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Suppose there is a fully functional space elevator built on Earth. The base is attached to coordinates

$ (\lambda, \varphi) = (0,0) $

e.g., on the equator on the zero-meridian.

What would happen if we were to suddenly remove the counterweight? At what speed would the tip hit the Earth, and at what coordinates (all approximate)?

I might be missing something, but I found this scenario exceptionally difficult to model properly.

I was trying to get general equations, so for arbitrary material, planet, etc. I suspect the equations involved will be quite messy, so please use abbreviations/substitutions/references to standard transformations etc. where possible. Ignore the atmosphere and any carts/stations/other masses attached to the elevator, assume spherical Earth, etc.

For completeness: it is a scenario described in the Mars Trilogy, by Kim Stanley Robinson. So this is purely a pet project :)

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At least 18,000km of space elevator cable falls on you? –  Martin Beckett Aug 17 '12 at 14:57
    
The Earth rotates; the cable will not fall straight down. –  Rody Oldenhuis Aug 17 '12 at 14:59
    
@RodyOldenhuis - I mean't 'you' to be the Earth. I suppose you (personaly) could step aside and watch it fall on the adjoining hemisphere! –  Martin Beckett Aug 17 '12 at 15:21
    
I've edited the link. –  Luboš Motl Aug 17 '12 at 15:33
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2 Answers

up vote 3 down vote accepted

The Earth is rotating but because the cable is attached to the Earth's surface, it's still more convenient to describe the behavior of the cable in the Earth's reference frame.

In that reference frame, there exists the attractive gravity force $GmM_E/r^2$ that points directly to the center of the Earth. This is partly compensated by the centrifugal force $mr\Omega^2$ going in the opposite direction. Note that the power laws are different. For objects at the geostationary orbit about 36,000 km above the surface, these two radial forces exactly cancel.

So of course, the upper portion of the cable isn't attracted by any force. Nevertheless, all the segments of the cable that are closer to the surface are attracted. The closer we get to the surface, the closer the acceleration acting on the cable is to the ordinary acceleration $g=9.81\,{\rm m/s}^2$. The lower parts of the cable are attached to the upper ones so they will drag the cable in the direction down, anyway.

Because the acceleration is larger at the bottom, the cable will remain stretched at the top, and only the portion of the cable that falls to the surface will bend and form loops.

However, there is another fictitious force in the reference frame, the Coriolis force, $-2m\Omega\times v$. Before the cable starts to fall/move, the Coriolis force is zero because it's proportional to the velocity $v$ of the segment of the cable. Once the cable starts to fall, $v$ goes towards the Earth's center. $\Omega$ is the Earth's spin, so the vector goes along the Earth's axis. The cross product is tangential to a circle concentric with the equator.

In other words, as the cable starts to increase the velocity in the downward direction, the Coriolis force will try to bend it in the shape of a "spiral" inside the equatorial plane. Some sign calculations would be needed to calculate the direction in which the cable would be bent. The precise shape of the bent cable would require one to solve a partial differential equation depending on two variables (time, spatial coordinate along the cable). It is probably hard enough – and would surely be affected the distribution of the mass in the cable.

Nevertheless, I am confident that the Coriolis force would only bend the cable and wouldn't change the qualitative behavior, namely the conclusion that the cable ultimately falls to the surface, distributed along a piece of the equator. Of course, the result would also depend on whether or not the cable may easily bend (like a rope) or whether it's more solid etc.

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Thanks. Could you perhaps be a bit more quantitative? Because this was indeed (one of) my general approach(es), but the devil's in the details here... –  Rody Oldenhuis Aug 20 '12 at 6:36
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Apologies, no, I think it's technically a very complicated exercise that depends on all the assumptions about the linear density and stiffness of the rope as a function of the position on its length and I believe that even for the simplest choices, it won't be analytically soluble. One may make order-of-magnitude estimates of the size of the rope on the surface. The fall from 18,000 km takes about five hours, or something of the sort, so it's clear that the rope won't have enough time to cover more than a continent, a fifth of the equator or so. –  Luboš Motl Aug 20 '12 at 8:29
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@RodyOldenhuis Probably the best idea would be to do a numerical simulation. You can see some preliminary results in Blaise Gassend's website. –  mmc Aug 20 '12 at 14:13
    
Agreed with mmc. –  Luboš Motl Aug 20 '12 at 14:15
    
@mmc It would appear the cable wraps around the Earth when it breaks near the counterweight. Interesting...Indeed, numerically it would be "easy". I was just wondering whether it was possible to keep everything non-specific and general. The problem is not solved, but the question answered satisfactorily. –  Rody Oldenhuis Aug 20 '12 at 19:34
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Blaise Gassend has created this simulation of "An elevator that breaks at the counterweight.":

"An elevator that breaks at the counterweight."

More discussion of various possible failure modes of a space elevator:

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+1, even though the animations were already mentioned in a comment to the accepted answer above. Thanks for the other links though! –  Rody Oldenhuis Jan 9 '13 at 7:43
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