If I had the material that had the tensile strength, would it be possible to have it dangle from the lowest most practically possible geostationary object down into the atmosphere?

Would it not experience the same issue as any object re-entering the earth's atmosphere and be subject to the same heating effects?


would it (because it is geo-stationary) not be affected by re-entry phenomenon? I would really appreciate some physics if anyone is up to it...my guess is that it would be subject to re-entry to the atmosphere and would burn up.

The reasoning being, that I would like to design a CO2 exhaust for Earth...hence my question

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    $\begingroup$ Heating while reentry requires a velocity between air and object. But why a geostationary orbit is called geostationary? $\endgroup$
    – Uwe
    Jul 11 '18 at 21:28
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    $\begingroup$ I dont know dude...that is why I am asking the question. I am not a Physicist and would like to understand if there is any such effect at all...i.e. if there is a re-entry price to pay, heat wise for such a dangler. In other words, intuitively, is not the same as asking a Physicist for the answer. If everything in the world had an intuitive answer, then we would not have any need for empiricism etc. $\endgroup$
    – Beezer
    Jul 11 '18 at 21:29
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    $\begingroup$ If the cable is "geostationary" - not moving relative to the surface of the Earth - it will only experience relative winds and not reentry velocities with their associated heating. The big problem with this concept is tensile strength. $\endgroup$ Jul 11 '18 at 21:30
  • $\begingroup$ Not the same thing, but this project is certainly food for thought, you may find it interesting to read about Could you enter this building when it was over New York City? $\endgroup$
    – uhoh
    Jul 12 '18 at 1:07
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    $\begingroup$ The latest XKCD what if is also related. (It discusses not a geostationary system, but one dangling from the moon, which means the is a strong differential motion.) $\endgroup$ Jul 12 '18 at 13:48

You're asking about an idea that's been around for a long time: the Space Elevator or the Sky Hook. Konstantin Tsiolkovsky wrote about a similar concept in 1895, though his concept was for a standing building, a compression structure supported by a foundation on Earth.

You can't just lower a cable from a geo bird. If you have a cable hanging from a geo station with nothing above it (ignoring the dynamics of getting that cable there), the weight of the cable, not fully offset by centrifugal force at the lower altitudes, would pull it down. You have to send a cable outward too, such that the mass and increased centrifugal acceleration of the outward segment balances the mass and decreased centrifugal acceleration of the inward segment. Once the inward segment reaches Earth (Sri Lanka was A.C. Clarke's favorite) it can be anchored and a counterweight attached to the outward end to keep the whole thing in tension, even when a payload is being hoisted up the cable.

That's the general attractiveness of the concept: if you could build such a thing, you could use it to "elevator" payloads up to GEO using only electric power. You could also return things from GEO to Earth for such things as repair or servicing.

Until fairly recently there were no known materials that could handle the tensile stress involved. But it appears that maybe carbon or boron nitride nanotubes, or diamond nanofibers, might be strong enough. But the static tensile stress is only part of the problem.

One problem is the "bullwhip" problem. A payload being sent upward places a sideward force on the cable because its tangential velocity is being increased, ultimately to GEO orbital velocity. This launches a displacement wave traveling both upward and downward in the cable. The wave traveling in the direction that the cable thickens isn't too much of a problem, but the wave traveling in the direction the cable thins grows in amplitude as it travels. This is the same phenomenon that allows someone to crack a bullwhip: the wave launched at the handle grows in amplitude as it approaches the skinny end, and that end can reach supersonic speeds. As the payload nears the GEO terminus, the wave launched from that thick part of the cable travels toward thinner sections, and the large displacements from equilibrium can result in large, sudden velocity changes that greatly increase the stress on the cable.

Another obvious problem is orbital debris and meteorites. Cables under extreme tension don't ike being partially severed. If a crack or gouge is big enough that it becomes a Griffith crack the whole cable severs, and that's a really bad day.

To answer the newly-edited and emphasised part of the question: no, it would not be subject to re-entry speeds and heating in the usual sense. Construction of the cable is a carefully monitored and controlled balance of sending new cable both upward and downward, to keep the net motion of the "station" at GEO constant, at GEO velocity. Even with new high-tensile-strength materials such as carbon or boron nitride nanotubes, at the GEO point, where stress is the maximum, the cable diameter required is really large, too large to "spool". It might be built up from spools of smaller-diameter cables, or even chemically fabricated in a continuous process. But either way, the net descent rate will be low, limited by the maximum fabrication rate that process allows.

