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Old 09-13-2017, 12:27 PM
HeyHomie HeyHomie is offline
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Space Elevator Question: If You're At The "Top," Are You In Zero-G?

Let's say the prefect the Space Elevator. You get on it and ride it up to the top, 400km above the surface of the Earth. Will you be "weightless" in the sense that astronauts aboard the ISS are?
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Old 09-13-2017, 12:50 PM
Weisshund Weisshund is offline
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Neat question.
I have no answer, but it definitely made me wonder, since you would be technically spinning pretty fast, would you experience false gravity and be drawn outwards?
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Old 09-13-2017, 12:51 PM
Darren Garrison Darren Garrison is offline
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No. You would be weightless only at the height of geosynchronous orbit.
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Old 09-13-2017, 12:53 PM
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You can't have a 400 km tall space elevator. A space elevator is basically a ridiculously long satellite in geostationary orbit. One end of the elevator reaches all the way to the ground, but the center of mass of the elevator is at 35,800 km altitude.
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Old 09-13-2017, 12:59 PM
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P.s. if you are just talking about a 400km tall tower, then no, you wouldn't be weightless at the top. If you step off the tower at the top, you'll fall straight down to the ground. Which is why it's not really a space elevator. A satellite at that altitude is "weightless" only because it's moving very fast, about 7 km/s. If you had such a tower and wanted to launch a satellite from the top, you'd still have to attach a rocket to it and accelerate it to 7 km/s.

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Old 09-13-2017, 01:04 PM
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Actually the center of mass would have to be well above the geostationary altitude, and the actual top with the counterweight much higher still, to counteract the weight of the cable and its cargo. Everything above geostationary altitude of course wants to have a longer orbital period than geostationary, so at geostationary velocity it tries to pull away to a higher altitude, producing tension on the cable.
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Old 09-13-2017, 01:06 PM
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You would be weightless after you stepped off the tower, just like the satellite, until you hit the atmosphere. The weightless satellite is falling as well, only it keeps missing the Earth.
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Old 09-13-2017, 01:11 PM
eburacum45 eburacum45 is offline
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Originally Posted by wolfpup View Post
Actually the center of mass would have to be well above the geostationary altitude, and the actual top with the counterweight much higher still, to counteract the weight of the cable and its cargo.
If you count the mass of the cable and its cargo together, the centre of mass needs to be at the geostationary point, or only a tiny bit above to cause some tension in the cable. If you put the centre of gravity too high the cable will fall upwards away from the Earth.
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Old 09-13-2017, 01:14 PM
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Originally Posted by eburacum45 View Post
If you put the centre of gravity too high the cable will fall upwards away from the Earth.
I thought you tied the Earth end to a very large rock or something.


Was Clarke's tethered in the Fountains of Paradise?
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Old 09-13-2017, 01:22 PM
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Originally Posted by eburacum45 View Post
If you count the mass of the cable and its cargo together, the centre of mass needs to be at the geostationary point, or only a tiny bit above to cause some tension in the cable. If you put the centre of gravity too high the cable will fall upwards away from the Earth.
Well, the cable is anchored at the equator, it's not going anywhere! If you make the center of mass too high you're going to needless expense and putting needless extra tension on the cable, but it does have to be above the geosynchronous point to a sufficient extent to support the payload, whose maximum weight will be at the surface.
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Old 09-13-2017, 01:28 PM
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Originally Posted by carnivorousplant View Post
I thought you tied the Earth end to a very large rock or something.


Was Clarke's tethered in the Fountains of Paradise?
Corrected link.

