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I've read this question, but most of its answers don't seem to apply to my question directly other than possibly this one, which suggests that in order to change a planet's orbit, you would have to change its rotational energy.

There's also this question, but the answers are more cataclysmic than anything.

What I'm imagining is the entire current population of earth (roughly 7.5 billion) being moved to the moon, with new buildings and roads to house and transport every individual both above and as far below the surface as possible. The moving of every person to the moon would be a steady and smooth stream until everyone has been transferred (let's say 100M/week).

So in the event of mass-colonization of our moon, at what point, if any, would it start moving out of its usual orbit? Would it fall or raise? I know there's no drag in space, but would the added height/weight/energy from the billions of bustling people cause its rotation to slow down/speed up at all? Assume all warning signs are ignored and people keep piling on.

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DCOPTimDowd
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No.

Mass of all of humanity: 4 x 10^11 kg

Mass of the moon: 7 × 10^22 kg

It's like asking if you could be knocked over by a bacterium (roughly a 11 order of magnitude mass difference too)

Matt Bowyer
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    But what if all the bacteria jumped at the same time. – wyldstallyns Dec 15 '16 at 17:39
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    What about all the added structures? If every germ had to have a house and a McDonalds in walking (feeler-ing?) distance, wouldn't that make them more noticeable? – DCOPTimDowd Dec 15 '16 at 17:41
  • If all those people were moved up to the moon in some survivable fashion, that implies that their velocities were matched with the moon's (soft landing.) In that case, the means of transportation would be supplying the required energy and angular momentum, person by person. And similarly for any earthly goods they were allowed to bring along. – Catalyst Dec 15 '16 at 17:44
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    Why would you bring a house with you? Build it out the moon.

    But even so, still no difference - even if everyone brings 100t with them, that's only a threefold difference in orders of magnitude; we're still not up to the equivalent of being knocked over by a mosquito.

     – Matt Bowyer Dec 15 '16 at 17:48
  • So the added weight and dimensions of the housing structures would not add any harmful effect? – DCOPTimDowd Dec 15 '16 at 17:48
  • Okay. I see what you're saying. – DCOPTimDowd Dec 15 '16 at 17:49
  • OK, just one more analogy - it's the effect of a mosquito (10^-5kg) landing on an Arliegh Burke-class destroyer (10^6kg) – Matt Bowyer Dec 15 '16 at 17:53
  • Would it be a separate question to ask, then, what human population would have any effect on the moon? – DCOPTimDowd Dec 15 '16 at 18:00
  • Actually, the factor is 13 orders of magnitude, not 11, since you need to compare against the mass of the Earth. See my Answer where I worked it out. So Matt needs another example. – JDługosz Dec 15 '16 at 18:01
  • @SonOfSam no that would be a stupid question. Look at my answer and adjust p yourself to see. How much of E could be people? – JDługosz Dec 15 '16 at 18:03
  • SonOfSam - think of it this way: the moon is 3,500km in diameter, now imagine humans standing shoulder-to-shoulder covering the whole thing. They're still barely perceptible as a fraction of the whole thing in terms of size, and they're much less dense than the moon itself. – Matt Bowyer Dec 15 '16 at 18:08
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    @JDługosz - good point, it's a mosquito on a Nmitz-class aircraft carrier. – Matt Bowyer Dec 15 '16 at 18:09
  • I think that my original idea didn't take into account all the water weight earth has, and because of that assumed the moon was much smaller than it is (or that earth is much bigger) – DCOPTimDowd Dec 15 '16 at 18:15
  • Water weight? I think water is a minor component. It might be a good exercise for you to draw a cross section of Earth with the ocean depth drawn to correct scale. Things like that will help you develop your intuition for orders of magnitude. – JDługosz Dec 15 '16 at 21:40
  • A bacterium (and some relatives) knocked me out the other week. – LosManos Dec 15 '16 at 22:11
  • @JDługosz Not the "weight" of the water, but it's total surface area of the planet. You're right though. I should get a better idea of scale for my next question. – DCOPTimDowd Dec 15 '16 at 22:19
  • @wyldstallyns If all of your bacteria are cooperating against you, you may have more problems than just whether they can knock you over... – Justin Time - Reinstate Monica Dec 15 '16 at 22:59
  • @MattBowyer Your comment raises a question: What if the moon saw Coruscant-like levels of development, being covered in massive skyscrapers full of people? It'd fit a lot more than everyone just standing on the surface, shoulder-to-shoulder, and could actually make the question plausible in a sufficiently advanced setting. – Justin Time - Reinstate Monica Dec 15 '16 at 23:02
  • Still not really - for starters there's the very important point that virtually everything involved in moon colonisation would come from either the Moon itself or from Earth.

    But EVEN so - covering the Moon entirely in Empire State Buildings would weigh around 2 x 10^15 kg - still 9 orders of magnitude short, or it's now moved up to a mosquito vs a lorry.

    The moon is REALLY, REALLY big - and the Earth (which is responsible for the gravitational pull) is EVEN BIGGER.

     – Matt Bowyer Dec 15 '16 at 23:16
  • @wyldstlyns when dealing with 11 orders of magnitude, one bacterium or all the bacteria are pretty much the same thing. :-) – SRM Dec 16 '16 at 00:33
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No. But let’s work it out:

The force of gravity will change if the product of the masses changes. Shifting the people of total mass p from Earth to Moon gives you a new product (E−p)(M+p) = EM + p(E−M) − p²

Since E = 5.97237×1024 kg
M = 7.342×1022 kg
and p is about 7.5 × 1011

So you can see that (E+p) and (E−p) can’t even be handled by a regular calculator, since p is insignificant. But the expanded form shows that the product will increase by a part that’s 13 orders of magnitude smaller than the original product. The difference between old and new will be like 1 vs 1.0000000000001. It won’t make any difference.

JDługosz
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Two important points to consider.

First, orbital velocity is a function of the body being orbited, not the body doing the orbiting. That is, the mass of the moon makes no real difference to its orbit; all that matters is the mass of the Earth. This is an extension of the famous principle that the speed of a falling object will be the same regardless of its mass - a feather and a hammer dropped in a vacuum from the same height will hit the ground at the same time.

Second, moving all of humanity and all of our infrastructure to the moon would increase the moon's mass (by a tiny fraction) - but it was also decrease the Earth's mass (by a tiny fraction). The mass of the Earth-Moon system will not change, just the barycentre (centre of mass) of the overall system - and that by only the tiniest amount.

Werrf
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  • Not, actually true, if rockets are used. Useful specific impulse only occurs with very high exhaust velocity. So virtually all the fuel will reach escape velocity. And since fuel mass fractions are quite high (like, 90% or so for current technology), much more mass will escape the Earth/Moon system than will be transferred to the Moon. – WhatRoughBeast Dec 15 '16 at 18:52
  • Good point, I hadn't considered the loss of mass from fuel exhaust - however if we're talking about mass migration from Earth to the moon, it's probably not going to be done with current technology, so it's hard to say what kind of propulsion to consider. – Werrf Dec 15 '16 at 19:44
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    Actually it depends on both, but when there are orders of magnitude difference in the sizes of the bodies the larger dominates. The Earth/Moon is “only” 2 orders of magnitude— unusually large for a moon. So it will be slightly faster than an apple. – JDługosz Dec 15 '16 at 21:36