# What payloads and launch speeds could a sling launcher get using modern materials on the Moon?

The only reference paper I have found on this subject is this paper by Landis, which is a great introduction to sling launchers - but the concept has been around for some time and there must be other treatments. At any rate, it is a simple device where a motor on a tower spins two cables - one bearing a payload and the other a counterweight - so fast that when released the payload flies into orbit or beyond. Modern motors are capable of doing this, and would even be highly efficient, if the cables used are very long - on the order of 50 km or more. The cables have to be incredibly strong, that's all.

The paper looks at the use of modern materials for such a device, but only briefly because the cables would be huge and extremely costly. It moves on to future possibilities using carbon nanotube cables.

If fullerene materials are not available, the concept could be implemented with existing materials. This increases the tether mass, and the sling launch becomes more difficult, but not impossible. The highest strength-to-weight ration for a currently available tether material are obtained with Poly(p-phenylene-2,6-benzobisoxazole), or "PBO" fibers, or with gel-spun polyethylene fibers. PBO (sold under the trade name "Zylon®") has a with tensile strength of 5.8 GPA, and density of 1.54 g/cm3 [12, 13]. High-strength polyethylene fiber (sold under the trade name Spectra-2000) has an ultimate strength of 4.0 GPa and a density of 0.97 g/cm3 [11]. Assuming an engineering factor of 2.5, the allowable load strength for the Spectra-2000 fiber is 1.6 GPa. For the example case of a launch to lunar orbit, 1.68 km/second, the required acceleration is 5.7 g (56 m/sec2 ). To carry a thousand kilogram payload, the force will be 56000 N. This will require a cable cross-section of 0.35 square centimeters at the tip. Since the cable must have additional cross-section to carry its own weight as well as the end mass, the cable must now be made to increase in cross-section from the tip to a wider cross-section toward the hub. This taper increases the cable mass. The cable mass is now about 2500 kg, no longer less than the mass of the launched object, but still a value which is feasible for an engineering system.

The reference paper made this calculation for escape velocity from the Moon. Could these materials take much more load without snapping under their own weight in such a scenario?

• You need to clarify a few things to make this question answerable: 1. What orbit do you want to achieve? This will determine the release velocity you have to achieve. 2. What's the mass of the object you're trying to sling? This, combined with the orbit question, will determine the requirements of the cables. – mLuby Apr 28 '15 at 0:46
• @MichaelLuby As I understand it a cable with current materials that could support its own weight and launch something into orbit is at the limit of what is possible, so I was pretty vague about specifics. But your point is well taken. Let me think about it and make an edit. – kim holder wants Monica back Apr 28 '15 at 2:00
• The was a Nasa Space Flight thread on slinging stuff from the moon: forum.nasaspaceflight.com/index.php?topic=5420.0 – HopDavid Apr 28 '15 at 5:32
• @HopDavid Just a note that they're NASASpaceFlight.com and it's just a name of a website. The way you wrote it suggests NASA associates with it. It doesn't. Misleading name doesn't really discredit it, it's IMO a good forum, I just thought to make that clear to anyone that doesn't know that. – TildalWave Apr 28 '15 at 7:19
• @MichaelLuby No. Feasibility of specific orbits, payload mass, even maximum radial acceleration and centrifugal force that would define suitability of such a launch system for humans is exactly what has yet to be established by analyzing performance of current high tensile strength cable technology. It's then pretty straightforward maths (numerical integration) and physics from then on. As in, you need this release speed (∆v) for that orbit, this cable length (ω²r) for that max acceleration, this much mass (M⋅a) for each release speed that the cable and payload could tolerate, and so on. – TildalWave Apr 28 '15 at 7:33

The paper does not describe how the calculations for the tether are done, but I can make a guess.

We take a small piece of the tether with mass $\delta m$ and length $\delta r$, at distance $r$ from the center. The tether is spinning at an angular rate $\omega$, and has an ultimate tensile strength $\sigma_\text{max}$ and density $\rho$. The area $A$ is a function of $r$.

We can write a relationship between the tension $T(r)$ on both sides of the piece:

$$T(r) - T(r+\delta r) = \delta m\ a_c = \delta m\ r \omega^2$$

We can substitute $\delta m=\rho A\delta r$, and recast this as a differential equation:

$$-\frac{\partial T}{\partial r}\delta r = \delta r A(r) \rho r\omega^2 \\ \frac{\partial T}{\partial r} = -\rho\omega^2rA$$

Next we can substitute $T=\sigma A$:

$$T = \sigma A \\ \frac{\partial T}{\partial r} = \frac{\partial \sigma}{\partial r}A + \sigma\frac{\partial A}{\partial r} = -\rho\omega^2rA$$

There are two types of solution we have to consider:

## Cable with Constant Cross-section

When the cable has a constant area, the differential equation becomes:

$$\frac{\partial \sigma}{\partial r} = -\rho\omega^2r \\ \sigma = -\rho\frac{\left(r\omega\right)^2}2 + C$$