The earth end of Clarke's space elevator was indeed anchored, with some tension, but just enough to keep the elevator stable. If I remember correctly, in the book, someone asks what happens if the cable were severed, and the engineer replies it'll just hang there - and someone else says that's not literally true, there's some tension to keep it stable.
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Old 09-13-2017, 01:47 PM
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someone asks what happens if the cable were severed, and the engineer replies it'll just hang there - and someone else says that's not literally true, there's some tension to keep it stable.
I remembered that, but I thought it was narration. Thanks!
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Old 09-13-2017, 01:55 PM
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If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth's surface) is in a stable orbit. So, at the top, you'd also be in a stable orbit and, therefore, weightless.
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Old 09-13-2017, 02:10 PM
iamthewalrus(:3= iamthewalrus(:3= is offline
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P.s. if you are just talking about a 400km tall tower, then no, you wouldn't be weightless at the top. If you step off the tower at the top, you'll fall straight down to the ground. Which is why it's not really a space elevator. A satellite at that altitude is "weightless" only because it's moving very fast, about 7 km/s. If you had such a tower and wanted to launch a satellite from the top, you'd still have to attach a rocket to it and accelerate it to 7 km/s.
This is wrong in many ways.

If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn't hold together at all.

Note that they don't have to put rockets at the top of skyscrapers to accelerate them to the appropriate speed after they're built, even though the tops are moving faster than the foundations.
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Old 09-13-2017, 02:16 PM
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Geosynchronous orbit is about 35,800 km above Earth's surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit. At every point, from the bottom to the middle to the top, your perceived gravity would be the difference between how fast Earth's gravity is pulling you downward and how fast the floor under your feet is moving away from you centripetally due to rotating once every 24 hours. At the equator, you're 6,371 km from Earth's center, rotating 40,030 km in 24 hours, which is 1668 kph or 463.3 meters per second, hence Earth is moving away from your feet at about .03 m/s2. Actual gravity there is 9.81 m/s2. Subtract the two and your apparent gravity is 9.78 m/s2 downward.

Now step in to the elevator and to up to geosynchronous orbit, 35,800 km up. Now you're 42,164 km from Earth's center. At that point, you're moving sideways at 3,066 m/s, so your centripetal acceleration is downward at 0.22 m/s2. You're nearly seven times farther away from Earth's center where actual gravity is also 0.22 m/s2. The difference between the two is zero, and you feel weightless. But you're only halfway up the elevator.

Go all the way to the far end of the elevator and you're 71,600 km from Earth's surface (78,000 km from the center). You're moving sideways at 5,672 m/s, so your centripetal acceleration is 0.41 m/s2 but your actual gravity is down to just 0.05 m/s2. The difference is 0.36 m/s2 NEGATIVE. If you step out of the elevator and let go of the hand rail, two seconds later you'll find your feet have drifted 36 cm off the floor. In ten seconds, you'll be 18 meters up and drifting further away at 3.6 m/s. You aren't really drifting upwards. Both you and the platform are moving downwards; it is moving faster than you are. The other end of the elevator is attached to Earth's surface and Earth is rotating, pulling the platform away from you, while gravity is pulling you downward at a much smaller magnitude. Sixty seconds after you let go, you'll be 648 meters away (nearly half a mile).

So, no, you aren't weightless at the top. You weight is NEGATIVE.
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Old 09-13-2017, 02:20 PM
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This is wrong in many ways.

If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn't hold together at all.

Note that they don't have to put rockets at the top of skyscrapers to accelerate them to the appropriate speed after they're built, even though the tops are moving faster than the foundations.
In that post, I was talking about a 400 km tall tower, in case what the OP meant by "space elevator" was just a tall tower, tall enough that the top of it was in space.

A 400 km tall tower is just a tower. It's just supported by compressive strength of the material. The top of the tower is nowhere near orbital speed for that height. If you dropped something from it, it would drop almost straight down.
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Old 09-13-2017, 02:24 PM
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I remembered that, but I thought it was narration. Thanks!
Actually you're right:

Quote:
[The visitor] reached out a cautious hand and stroked the narrow ribbon linking the planet with its new moon.

"What would happen," he asked, "if it broke?"

That was an old question. Most people were surprised at the answer.

[Morgan answered,] "Very little. At this point, it's under practically no tension. If you cut the tape, it would just hang there, waving in the breeze."

