning you need yet more fuel (and, incidentally, more containers to hold it)... Any possible 'free' source of delta-v, once you're already off the planet, is therefore extremely valuable. One good example is the gravitational slingshot maneuver, where you fly really close to the moon and get a lot of free acceleration.
The proposal here would effectively make it much, much, much cheaper to get back and forth from the moon, or to higher orbits between geostationary and the moon.
(It seems like a great road to jumpstarting lunar mining, honestly... which would be great, because it's way cheaper to lift things off the moon than off of earth.)
posted by kaibutsu at 2:25 PM on September 12, 2019 [1 favorite]
The basic currency of space flight is delta-v, which is a really funny way to say 'acceleration'. If you want to change orbits, you need to accelerate. If you want to land on the moon without turning into a pancake, you need to accelerate. If you want to get off the moon after your visit? More acceleration.
This acceleration is usually provided by rocket fuel, and therefore needs to be hauled at great expense into space. Meaning the weight of the craft is heavier, meaning you need yet more fuel (and, incidentally, more containers to hold it)... Any possible 'free' source of delta-v, once you're already off the planet, is therefore extremely valuable. One good example is the gravitational slingshot maneuver, where you fly really close to the moon and get a lot of free acceleration.
The proposal here would effectively make it much, much, much cheaper to get back and forth from the moon, or to higher orbits between geostationary and the moon.
(It seems like a great road to jumpstarting lunar mining, honestly... which would be great, because it's way cheaper to lift things off the moon than off of earth.)
posted by kaibutsu at 2:25 PM on September 12, 2019 [1 favorite]
also imma just say that the overall quality of space discussions on the internet has boomed since kerbal space program was released.
posted by kaibutsu at 2:28 PM on September 12, 2019 [19 favorites]
posted by kaibutsu at 2:28 PM on September 12, 2019 [19 favorites]
It looks to me like they calculated the cost of first getting into LEO and *then* getting to the cable, whereas you could almost always do better by taking a suborbital trajectory to the cable.
I was wondering about that, too... I suppose that if you do the suborbital trajectory and catch a hook at the end of the line, you're effectively placing extra stress on the line to change your velocity (which in turn lowers the maximum load you can place on the line). The situation may be worse if catching the end adds jerk to the system...
posted by kaibutsu at 2:36 PM on September 12, 2019
I was wondering about that, too... I suppose that if you do the suborbital trajectory and catch a hook at the end of the line, you're effectively placing extra stress on the line to change your velocity (which in turn lowers the maximum load you can place on the line). The situation may be worse if catching the end adds jerk to the system...
posted by kaibutsu at 2:36 PM on September 12, 2019
Jolt or jounce, technically.
...he said while trying not to sound like a jerk, and failing.
posted by Ryvar at 2:42 PM on September 12, 2019 [3 favorites]
...he said while trying not to sound like a jerk, and failing.
posted by Ryvar at 2:42 PM on September 12, 2019 [3 favorites]
How does it work that one end is fixed on the moon, which most definitely is not geostationary, but the other end is hovering in one spot in a geostationary orbit?
Thanks for everyone who tried to explain this, but I'm still confused. Is the answer that the end of the cable is at geostationary altitude, and centered on Earth's equator. But it's not moving fast enough to be geostationary and instead is rotating around the earth every 29ish days? If the cable end were a stationary satellite going that slow it'd fall immediately to a lower orbit, but the cable tethered to the moon is in effect holding it up?
posted by Nelson at 2:58 PM on September 12, 2019
Thanks for everyone who tried to explain this, but I'm still confused. Is the answer that the end of the cable is at geostationary altitude, and centered on Earth's equator. But it's not moving fast enough to be geostationary and instead is rotating around the earth every 29ish days? If the cable end were a stationary satellite going that slow it'd fall immediately to a lower orbit, but the cable tethered to the moon is in effect holding it up?
posted by Nelson at 2:58 PM on September 12, 2019
Jerk is d/dt of acceleration ... jounce is d/dt of jerk; I'd forgotten that one. What's jolt?
