2
$\begingroup$

Which solar system objects have the right combination of low gravity and fast spinning rate, that an object tall enough to be a space elevator could be built on with modern construction tech?

By "modern construction tech", I'm referring to readily available reinforced concrete of the sort they used to build the Burj Khalifa, not carbon nanotubes or any other 'theoretical' building materials.

Are there any other considerations, such as the presence of moons, that need to be accounted for?

$\endgroup$
11
  • 1
    $\begingroup$ Since construction in space is a largely unexplored field, I think the answer is "none". It might be useful to reframe this in strict terms of current materials science instead. $\endgroup$ Sep 21, 2019 at 5:55
  • 1
    $\begingroup$ @ChrisB.Behrens I mean yes, ignoring how one would actually build such a thing, where one would get materials, etc. If you could place a space elevator made of reinforced concrete on a solar system object, would the material be strong enough? is the main question. $\endgroup$
    – Ingolifs
    Sep 21, 2019 at 6:00
  • 1
    $\begingroup$ Why reinforced concreet and not steel cables? Is there a reason to restrict us to some modern building materials? $\endgroup$
    – lijat
    Sep 21, 2019 at 6:18
  • 1
    $\begingroup$ Lower gravity objects are a bet since you wouldn't need such exotic materials with extreme strenghts. A primary target for this is in fact Phobos: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20030065879.pdf but even if promissing this is still far far away from what we could possibly start to build now. $\endgroup$
    – Swike
    Sep 21, 2019 at 10:00
  • 1
    $\begingroup$ @Swike - I think that pdf you're linking to is probably a good answer. $\endgroup$ Sep 21, 2019 at 16:50

1 Answer 1

3
$\begingroup$

Here's how you can calculate this for any spinning body.

First, you need its stationary orbital radius

$$r_{stationary} = \sqrt[^3]{\frac{\mu}{\omega}}$$

  • $\mu$ is the body's mass parameter, a shorthand for its mass times the gravitational constant.
  • $\omega$ is it's rotational angular velocity. If you have it's rotational period, you can obtain it as $\omega = \frac{2\pi}{T}$

Secondly, we need to find how many Yuris (m²/s²) the space elevator is.

To obtain that, we can first look at what acceleration is acting on the elevator in a co-rotating frame of reference at some radius.

$$a(r) = \omega ^2 r - \frac{\mu}{r^2}$$

Integrating that over distance, we get the specific strength requirements:

$$\int_{r_{surface}}^{r_{stationary}} a(r) dr = \frac{\omega^2 r_{stationary}^2}{2} + \frac{\mu}{r_{stationary}} - \frac{\omega^2 r_{surface}^2}{2} - \frac{\mu}{r_{surface}}$$

Or somewhat more compact:

$$Y_{elevator} = \frac{\omega^2}{2} \left(r_{stationary}^2 - r_{surface}^2\right) + \frac{\mu}{r_{stationary}} - \frac{\mu}{r_{surface}}$$

This can be used to calculate the cross section ratio "taper ratio" between the upper part and the lower part of the elevator:

$$A = e^{\rho \cdot Y_{elevator} /T_s}$$

Where $\rho$ is the density of your material of choice, and $T_s$ is its tensile strength

As always with exponential equations, they grow very quickly. A cross section ratio of 10 is probably okay, while 1000 is out. And that's a difference in material strength of only a factor of 3.


Are there any other considerations, such as the presence of moons, that need to be accounted for?

Very few solar system bodies have moons below their stationary orbital altitudes that can get in the way. The Mars moon Phobos is one of the few examples.

Another consideration to make is that almost all moons in the solar system are tidally locked, and thus don't really have a stationary altitude. In those cases, you can build an elevator to their L1 and L2 Lagrangian points instead. The equation is a bit longer (not too hard! just do the integral above with another mass added), however you can use the following approximation, which is only a tiny bit too pessimistic:

$$Y_{Lpoint} = \frac{\mu}{r_{surface}}$$


Solar system space elevator requirements: (MY = megayuris)

  • Jupiter: 660 MY (not possible within current physics)
  • Venus: 53.4 MY (due to very slow rotational period)
  • Earth: 48.4 MY (only possible with currently impractical materials)
  • Mars: 9.5 MY (within current capabilities, except logistics)
  • Mercury: 8.9 MY
  • The moon: 2.7 MY (To EML1 or EML2)
  • Ceres: 50.6 kY (possible with any normal rope)
$\endgroup$
2
  • $\begingroup$ What level of physically impossible is Jupiter? (Chemistry or physics?) $\endgroup$
    – ikrase
    Oct 2, 2020 at 3:10
  • $\begingroup$ @ikrase Chemistry with a huge margin, physics very likely. $\endgroup$ Oct 2, 2020 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.