# Space Elevator on Mars with Today's Technology Possible?

I know it's not possible with Earth today, but With today's level of material science technology, would it be possible to make cable strong and light enough to make a space elevator system connecting a Mars-synchronous satellite to an anchor on Mars' surface, ignoring any load at all by an elevator car? Or can someone say with a high degree of certainty that the level of strength to weight possible in cables today is still far from possible even with Mars' lower mass compared to Earth?

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More on space elevators: physics.stackexchange.com/q/15052/2451 –  Qmechanic Aug 6 '12 at 12:44

the simple answer is no. Materials for an Earth elevator are at least one order of strength too weak at this time. Mars gravity is around 0.378 of earth, so materials are still too weak.

The long answer is much more complicated: 1. Taper of the tether plays a role as much as the safety factor you want to engineer into your elevator, how much a tether can hold, and how much material you can put into space to construct it. Strength, taper, and tether mass are related. You can check for the Space Elevator feasibility condition via isec.org or Google. It is basically some form of decay equation: how much do you need to keep lifting to maintain or grow your elevator, basically taking away from transport payload for maintenance/repairs. 2. Mars has some interesting options with Deimos being made of mostly Carbon. Material for a tether could be refined on site, e.g., when Planetary Resources engineers related technology for asteroids. Deimos' orbital period is with 30.30h is close to the sidereal rotation period of Mars with 24.6229 h. You could work a tether that drops from Deimos part way into the Mars atmosphere and get something that you could attach to with much smaller Delta V than getting to orbit, thus limiting tether strength and mass requirements substantially further. I strongly believe that that the Deimos model is probably close to being within reach of today's advanced materials. Of course you need to work out some control to avoid Phobos dropping by every so often below Deimos. 3. The best anchor point for a Mars Elevator, once material becomes sufficently strong for surface attachment would be on Olympus mons. That puts it above sand storms and other Mars weather.

Finally: I believe that Mars is destined to become the shipyard for the solar system. It is sufficiently friendly for construction sites and the lower gravity makes it much more amenable to lift things to space than Earth.

Hope this helps. Let me know if you have any more questions about the material strength.

Martin

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You can convert the exponential falloff into a linear material strength required, as was done in the NIAC study by Brad Edwards. You can then also ask the question if today's materials are the ~3 GPa strength of Kevlar or maybe the ~5-10GPa of M5, should you be able to acquire it. Or how much taper you are willing to deploy with. That is why this answer is not as simple, not even with your assumption that I just used a linear factor, which is not stated. With clearer boundary conditions we can give a clearer answer, i.e., calculate it. :) –  Martin Aug 6 '12 at 20:46
Sorry for deleting my comment just before you posted your...I had concluded that I was wrong on roughly the basis you give. –  dmckee Aug 6 '12 at 20:47

The calculations for the tapered space elevator are a little nasty, but apparently fairly workable as a closed form. I feel like it's rather unsatisfactory to leave this question without a quantitative sense, considering that it's possible.

To start off with, we need to formalize the fact that we're interested in the change in area, tension, or linear mass-thickness over the length of it. All 3 of these are directly proportional to each other. What's more, the differential equation for tapering doesn't care which. That is:

$$\frac{ \lambda'(r) }{ \lambda(r) } = \frac{ \frac{ G M }{ r^2 } - \omega^2 r }{ \left( \frac{ \sigma}{ \rho} \right) }$$

The (sigma/rho) quantity is the material specific strength. I'm using lambda to indicate kg/m linear mass thickness, but you can easily replace it with any of the other proxies. This equation has a solution. Here it is in terms of simple values, and then in terms of the pseudo-potential.

$$\lambda(r) \propto \exp{ \left( \frac{ - \frac{ GM}{r } - \frac{1}{2} \omega^2 r^2 } { \left( \frac{ \sigma}{ \rho} \right) } \right) } \rightarrow \exp{ \left( \frac{ U(r) }{\left( \frac{ \sigma}{ \rho} \right) } \right) }$$

To get the "taper ratio", we're interested in the area at geosynchronous orbit divided by the areas at the surface. This just consists of plugging in values.

$$\frac{ \lambda(r_{GEO}) }{ \lambda(r) } = \exp{ \left( \frac{ U(r_{GEO}) - U(r) } { \left( \frac{ \sigma}{ \rho} \right) } \right) }$$

Most people reading this probably have the requisite knowledge to do this calculation. Here are some, just to look at. I divided the specific strength by a factor of 3 to give a sense of the "engineering margin"

   mat        m^2/s^2     ratio       3x ratio
Nanotubes   46268000      1.228         1.851
Zylon        3766000     12.449      1929.560
Kevlar       2514000     43.707     83493.590
Stainless     254000      1.728E+16     5.16E+48
seatbelt      108693.     8.806E+37


You can see that in a pedantic way, yes, kevlar is strong enough to be a Mars elevator going by yield strength. But that's only if it is literally stressed to its yield strength. Even then, you're at a taper ratio of 43. Add in some margin, and this explodes to factors which are a good fraction of a million.

I didn't include the 3x ratio for seatbelt material, because I took the value which is the actual "working load". It's a good sample for an "everyday" material, even if depressing.

But still, I'll have to concede that it is possible, since Zylon is a real material, even if I am skeptical the industrial production capability and quality control. If you'll just accept the ratio of 1,929 then we can agree it is possible. It might be promptly broken by a meteorite, but that's not the question. Since this is an area ratio, the diameter ratio will only be about 43. That is, if the diameter on the surface is 1 mm, then the diameter at GEO will be 4.3 cm. You are free to interpret the consequences of that result as you see fit.

The sci-fi book Ouroboros Wave entertains this exact idea. In that book, they used engineered systems to flex a Martian space elevator made of carbon nanotubes to avoid Phobos when it passed by, which is an important point.

If you were going to build a Martian space elevator, you would either have to do something like this to avoid Phobos, or just crash it into the planet before you start. There is a distinct proposal to put a space elevator on the dark side of Phobos, since it is tidally locked. Phobos orbits almost 3 times every Martian day. That means that $\omega$ is larger for that system, which means that it will be much easier to build.

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It is actually an old ladies myth that the space elevator cannot be built with current technology. It would just be extraordinarily more expensive with current materials.

The elevator needs to be built from geostationary orbit downward. This means that all the material needs to be pulled up by conventional propulsion, this is the principal factor affecting cost. The conventional approach to build an elevator without materials satisfying the strong tensile requirements that are required for the single-tether design would have to be an exponential structure; a single tether connected to earth, supported by two tether, supported themselves by 4 tethers,.. until reaching above geo orbit, where we'll have to attach $2^n$ tethers to the counterweight, where $n$ is given by the tensile properties of the chosen material.

If you have already one space elevator, then it is easier/cheaper to build a second one, because the expensive step of moving the mass up has already been bootstrapped. In the case of mars, it would make more economical sense to actually send the building materials from earth than to uplift them from mars surface, assuming of course we already have a working elevator on earth

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Surely you'd start by putting a mass-driver on some suitable asteroid to get it into orbit, and then use the remaining mass of it as material for both the tether and counterbalance. Much cheaper than lifting billions of tons by rocket. We don't yet have the technology for the mass driver, but neither can we build all the rockets... –  hdhondt Aug 7 '12 at 4:00
@hdhondt i doubt the tradeoff of extracting the raw materials from an asteroid and processing them (say, into graphene nanotubes) will make it significantly more economically viable, but there is a lot of room for improvement in the next decades in that area –  lurscher Aug 7 '12 at 15:57

## protected by Qmechanic♦Oct 4 '14 at 7:18

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