I think jerk is the important thing here, since that's what's going to send waves all the way up the cable, but under their assumptions it's independent of whether you do a suborbital or an intermediate orbital trajectory, because you're going to do a maneuver at the cable tip to match it's velocity either way, not try to get it to suddenly yank you to a stop. They explicitly calculate how much of a maneuver you need at the end of a trajectory starting from LEO, they just fail to see that (for alpha less than 25) you could get away with a smaller final maneuver by just not starting from LEO in the first place.
posted by roystgnr at 2:58 PM on September 12, 2019
I think jerk is the important thing here, since that's what's going to send waves all the way up the cable, but under their assumptions it's independent of whether you do a suborbital or an intermediate orbital trajectory, because you're going to do a maneuver at the cable tip to match it's velocity either way, not try to get it to suddenly yank you to a stop. They explicitly calculate how much of a maneuver you need at the end of a trajectory starting from LEO, they just fail to see that (for alpha less than 25) you could get away with a smaller final maneuver by just not starting from LEO in the first place.
posted by roystgnr at 2:58 PM on September 12, 2019
Sounds like you've pretty much got it, Nelson. The structure in question isn't actually orbiting the Earth.
posted by tobascodagama at 3:03 PM on September 12, 2019 [1 favorite]
posted by tobascodagama at 3:03 PM on September 12, 2019 [1 favorite]
Re: jolt vs jerk - apparently this is a British/American English terminology disagreement for the same thing (third derivative of position), but jolt is so anachronistic these days that it could also be considered a faceplant in a Metafilter physics thread.
posted by Ryvar at 3:09 PM on September 12, 2019
posted by Ryvar at 3:09 PM on September 12, 2019
It might make more sense to call this a Space Descender, but space descent isn’t all that special, I mean, any old thing can fall into a gravity well, so I propose Space Funicular.
posted by rodlymight at 4:09 PM on September 12, 2019 [6 favorites]
posted by rodlymight at 4:09 PM on September 12, 2019 [6 favorites]
What's jolt?
I didn't think it was still being made, but apparently I'm wrong.
posted by Greg_Ace at 4:12 PM on September 12, 2019
I didn't think it was still being made, but apparently I'm wrong.
posted by Greg_Ace at 4:12 PM on September 12, 2019
Funicular is an underused word, so I absolutely support renaming all of these long, stringy things for moving objects up and down gravity wells "space funiculars".
posted by tobascodagama at 4:22 PM on September 12, 2019 [4 favorites]
posted by tobascodagama at 4:22 PM on September 12, 2019 [4 favorites]
I am still confused. My understanding is that this cable will enter Earth's atmosphere or orbit once a month, and we'd still need to rocket up a payload to attach to the cable. That sounds like a tricky maneuver, to be sure.
But I still can't find an answer to the basic question: where would the Earth end of this cable be, exactly? Between, say, the ISS and Earth? Dangling 10 feet from the ground? Where??
posted by zardoz at 4:26 PM on September 12, 2019
But I still can't find an answer to the basic question: where would the Earth end of this cable be, exactly? Between, say, the ISS and Earth? Dangling 10 feet from the ground? Where??
posted by zardoz at 4:26 PM on September 12, 2019
zardoz: The earth-end would dangle at the geosynchronous orbit, about 35km above earth. Then to use it, you would rocket up and attach stuff at that orbit. Imagine the moon as a big round popsicle whose stick always points towards the earth (because the moon is tidally locked, which is why we always see the same face). The moon's orbit is pretty circular, so the end of the "stick" is always at roughly the same distance from earth.
posted by kaibutsu at 4:40 PM on September 12, 2019 [2 favorites]
posted by kaibutsu at 4:40 PM on September 12, 2019 [2 favorites]
Ok, thanks. That's about what I was guessing but with that confirmation I am just confused about the concept itself. We would still need to use fuel to get 35km out of Earth's gravity well, so I guess that makes it...cheaper? But still not cheap. Seems like an actual Earth-based elevator or launch loop would make more sense.
posted by zardoz at 4:56 PM on September 12, 2019
posted by zardoz at 4:56 PM on September 12, 2019
The earth-end would dangle at the geosynchronous orbit, about 35km above earth.
(35 thousand km)
posted by musicinmybrain at 5:26 PM on September 12, 2019 [4 favorites]
(35 thousand km)
posted by musicinmybrain at 5:26 PM on September 12, 2019 [4 favorites]
whats a few orders of magnitude amongst friends
posted by kaibutsu at 5:29 PM on September 12, 2019 [5 favorites]
posted by kaibutsu at 5:29 PM on September 12, 2019 [5 favorites]
If we're going to build a 380,000 km long structure from the earth to the moon, we might as well build a 264,000 km circumference ring around the earth at geosynchronous height*, then you just need enough delta-v to reach a 35,000 km apogee, grab onto the ring, and you're done.