And because the upward and downward segments cancel net tangential forces as well as radial forces, this prevents the usual acceleration to high tangential speeds that you usually get when an orbiting object descends. That acceleration occurs when the orbiting object is in freefall (i.e., subject only to gravitational forces from the primary, the large body the object is orbiting), but the end of the cable isn't in freefall: it has large non-gravitational forces on it from the rest of the cable.

The net result: you don't have to worry about re-entry heating.

EDIT 2018 July 15

I wrote software to calculate the required cable diameter as a function of altitude, only for a static cable in position, so no installation dynamics, no propagating waves, and no design margin! so it gives the minimum diameter the cable could be under perfect conditions, at each point along its length. Step-by-step upward from the anchor point (and the specified bias tension there) it calculates the net acceleration due to gravity and centrifugal acceleration, calculates the differential tensile force with altitude (integrating upward from the anchor point as it goes), and then at each step uses the material's tensile strength to calculate the cable diameter. Putting in different material properties yields interesting results.

Using Kevlar, with a tensile strength of 3.62 GigaNewtons (GN) per square meter and mass density of 1,440 kg per cubic meter, and a bias tension of 1 MegaNewton (MN), the cable diameter at the anchor is ~1.87 cm, but a whopping 285 m at the GEO point. The cable mass from GEO down is ~1.2E12 metric tons!

But using carbon nanotubes, assuming you can get large-scale performance as good as the small-scale performance demonstrated in the laboratory (63 GN per square meter and mass density of 1,400 kg per cubic meter), the cable diameter at the anchor is only 0.45 cm and only 0.77 cm at GEO, with a down-segment mass of only ~2,000 tons. But this is an extremely optimistic estimate. It assumes that any individual carbon nanotube that starts out at the anchor continues unbroken all the way to GEO, and that any nanotubes added as you go upward also extend all the way to GEO, and bind so strongly to their neighbors that slippage doesn't occur.

Even if the large-scale performance is only 1/10 of the lab performance, that still reduces the cable diameter at GEO (compared to Kevlar) to ~3.1 m and total down-segment mass of ~1.7E8 tons.

Again, this is under ideal conditions. When you start adding the strength to handle orbital debris and meteorite damage, dynamics such as propagating waves, systems to control the dynamics, and the ever-present design margin (if you've ever driven across a bridge, you owe your life to that!), cable sizes get significantly larger, along with the masses.

  • $\begingroup$ The fall of the space elevator is described in Red Mars. It is indeed a bad day. The cable for Earth would go 9 times around the world, with enormous mass and terrific speed... $\endgroup$
    – njzk2
    Jul 12 '18 at 3:58
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    $\begingroup$ @njzk2 A real space elevator would be thinner-than-paper ribbons weighing about a gram per meter. Most of the length would burn up in the atmosphere and the lowest few hundreds of km would flutter down at the speed of falling leaves. $\endgroup$
    – JollyJoker
    Jul 12 '18 at 7:53
  • $\begingroup$ This does not answer the question which is about heat. The setup of the question is wrong, OK, but assume that was fixed. Besides, Organic Marble answered the question in a comment. $\endgroup$ Jul 12 '18 at 8:19
  • $\begingroup$ @njzk2 Except Kim Stanley Robinson was more concerned about exciting fiction than laws of physics. $\endgroup$
    – Graham
    Jul 12 '18 at 10:36
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    $\begingroup$ @tim i'm afraid people have been "putting it in the ocean" for too long, too much for it to be a viable option. I doubt your space elevator crashing down on an endangered whale family and causing a tsunami... (oh, wait, let me post a question to worldbuilding.se) $\endgroup$
    – Mindwin
    Jul 12 '18 at 19:19

Material strength/weight issues aside, a space elevator wouldn't suffer destructive re-entry heating if you lowered the cable (as opposed to dropping it in an uncontrolled manner), which would be a good idea for a lot of reasons.

If you drop the cable, it's going to accumulate a lot of kinetic energy - exactly how much depends on its mass and cross-sectional area. So I'll hand-wave with "a lot". When it stops dropping, bad things will happen regardless of how long the cable is.