Kingsley made an expression of distaste; both knew that this was a considerable oversimplification. At the moment, each of the four tapes was stressed at about a hundred tons, but that was negligible compared to the design loads they would be handling when the system was in operation and they had been integrated into the structure of the Tower. There was no point, however, in confusing the boy with such details.
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Old 09-13-2017, 02:26 PM
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Geosynchronous orbit is about 35,800 km above Earth's surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit.
Place a massive object at the top; would that not move the center of mass upwards? We move an asteroid to synchronous orbit, and lower a cable from it.
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Old 09-13-2017, 02:27 PM
Darren Garrison Darren Garrison is offline
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If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn't hold together at all.
Nope, a rigid tower can't be at proper orbital velocity at every height. Look at this calculator. At a point 50 km up, orbital period is 1.42 hours. At 100 km, 1.44 hours. At 200 km, 1.47 hours. At 400 km, 1.54 hours. (All figures rounded.) But if it is a rigid tower rising off of Earth, each section would have to have an orbital period of 24 hours.
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Old 09-13-2017, 02:35 PM
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There is a way to put a 400km high space elevator up that would just hang there; build an orbital ring first. This link explains how that could work.
https://en.wikipedia.org/wiki/Orbital_ring
Note that an orbital ring is even more speculative than a geostationary space elevator. If you stepped off a 400km orbital ring you'd just fall towards the Earth, so I wouldn't advise it.
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Old 09-13-2017, 02:35 PM
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If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth's surface) is in a stable orbit. So, at the top, you'd also be in a stable orbit and, therefore, weightless.
Not correct.

A space elevator, as a whole, is in a stable orbit, but not every part of it is. The center of mass is at 22,200 miles high (or close to it). And because there's 22,200 miles of cable below that point, there needs to be a hell of a lot of mass above that point, too. If you were in an elevator climbing up this cable, at 22,200 miles you'd be weightless. Your height and speed would be just right to stay in orbit, so it doesn't matter if the elevator cab is around you or not. But below that point you'd be going slower than a satellite, so you'd still feel some sense of gravity inside the elevator. Above that, you'd be going faster than a satellite and you'd be pressed, slightly, against the ceiling of the elevator.
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Old 09-13-2017, 02:37 PM
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Originally Posted by dstarfire View Post
If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth's surface) is in a stable orbit. So, at the top, you'd also be in a stable orbit and, therefore, weightless.
I'm not sure what you mean by "stable orbit" but that's not correct by most definitions, and you certainly would not be weightless at the top of a space elevator if by "top" you mean the maximum extent of the cable, where the counterweight is. The counterweight is in an artificially constrained orbit, orbiting with a geosychronous period but at a higher altitude where its velocity is higher than it should be for a stable orbit. In effect everything at that speed and altitude has negative weight; if the counterweight were released, it would climb to a higher orbit, just as a satellite would do if it got a rocket boost. That's what gives the cable its tension. It would keep climbing until the earth starting pulling it back, and that point would become the apogee of a new stable orbit.

The only point on a space elevator where you would be weightless would be at the geosynchronous point. You would get lighter and lighter until you weighed zero at that point. Climbing further, toward the center of mass and the counterweight, your weight would become increasingly negative, pointing away from the earth. If you jumped out into space you would of course immediately become weightless since you would be on a ballistic trajectory, but you'd end up rising higher and then settling into a stable higher orbit.
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Old 09-13-2017, 03:08 PM
RedSwinglineOne RedSwinglineOne is offline
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So if I understand correctly, a space elevator would be of little use getting to low earth orbit because at that altitude, whatever you are lifting is not moving fast enough to maintain orbit.
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Old 09-13-2017, 03:10 PM
Marvin the Martian Marvin the Martian is offline
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If you jumped out into space you would of course immediately become weightless since you would be on a ballistic trajectory, but you'd end up rising higher and then settling into a stable higher orbit.
Plus, if my quick calculations are correct, if you went past about 11km above the geosynchronous point and let go you would be moving faster than escape velocity - you would go not just a higher orbit but out of the earth's gravitational field completely. I believe Clarke made this point also in Fountains of Paradise - the space elevator could also be used as a slingshot to launch interstellar payloads.