* Note: Currently available materials may not be able to support the construction of such a ring.
posted by Mr.Encyclopedia at 5:39 PM on September 12, 2019 [3 favorites]
* Note: Currently available materials may not be able to support the construction of such a ring.
posted by Mr.Encyclopedia at 5:39 PM on September 12, 2019 [3 favorites]
A huge issue with space elevators, regardless of the availability of materials of sufficient strength-to-weight ratio will be very lightly damped longitudinal and lateral vibration modes. (These are really not elevators but cranes, because they are not guided by tracks or shafts.) The amplitude of these modes could be many miles, although the frequencies would be extremely low. I have never seen this issue discussed never mind addressed. The authors mention this problem only briefly near the end of the appendices (C.3.2). Does anyone here know of any deeper consideration of vibration or its mitigation for these systems?
posted by haiku warrior at 5:51 PM on September 12, 2019 [2 favorites]
posted by haiku warrior at 5:51 PM on September 12, 2019 [2 favorites]
Ryvar: "there is no realistic amount of force we are going to impart that will palpably alter the mechanics of the Earth-Moon system. The masses involved are so far beyond anything we will be capable of affecting in the next few centuries that it's not worth spending further words on."
The practical ways of moving a moon- or planet-sized body are left to the reader as an exercise
posted by andycyca at 6:05 PM on September 12, 2019 [3 favorites]
The practical ways of moving a moon- or planet-sized body are left to the reader as an exercise
posted by andycyca at 6:05 PM on September 12, 2019 [3 favorites]
zardoz: An actual Earth-based elevator would be a *lot* better. We can't do that now. The only news about this (besides the surprising structure) is we theoretically *could* do this now with the tech we have. The only material I've heard of that has a chance at being strong enough to do the Earth based one is carbon nanotubes. And we don't know how to make them more than very small lengths as yet.
posted by aleph at 6:06 PM on September 12, 2019 [1 favorite]
posted by aleph at 6:06 PM on September 12, 2019 [1 favorite]
You don't need to build the ring that high, LEO will do. The trick is that the ring is not in orbit, it's stationary with respect to the earth below it. Inside the ring is a wire, spinning at orbital speed. Magnets on the ring provide lift. Insane, but theoretically buildable with standard materials.
Isaac Arthur explains
(Be sure to watch to the end where he explains how these structures could allow humans to walk (or take a train) from the earth to the moon.)
posted by joeyh at 6:12 PM on September 12, 2019 [1 favorite]
Isaac Arthur explains
(Be sure to watch to the end where he explains how these structures could allow humans to walk (or take a train) from the earth to the moon.)
posted by joeyh at 6:12 PM on September 12, 2019 [1 favorite]
It seems like a great road to jumpstarting lunar mining, honestly... which would be great, because it's way cheaper to lift things off the moon than off of earth.
Mine the moon for what? There's nothing on the moon that can't be mined a thousand times more easily on the earth. All of the precious metals on earth - gold, silver, copper, platinum, etc -are the result of geotectonic processes involving flowing magma and water that concentrate these rare metals into mineable ores. There are no tectonics on the moon, there is no mountain building on the moon, because it has been cold and dead for almost 4 billion years. There are no fluids on the moon. Therefore there are no likely mineable ores on the moon. It just a bunch of boring rocks of no more use than in your backyard.
posted by JackFlash at 6:17 PM on September 12, 2019 [2 favorites]
Mine the moon for what? There's nothing on the moon that can't be mined a thousand times more easily on the earth. All of the precious metals on earth - gold, silver, copper, platinum, etc -are the result of geotectonic processes involving flowing magma and water that concentrate these rare metals into mineable ores. There are no tectonics on the moon, there is no mountain building on the moon, because it has been cold and dead for almost 4 billion years. There are no fluids on the moon. Therefore there are no likely mineable ores on the moon. It just a bunch of boring rocks of no more use than in your backyard.
posted by JackFlash at 6:17 PM on September 12, 2019 [2 favorites]
Yes, a thousand times easier on earth but *not* that easy/cheap to get it to orbit. And since moon mining is the vaporist of vaporware right now, not worth arguing about.
posted by aleph at 6:20 PM on September 12, 2019
posted by aleph at 6:20 PM on September 12, 2019
Mine the moon for what?
