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I'm a Sci-Fi author, trying to keep the physics real in my books. I am in the process of moving a space station to the Earth-Moon $L_1$. I've read that a vessel can't just sit at that point, but has to orbit it. Can anyone tell me the characteristics of that 'orbit' relative to an Earth-Moon frame of reference? What period, velocity? The usual equations don't seem to apply. Coming 'up' from LEO, is there a particular orbit which would be easier to get into, or more favourable for customers, who will be using the station as a waypoint on trips to the Moon?

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    $\begingroup$ Have you searched our site for other questions tagged as lagrange-points? Some other keywords you're after are halo orbit and Lissajous orbit, but we do have actual examples on the site too. One that comes to mind is ARTEMIS, more links and explanations e.g. here (includes data on EML1). $\endgroup$ – TildalWave Jan 11 '15 at 16:56
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    $\begingroup$ Kudos for trying to keep the science sound. You may be interested in the Worldbuilding and Physics Stack Exchange sites as well. $\endgroup$ – a CVn Jan 12 '15 at 10:30
  • $\begingroup$ When you want a trip to the moon, a stop at L1 would be an unnecessary detour. When you want to do trips to the surface in dedicated landing crafts, a base in a low moon orbit would be the ideal choice. That's how the Apollo missions were doing it. $\endgroup$ – Philipp Jan 12 '15 at 16:19
  • $\begingroup$ The only reasons I could think of which would make a base at Earth-Moon-L1 useful would be a) observing the earth-facing side of the moon from outside Earths atmosphere, b) observing earth from the perspective of the moon but without being on the moon surface... or c) creating a permanent artificial moon eclipse, but that would require a satellite so gigantic in diameter that its mass would no longer be negligible. $\endgroup$ – Philipp Jan 12 '15 at 16:24
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    $\begingroup$ @RichardPenn I'd advise against moving it to LLO (Low Lunar Orbit, I guess that's what you meant with LMO?) because they're not really long term stable due to lunar mascons (mass concentrations). Perhaps considering some form of a cycler (there's some good cyclers with 5 or 6 visits per lunar orbit explained online) would be more suitable to your needs? Also, FWIW, I'd keep your question as is (perhaps edit in new info you now add) and rather ask new ones if needed, because we could use a few good answers explaining why is it so darn complicated to stay stationary w.r.t. libration points. $\endgroup$ – TildalWave Jan 12 '15 at 18:26
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All first three Lagrange points are unstable, namely the effective potential near these points are saddle points. The L4 and L5 points are stable when the mass ratios are big enough. The unstable directions of those saddle points are inline with the vector connecting the two celestial bodies, however the forces in the plane normal to this direction will be directed towards the respective Lagrange point. So if you move in this plane away from such a point and give yourself a relative velocity perpendicular to the attractive force, you could orbit that point, but you would still have to correct the unstable direction, because any perturbation will build op exponentially over time if you don't.

If you want to know the periods of these orbits as a function of the radius (limiting to circular orbits) you have to look at the forces involved, which should act as the centripetal force. I have to make the assumption that the velocity, with which the L1 point is being orbited, is small compared to the rotational velocity of the rotating reference frame, such that the fictitious centripetal force would only be a function of the position. The variables that I will be using are illustrated in the following figure.

enter image description here

Where $m$ is the Moon, $M$ is the Earth, the $\times$ illustrates the center of mass of the two and $L_1^*$ is a translation of distance $r$ of the $L_1$ point in the perpendicular plane.

The resulting acceleration, towards $L_1$, can be calculated as follows,

$$ a = \frac{\mu_m r}{\left(R_m^2+r^2\right)^{3/2}} + \frac{\mu_M r}{\left((R_C+R_M)^2+r^2\right)^{3/2}} - \omega^2r, $$

where $\mu_i$ is the gravitational parameter of body $i$ and $\omega$ is equal to the angular velocity with which the two bodies rotate around each other. If $r$ is small compared to the other distances, then this equation can be approximated with,

$$ a \approx r \left(\frac{\mu_m}{R_m^3} + \frac{\mu_M}{(R_C+R_M)^3} - \omega^2\right). $$

This should be equal to the centripetal acceleration, thus the required velocity, $v$, and corresponding period, $T$, can be expressed as,

$$ v = r \sqrt{\frac{\mu_m}{R_m^3} + \frac{\mu_M}{(R_C+R_M)^3} - \omega^2}, $$

$$ T = \frac{2\pi}{\sqrt{\frac{\mu_m}{R_m^3} + \frac{\mu_M}{(R_C+R_M)^3} - \omega^2}}. $$

So the period would be independent of $r$ (if sufficiently small) and for the Earth-Moon L1 point would be roughly equal to 13.4 days.

I am not an expert in this field and did a few simplifications, so I also looked for a reference and according to this paper for a small $r$ the period would be closer to 11.94 days, so I might have used wrong values for the involved parameters and/or simplified the problem to much, but hopefully this does give some insight into the physics involved, or the low eccentricity of the Moon also has an influence. I also tried to calculate it for the L2 point and for both points in the Sun-Earth system, but kept getting about 1.2 times too high periods compared to my reference.

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    $\begingroup$ That's a great response, Fibonatic, thanks for all the effort. I'll stick with the EML1 halo orbit, keeps my station in continuous comms with Earth and whole front of the Moon, continuous sunshine, gives low-cost access to both GEO and interplanetary transfers. The station will have to de-spin every 12 (or 13?) days so they can apply a correction burn, but that seems reasonable and the delta-V's involved seem sustainable. That paper is awesome, what I can understand. Clearly a Spirograph lover! Need to remember that cargo vessels take much longer to fly, while crewed ones burn more fuel. $\endgroup$ – Richard Penn Jan 13 '15 at 12:24
  • $\begingroup$ Yep, your 1.2 seems right! In this question I show an historic calculation of halo orbits. In the reduced units the synodic period is 2π, and the half_periods of the halo orbits would be π/2 or about 1.57 if your calculations were used there. The numerical values for periods also turn out to be about 20% smaller than this, and the difference increases for larger amplitude. $\endgroup$ – uhoh Apr 12 '18 at 12:35

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