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A space elevator from the Moon's equator, slightly past the L1 Lagrange point and extending into cis-lunar space would have a substantially lower tensile strength requirement than an Earth-based elevator.

But just what tensile strength would be needed for the cable?

lunar space elevator Source

The counterweight is slightly on the Earth side of L1, because L1 is unstable to perturbations. The closer the counterweight is brought to Earth, the harder it will be to perturb, but there will also be more tension needed in the cable. Thus, one picks a point as close to L1 as feasible. The answer will be approximately the same whether you pick exactly at L1, or slightly on the Earth side of L1. Therefore, you may use the distance to L1 for calculations, even though the actual distance would be slightly more than that.

I am looking for a numerical answer, not a list of possible materials or a comparison to other space elevators. Neither the Wikipedia article nor answers to the following two questions provide an actual number.

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    $\begingroup$ Does a space elevator to an L1 point even work? I don’t think there’s anything to keep it up. $\endgroup$ – Bob Jacobsen Jun 8 at 6:08
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    $\begingroup$ It might as the moon is tidaly locked. I was looking into this question trying to quickly debunk the premis but it was more involved than expected. I guess it was a better question than I initially thought. $\endgroup$ – lijat Jun 8 at 12:27
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    $\begingroup$ I made an edit, perhaps you can add some of your last comment back into the question as well? I think that the more you explain the issues, the question becomes much more interesting and people may spend more time looking at the PDF linked in the Wikipedia article to see if there's anything helpful there. $\endgroup$ – uhoh Jun 10 at 23:59
  • $\begingroup$ * Not to scale - I'd hope not haha! Else your counterweight is 50-100 Deimos. $\endgroup$ – Magic Octopus Urn Jun 11 at 14:26
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    $\begingroup$ @BobJacobsen: The elevator would need to reach past L1 and have a counter-weight to keep it taut. $\endgroup$ – SF. Jun 13 at 10:00
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An exact value isn't available, but there are some proposed designs in existence that we can look at. The materials exist today. A breaking height of about 300 km is required. Kevlar could theoretically be used, but other materials that allow for a larger tensile strength would be preferred, such as Zylon or M5, with a total tensile strength around 4-6 GPa.

The bottom line, it depends on how you design the elevator, what amount of cargo and its distribution you need to support, and what the taper is, but anything in the range of 4-6 GPa will be good enough to create a design that works, higher allowing for a less tapered solution.

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Your question in its present form is unanswerable. Tensile strength is not the limiting factor of how space elevators are designed. Space elevators are designed to have an arbitrary constant amount of tension. This is accomplished by varying the thickness of the cable as you move up to either L1 or a GEO orbit for Earth.

Imagine it this way, when you start at the bottom where the cable touches the ground there is no force so the cable can be as small as you want. Go up one mile and now the cable has to be at least as thick as it was at the bottom plus additional thickness to carry all of the weight of the cable below it. This sort of design is called a constant tension space elevator because the thickness is varied to keep the tension constant. The first designs of space elevators used constant cross-sections, but that design has long since been abandoned because it is mass-inefficient to the point that no material could work (even carbon nano-tubes on Earth).

So, I think what you really want is a comparison of taper ratios on Earth versus a Moon L1 elevator. The taper should start at a minimum and reach its maximum at either GEO or L1 and then taper down to a counterweight. Derivation time.

For Earth, where T is the tension in the cable, $\rho$ is the density of the cable, $\mu_e$ is the standard gravitational parameter of Earth and $\omega$ is the angular velocity of Earth's rotational period. An intuitive explanation of the terms: The terms on the left side is the tension holding the cable up at some point dr, the terms on the right are first gravity dragging you down and then centripetal acceleration slinging you up.

$T \space dA = \rho A(\mu_e/r^2 - \omega^2 r ) \space dr $

$ dA/A = (\rho /T )(\mu_e/r^2 - \omega^2 r ) \space dr $

Re-arranging and integrating from the surface $R_s$ to GEO $R_g$.

$ln(A_g/A_s) = (\rho /T )(\mu_e(1/R_s -1/R_g) + \omega^2R_g^2/2 -\omega²R_s^2/2) $

$TaperRatio =A_g/A_s =exp[ (\rho /T )(\mu_e(1/R_s -1/R_g) + \omega^2R_g^2/2 -\omega²R_s^2/2)] $

I am going to use kevlar as an example because it has one of the best strength to weight ratios of any material.