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Old 09-13-2017, 03:19 PM
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There is a way to put a 400km high space elevator up that would just hang there; build an orbital ring first. This link explains how that could work.
https://en.wikipedia.org/wiki/Orbital_ring
Note that an orbital ring is even more speculative than a geostationary space elevator.
Also touched on in this thread.
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Old 09-13-2017, 03:23 PM
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I believe Clarke made this point also in Fountains of Paradise - the space elevator could also be used as a slingshot to launch interstellar payloads.
Interstellar?
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Old 09-13-2017, 03:28 PM
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With a space elevator, getting to LEO would be cheaper than it is now, but it'd still be awkward. But that's OK, because there's very little value in LEO in its own right. The only reason we launch so much stuff to LEO right now is because it's the cheapest orbit (with our current technology). A space elevator would make GEO orbits and highly-eccentric orbits much cheaper, so most of what we launch to LEO now would just be launched to some other orbit instead.

EDIT: And yes, you could do interstellar launches from a space elevator if you wanted. But you'd still have to solve all of the other problems with an interstellar mission, like providing power for thousands of years. Much more practical would be interplanetary missions.
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Old 09-13-2017, 03:33 PM
Marvin the Martian Marvin the Martian is offline
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Interstellar?
Meant interplanetary.
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Old 09-13-2017, 03:56 PM
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So if I understand correctly, a space elevator would be of little use getting to low earth orbit because at that altitude, whatever you are lifting is not moving fast enough to maintain orbit.
Except that (as Chronos already said) getting up to an orbital altitude and with a good fraction of the needed orbital speed (even if insufficient) already saves you a very, very high percentage of the cost of trying to lift it from the ground in a rocket at standstill! Plus, if you go higher than you need to be, you have free energy to play with, though I confess I don't have a sufficient intuitive sense of the orbital mechanics to know how you would effectively use that -- I believe you would inevitably need some amount of rocket thrust to maneuver into the desired orbit. The international consortium working on this proposes a "LEO gate" for low earth orbits that would release the payload at around 24,000 km, IIRC, as opposed to the GEO (geosynchronous point) at 35,786 km. They also propose a "lunar gate" and a "Mars gate" at higher altitudes above GEO.
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Plus, if my quick calculations are correct, if you went past about 11km above the geosynchronous point and let go you would be moving faster than escape velocity - you would go not just a higher orbit but out of the earth's gravitational field completely. I believe Clarke made this point also in Fountains of Paradise - the space elevator could also be used as a slingshot to launch interstellar payloads.
That intuitively seems far too little. The Wikipedia article on space elevators states that escape velocity is reached at 53,100 km, as compared to GEO at 35,786.
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Old 09-13-2017, 04:17 PM
Marvin the Martian Marvin the Martian is offline
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That intuitively seems far too little. The Wikipedia article on space elevators states that escape velocity is reached at 53,100 km, as compared to GEO at 35,786.
The Wikipedia article says, "An object attached to a space elevator at a radius of approximately 53,100 km would be at escape velocity when released." Radius, not altitude. The radius of a geosynchronous orbit is about 42,000 km.
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Old 09-13-2017, 04:29 PM
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The Wikipedia article says, "An object attached to a space elevator at a radius of approximately 53,100 km would be at escape velocity when released." Radius, not altitude. The radius of a geosynchronous orbit is about 42,000 km.
OK, thanks, but what caught my attention was that you wrote "11km above the geosynchronous point", which seemed wrong. I wasn't trying to be snarky, it actually didn't occur to me that this was probably a typo and that you probably meant 11K km! If so, then apologies and all is well -- you were indeed in the ballbark.
