# What are the calculations for gas transport to the Moon with a pipeline?

This paper, acquired from this question about tethers on the Moon, describes LADDER, a mission to deploy a Lunar Space Elevator (LSE).

The space tether, made of Zylon fiber, would be 264,000 km long, erected from the lunar surface, passing the L1 lagrange point with a counterweight in deep cislunar space.

Now imagine a 356,000 km long Zylon pipeline, extending from Sinus Medii at the Moon, through the L1 point to a compressor station at its end, hanging just above Earth's atmosphere.

Because the Moon's distance to the Earth varies from a minimum of 356,400 km to a maximum of 406,700 km, the height of the compressor station above the Earth would vary between 400 km and 50,700 km during one Moon's orbit of 27.3 days.

Consequently the distance of the compressor station to the center of the Earth would vary by a factor of over 7, and the Earth's gravity for that station at maximum distance would be a factor 50 less !

So if a tank with liquid gas could be delivered at the compressor station every 27 days at minimum distance from Earth, and the gas could be released into the pipeline at maximum distance, much less energy would be needed to get the gas to the Moon.

Question: Would it be possible for one compressor station at the end of the pipeline to deliver gas all the way to the Moon using the much less gravity, and can it be calculated how much pressure is needed because Earth's gravity varies wlth distance ?

• Out of curiosity, what gas would you want to transport and why? – gerrit Oct 2 at 12:28
• @gerrit Of course H$_2$ and O$_2$, giving energy and water on the Moon, but also CO$_2$ to combat the global warming problem on Earth ? – Cornelisinspace Oct 2 at 12:38
• @Conelisinspace, Why do you want to transport hydrogen and oxygen to the moon, since tons and tons of water ice are in the polar region of the moon itself? Just do electrolysis with solar power! – M. Guru Vishnu Oct 4 at 14:52
• In the long run much more water is needed , it s to precious to use it for electrolysis. But O$_2$ and N$_2$ will also be needed , of course. – Cornelisinspace Oct 4 at 16:33

Pipelines two orders of magnitude shorter on Earth are usually divided into many compressor stations.

But let's take a look at how feasible it is. At the very least, a compressor station would have to provide a pressure greater than the pressure at the bottom of the pipeline.

Fortunately, the pressure calculation needed is unusually simple for a CR3BP ( Circular Restricted Three Body Problem ) system:

$$P=\rho\left(\mu_{earth}\left(\frac{1}{r_0}-\frac{1}{r_1}\right) - \mu_{moon}\left(\frac{1}{r_{moon} - r_1}-\frac{1}{r_{moon} -r_0}\right) + \frac{\omega^2r_0^2}{2} - \frac{\omega^2r_1^2}{2}\right)$$ ($$r_0$$ is the distance from compressor station to Earth and $$r_{moon}$$ the orbital radius of the Moon.)

While it can account for the altitude variations in the question, it doesn't quite capture the oscillation effects. I presume they are at least an order of magnitude smaller than the numerical result. (the $$r_0$$ and $$r_1$$ in the last two terms should technically be form the barycentre, but this is close enough)

This formula can also not be used across L1, but that's not an issue since the gas can just flow down from there. The important distance is base - L1

Filling in the numbers, this is a pressure of around 4 GPa when farthest from Earth. That's already higher than any compressor pressures I can find, and that's before even considering the friction in a 350,000 km long pipe.

You will need multiple compressor stations.

• It's the usual tether strength requirement formula, modified to yield a pressure instead. – Hohmannfan Oct 2 at 12:49
• It absolutely makes a very large difference. The 4 GPa number is for the maximum distance, for the minimum distance it's about 40 GPa – Hohmannfan Oct 2 at 13:03
• 4 GPa? I suspect many gases will be solids at that pressure. – Hobbes Oct 2 at 17:57
• You are using the wrong angular velocity. You're supposed to use the angular velocity of your tether, and since it's co-rotating with the Earth-Moon system, it has the same orbital period as the Moon. That's 0.0096 radians per hour. – Hohmannfan Oct 5 at 13:54
• The tether strength formula assumes constant density and strength. A gas column under gravity has a decreasing density higher up. All of Earth’s atmosphere only provides the equivalent of < 10 km pressure from a constant-density atmosphere. So you have to start with a desired pressure at the L1 turnover and integrate down. – Bob Jacobsen Oct 6 at 3:34

In this answer on Physics stackexchange it is stated that a total of 368 compressor stations along the pipe line should be installed in the case of real gas transport with a given flow velocity of < 10 m/sec.

The pressure drop between the compressors would be from 1 bar to about 0,4 bar.

• You shouldn't cross-post the same question to multiple stacks. – Organic Marble Oct 8 at 12:18
• @OrganicMarble Why not ? – Cornelisinspace Oct 8 at 12:55
• meta.stackexchange.com/questions/64068/… – Organic Marble Oct 8 at 12:56
• @OrganicMarble In the question here, i asked for calculations for only one compressor station, and because it appeared from the answer above this would not be feasible, i wanted to know what the calculations would be with multiple stations. Because another question here woulld be a duplicate, and because at Physiics .SE the chances for a good answer would be better, i asked there. – Cornelisinspace Oct 8 at 13:18