Obviously the greater a planet's gravity, the stronger the material of the space elevator's cable would need to be in order to support its own weight. However, doesn't greater rotation speed reduce an elevator's required length by increasing centrifugal force?

Absolutely massive super-Earths are probably out of the question with regard to space elevator construction (but correct me if I'm wrong!), but what about a planet of two Earth masses? According to this planet calculator I found through Google, a planet of Earth's density but twice the mass would have 1.26 times the radius and 1.27 times Earth's surface gravity (I'd also love to be corrected here if this is wrong).

Forgive me for being unable to do the math myself, but given a surface gravity of 1.27g:

  1. Would it even be possible for conceivable materials (such as carbon nanotubes or graphene ribbons or something else I've never heard of) to support a space elevator if the planet had an Earth-like ~24 or ~25 hour day?
  2. If an elevator at such a rotation speed is in fact feasible, how long could the days be before a space elevator is impossible using conceivable materials?
  3. If a shorter day is required to build a space elevator on such a planet, how short would the day need to be?

I'm not a physicist or chemist and admit I don't know the bounds of "conceivable materials". I don't want to use unobtainium.

In case it affects any answers, FYI my primary interest in asking this question regards a planet I'm trying to design that is the homeworld of an alien civilization, not humans, so "find a better candidate planet" isn't really an option for them. I'm pretty married to the planet's gravity, so I'm willing to accept the suggestion of discarding the whole space elevator idea if it turns out to be basically impossible to execute.

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    $\begingroup$ Your asking for more detail than I can give, but I will point out that centrifugal forces don't exist. I believe you want to look up centripetal. $\endgroup$ – bowlturner Jan 15 '15 at 14:11
  • $\begingroup$ you want enough spin that a geostationary orbit is possible, that's all $\endgroup$ – ratchet freak Jan 15 '15 at 16:54
  • $\begingroup$ No need to use unobtainium. Just use interlaced carbon nanotubes reinforced with an alloy of mythril and adamantium on graphene sheets and you are done. Oh wait... This material is just called "unobtainium". $\endgroup$ – Victor Stafusa Jan 15 '15 at 18:16
  • $\begingroup$ @bowlturner: xkcd.com/123 $\endgroup$ – Joshua Snider Mar 1 '16 at 23:52
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    $\begingroup$ @bowlturner Actually centrifugal forces do exist if you assume the rotating system itself as the reference. And in this case — where we are sitting on a rotating planet — this becomes very relevant. $\endgroup$ – MichaelK Jun 25 '16 at 19:04

It turns out that it doesn't matter all that much how fast a planet spins for the purpose of creating a space elevator, unless it spins significantly faster than Earth. Let's run through the calculations:

The critical point of a space elevator, stress-wise, is the point at geostationary orbit. Everything below this point effectively hangs down from it, and everything above this point pulls upwards. (this isn't quite accurate, since we want a bit more upward pull for stability, but it's true within some small factor of safety.) Based on this, we can compute the maximum strain experienced by the space elevator by integrating the weight of the hanging portion of the elevator from the ground up to the geosynchronous point. we start with the equation for acceleration due to gravity:

$$|g| = \frac{GM}{r^2}$$

We can use this to obtain $\frac{d \sigma}{dr}$ (derivative of stress against radial distance from the planet's center of gravity) by multiplying by $\frac{dm}{dr} = A \rho dr$ (our differential force) and then dividing by area, $A$, to get our differential stress equation:

$$\frac{d \sigma}{dr} = \frac{GM \rho}{r^2} dr$$

We then integrate this from $r_0$, the radius of the planet, to $r_1$, the radius of geosynchronous orbit, to get total stress at our maximum stress point. (Approximately. Our actual stress will be a bit higher.)

$$\sigma = \int_{r_0}^{r_1} \frac{GMA \rho}{r^2} dr = \frac{GMA \rho}{r_0} - \frac{GMA \rho}{r_1}$$

For $r_1 \gg r_0$, we can effectively ignore the second term.

On Earth, $M = 5.97 \times 10^{24}$ and $r_0 = 6.37 \times 10^6$. This yields, with $\rho = 1400 \text{ kg/m}^2$ for carbon nanotubes, a stress of 87.6 GPa, which is below the maximum stress for our nanotubes.

On your planet, this will equal about 139 GPa, which is within the realm of what's been proposed as the upper end for multi-walled carbon nanotubes (150 GPa, according to this source.).

  • $\begingroup$ I think it's worth mentioning what the terms in these equations stand for. $\endgroup$ – bendl Dec 13 '17 at 18:38

Since you've got these tags (science-based, reality-check, technology, physics), I leave you with this YT video, which pretty much decimates any possibility of a space elevator: https://www.youtube.com/watch?v=iAXGUQ_ewcg

On Earth...

  • Just to lift a thin film solar panel (no lifter structure, motors, payload, friction or drag, just the thin film panel) would take a week.
  • Lasers would be better, but not much:
    • Atmospheric distortion would reduce the efficiency to 0.25% (That's not 25%.)
    • That's the same as requiring a nuclear reactor for a single mid-sized office building building.
    • Adaptive optics would increase the efficiency to 2.5%. Like a nuclear reactor powering 10 mid-sized office building buildings. Still really bad.
    • When you add in the structure, motors, payload, friction and drag, we're talking a stupendous amount of energy.
  • Carbon nanotubes are still in the small-scale laboratory scale, and have been there for quite some time.

Thing on your planet wouldn't be any better.

  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ – L.Dutch Oct 28 '17 at 11:29
  • $\begingroup$ @L.Dutch "essential parts" added. $\endgroup$ – RonJohn Oct 28 '17 at 13:36
  • $\begingroup$ Earth can spin up to 1.4 hour days and still hold an atmosphere. $\endgroup$ – Muze Mar 10 '18 at 23:28

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