Suppose we begin to colonize the moon. Assuming the building materials and biological material are all imported (not taken on-site), would it be possible for the moon to become massive enough to fall out of its orbital and come crashing into the Earth?

We should assume the materials are taken from Earth, meaning the Earth becomes lighter as the moon gets heavier.

How much mass does that represent? What scale is necessary for such an event? Would the moon need to be covered in tall buildings? Would every structure need to be built out of very dense materials?

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    $\begingroup$ Huh. This is actually very interesting. The gravitational attraction between the Earth and the moon can be increased by making their masses more equal (recall F is proportional to m*M), but you would also change the location of the Centre of Mass about which the objects rotate and thus change the centrifugal forces. I suspect that it all balances out, but I don't have time to do the barycentric maths right now. It is probably a proven result somewhere online though... $\endgroup$ – Mark_Anderson Dec 7 '18 at 16:39
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    $\begingroup$ Simply, no. Every mass transfer is going to be accelerated to lunar orbital speed so it won't slow the Moon down. If the Earth becomes significantly lighter, the Moon will orbit further away, or leave entirely. $\endgroup$ – David Thornley Dec 7 '18 at 16:43
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    $\begingroup$ @DavidThornley Perhaps? Do you have calculations? We need to check the maths as the net gravitational force will change with the mass transfer (which surprised me when I actually did some maths to check intuition). The mass transfer will, initially, increase the net attraction so we are at risk of the moon crashing into the earth if anything... $\endgroup$ – Mark_Anderson Dec 7 '18 at 16:51
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    $\begingroup$ @Mark_Anderson: No, because the Earth-Moon system will continue to revolve around its common center of gravity. So if for instance we moved enough of Earth's mass to the Moon to make their masses equal, then they'd revolve around a point midway between them. And as we kept moving mass to the Moon, the remnant Earth would now orbit it. (Note that while this mass transfer is pretty much impossible for the Earth-Moon system, something similar can happen with binary star systems.) $\endgroup$ – jamesqf Dec 7 '18 at 18:39
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    $\begingroup$ What if it wasn't mass-transfer? Would landing asteroids on the moon slow it down? $\endgroup$ – Mazura Dec 7 '18 at 22:36

As mentioned in comments, I don't have the full stability answer to hand (although see edit below). But I do have a practical answer.

The practical answer is that no feasible human effort could change the orbit of the moon via Earth-mass-transfer.

Here is how the net attraction between the earth and the moon would change as you moved Earth mass to the moon (Moon is approx 1.2% Earth mass), using Newtonian gravity

F= G(M*m)/r^2 enter image description here

Wow! Peak 12 times increase in gravitational force?! Okay, sure. I'm still not sure that the change in barycentric rotation won't perfectly compensate for that to leave us in a stable orbit but yeah, there's a big potential change in net attraction.

Let's zoom in on the left of the chart: enter image description here

See that red marker? Waaaaaay down there on the left? That represents moving the ENTIRETY OF THE OCEANS TO THE MOON. It changes the net gravitational attraction by 0.01%.

Therefore, even if mass transfer could destabilise orbits (and I am not sure that is true), the amount of mass transfer that could be achieved is incredibly insignificant to orbital dynamics.

EDIT: Having thought about the general stability of mass transfer from the moon to the Earth, I think the question isn't properly defined. To define stability we need to know both the change of mass and the change of velocity of the Earth and the Moon during the mass transfer. Keeping constant velocity sounds like the simplest option, but it is clearly unphysical: imagine we transfer mass such that the Earth and Moon are the same mass; they now have a barycentre [the point about which they rotate] exactly halfway between them --- meaning both must have equivalent tangental velocity around this centre. But we have also assumed that the Earth's and Moon's tangental velocities don't change. Therefore we have a contradiction. We can only solve this by making explicit assumptions about momentum transfer.

EDIT2: A practical consideration I forgot to mention previously. Gravity is a harsh mistress that makes moving mass very expensive. If you wanted to build extensively on the moon it would rapidly become preferable to mine materials from the asteroid belt (send a small set of mining materials up-gravity, send lots of mined mass down-gravity) rather than sending up a full set of mass from Earth to the Moon. See https://xkcd.com/681/ for a rough idea of the Gravity Well energy problem

