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Assume the observer was on the ground, 4 km from the base (bottom) of the Space Elevator.

Assume a "standard" Earth Elevator "tethered" at <= 20 degrees of equator, made of diamond nanothread, with an elevator "top" in LEO and a final tether in geosynchronous orbit.

Please take into account oscillation and differing rotational momentum between Earth and the geosynch tether. How long would it take for the effects of the severing to be noticed, as well as what the observer on the ground would see?

I add this to "world-building" as I need specific numbers to play around with for both Earth and a different world for a piece I'm writing.

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    $\begingroup$ The point where the orbital speed matches ground speed is in geosynchronous orbit. The balance point extends beyond this point. Compare for example space.stackexchange.com/a/5165/415 and space.stackexchange.com/questions/5163/…. $\endgroup$ – a CVn Feb 15 '17 at 20:34
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    $\begingroup$ Is the observer near the base? At what distance are they? I mean, if the base is severed, the people at the base would notice... $\endgroup$ – PatJ Feb 15 '17 at 20:49
  • $\begingroup$ 1 second for every 186,000,000 miles $\endgroup$ – Richard U Feb 15 '17 at 22:42
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    $\begingroup$ No "standard" design i know of assume LEO as elevator's top. Typically elevator is supposed to extend beyond geostationary orbit, @Michael is right. So first, make sure your design can exist. And describe it. Then and only then this will be really answerable. $\endgroup$ – Mołot Feb 16 '17 at 23:40
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    $\begingroup$ Indeed. And it is that you need the end far beyond geosynchronous orbit. Yet your question places end at that orbit. That's my point. $\endgroup$ – Mołot Feb 17 '17 at 16:37
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Breaking at the bottom would be the best way for it to break.

A couple of notes: The space elevator needs to have a top at geosync orbit, and a counter balance extending above geosync orbit. You can't use a space elevator to get into low Earth orbit, because at low Earth orbit heights you aren't going fast enough to be in orbit.

Not having the elevator on the Equator is possible, but adds to stresses as there is now bending in the elevator cable.

Now the elevator will be under a lot of tension. If it breaks at the base, the whole thing will lift off and be flung off the Earth, and probably end up in solar orbit. The bottom part will be dragged at high velocity through the atmosphere, probably fragmenting and the fragments will fall in the neighbouring districts.

If the elevator breaks higher up, the lower part will fall down, wrapping itself around the equator. God help anyone living on the Equator. Although the broken elevator won't hit the ground at very high speeds (thank you atmosphere) it will cause a line of destruction through Brazil, Congo, Indonesia and other countries on the Equator.

These breaks can be simulated: Blaise Gassend has done some animated gifs of an elevator break at various altitudes. http://gassend.net/spaceelevator/breaks/index.html

His model assumes "The elevator that is simulated is an equatorial uniform stress elevator with Brad Edwards' standard parameters. Length is 91000 km, density is 1300 kg/m^3, strength is 130 GPa with a factor of safety of 2, Young's modulus is 1 TPa."

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  • $\begingroup$ One nitpick: You could use a space elevator to get to low Earth orbit height. You are correct, however, that this would not actually be an orbit. Also, building a space elevator that only carries you up to LEO height and wastes the rest of the structure would be nearly useless; possible but useless. $\endgroup$ – Loduwijk Apr 17 '17 at 19:12
  • $\begingroup$ And of course you could use the space elevator to get into LEO: First use the elevator for the required height, and then use your rocket engine to get the required speed. $\endgroup$ – celtschk Apr 17 '17 at 22:03
  • $\begingroup$ If you still need a rocket then you have wasted a lot of money on an elevator. The hard part about getting into orbit isn't the height, it is the speed. What I say in paragraph 2 is still basically right. $\endgroup$ – James K Apr 17 '17 at 22:23
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Visually, 1 second for every 186,000 miles, because that's the speed of light.

A geostationary equatorial orbit (GEO) is a circular geosynchronous orbit in the plane of the Earth's equator with a radius of approximately 42,164 km (26,199 mi) (measured from the center of the Earth). A satellite in such an orbit is at an altitude of approximately 35,786 km (22,236 mi) above mean sea level.

Because it's only 22,236 high, even if you severed it at the tippy top, they would notice pretty much instantly (or near to) if they have a clear view. (I assume they'd be able to look through a telescope). (source)

But, Because of the slow rate of rotation of the Earth (only one revolution per 24 hours) the cable has to be very long - theoretically at least 25,000 miles, and in practice closer to 60,000 miles. (source)

Still, even at 60,000 miles, it will take less than a second for the light to travel and show us that it has happened.