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Old 09-13-2017, 04:38 PM
Marvin the Martian Marvin the Martian is offline
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OK, thanks, but what caught my attention was that you wrote "11km above the geosynchronous point", which seemed wrong. I wasn't trying to be snarky, it actually didn't occur to me that this was probably a typo and that you probably meant 11K km! If so, then apologies and all is well -- you were indeed in the ballbark.
Yes, another typo . Meant 11,000 km. Really wish Mm (megameter) were in common usage.
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Old 09-13-2017, 04:48 PM
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Yes, another typo . Meant 11,000 km. Really wish Mm (megameter) were in common usage.
From your username, I should have realized that you'd be closely familiar with the velocities necessary to return home!
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Old 09-13-2017, 05:18 PM
RedSwinglineOne RedSwinglineOne is offline
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Except that (as Chronos already said) getting up to an orbital altitude and with a good fraction of the needed orbital speed (even if insufficient) already saves you a very, very high percentage of the cost of trying to lift it from the ground in a rocket at standstill! ...
But on the surface of the earth, you are not at a standstill. At the equator you are moving east at over 1000mph. 300 miles up you are only going very slightly faster. 1100mph or so if my math is correct.
It is a great help to be above the atmosphere of course, but speed seems still to be an issue.
You are probably right that something may be gained by going higher (and therefore faster) and changing orbit later on.
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Old 09-13-2017, 06:11 PM
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I'm certainly not an expert on orbital mechanics, but I think that's the general idea. Tremendous cost and energy is saved by gaining altitude, because a rocket has to lift not only the payload, but also itself and all its fuel, so it all snowballs exponentially. Plus, you're not talking about hundreds of miles in altitude but potentially thousands -- as mentioned, the LEO gate proposed by the elevator consortium to the best of my knowledge is at 24,000 km, about two-thirds of the way to GEO -- which gives the payload lots of orbital speed at release. I would imagine relatively minimal thrusters could put it into a nice circular LEO.
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Old 09-13-2017, 07:58 PM
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I'm certainly not an expert on orbital mechanics, but I think that's the general idea. Tremendous cost and energy is saved by gaining altitude, because a rocket has to lift not only the payload, but also itself and all its fuel, so it all snowballs exponentially. Plus, you're not talking about hundreds of miles in altitude but potentially thousands -- as mentioned, the LEO gate proposed by the elevator consortium to the best of my knowledge is at 24,000 km, about two-thirds of the way to GEO -- which gives the payload lots of orbital speed at release. I would imagine relatively minimal thrusters could put it into a nice circular LEO.
This is not correct. It takes much more energy to achieve orbital speed than it does to achieve orbital height.
XKCD has an article about this.
https://what-if.xkcd.com/58/
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Old 09-13-2017, 08:22 PM
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Ummm.... Perhaps this is too simple of a view. But thinking about the OP's question. - Wouldn't the anchor in space have to have 'negative' gravity to hold up the climbing ribbon/cable to keep it taught? Not even considering if the payload climbs or is pushed up with lasers.

So, if you 'stepped off', you would fly into space away from earth.
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Old 09-13-2017, 08:30 PM
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Well, getting above the atmosphere does still save you something. Still, I imagine that if, for some reason, you really must have LEO, then the most efficient way to get there is probably to lift to a higher height and release, to go into an orbit with its perigee at LEO height. Then, when you're down there, you do a rocket burn to circularize. Or maybe a combination of rocket burns and aerobraking. I haven't done the calculations on how high you'd need to go for your original release, but the 24 Mm cited by Wikipedia seems plausible.

