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If the space elevator platform (The very top) is orbiting a planet in a geosynchronous orbit, wouldn't the forces on the elevator "rope" be minimal?

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    $\begingroup$ Only the top is in a geosynchronous orbit. This shouldn't be too surprising, since there's only one geosynchronous orbit for a given spot - all the "rope" lower or higher than that necessarily isn't in a geosynchronous orbit, and will be dragged around, which introduces/requires tension on the cable. Not to mention that the force of the wind alone is well in excess of any materials we can currently produce, the last time I checked :) $\endgroup$ – Luaan Oct 14 '16 at 14:00
  • $\begingroup$ @Luaan The top needs to be above geosynchronous orbit (though by using a counter weight you can avoid putting it terribly far above) to cancel the tension from the inner part of the cable. In most science fictional schemes they put a station at the geosyncronouos altitude and the cable has it's maximum thickness there, but the cable must extend beyond. $\endgroup$ – dmckee Oct 14 '16 at 21:14
  • $\begingroup$ @dmckee Yup, that's the image I see as well. The counterweight is a cheap way to make everything stable (some works put a rock on the end to make the cable shorter, but it really isn't much of a deal), and it gives you a nice way to boost your spaceships on their way out. $\endgroup$ – Luaan Oct 14 '16 at 23:13
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Supplemental to the other answers; you are correct that the net force on the tether would be minimal, since the rotation of the counterweight would counteract the force of gravity. But, the individual components of this net force aren't being distributed evenly.

Consider, say, the first kilometer of tether from the ground. This is being pulled down by gravity and "held up" by the counterweight, but the counterweight is many many kilometers farther away whereas gravity happens right there. So, the rest of the tether is being "pulled apart" by these two forces.

(Incidentally, the counterweight at the top of the tether is going to be well above what "geosynchronous orbit" would be for an ordinary satellite. Instead the center of gravity for the entire counterweight-tether system would be at geosynchronous orbit altitude, or a bit beyond.)

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    $\begingroup$ Correction: The center of gravity of the entire system needs to be at geosynchronous altitude, which means the center of mass needs to be well beyond geosynchronous altitude. A space elevator of uniform thickness and no counterweight would need to reach to about 6 times geosynchronous altitude to have the center of gravity at geosynchronous altitude. $\endgroup$ – David Hammen Oct 14 '16 at 19:57
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    $\begingroup$ @DavidHammen Good point! In my head center of mass and center of gravity meant the same thing, but I don't spend much time contemplating the physics of space elevators. Edited appropriately. $\endgroup$ – Maxander Oct 14 '16 at 22:42
  • $\begingroup$ Great answer. Personally, I don't quite see how (and where) a counterweight could be used in this case. It's been mentioned in the comments, but not by OP. Could you explain further on that, or maybe do a cheap paint drawing? I understand this is borderline of asking a new question... $\endgroup$ – Marc.2377 Oct 15 '16 at 1:12
  • $\begingroup$ @Marc.2377 The easiest way to understand why the counter weight is needed is to imagine building the elevator by unreeling it from the geostationary station. As you let 'down' the in-gong cable the center of gravity of the whole thing shifts inward resulting in a faster orbit unless you also let 'up' a second mass to balance things. As a bonus, you get tidal stabilization of the orientation of the structure. By using a heavy mass the out-going end can be relatively short, or you can make it long and get a 'free' inter-planetary boost by climbing to the far end and letting go. $\endgroup$ – dmckee Oct 15 '16 at 20:54
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The force on the rope is due to the weight of the rope. You can imagine a rope 36,000 km long weighs a lot.

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    $\begingroup$ specifically to the question: the very top might be in a geosynchronous orbit, but the rope is not. $\endgroup$ – null Oct 13 '16 at 19:58
  • $\begingroup$ And since for the elevator to be in orbit and steady with respect to earth's surface its center of mass needs to be in GEO, thus it's a much longer rope. $\endgroup$ – Daniel Jour Oct 14 '16 at 13:49
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Imagine building a stack of dirt. For a short period of time, it might stick straight up, but eventually it will fall down, forming something of a pyramid. Every material has an amount at which you can build with it and not have it collapse. Going to a pyramidal shape helps somewhat, but it doesn't solve everything. Specifically what is important is the Breaking Length, which can be defined as the maximum weight of an equal area vertical structure that can be supported for the given material. If a material can sustain a 5000 km vertical tower (As measured at sea level), it is strong enough to use for building a space elevator.

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    $\begingroup$ You seem to talk about compression while the space elevator concepts involve a tether whose critical property is its tensile strength. $\endgroup$ – Peter - Reinstate Monica Oct 15 '16 at 13:18
  • $\begingroup$ @PeterA.Schneider 's comment is pretty poignant (evoking a keen sense of sadness or regret.). Can you add something to help address how/why a compressional argument about a stack of dirt on the ground really applies to a space tether? $\endgroup$ – uhoh Feb 7 '18 at 3:02
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    $\begingroup$ @uhoh no regret here ;-)... see merriam-webster.com/dictionary/poignant $\endgroup$ – Peter - Reinstate Monica Feb 7 '18 at 6:39
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A space elavtor works by balancing gravity with centrifugal force.

As you get further from the center of the earth (which is both the center fo gravity and the center of rotation) gravity decreases and centrifugal force increases.

Each segment of the "rope" can be assigned an "effective weight" consisting of weight minus centrifugal force (both weight and centrifugal force are proportional to mass). For segments below geostatoinary orbit the effective weight will be positive, for segments above geostationary orbit the effective weight will be negative.

Tension in the rope is highest at geostationary orbit where all the rope with negative effective weight is pulling on all the rope with postive effective weight.

While the effective weight of the rope below geostationary orbit will be lower than it's actual weight it will still be a substantial fraction of it.

We can to some extent work arround this by making the rope non-uniform but just as with a rocket you need high ISP to avoid ludicrous mass ratios with a space elevator cable you need high strength to weight ratio to avoid ludicrous thickness ratios.

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The cable is under tension - maximum in the center - because every foot you go up the cable, that's one more foot of cable hanging from it.

This can be addressed by tapering the cable: the cable cross-section increases as you approach the center in order to support the increasing amount of cable hanging from it. There is a 'taper ratio' equation (I read an article about all this, IIRC published in either "New Destinies" or "Analog" at least ten years ago). Given carbon fiber, the taper ratios start getting down to (again, IIRC) about 4.5.

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