The tension on the end of the cable, where a payload of mass $M$ is supported, is:

$$T(R) = M\omega^2 R \\ \sigma(r)A = M\omega^2 R$$

Setting $\sigma = M\omega^2 R/A$ at $r=R$ and $\sigma=\sigma_\text{max}$ at $r=0$, we get:

$$\sigma_\text{max} - \frac{(R\omega)^2}{2\rho} = \frac{M \omega^2 R}{A}$$

Substituting in the edge velocity $V=\omega R$ and solving for $A$ we get a required cable area of:

$$A = \frac{MV^2}{R}\left(\sigma_\text{max} - \rho\frac{V^2}{2}\right)^{-1}$$

We can rewrite this in terms of the critical velocity $v_\text{crit}=\sqrt{2\sigma_\text{max}/\rho}$:

$$A = \frac{MV^2}{R\sigma_\text{max}\left(1-(V/v_\text{crit})^2\right)}$$

Now for some examples. I'll use the three speeds given in the paper:

• Lunar Orbit: $1.68~\text{km}/\text{s}$
• Lunar Escape Trajectory: $2.40~\text{km}/\text{s}$
• Mars Injection Orbit: $3.84~\text{km}/\text{s}$

And I'll assume a Spectra cable:

• $\sigma_\text{max}=3.6~\text{GPa}$
• $\rho = 0.97~\text{g}/\text{cm}^3$
• $v_\text{crit}=2.7~\text{km}/\text{s}$

The dashed lines represent the required area with a margin of $25\%$. Note that the fastest speed doesn't even appear on the plot: since it's faster than $v_\text{crit}$ the cable would snap under its own weight.

## Tapered Cable (Varying Cross-section)

When the cable is operating at maximum performance, $\sigma=\sigma_\text{max}$ along its whole length:

$$\frac{\partial A}{\partial r} = -\frac{\rho\omega^2}{\sigma_\text{max}}rA \\ A(r) = A(0) \exp\left\{-\frac{\rho \omega^2}{2\sigma_\text{max}}r^2\right\} \\ A(r) = A(0) \exp\left\{-\left(\frac V{v_\text{crit}}\right)^2\right\}$$

Thus, to support a mass $M$ at the end of the cable ($r=R$) we have:

$$\sigma_\text{max} A(r) = M\omega^2 R \\ \sigma_\text{max} A(0)\exp\left\{-\left(\frac V{v_\text{crit}}\right)^2\right\} = M\frac{V^2}{R} \\ A(0) = \frac{MV^2}{R\sigma_\text{max}}\exp\left\{\left(\frac V{v_\text{crit}}\right)^2\right\}$$

Our plot now looks something like:

We can find the total mass of the cable by integrating $\delta m$:

$$\int \delta m = \int \rho A\ \delta r\\ = M\sqrt{\pi}\left(\frac{V}{v_\text{crit}}\right)e^{\left(\frac{V}{v_\text{crit}}\right)^2}\text{erf}\left(\frac{V}{v_\text{crit}}\right)$$

The ratio between cable mass and payload mass turns out to be purely a function of the ratio between tip velocity and the critical velocity. It looks like this:

The tether works on the Moon, but we can see that on Mars or Earth where the escape speeds are two to five times higher, the tether system quickly becomes impractical with current materials. Even carbon nanotubes, with $v_\text{crit}=9.6~\text{km}/\text{s}$, would have a cable-to-payload mass ratio of over 7 launching from Earth.

• Your calculus skills surpass mine, but so far as I can see the math is sound. Hoping someone like Mark Adler will check it. Some very useful concepts, first time I'd seen Vcrit for tethers. – HopDavid Apr 28 '15 at 15:19
• I have to admit your answer is unpleasant to me -- Some of my favorite day dreams are Clarke towers from asteroids. For these shallow gravity gravity wells, centrifugal force quickly becomes the dominant force as radius grows and thus are close to your sling model. I will have to revisit some of my sci fi scenarios set on Ceres and Vesta. – HopDavid Apr 28 '15 at 15:22
• @Hop Clarke tower == space elevator? Sorry, I'm not hip with the lingo here... – 2012rcampion Apr 28 '15 at 16:11
• I do have an odd manner of speech that's sometimes opaque. Yes, Clarke tower == space elevator. hopsblog-hop.blogspot.com/2012/09/… In my spreadsheets lowering the tether foot will raise the tether top. (A spreadsheet based on equations in P. K. Aravind's paper). I think this may be a way to look at taper ratios for very tall elevators from Vesta or Ceres. – HopDavid Apr 28 '15 at 16:28
• A relatively high launch cable mass works to your advantage in one way: When you release your load the system immediately goes out of balance and shifts its rotation to the new center of mass. This can be managed by (for example) supporting your central hub on a sliding or articulating base, or by allowing the cable to slide through the base a certain distance. The higher your cable mass fraction the smaller the displacement in CofM and therefore the easier/cheaper your post-launch displacement issue. – Kengineer Oct 19 '17 at 18:03