Now, non-visually, the effects are going to depend on HOW you handle the severing. Is it a large explosion? Is a line mechanism just cut?

Let's also talk about how elevators are handled in large buildings. Most of the time, the elevators go to certain floor, then you have to change elevators to go up the rest of the way. For safety's sake, I would expect it would be handled this way. So maybe 1/2 way or a 1/4 of the way up, the doors open and you go to another elevator in the same bay, switching sides as you go.

If the elevator doesn't work, you'd be stuck at the second to last elevator up. Now, some of these proposed space elevators don't work this way--they are a single elevator that travels up at a high speed, but this is the way I'd build it, mainly because segmented systems can be easier to maintain. There are lots and lots of models for this, so I can't answer this question fully without knowing the method of design (there's the ski-lift version that involves multiple cars, and I'd expect that this would be segmented in some way as well).

I also think that the speed is important to the design as well. See this answer as to the top speed of the elevator. That's because if you have people travelling in the elevator for 2-3 days, those people are going to have NEEDS. Like eating and going to the bathroom. Will there be a bathroom on the elevator? Will the elevator(s) have stops on tiny platform to allow for eating and drinking? How large will each one be? Is it one single car, or several sent in a row? If it's several, the car that doesn't make it might procedurally send a message down on arrival, and that will be noticed, or the system simply won't take them up any further and they might radio for help.

Answering this question can only be fully accomplished once you've done a lot more research on how these things work and how your operating system is going to work. This determines how communication works, how many stops there are and all that--since I am sure the ground likely communicates with the elevator(s) along the way. If many are being sent up, one right after another, that changes things vs. one single car, (because that will take 2-3 days to reach the top) as will stops and communication with people on the top.

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  • $\begingroup$ My apologies if I wasn't clear enough: I assume a space elevator with a top of the elevator 22,000 miles high in geosynchronous orbit with an additional 40,000 miles to the rotational counterweight. Also, the assumed severing is not "at the tippy top" but at the base: if this were to occur, I'd like to know what an observer on the ground, ~4 km away from the base, would see: How long would it take for anything observable to happen viz. this cable now dangling from the sky, with suddenly altered center of gravity, angular momentum, etc., and what exactly would be seen? $\endgroup$ – Ozymandias Feb 17 '17 at 22:42
  • $\begingroup$ @Ozymandias Depends on how it's designed, greatly. $\endgroup$ – Erin Thursby Feb 18 '17 at 22:36
  • $\begingroup$ @Ozymandias Is the base on the ground? base1 bās/ noun 1. the lowest part or edge of something, especially the part on which it rests or is supported. "she sat down at the base of a tree" synonyms: foundation, bottom, foot, support, stand, pedestal, plinth "the base of the tower" $\endgroup$ – Erin Thursby Feb 18 '17 at 22:39
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    $\begingroup$ @Ozymandias if the severing is done at the base, which is the BOTTOM by most definitions, I would guess that an observer on the ground, at the bottom would notice. If they are only 4 KM above that, then...you know, they might notice, if they are going up. But you say the observer is on the ground, and the "base" is 4KM away. Are they IN the elevator? On the ground next to it? Or directly underneath it? I think you should definitely edit the question for clarity. By most definitions the base of any tall structure is on the very bottom. $\endgroup$ – Erin Thursby Feb 18 '17 at 22:44
  • $\begingroup$ 186 thousand miles per second. You have million. I tried to edit it for you, but it demanded I change a minimum 6 characters and that is only 4. $\endgroup$ – Loduwijk Apr 17 '17 at 19:17
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Look online for videos of a slinky toy being dropped. The top starts falling when it is released, but the bottom does not move yet! The bottom is still supported by material that is not itself supported. The loss of tension is noticed by other parts of the system at the speed of sound in that material.

The space elevator, assuming it’s under tension and cut at the bottom, is the same thing. Various cars gripping it will notice the lack of tension after a delay ed on the speed of sound. Likewise, this is the time scale at which it will start to bend and whip around.

A ground observer can’t see the tension. He won’t notice anything until it moves far enough to resolve. How fast the cut end snaps up depends on the specific design and tension, and can vary by orders of magnitude! Newer concepts don’t need an anchor at all, but hang in the air and normally don’t touch the ground. So at one extreme you won’t see anything happen; at the other it will shoot up at supersonic (in air) speed.

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