  • If the cable is long enough to hit the ground, it's got the geostationary object's parallel velocity of ~3km/s and a perpendicular velocity of up to 9km/s (masses cancel for KE = GPE, leaving 1/2 v^2 = gh. Plug in 9.8m/s^2, ~36000km, convert and solve), both of which will suffer from a lot of aerobraking (which will, again, depend on the cross-sectional area). Congratulations, you've just orbitally bombarded the Earth
  • If the cable isn't long enough to hit the ground, the spool is going to rip out of whatever it's attached to on the geostationary object. I'm not even going to provide some back-of-the-napkin math on this

The Concorde experiences some heating at Mach 2-ish, but probably not enough to be terribly destructive to something with the kind of strength:weight ratio you need for a space elevator

Bear in mind that as you lower a cable from a geostationary object, you'll need to unspool a counterweight to keep the center of mass from shifting out of its geostationary slot.

Edit: Whoops, I was beaten to the Space Elevator link. But FWIW, there'll still be some heating on the cable, which is moving, regardless of whether the center of mass moves.

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    $\begingroup$ For an object to accelerate to those speeds it must be in freefall, which the cable is not. The tendency of the inward segment to accelerate tangentially as it goes down is offset by the tendency of the outward segment to decelerate as it goes up. But there is a consequence: if you "just lower" the cable, it doesn't head straight for Earth, it gets out in front of the GEO platform! Just as a satellite initially co-orbiting with another will, if it goes lower in altitude, get out in front of the other, the cable does the same. This makes it like a *gravity gradient stabilized" satellite... $\endgroup$ Jul 11 '18 at 22:00
  • $\begingroup$ I also, must say, for a first answer on this site-- that your answer is wonderfully sourced. $\endgroup$ Jul 11 '18 at 22:01
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    $\begingroup$ ...which is not in equilibrium straight up and down. It's slanted. Getting it to "straighten out" and descend to the intended anchor point will require propulsion or some other kind of tangential force. $\endgroup$ Jul 11 '18 at 22:04
  • $\begingroup$ With the exception of diamond, there is a strong overlap between materials that can handle high compression loads and materials that are heat resistant. With the exception of light bulb filament metals, materials that can handle high tension loads tend to not be resistant to high temperatures. $\endgroup$
    – Jasper
    Jul 12 '18 at 6:55
  • $\begingroup$ To make this answer complete, you could add a remark on atmospheric temperature en.wikipedia.org/wiki/… see the graph temperature vs height $\endgroup$ Jul 12 '18 at 8:23

This is essentially Tsiolkovsky's 1895 "Space Elevator" idea. You'd need an adjustable counterweight extending further out to keep the center of mass at the geostationary point. If done correctly there should be no movement at the lower end. No movement means no air friction and thus no heating...


For something to be in geostationary orbit it must be at a distance of about 36,000 km from Earth. That's a lot of material.

As something gets closer to the Earth it must travel faster to maintain a stable orbit (angular momentum applies), and within the atmosphere it must travel slower to avoid problems with heating due to friction etc.

I imagine that any such cable or space elevator guide rail would bow outwards (in the direction of Earth's spin) between its end points. Assuming the material has the strength to not break, I don't have any idea what effect the stresses and forces would have on the endpoints, but I can guess that there would be a problem with securely anchoring the endpoint on Earth so that it is not ripped free, and the space end would need to have additional sufficient mass at a suitable distance beyond the geostationary end to prevent that end being dragged around faster than a natural orbit speed.

I also imagine that moving any mass along it will add complications to the stability, too fast or too slow at any point will quickly destabilise it, creating oscillations along its length.


Satellites in that orbit tend to move in a figure 8 shape as seen from earth, so that needs to addressed somehow also.

From http://www.satobs.org/geosats.html :

Strictly speaking, a geostationary satellite would be in an orbit of 0 degrees inclination, zero eccentricity and a mean motion of 1.002701 revolutions per day or a period of 1436 minutes per revolution. The Earth rotates once in about 23 hours and 56 minutes (1436 minutes); the remaining 4 minutes allow the Earth to rotate further, compensating for the apparent change in position of the Sun. This arises from the movement of the Earth in it's orbit about the Sun. In fact most geostationary satellites are really geosynchronous. Having mean motions between 0.9 to 1.1 revolutions per day they are allowed to drift across a box before corrections are made by on board thrusters. The size of this box is dictated by mission requirements. For example the box for a TV broadcast satellite is determined by the beamwidth of the reception dishes used.