Bitcoin, obviously. Blanket the lunar surface with solar panels and mining clusters, bankrolled by Silicon Valley VC. Sell access to mining consortiums on Earth. Create the biggest, dumbest entropy generator in history. A monument of sorts to mankind’s idiocy.
posted by dephlogisticated at 7:10 PM on September 12, 2019 [12 favorites]
Bitcoin, obviously. Blanket the lunar surface with solar panels and mining clusters, bankrolled by Silicon Valley VC. Sell access to mining consortiums on Earth. Create the biggest, dumbest entropy generator in history. A monument of sorts to mankind’s idiocy.
posted by dephlogisticated at 7:10 PM on September 12, 2019 [12 favorites]
Does anyone here know of any deeper consideration of vibration or its mitigation for these systems?
posted by haiku warrior
Specific to this case: no. In general? Yes, everyone debating space elevators or variations thereof thinks about this, as it's a minor but significant plot point in Clarke's The Fountains of Paradise, which is what every would-be orbital megastructure engineer cuts their teeth on.
[SPOILERS]
Specifically it turns out to be the solution to dealing with one of the moons of Mars regularly passing through the equatorial plane - stagger the timing of the elevators to control the natural vibration of the "string" such that every time the moon would plow through the average position of the elevator at that height, the elevator happens to be dozens of miles to one side or another. Charge thrillseekers a massive premium for rides that stop at the exact right altitude on particularly near approaches. It comes up again later during the rekindled Earth elevator project, but only as an item on the mid-development checklist of problems they're fine-tuning solutions for.
posted by Ryvar at 7:15 PM on September 12, 2019 [1 favorite]
posted by haiku warrior
Specific to this case: no. In general? Yes, everyone debating space elevators or variations thereof thinks about this, as it's a minor but significant plot point in Clarke's The Fountains of Paradise, which is what every would-be orbital megastructure engineer cuts their teeth on.
[SPOILERS]
Specifically it turns out to be the solution to dealing with one of the moons of Mars regularly passing through the equatorial plane - stagger the timing of the elevators to control the natural vibration of the "string" such that every time the moon would plow through the average position of the elevator at that height, the elevator happens to be dozens of miles to one side or another. Charge thrillseekers a massive premium for rides that stop at the exact right altitude on particularly near approaches. It comes up again later during the rekindled Earth elevator project, but only as an item on the mid-development checklist of problems they're fine-tuning solutions for.
posted by Ryvar at 7:15 PM on September 12, 2019 [1 favorite]
Ok so like not a space elevator but how about a space dumbwaiter.
posted by glonous keming at 7:16 PM on September 12, 2019 [3 favorites]
posted by glonous keming at 7:16 PM on September 12, 2019 [3 favorites]
The practical ways of moving a moon- or planet-sized body are left to the reader as an exercise
One spoonful at a time?
posted by rhizome at 7:41 PM on September 12, 2019
One spoonful at a time?
posted by rhizome at 7:41 PM on September 12, 2019
It is absolutely vital that we leave something else on the moon that we never go back to.
posted by rum-soaked space hobo at 5:03 AM on September 13, 2019
posted by rum-soaked space hobo at 5:03 AM on September 13, 2019
The moon's orbit is pretty circular, so the end of the "stick" is always at roughly the same distance from earth.
Apogee is 362,600 km and perigee is 405,400 km, so there would be a variance of *only* about 43,000 km.
posted by rocket88 at 8:02 AM on September 13, 2019
Apogee is 362,600 km and perigee is 405,400 km, so there would be a variance of *only* about 43,000 km.
posted by rocket88 at 8:02 AM on September 13, 2019
a variance of *only* about 43,000 km.
And I thought it was a long way to the chemist.
posted by nubs at 8:21 AM on September 13, 2019 [3 favorites]
And I thought it was a long way to the chemist.
posted by nubs at 8:21 AM on September 13, 2019 [3 favorites]
Ryvar: "there is no realistic amount of force we are going to impart that will palpably alter the mechanics of the Earth-Moon system."
Bummer.
*stops jumping up and down*
posted by Hairy Lobster at 9:32 AM on September 13, 2019
Bummer.
*stops jumping up and down*
posted by Hairy Lobster at 9:32 AM on September 13, 2019
“If you drop a tool from the International Space Station it will seem to rapidly accelerate away from you,” point out Penoyre and Sandford.Uhh, what?
posted by lucidium at 12:24 PM on September 13, 2019
I thought they were talking about the gravity gradient at that orbit height. That would make things eventually move relative to being "dropped" motionless. But wouldn't have thought it'd be that fast.
posted by aleph at 1:14 PM on September 13, 2019
posted by aleph at 1:14 PM on September 13, 2019
Well, you know, "rapidly" is relative.
Actually, I think they're just being imprecise while trying to avoid technical language. The phenomenon they're trying to explain is a feature of orbital dynamics where lower-altitude orbits have less energy than higher orbits but also have more angular velocity.
