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I guess the concept of a space elevator is pretty well known. The idea, first published by Konstantin Tsiolkovsky in 1895, and popularized (among others) by Arthur C. Clarke in The Fountains of Paradise, is to have a geosynchronous satellite with a very strong and light cable hanging down to the ground at some anchoring point on the equator circle, and a counterweight extending outward in space to keep the center of mass at the rigth geosynchronous altitude (this variant is due to Yuri N. Artsutanov). Whether it can be done for Earth remains doubtful, as we do not (yet?) know of materials that can sustain the strain (afaik), but the concept is interesting and has been reused in several science-fiction stories, for Earth and other planets.

In at least two of the novels I read that use the concept, the elevator cable breaks, or is broken, below the center of mass and the cable falls back to the ground (with or without the car). But these two novels do not seem to agree on how it falls.

In one novel, the cable falls in a heap on its ground anchoring place. In another, it falls on a big circle (the equator), making a "straight" line across the planet surface, though I do not recall whether it is forward (ahead of its anchor, with respect to the planet rotation) or backward (behind its anchor, with respect to the planet rotation). The planet may not be Earth.

My knowledge of mechanics no longer being was it may once have been, I am not sure I can analyze the problem correctly. I seriously doubt the cable would fall as a heap onto its anchor (or that the car would do that if it were to get loose from the cable, as suggested in one novel).

So my question is what are the machanical laws of the phenomenon. and where do the car and the cable fall and how.

They could fall ahead of the anchoring point, or behind the anchoring point (with respect to earth rotation). The cable could be taut on the equator, or zigzaging because it fell too fast with respect to earth rotation. It could start falling in one direction (forward or backward) and later reverse the other way for the remaining span. I just have no real idea of what might happen, and I wonder how it is to be analyzed.

I tried to get a first understanding by considering the car alone getting loose from the cable, and I am putting it in a first tentative partial answer, so that this question does not get too long. My conclusion for the car felt counter-intuitive at first, becoming obvious in retrospect. But what of the cable?

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You could use the Coriolis force to analyze this, or just a little common sense.

Assuming the cable is anchored at the equator, the linear velocity of a point "high up" will be greater than the velocity of the anchor point. When each point of the cable is in free fall, that velocity will carry the higher parts of the cable "ahead of" the anchor point. It will land to the East, and not in a heap (although elastic forces may further complicate things, it is unlikely to completely reverse this).

You can reach the same conclusion by thinking about conservation of angular momentum.

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    $\begingroup$ You are right, but ... Coriolis forces are at the edge of my remaining knowledge. Angular momentum gives a general idea of where things go, but an imprecise intuition (at least for me). In my answer, about the car alone, I thought orbit analysis might give me a better idea of trajectories, so as to better understand what happen to the cable. A basic issue is whether two neighboring segments of the cable will tend to get closer or further apart. If it is closer, we get essentially the same result as for the car alone. If they diverge, then forces and energy will propagate along the cable. How? $\endgroup$ – babou Feb 23 '16 at 13:27
  • $\begingroup$ physics.stackexchange.com/questions/277688/… $\endgroup$ – Muze the good Troll. Sep 2 '16 at 15:40
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    $\begingroup$ @Troll - really Jen? You changed your name to that? $\endgroup$ – Floris Sep 2 '16 at 16:08
  • $\begingroup$ @Floris I like it... not so much for the definition. $\endgroup$ – Muze the good Troll. Sep 2 '16 at 17:11
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The car alone falls ahead of the anchoring point

This is a partial answer to illustrate the question in a simpler case. It is separated from the question to keep it short enough, and because it is more an answer than a question.

If one considers the car alone, getting suddenly loose, its angular speed would be to low to keep it in a circular orbit, at its current altitude, since the whole elevator is rotating at planet-synchronous speed, which corresponds to a satellite above the car altitude. Thus the car would dive into an elliptical orbit that might, or might not intersect the ground level (I am ignoring the atmosphere for simplification, though the elevator is not really needed when there is none).

Car touching ground at perigee

The orbital period is actually determined by the length $a$ of the semi-major axis of the orbit, according to the formula $T = 2\pi\sqrt{a^3/GM}$, where $M$ is the mass of the planet. That is $a$ is the initial height of the car above the planet center. If it touches ground just at perigee, it will have taken half a period of that elliptical orbit, a time shorter than half a period of planet-synchronous orbit (which has a longer semi-major axis). Hence the car will reach the ground in less than half of a planet rotation period, at a point that would be half the equator from the anchor, if Earth were not rotating. Thus the car will fall ahead of the anchoring point. This reasonning can be done without the exact formula, using only Kepler third law.

The general case for the car alone

I have not done the precise calculation for all cases, and would not be very good at it, but it seems that qualitative reasonning is enough to show that, if the car hits the ground (i.e. its orbits intersect the planet surface), then it is always ahead of the anchoring point.

A first remark is that, if the car crashes on the ground, it does so somewhere along its first half orbit from apogee (when it gets loose) to perigee (when it is closest to the planet center, if it can). According to Kepler second law of equal areas (using a differential form of it) its angular speed increases continuously from apogee to perigee, as its distance to the planet decreases.

The car's angular speed at apogee is the same as the planet's, i.e., when the car gets loose. From then on, when it crashes, its angular speed will only increase until it crashes. Hence it will continuously get further ahead of the anchoring points as its follows its orbit to the ground, and will necessarily crash on the equator ahead of the elevator anchoring point.

The cable case

Thus I would tend to believe that a ruptured cable will fall similarly ahead of the anchoring point. But I have no idea how forces propagating along the cable might affect its motion, and whether it will be taut on the ground.

Examining the above reasonning, it is clear that the car alone falls on the first half of the equator great circle, ahead of the anchoring point. Now, it we consider a space elevator for Earth, if the cable is broken right under the geosynchronous satellite, its length is about 35,786 km, while half of the equator great circle is only about 20,000 km. Hence there is no way the cable can be taut along the equator while falling only on its first half, which is shorter.

Could it be that some extra energy is propagated to the end (highest part) of the cable allowing it to stay "in orbit" beyond the half great circle limit?

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