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

Question

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.

How was this analysis of the cable done?

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?

Answer 1

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.