EDIT: enipla, the top point must be at least some amount above GEO in order to balance it, so if you went to the top and let go, you would certainly "fall" upwards, at least initially. But depending on the design of the elevator, you might end up still in an orbit around the Earth and come back down to that same height again a little over a day later (and then up again and so on), or you might end up escaping completely. How high up the end needs to be depends on how massive the counterweight is, and with a really big counterweight, it might be only a little above GEO height. Still, it'd be really nice to be able to launch things on escape trajectories for free, so in practice, I expect that any space elevator would be at least tall enough for that.
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Last edited by Chronos; 09-13-2017 at 08:33 PM.
  #39  
Old 09-13-2017, 09:09 PM
wolfpup wolfpup is online now
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Originally Posted by gazpacho View Post
This is not correct. It takes much more energy to achieve orbital speed than it does to achieve orbital height.
XKCD has an article about this.
https://what-if.xkcd.com/58/
That's quite true, and worth pointing out. My comment was badly worded. Satellite launches tend to go fairly straight up only to clear the worst of the atmosphere, and then they heel over and go for speed. The thing is, an enormous proportion of the fuel is consumed in those early seconds and minutes, which is why experimental airlifted spacecraft ("air launch to orbit") have had some success, even though the predominant advantage offered by air launch is high altitude and thin atmosphere rather than any substantial velocity. But certainly your statement about where most of the energy goes is correct. And the beauty of space elevators is that enormous altitude and huge orbital velocity go hand in hand.
  #40  
Old 09-13-2017, 10:27 PM
JWT Kottekoe JWT Kottekoe is offline
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Last edited by JWT Kottekoe; 09-13-2017 at 10:29 PM.
  #41  
Old 09-14-2017, 12:28 PM
iamthewalrus(:3= iamthewalrus(:3= is offline
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Originally Posted by scr4 View Post
In that post, I was talking about a 400 km tall tower, in case what the OP meant by "space elevator" was just a tall tower, tall enough that the top of it was in space.

A 400 km tall tower is just a tower. It's just supported by compressive strength of the material. The top of the tower is nowhere near orbital speed for that height. If you dropped something from it, it would drop almost straight down.
My mistake. I took the OP's "400km" as a wild-ass guess about the proper height for an actual geosynchronous elevator, and was responding as such, but you're correct, a 400km tower is just a tall tower.

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Originally Posted by Darren Garrison View Post
Nope, a rigid tower can't be at proper orbital velocity at every height. Look at this calculator. At a point 50 km up, orbital period is 1.42 hours. At 100 km, 1.44 hours. At 200 km, 1.47 hours. At 400 km, 1.54 hours. (All figures rounded.) But if it is a rigid tower rising off of Earth, each section would have to have an orbital period of 24 hours.
Sorry, I meant "rotational velocity", not orbital velocity. It sounded like scr4 was saying that the top of the tower wouldn't already be rotating at the proper speed (to remain a tower, not to be in orbit).

My point was that if you are building a rigid tower, at no point do you have to do something special to accelerate the higher bits, because they are accelerated as you raise them up to build the tower. But I used some pretty imprecise language to get there.
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  #42  
Old 09-14-2017, 01:03 PM
Elendil's Heir Elendil's Heir is offline
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Originally Posted by carnivorousplant View Post
...Was Clarke's tethered in the Fountains of Paradise?
Fascinating book - I just reread it a year or so ago. Star Trek: Voyager had a space-elevator episode, too: http://memory-alpha.wikia.com/wiki/Rise_(episode)
  #43  
Old 09-14-2017, 03:55 PM
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Originally Posted by HeyHomie View Post
Let's say the prefect the Space Elevator. You get on it and ride it up to the top, 400km above the surface of the Earth. Will you be "weightless" in the sense that astronauts aboard the ISS are?
Gravity scales with 1/r^2 and so called centrifugal force scales with r. Earth's radius is about 6378 kilometers.

If you're a 150 lb guy, climbing to the top of a 400 km tower on the equator would lose you about 3 lbs.

Centrifugal force doesn't cancel gravity until you reach an altitude of about 36,000 km. This is the altitude of geosynchronous satellites.

To reach an orbit with a perigee above earth's surface you need to climb to an altitude of around 30,000 kilometers.

