I'm slightly confused about where exactly the forces involved in a space elevator are being applied, and my lack of a degree in physics makes it difficult for me to understand the sources that google gives me when I try to search for answers.

I'll do my best to explain what I think I know or at least know about, and would like for people to correct me when I inevitably say something that is wrong and/or fill in gaps of things that are relevant which I failed mention (likely due to not knowing about them).

As far as I know, there are only 4 sources of force in a space elevator system for Earth that are particularly relevant. Earth's Gravity, Moon's Gravity, Centrifugal Force, and the orbital velocity of any given part of the elevator.

For an elevator going up to a geosynchronous point from earth's surface, the effect of the Moon's gravity will fluctuate because the moon isn't in a geosynchronous orbit- so as the moon orbits, the effect of gravity from it will increase or decrease and probably have pretty significant effects on the elevator itself that could destabilize it and cause it to collapse or snap somehow.

Earth's gravity, of course, is relevant all the way up to the Lagrange point and most ideas for elevators involve going beyond the lagrange point so that the weight of the elevator beyond that point will help to counterbalance Earth's gravity with centrifugal force. It is believed currently that there are no materials capable of withstanding those forces- at least none of which we are certain and can industrialize/mass produce to the degree necessary for such a project.

However, this part confuses me somewhat for two reasons- 1 - The Lagrange point is not geosynchronous, as far as I am aware it is a point which always rests at some point where the gravitational forces of two bodies are equal. As stated, the Moon is not orbiting geosynchronously as rather obviously it moves through the sky from the perspective of anyone on the ground whereas a geosynchronous moon would seem to hang perfectly still at all times, so for an elevator that is geosynchronous (that is, fixed to a particular spot on the Earth's surface) the Lagrange point for the Earth and moon would be constantly shifting around and wreaking havoc on the elevator itself. That is, for a geosynchronous elevator, I do not see how there is ever a point where the gravitational forces at play are consistent. The solar lagrange obviously must be further than the Moon's orbit because if it was inside the moon's orbit the moon wouldn't orbit earth at all, as soon as it's path crossed that point it'd be captured by the Sun and go off on its own. Any elevator going to the solar lagrange would just get slapped by the moon itself or its gravity as it passes. Where are proposals placing space elevators that there is some form of lagrange point they are working with to keep gravitional forces consistent?

2 - The idea that materials cannot withstand the forces at play confuses me as well. Is not the entire purpose of the centrifugal portion of the elevator to negate the gravitional forces and thereby achieve net zero force? IE- for a normal centrifugal system the only force applying tension to the rope is the centrifugal force being resisted by the material's attachment to some fixed point. In this case, however, there is also Earth's gravity pulling along essentially the entire length of the elevator up until the Lagrange point where the gravity of.. some other body (and clearly not the moon or sun, for reasons above) kicks in and starts applying to the entire other half of the rope. But, if the force of Earth's Gravity and the force of whatever body is providing the stable lagrange point are negating each other, then where is the centrifugal force here? Objects can simply be in a geosynchronous orbit without any need for a rope at all whatsoever, and portions beyond the lagrange could be orbiting earth at whichever speed keeps them stationary with the lagrange. If every given point on the elevator is orbiting geosynchronously- then why would there be any centrifugal force at all? The only thing I can think of is something like air resistance from Earth's atmosphere making geosynchronous orbit at low altitudes functionally impossible, thereby causing those portions of the elevator to pull on the entire rest of it as they slow down and wobble to the side. But that isn't a centrifugal force at all! The way I think about it, the "source" of a centrifugal force comes from the resistance of an object to its path being changed via attachment to some rotating point because the object wants to move further away due to its path but the physical attachment is pulling it to keep it at a fixed distance. But if an object's path is being changed by something else entirely- in this case gravitational forces keeping that distance fixed- then I do not see where centrifugal forces come in?

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    $\begingroup$ Orbital velocity isn't a force, it's part of the centripetal force required to keep massive objects in a circular trajectory $\endgroup$
    – Triatticus
    Jun 1 at 3:48
  • $\begingroup$ There's a section on space elevators in Soonish $\endgroup$
    – Barmar
    Jun 1 at 14:31
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    $\begingroup$ There is a difference between geosynchronous and geostationary. Geosynchronous return to the same point in the sky every 24 hours, but may have inclination and eccentricity to their orbits, so they may drift about through the 24 hours, moving up and down or drawing figure 8s in the sky, but return to the same spot at the 24th hour (23hrs, 56min, 4sec). Geostationary appear as a fixed point in the sky and are a circular orbit along the equatorial plane. Geostationary is the only orbit that has an apparent ground speed of zero. The ISS, at ~90% of earth gravity, has a ground speed of ~17,500mph. $\endgroup$
    – David S
    Jun 1 at 17:21
  • $\begingroup$ As the answers say, Lagrange points don't really have anything to do with the calculations for an Earth-based space elevator. They are much further away than a space elevator would reach. However, for a lunar space elevator - that is, one built on the Moon rather than the Earth - the Earth-Moon Lagrange points are relevant. It could be that your confusion has come from reading about that without realising it wasn't talking about an Earth-based space elevator. $\endgroup$
    – N. Virgo
    Jun 2 at 5:31

3 Answers 3


Rather than trying to answer your questions directly, I'll try to explain how space elevators could theoretically work, and that may clear up your confusion.