$\rho $ = 1400 Kg/m^3

T = Max tensile strength = 3.6 GPa = 3.6 *10^9 Pa

*note if doing the calculation yourself, $\rho/T$ needs to be in s^2/km^2 units for everything to cancel properly.

I am going to incorporate a safety factor of 1.1, which makes the max T = 3.27 GPa.

Assuming $R_s$ = 6371 Km, $R_g$ = 42164 Km, $\mu_e$ = 398600 Km^3/s^2, $\omega$ = 7.272e-5 rad/s, and $\rho/T$ = .3889 s^2/km^2, The taper ratio for a Kevlar based Earth elevator would be 5.2577e+10. This large taper ratio is why the elevator would be impractical on Earth.

Now for the derivation of the taper ratio on a Lunar L1 elevator. This answer does not apply to an L2 elevator. The basic force balance equation is the same except now there is a term for the effect of Earth's gravity. The Earth also pulls up on the elevator. The constant c is just the radius of the orbit between the Earth and the Moon and $\omega$ is now the angular velocity of the Moon instead of the Earth. Since the the Moons rotational period and orbital period are very similar, the Moon is tidally locked. I assume a perfect tidal lock, if this was a real design you would have to allow the tether to move around on the surface of the Moon a little bit.

$T \space dA = \rho A(\mu_m/r^2 -\mu_e/(c-r)^2 - \omega^2 r ) \space dr $

Re-arranging and integrating from the surface of the Moon $L_s$ to Lagrange point 1 $L_1$.

$A_1/A_s =exp[ (\rho /T )(\mu_m(1/L_s -1/L_1) + \omega_m^2L_s^2/2 -\omega_m²L_1^2/2) + \mu_e(1/(c-L_s) -1/(c-L_1))] $

Here is the simplified equation for the L1 elevator if you want to try it yourself for various materials. $TR = exp((\rho\space 2.5289)/T)$

Plugging in using the same safety factor and Kevlar, you get a taper ratio of 2.95 for an L1 elevator where $L_s$ = 1738 Km, $L_1$ = 63008 Km, C = 384400 Km, $\mu_m$ = 4900 Km^3/s^2, and $\omega_m$ = 2.6617e-06 rad/s.

To sum it all, taper ratios are how you compare different elevators not tensile strength requirements. For Earth with Kevlar, the TR is 5.2577e+10 and for a Moon-L1 elevator it is 2.95. This is what I believe @LaurenPechtel's answer is talking about for blowing up exponentials.

Once you have found your taper ratio, you can then begin looking at how much mass you want to lift. Lets say you want your elevator to be able to lift 100 thousand kilos, you just do a force balance at the bottom and that sets your starting area.

$A_s T = mg => A_s = mg/T $

On the Moon that means you need an initial cross section of .45 cm^2 and a max cross section at L1 of 1.35 cm^2 to lift 100,000 Kg. On Earth, it would be 2.72 cm^2 starting cross section and 14.3125 Km^2 max cross section at GEO.

Bringing the counterweight closer to L1 from the Earth side will not change the tension requirements at all. It just changes where you decide to stop tapering down. Read this paper to understand why.

Last thought, the image you show for an L1 elevator can be somewhat confusing. An L1 elevator does not need a section to the pole for engineering reasons. Everyone who has looked into L1 elevator designs, generally has thought it would be nice to go straight to the poles. It is possible to attach a secondary string from the main elevator to the poles, but this is done for mission reasons.

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  • $\begingroup$ You're right about what I was saying about the exponentials blowing up. A small decrease in the strength can cause a big increase in the taper. You're fired for designing a cable with a safety factor of only 1.1, though! $\endgroup$ – Loren Pechtel Jun 13 at 3:50
  • $\begingroup$ For the Lunar elevator, the angular velocity is around the barycentre, so should not the radius r be the distance to that centre ? $\endgroup$ – Cornelisinspace Oct 6 at 9:49
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The problem is you are asking for an exact value that does not exist.

The biggest load on an elevator cable is the cable itself, the cable near the center needs more strength than the cable near the ends. In practice this means you would build a cable that is thicker in the middle.

The greater the increase in thickness you are willing to accept the lower the strength requirement for the cable material. Thus you get a range of answers--material strength and taper ratio.

You still have a minimum strength because the equation is exponential and blows up very badly if your cable is too weak.

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