Here I have some drawings of the paths payloads would follow if released from different parts of an elevator.
  #44  
Old 09-14-2017, 04:03 PM
k9bfriender k9bfriender is offline
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Originally Posted by HopDavid View Post

If you're a 150 lb guy, climbing to the top of a 400 km tower on the equator would lose you about 3 lbs.
I think I'd lose more than just 3 lbs if I climbed a 400km tower.
  #45  
Old 09-14-2017, 04:04 PM
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Originally Posted by sbunny8 View Post
Geosynchronous orbit is about 35,800 km above Earth's surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit.
As you go outward gravity falls faster than centrifugal force climbs. So the acceleration gradient isn't symmetrical about geosynchronous orbit.

To balance and maintain tension, the tether above geosynchronous would need to be around twice as long as the tether below geosynchronous orbit.
  #46  
Old 09-14-2017, 04:20 PM
carnivorousplant carnivorousplant is online now
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Originally Posted by k9bfriender View Post
I think I'd lose more than just 3 lbs if I climbed a 400km tower.
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  #47  
Old 09-14-2017, 06:26 PM
K364 K364 is offline
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Originally Posted by HopDavid View Post
As you go outward gravity falls faster than centrifugal force climbs. So the acceleration gradient isn't symmetrical about geosynchronous orbit.

To balance and maintain tension, the tether above geosynchronous would need to be around twice as long as the tether below geosynchronous orbit.
I'm thinking that below geosynchronous the tower wants to drop, and above geosynchronous it wants to fly away to a higher orbit.

So, centripetal force depends on the speed and mass. In an unconstrained orbit these are equal to gravity. In the tower above geosynchronous it is more than gravity and pulls the tower up. So, you don't need any special length for the upper tether, just more mass.

BTW, in the category of "Possible, just needs more technology than we have today" I think this engineering will never happen. The scale of this thing is immense and any defect or mishap will be catastrophic.

Last edited by K364; 09-14-2017 at 06:26 PM.
  #48  
Old 09-14-2017, 08:57 PM
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I'm thinking that below geosynchronous the tower wants to drop, and above geosynchronous it wants to fly away to a higher orbit.
Correct. Gravity is GM/r^2 . And so called centrifugal force is ω^2 r. ω is angular speed in radians.

Net acceleration is GM/r^2 - ω^2 r.

If r > than geosynchronous, centrifugal force is greater and net acceleration is up. If r is below geosynchronous, gravity is greater and net acceleration is down.

Quote:
Originally Posted by K364 View Post
So, you don't need any special length for the upper tether, just more mass.
Net acceleration is small if you're close to geosynchronous orbit. So a counterweight near geosynchronous would provide only a small amount of upward newtons unless it was many tonnes.

Having the tether length extend to an altitude of about 144,000 kilometers would be the minimum mass way to counterbalance the downward newtons.

And if your tower is that tall, it could fling payloads most the way to Neptune.
  #49  
Old 09-15-2017, 07:41 AM
eburacum45 eburacum45 is offline
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I am pretty sure that you could use a space elevator to leave the Solar System if you aimed the payload just right- using a flyby manoeuvre at Jupiter, for instance, like Voyager.

It is a remarkable thing that once you get past geostationary orbit all that acceleration would come from the rotation of the Earth. But launching payloads towards the outer solar system would have an effect - there ain't no such thing as a free lunch. It would cause a drag on the Earth, and slow it down, That's why you need to attach the elevator firmly at the bottom - otherwise the bottom end of elevator would career off through the atmosphere, and extract angular momentum by friction.
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  #50  
Old 09-15-2017, 08:06 AM
Chronos Chronos is offline
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On the other hand, you could also set up an exchange of material: Say, put another elevator on Mars, and ship equal tonnages of Earthly seawater to Mars, and Martian iron ore to Earth. In that case, you'd have no net effect on the rotation or orbit of Earth or of Mars. And you could set it up so the energy cost per ton (both ways) was arbitrarily small.

(nitpicking myself: You couldn't actually do arbitrarily small, since you'd need computers to calculate the proper trajectories, and it costs energy to do those calculations. But that's so small compared to the energies involved in the movement of the matter that it might as well be zero.)
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