The drift from the ideal position arises due to anomalies in the Earth's gravitational field, at this altitude atmospheric drag is not a consideration. The gravitational influence of the Moon provides an out-of-plane force too, which gradually increases the orbital inclination towards that of the Moon about the Earth (which itself varies between 18 and 29 degrees). The satellite now tends to describe a figure-of-eight ground track; ground controllers aim to restrict this to the box mentioned earlier given that enough orbit-keeping fuel remains. This wandering has been allowed to grow unchecked in the case of a few communications satellites in order to provide better coverage of the polar regions which is otherwise poor (from the poles a geostationary satellite would almost graze the horizon). Net connectivity to US research stations in the Antarctic was achieved in this manner.

  • $\begingroup$ If you're quoting someone, you should attribute the quote to them. $\endgroup$ Jul 12 '18 at 16:56
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    $\begingroup$ I put the link where I got it from, isn't that enough? $\endgroup$
    – CrossRoads
    Jul 12 '18 at 16:59
  • $\begingroup$ Looks fine to me! $\endgroup$
    – uhoh
    Nov 7 '18 at 15:51
  • $\begingroup$ I noticed that you participate in Aviation SE. Here's a tough question, but if you have a simulator program, you might want to give it a try! Could an aircraft ever simulate Martian gravity perpendicular to the aircraft's floor? $\endgroup$
    – uhoh
    Nov 7 '18 at 15:53
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    $\begingroup$ I don't play with simulators, just fly real airplanes. Home simulators are only good for practicing approaches. $\endgroup$
    – CrossRoads
    Nov 7 '18 at 18:52

If the spacecraft is in geosynchronous orbit, then by definition it is rotating with the Earth, so if you have a cable rotating at the same angular velocity as the spacecraft, then it will also be moving with the Earth, and also with the Earth's atmosphere. Since there's no relative motion between the cable and its surroundings, there wouldn't be any reason for it to "burn up".

Note, however, that for geosynchronous orbit to be maintained, the center of mass must be at geosynchronous altitude. If the spacecraft has a cable dangling below it, then the entire spacecraft+cable system must be in geosynchronous orbit, so the spacecraft would need an altitude higher than that what it would have without the cable, or to have cable extending above it to counteract the weight of the cable below it.

Another issue is that "dangling" implies that you're releasing the cable from orbit. If you're in freefall, then by definition there isn't local gravity in your reference frame, so you can't "drop" a cable: if you put a cable outside your spacecraft, it will just float there.

If you do get it significantly below you, it won't stay there. Tom Spilker claimed that it will pull the spacecraft down, but this is mistaken. A larger radius means a lower angular velocity, but higher linear velocity. If you were to somehow arrange for the cable to be at the same angular velocity as the spacecraft, then it would indeed pull the spacecraft down. But if the cable starts at the linear, rather than angular, velocity of the spacecraft, and is placed in a lower altitude, its velocity would be too high to maintain circular orbit at that altitude. Instead, it will enter into an elliptical orbit, and in the frame of reference of people on the spacecraft, the cable will appear to orbit around them.

  • $\begingroup$ @Accumulation, Look again. If you lower a cable straight toward Earth without moving the GEO station out of its radial position, the center of mass moves closer to Earth, while moving at less than orbital velocity for that altitude. In that case the gravitational force is greater than the centrifugal force, and the net movement of the entire system is downward. The key is the force required to lower the cable. If an attempt is made to lower it just by releasing it, an infinitely flexible cable will remain at GEO altitude; you have to push it downward, and that pushes the station upward... $\endgroup$ Jul 12 '18 at 21:22
  • $\begingroup$ ...If you simultaneously push an outward segment you can balance the radial forces such that the GEO station remains at GEO, but you do get rotation around an axis perpendicular to both the orbital velocity and the direction to Earth. I briefly describe the dynamics in a comment to @Punintended's answer. The Coriolis force on the moving (downward) cable is the -2(Omega X V) term of the Coriolis equation. $\endgroup$ Jul 12 '18 at 21:35
  • $\begingroup$ If the force comes from the spacecraft, then the center of the mass won't move. Only if there is some external force moving the cable will the center of mass move. Moreover, you simply ignored my claim that objects in higher orbits have higher linear velocity. If you wish to dispute that claim, you should do so explicitly. If you wish to dispute that the linear, rather than angular, velocity is what will be conserved, you should say that explicitly. And the rotation you speak of is, I think, referring to the same thing as me discussing the cable rotating around the spacecraft. $\endgroup$ Jul 13 '18 at 1:15
  • $\begingroup$ The acceleration gradient is much steeper earth side of GEO. To balance you need more tether above geosynchronous than below. The center of mass would be above GEO. $\endgroup$
    – HopDavid
    Jul 15 '18 at 13:06

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