To start, remember that, the closer a satellite is to Earth, the faster it moves around it. That's why the ISS takes 90 minutes per orbit, and the Moon takes 29 days. In GEO (35,786 km), it takes 24 hours, so a satellite directly above the equator in GEO will seem to sit at a fixed point in the sky.

If an object were going around the Earth in 24 hours, but below GEO, it would not have enough speed to stay in orbit, and it would fall back to earth. Similarly, any 24-hour object above GEO will fly away from Earth altogether.

If we build an elevator to GEO, there are two ways we could do it. First, we could build a tower. Unfortunately, most materials would collapse under their own weight. It could possibly be done by making the tower exponentially larger as you go down, but then you end up with a continent spanning structure (I have not done any calculations to decide how large it would be!)

Alternatively, we could lower a "rope" from GEO orbit. Now we have 2 problems. The first is that we do not know of any material that is strong enough to support itself. In other words, the rope would snap under its own weight. But, perhaps one day we'll discover a material that is strong enough. More on this later.

Now we come to the second problem. If we hang a cable from GEO, all its mass is below GEO, and as discussed above, it will fall down unless it goes around faster than once every 24 hours. In other words, it will drag the cable down to Earth. The way to fix this problem is to extend the cable beyond GEO. The mass above GEO wants to fly away and will counteract the mass below GEO. If we do it correctly, with the cable extending to about 100,000 km, the entire structure will hang at GEO. We could even attach a small asteroid to the far end of the cable, to minimise the total length needed.

Remember that any cable material below GEO wants to fall down, any material above GEO want to fly away. Both parts of the cable literally "hang" from its GEO anchor, downwards in the lower part, and the upper part "hangs" outward. That is why we need to be concerned about its strength. The stresses in the cable are caused by the cable material, both above and below GEO, wanting to leave that orbit.

As the end of the cable would travel at significantly more than escape speed from Earth, spacecraft can be launched to other planets, simply by ferrying them up and letting go at the correct moment and altitude.

Note that nowhere have I mentioned any Lagrange points. Lagrange points do not come in to this discussion. I assume you mean the L1 point, the Lagrange point between Earth and Moon. That is about 320,000 km away, much further than the end of a space elevator.

A couple more comments. There is no "geosynchronous orbit at low altitudes". GEO is at the fixed altitude of 35,786 km. Only at that distance can a satellite stay precisely above one spot.

"gravitational forces keeping that distance fixed". In orbit, as everywhere else, object will fly in a straight line unless a force acts on them. The gravitational force pulls the satellite straight down to the earth's centre, but its orbital speed means that, while it falls straight down, its sideways motion means that it keeps missing the Earth. Newton was the one who first explained this by imagining a cannon on a hill. As the starting velocity of the bullet increases, it flies further and further until, when the speed is high enough, it achieves orbit.

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    $\begingroup$ This does answer alot of my questions, and clears up many of the misunderstandings I now clearly see I had! Much of my misunderstanding came from what I now recognize as a rather awful source in the form of an article that was speaking about an advancement towards space elevators, which mentioned things like the lunar lagrange. As a followup question- Are the forces involved applied to the entire length of the theoretical cable? I suppose a more direct question is how much of the theoretical cable's length is at risk of snapping? $\endgroup$
    – Dicerson
    Jun 1 at 8:50
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    $\begingroup$ @Dicerson You can imagine an analogous situation without any of the complexities of orbital mechanics. The whole thing is no different from suspending a rope from, say, a branch. What are the forces acting on different parts of the rope? Where does the rope need to be thickest to support the greatest load/length? How much thickness (and thus mass) do you need to add to keep extending the rope? How thick would the thickest part need to be to reach, say, 36 thousand kilometres? :P $\endgroup$
    – Luaan
    Jun 1 at 13:08
  • $\begingroup$ So you can stick the cable anywhere as long as the center of mass is at GEO? $\endgroup$
    – Seggan
    Jun 1 at 21:27
  • $\begingroup$ The thing that's confused me is the "ferrying" up the rope. Wouldn't conservation of momentum mean that, by having something pull its way up the rope, part of the rope also gets pulled down? And wouldn't that shift the center of mass of the elevator towards the earth, so the whole thing would start to collapse? $\endgroup$
    – Dan Staley
    Jun 1 at 21:30
  • $\begingroup$ @DanStaley Without a counteracting force, yes. However, that would be a separate engineering challenge. $\endgroup$
    – David S
    Jun 1 at 22:20

I'll try and address your main questions, as I can identify them:

4 sources of force ...Earth's Gravity, Moon's Gravity, Centrifugal Force, and the orbital velocity of any given part of the elevator.

