What would be the distances between compressors part of a gas pipeline from Earth to the Moon?

Question

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 the surface of the Moon through the L1 point to a compressor station at its end, hanging just above Earth's atmosphere.

To be able to transport gas through the pipeline towards the Moon, it would have to be compressed at certain intervals because of the gravity of the Earth.

Supposing that the compressors, being part of the pipeline, would each compress the gas from 1 to 10 bars, what would be the distances of those compressors between each other, accounting for both the gravities of the Earth and the Moon ?

To make the calculation somewhat easier, we can suppose that between two successive compressors the gravities of both the Earth and the Moon don't change with distance and that the pipeline is frictionless for the gas.

Answer 1

We will consider the case of real transport with a given flow rate $J=\rho u$. We assume that all compressors are identical, capable of maintaining a given air flow. Specifications: the flow velocity is limited by the condition $1\le u\le 10$ m/s. It is necessary to determine how many compressors are needed and how to optimally position them when transporting air from the surface of the earth to the surface of the moon. For the problem to have an unambiguous solution, it is necessary to set the equation of state, for example, the adiabatic flow between compressor stations $p=p_0(\rho /\rho _0)^\kappa$. Define the parameters for the Earth-Moon system: $G = 6.67408*10^{-11}; R_E = 6.3710*10^6m; R_M = 1.7381 10^6m;M_E = 5.9736 10^{24} kg; M_M = 7.342 10^{22} kg; L=356400000m$.

The acceleration of gravity at the Earth-Moon distance is

$$g=-\frac {GM_E}{x^2}+\frac {GM_M}{(L-x)^2}$$ Here $x$ is counted from the center of the Earth to the Moon. Define the gas parameters:$\rho _0=1.29 kg/m^3, P_0=10^5 Pa,\kappa =1.4,J=1.29 kg/sm^2$. The gas motion equation has the form $$u\frac {\partial u}{\partial x}+\frac {1}{\rho}\frac {\partial p}{\partial x}=-ku+g$$ Here $k$ is the coefficient of viscous friction. Put $k=1/100$, then the numerical solution for the flow velocity is shown in Fig. 1 on a large (left) and small scale (right) in the radii of the Earth

In Fig. 1, each peak corresponds to the position of the compressor. The first compressor must be installed at a height of $h_1=16884.25m$ above the surface of the earth, the second at a height $h_2=33858.7m$. A total of 370 compressors should be installed.The last compressor is located at a distance of about $7R_E$ from the center of the moon. Figure 2 shows the position of all compressors relative to the surface of the earth. At a distance of $R_E$ from the surface, 190 compressors should be installed. The rest are unevenly distributed, the latter is at a distance of 272835 km.

If we take into account the rotation of the Earth-Moon system with a period of $T_M = 27.321661 d$, then the number of stations decreases to 368. If we increase the pressure to 10 bar, then the number of stations decreases to 279.