Like say you're in a spacecraft on the same orbit as the ISS but 500 km "behind" it. If you point along your current direction of travel (i.e., facing roughly towards the ISS) and fire your engine, the ISS will seem to accelerate away from you. This is because firing your engine added to your orbital energy, but that energy goes into increasing your orbital altitude rather than increasing your angular velocity -- your new orbital altitude has a lower angular velocity, so you seem to "slow down" relative to the ISS. If you want to "catch up" to the ISS, you actually have to fire your engine in "reverse", lowering your orbital altitude, which in turn increases your angular velocity.
(Even that paragraph is a loose simplification of orbital rendezvous maneuvers.)
So I think what they're saying is that if you "dropped" -- i.e., lowered the orbital altitude of -- a tool from the ISS, it will rapidly gain angular velocity because of its new orbit.
The thing about the gravity gradient is absolutely true, and it's why tools and astronauts are always tethered while working outside of a spacecraft, but nothing about that process is "rapid". It'd take a couple of orbital cycles to become obvious, I think.
posted by tobascodagama at 1:37 PM on September 13, 2019
Actually, I think they're just being imprecise while trying to avoid technical language. The phenomenon they're trying to explain is a feature of orbital dynamics where lower-altitude orbits have less energy than higher orbits but also have more angular velocity.
Like say you're in a spacecraft on the same orbit as the ISS but 500 km "behind" it. If you point along your current direction of travel (i.e., facing roughly towards the ISS) and fire your engine, the ISS will seem to accelerate away from you. This is because firing your engine added to your orbital energy, but that energy goes into increasing your orbital altitude rather than increasing your angular velocity -- your new orbital altitude has a lower angular velocity, so you seem to "slow down" relative to the ISS. If you want to "catch up" to the ISS, you actually have to fire your engine in "reverse", lowering your orbital altitude, which in turn increases your angular velocity.
(Even that paragraph is a loose simplification of orbital rendezvous maneuvers.)
So I think what they're saying is that if you "dropped" -- i.e., lowered the orbital altitude of -- a tool from the ISS, it will rapidly gain angular velocity because of its new orbit.
The thing about the gravity gradient is absolutely true, and it's why tools and astronauts are always tethered while working outside of a spacecraft, but nothing about that process is "rapid". It'd take a couple of orbital cycles to become obvious, I think.
posted by tobascodagama at 1:37 PM on September 13, 2019
"It'd take a couple of orbital cycles to become obvious, I think." Would think it would take a while.
I think it's cool that they use that gravity gradient thing to hang weight "lower" in orbit and use that effect to stabilize the satellite.
posted by aleph at 2:15 PM on September 13, 2019
I think it's cool that they use that gravity gradient thing to hang weight "lower" in orbit and use that effect to stabilize the satellite.
posted by aleph at 2:15 PM on September 13, 2019
I'm dumb so I am simply not getting this at all. A cable based on the moon, with the other end in "geostationary orbit." But we'd still need rocket fuel to get to the Earth end of the cable, which is still very far out of the atmosphere, so what's the point? What am I missing?So you need 1/3 the fuel to get there that you need to get to the moon. Also, something that seems obvious but isn't mentioned--if you send stuff back from the moon to the earth, you might be able to have a net energy gain depending on how far into the earth's gravity well the thing was dangling (since you have to slow it down the last part of the trip which could generate energy).
I think the "dropped" thing implies "if you were stationary and suspended at that height." Obviously, if you dropped a tool while you were in orbit it wouldn't fall away from you. But what do I know about rocket science?
posted by Gilgamesh's Chauffeur at 2:20 PM on September 13, 2019 [1 favorite]
This report ("On the Gravity Gradient at Satellite Altitudes" from Harvard) has way too much math for me to tackle now but for anybody wanting to know the details of the gravity gradient at orbital heights (and I mean *all* the details) they can wade through it. I mostly skimmed it but they were tossing around factors of (10^-11 )/(sec^2) for the horizontal and vertical components of the gravity gradient so I doubt it matters for dropped tools. Don't know if link will resolve but can google the title.
posted by aleph at 2:57 PM on September 13, 2019 [1 favorite]
posted by aleph at 2:57 PM on September 13, 2019 [1 favorite]
Stronger Than Steel: Synthetic Spider Silk is Real, just sayin'! like if graphene or some other diamond-age-like material doesn't work out, protein may have a way :P
posted by kliuless at 2:34 AM on September 14, 2019
posted by kliuless at 2:34 AM on September 14, 2019
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posted by tobascodagama at 10:27 AM on September 12, 2019 [8 favorites]