5
$\begingroup$

A sling launcher (discussed in this write-up by Landis) is a tower with a motor that spins a hub with two or more cables attached. Payload(s) are attached to the end of one or more cables and counterweights to others. Once the hub is spinning, the cables are reeled out a number of kilometers until the angular momentum at their tips is so high the payload reaches orbit after it is released.

schematic of sling launcher

The paper linked to discusses its use on the Moon. I can see why drag makes the system unusable on Earth, but the atmosphere of Mars is only 1% that of Earth. There must be a density below which the atmospheric drag is manageable. Actually, considering the high efficiency of sling launchers as a launch system, and their relative simplicity, perhaps it could be worth dealing with quite a bit of drag, considering the savings on propellant.

Is the atmosphere of Mars too thick to make sling launchers practical?

Edit: Sensibly, several people have mentioned that the cable for this launcher is impossible using current materials, including 2012rcampion's answer, which lays out the calculations that show that. The paper referenced based most calculations on carbon nanotube cables. This technology seems feasible if still distant, so let us assume it would be used in this case. From the paper:

The ultimate tensile strength of fullerene nanotubes is predicted by theory to be well over 100 GPa [9], with measured values on individual tubes approaching this value. Allowing an engineering factor of 5 (including the added strands for cross-connections, discussed below), the material should allow a working stress of 20 GPa. A mass of one thousand kg at 11.5 gravities (110 m/sec2 ) results in a force on the cable of 112,000 N, as shown in table 1, so the required cross-section of the cable is 0.00389 cm2 (0.389 square millimeters) per ton of end-mass. (More likely, this will be in the form of a number of separate cables which sum to a total cross-sectional area .389 mm2 ). The density of fullerene nanotubes is 1.3 gr/cm3 . The mass of the 50 km cable itself is then about 25 kg, and it is clear that neglecting the mass of the cable itself in the calculation was justifiable. The counterweight cable carries half the mass, but at half the acceleration, and half the length, so the counterweight cable has a total mass of about 13kg. An ultimate strength of 20 GPa may be optimistic for a realistic material. If the working stress is reduced to 10 GPa, the cable cross-section is doubled, and the mass increases to 51 kg and 15 kg for the launch cable and the counterweight respectively

Improve this question
$\endgroup$
6
  • $\begingroup$ I think the major issue would be the launch cable. The analysis shows that with existing materials, a cable mass of 2,500 kg would be sufficient to launch a 1,000 kg object; only with fullerene nanotubes does this turn into a more practical 50 kg. Remember, Mars escape velocity is about double that of the moon, requiring a cable four times as strong. I think that makes existing materials entirely infeasible. $\endgroup$
    – raptortech97
    Commented Apr 25, 2015 at 16:11
  • 1
    $\begingroup$ @raptortech97 this is true, and I think it is necessary to assume carbon nanotube cables in order to usefully answer the question. I'm about to put in the relevant quote from the paper. But that doesn't seem out of order for a project that couldn't be realized for several decades anyhow. Rice University is spinning nanotube thread now, after all. $\endgroup$
    – kim holder
    Commented Apr 25, 2015 at 22:46
  • $\begingroup$ @briligg Rice only reports tensile strengths of 0.2 GPa, giving them a specific strength of less than 200 kN.m/kg. $\endgroup$
    – 2012rcampion
    Commented Apr 25, 2015 at 22:58
  • $\begingroup$ @2012rcampion well sure, but it's still early days. $\endgroup$
    – kim holder
    Commented Apr 25, 2015 at 23:00
  • $\begingroup$ @briligg 2012rcampion's answer identifies that cable strength isn't the only problem. Peak black body temperature at 939 MW/m² (my calculations for 0 altitude standard datum) is 11,345 K, or about the same flux as for laser material hardening. Even on Olympus Mons, black body temperature would be 7,200 K. Kevlar would melt and conductive CNT would generate a whole lot of triboelectric charge. Point is, that even if we could make such a cable, we couldn't attach anything of much use at its end. Well, maybe diamonds. I'd be fine with those. :) $\endgroup$
    – TildalWave
    Commented Apr 25, 2015 at 23:47

1 Answer 1

Reset to default
9
$\begingroup$

We can compute the power required to maintain speed as:

$$ P=\frac{C_D}2\rho A v^3 $$

Assuming the hypersonic drag coefficient is around $1$ and that the atmospheric density is $1\%$ of Earth's, we get:

$$ \frac P A=\frac 1 2\times 0.01225~\text{kg}/\text{m}^3\times\left(5.0~\text{km}/\text{s}\right)^3 = 780~\text{MW}/\text{m}^2 $$

Even on Mars the atmosphere is far too thick to achieve escape velocity inside the atmosphere.

The linked paper uses a power estimate of $100~\text{kW}$, meaning the largest payload you could launch would have a fairing diameter of around $1~\text{cm}$. That's not even accounting for the drag from the cable itself!

The big attraction of this method is that it uses very little power, unlike electromagnetic launch systems, but even in a very thin atmosphere this benefit disappears.

Improve this answer
$\endgroup$
12
  • $\begingroup$ Actually, you were too forgiving for the mass of Martian atmosphere, because you seem to have assumed same specific gas constant than for Earth's atmosphere. At standard datum, I get 0.01503 kg/m³ or 939 MW/m² (replace those two 0's in the link with any altitude in meters). It gets a lot better at the top of Olympus Mons (21,229 m) tho, about 152 MW/m². I used this atmospheric model as input. $\endgroup$
    – TildalWave
    Commented Apr 25, 2015 at 22:42
  • $\begingroup$ Randall Munroe's model is a disk. I believe this model is wrong for a tether. A tether would be tapered, the thickest portion at the hub and thinnest at the fast moving end. Exactly the opposite of a disk. $\endgroup$
    – HopDavid
    Commented Apr 27, 2015 at 19:45
  • $\begingroup$ @HopDavid After thinking about it for a little, I realized you're right. He uses the equation to calculate the strength of a tether, but the equation isn't valid anymore. For a cable of constant cross-section, the maximum speed is raised by a factor of $\sqrt{2}$. A tapered cable can achieve any speed, but the required thickness increases too quickly with any real material (just like a space elevator cable). I've added my derivations to the question. $\endgroup$
    – 2012rcampion
    Commented Apr 27, 2015 at 21:34
  • $\begingroup$ This is an excellent treatment of the cable issue, one I will have to pore over in detail. But I sort of feel it could get overlooked due to the fact the question didn´t focus on the cable at all. I have an itch to ask another question that specifically focuses on sling launcher cables so this gets the attention it deserves. What do you think? You´d need to move your answer. It might bear discussing in chat. $\endgroup$
    – kim holder
    Commented Apr 27, 2015 at 21:39
  • $\begingroup$ @briligg I'd be fine with moving the cable calculations to a new question while leaving the drag calculations here. $\endgroup$
    – 2012rcampion
    Commented Apr 27, 2015 at 21:43

Your Answer

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

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