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    $\begingroup$ Yeah, normally when sending a rocket from the earth to the moon, you don't factor in the the change in the earth or the moon's velocity or angular momentum because of the rocket. But if you were doing this quadrillions of times and taking mass with you, the cumulative effects would be significant. But, it could probably be controlled for. One could take off and land on alternate sides of the planet to counteract these effects probably. $\endgroup$ – Shufflepants Dec 7 '18 at 22:06
  • $\begingroup$ Well, there's the rub. How do you control them? You will want to pick one of the family of momentum-transfer solutions that lead to stable orbits of the Earth and Moon. Not sure what those are off the top of my head... $\endgroup$ – Mark_Anderson Dec 7 '18 at 22:09
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    $\begingroup$ Re: "Gravity is a harsh mistress" -- yes, but so is the moon. $\endgroup$ – Pete Becker Dec 8 '18 at 13:52
  • $\begingroup$ @Mark_Anderson I'm not sure what you mean, but I'm not sure you know what I mean. Taking off from one side of the planet pushes it in the opposite direction a tiny amount and same with landing on a planet (at least if you brake primarily via collision with the atmosphere). And you'll have control over when and where to take off and land from. So whoever is in control of the rockets will have a lot of control over the change in momentum and orbits of the two planets. $\endgroup$ – Shufflepants Dec 10 '18 at 7:30
  • $\begingroup$ @Shufflepants Yeah, that's exactly it. You can control the momentum imparted to both the moon and the Earth during the mass transfer. Depending on how that is done, you will either be pushing towards a stable or unstable orbit regime. (Sorry, I should have been clearer before; by "how do you control them" I meant 'how do you choose to do the orbital transfer to ensure stability', not 'how is it physically possible to cause change') $\endgroup$ – Mark_Anderson Dec 10 '18 at 13:57


  1. One has to increase one's momentum to the moon's level in order to actually get to the moon. If you're planning to land on the moon instead of crashing into it; you're basically performing a rendezvous with the moon.

    Consider normal rendezvous in LEO. The ISS doesn't fall out of the sky when its mass is doubled by a docking space shuttle, because the space shuttle's mass has the same momentum. It's only when these have different momentum and still connect with each other (collide, more like) that the apsides change.

  2. By adding mass to Moon, you're removing it from Earth. Therefore, not only do the orbital parameters stay the same, but the gravitational force felt between these two should stay constant.

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    $\begingroup$ Re: point 2. That's incorrect. Do the maths. G(M-x)(m+x)/r^2 for varying x. You change the net gravitational force by moving mass between bodies in a paired system. It's impossible not to, as the exchange is moving mass to a different GPE and you need to conserve energy. We don't care about this correction for the ISS because the shuttle is a negligible fraction of the Earth's mass, thus x~0. $\endgroup$ – Mark_Anderson Dec 8 '18 at 16:21

No. Just no.

1.6 kilometers per second are required to achieve low lunar orbit. From there, another 0.7 kilometers per second are required to get from low lunar orbit to touch the earth surface. Therefore, our work on the moon would require imparting 2.3 kilometers per second of delta v.

If you tried to deorbit the moon intentionally you would have to come up with 1.18 kg km / s of change. If we used the nuclear thermal engine for its ~800 second impulse, we need 1.2 * 10^23 kilograms of fuel to intentionally cause the moon to deorbit.

Since you're trying to colonize, you're soft-landing stuff on the moon. There is no impulse provided to the moon. On the other hand, the mass came from earth so earth gets a little bit lighter, so the moon ends up a little bit further away. Considering the size of the other number, we can assume this is negligible for any reasonable purpose including terraforming the entire lunar surface.

Surface area of the moon: 3.8 * 10^7 km

Amount of dirt to coat the moon 10 feet deep: 1.7 * 10^17 kg

Mass of earth: 5.9 * 10^24 kg.

Earth's escape velocity before mass transfer sqrt(2G * (5.9*10^24 kg) / 405400 km): 1394m/s

Earth's escape velocity after mass transfer sqrt(2G * (5.9*10^24-1.7 * 10 ^17 kg) / 405400 km): 1394m/s

No effective change.

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Based on this article it seems that the increase of the mass of the moon would disrupt the earths system of time. We would become locked in an entirely different length of days and years etc. not to mention the ocean tides would be different.

And if the mass grows ina region of the moon instead of uniform, the moon might either spin a little or a lot considering the center of its mass would change. As opposed to it not spinning at all right now. But I can’t find any evidence of the moon crashing into the earth, only changing its relationship.

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  • $\begingroup$ Of course any changes in the day length would be irrelevant to the question, since if you transferred more than the tiniest fraction - say a layer 0.1% of its radius thick, or 6 km - you would render the earth uninhabitable. $\endgroup$ – jamesqf Dec 8 '18 at 5:16

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