Centrifugal Force and "orbital velocity" are the same thing. So there are only 3 primary forces. The Moon's gravity is much less relevant than Earth's, so to a first-order model we need only consider Earth's Gravity and Centrifugal Force (CF). CF must be at least equal to or somewhat greater than the weight of the whole structure for it to work, because the structure is supported by tension (hence the need to extend beyond GEO).

Next regarding the forces necessary: centrifugal force does not simply cancel the weight of the tether, it supports the weight of the tether, in the same way a hook on the ceiling supports a chandelier. But the hook and the cable connecting it to the chandelier still have to withstand the weight.

As for the magnitude of that weight, it's hard to state just how much the size of this tether would dwarf any structure humanity has built. The tether length would be a minimum of $2.8$ times the diameter of the Earth. The Golden Gate Bridge, The Great Wall of China, and the Burj Khalifa are all absolutely miniscule by comparison. For instance, if the tether were only $1$ meter in diameter, it would require about $112,000,000$ cubic meters of material. This represents a cube of material whose edge is nearly as tall as the Burj Khalifa (world's tallest skyscraper). That is the weight of material that would need to be supported (granted, not all of the tether experiences $1 \, \rm{g}$ of normal Earth gravity).

Hopefully that gives you a basic idea of the scale of the problem.

This great video by Practical Engineering also explains it in greater detail:


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    $\begingroup$ "That is the weight of material that would need to be supported" - I'm not sure "weight" is a good way to think about it. The tension force in the cable needs to hold the apparent weight of the cable, which has more to do with the orbital characteristics than the dead weight. The cable mass at GEO, for example, takes no force whatsoever to remain there since it's in freefall orbit, while the mass just below it stays there with very little force to correct its slightly too-low orbit. I'm actually not sure if the tension force would be more or less than the dead weight of the cable. $\endgroup$ Jun 1 at 15:55
  • $\begingroup$ It's true the weight of the cable that needs to be supported is quite a bit less than the weight of the cable at sea level. It would be found by $\int \frac {G}{R^2}dm$ rather than $mg$. But my goal was simply to illustrate the rough scale of the forces involved and show why no known material is strong enough. Strength to weight ratio is the more relevant quantity. $\endgroup$
    – RC_23
    Jun 1 at 19:07
  • $\begingroup$ No, that would be the weight if you just hung a long string through a gravitational gradient, but the space elevator is held up by its own orbital inertia. Suppose you had a space elevator with all its mass concentrated at GEO - the cable feels no tension whatsoever, not 3% of the dead weight because it feels 0.03G. The cable is thickest at GEO, and none of that mass takes any force whatsoever to keep in orbit. $\endgroup$ Jun 1 at 19:17
  • $\begingroup$ The weight is $\int \frac {G}{R^2} dm$. The tension in the cable at a given point is $\int \frac {GM_E}{R^2} dm - \int r\omega^2 dm$ (gravity minus centrifugal), yes. Fair point I suppose. But the 2nd term is fairly negligible until you get to a significant orbital altitude. Again I was just giving a rough idea of how massive this would be, but it pays to be precise. $\endgroup$
    – RC_23
    Jun 1 at 20:28

You have a bad or misunderstood source. Lagrange points are not necessary.

The point is that you boost the center of mass to a height at which a geosynchronous tangent velocity is faster than the orbital speed at that height. Such an object will rise to a higher orbit if not constrained. Constraining it requires / provides a tension force.

If we have an object moving in a circle, and the only forces acting on it are some force $T$ and gravity, then the sum (taking downward forces as positive) must be the force required for uniform circular motion

1: $$mv^2/r = T + GMm/r^2$$

Define $v$ to be the velocity so that you get a geosynchronous motion. Given "ground" velocity $v_s$ at Earth's surface $r_0$...

$v = v_sr/r_0$

plug in and evaluate for $T=0$ to get the natural geosynchronous orbit

2: $$r = \sqrt[3]{GMr_0^2/v_s^2}$$

Solve 1 for $T$ and plug in $v$ to get a function only in $r$:

3: $$T(r) = mv_s^2r/r_0^2 - GMm/r^2$$

set all the constants equal to $1$ instead of looking them up because I'm lazy, and plot

enter image description here

Points lower have a negative force. Recall I defined downward forces positive, so negative forces point up: that is, the center of mass would require support from below to have that height. This is our common experience of unsupported objects moving with the surface of the earth - they fall.

Points higher have positive force. Again recall that downward forces are positive, so this is the tension that is required to keep the center of mass from rising to its natural, higher orbit.

2: If we input $v_s = 2 \pi r_0 / t$ where $t$ is the orbital period (1 day), we can re-arrange to obtain the more familiar equation for geosynchronous orbit,

$r = \sqrt[3]{GMt^2/4 \pi^2}$


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