Imagine that humanity builds mining facilities on the moon. We know that the moon has Helium-3 which could be used as fuel for fusion reactors. Also, there are a lot of other resources which could be used for building ships and buildings instead of transporting them from Earth.

Thousands of tons of metals will be mined from the moon in order to sustain a giant fleet of military, civilian and scientific ships for further colonization of our Solar System.

Because all of these actions, the moon will start to lose its own mass. Would this have some kind of impact on its orbit? If it is, how severe would the changes be?

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    $\begingroup$ Thousands of tons... that's like "does the gravitational pull on a spoon of sugar change if I remove one grain?". Try something like 10% of the moons mass. $\endgroup$
    – Erik
    Commented Mar 18, 2019 at 8:29
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    $\begingroup$ I have to agree with @Erik . Thousands of tons are not even noticable in comparison to the mass of the moon, which is about 7,349*10^19 tons. The factor difference is about 10^16. Sadly, I cannot calculate everything, hence no answer and just a comment. $\endgroup$
    – DarthDonut
    Commented Mar 18, 2019 at 8:34
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    $\begingroup$ @Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul. $\endgroup$
    – DarthDonut
    Commented Mar 18, 2019 at 8:51
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    $\begingroup$ Bringing stuff from the Moon to the Earth may change the Moon's orbit, depending on how it's done. No, the changes will not be measurable unless we somehow develop the ability to move vastly larger quantities of stuff. To give you an example, the total worldwide extraction of iron ore is about 2.4 million tonnes per year; this is about one thirtieth of one thousandth of one billionth part of the mass of the Moon. Shifting our entire iron ore extraction from the Earth to the Moon, and keeping at it for one billion years, would extract one thirtieth of one thousandth of the mass of the Moon. $\endgroup$
    – AlexP
    Commented Mar 18, 2019 at 9:16
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    $\begingroup$ @AlexP And you'd be transporting much of it to Earth so the gravity between the two wouldn't really change. I am guessing the biggest effect would actually come from launching and landing the stuff causing "recoil". Although, now that I think of it "big" might not be the correct root word to use. $\endgroup$ Commented Mar 18, 2019 at 9:20

4 Answers 4


The mass of the Moon is 7.342×1022 kg. One ton is 103 kg. How much is thousands of tons? Let's say you have thousands of thousands of tons. That's one million tons, or 109 kg. This is still ten thousands times a billion less than the mass of the moon.

enter image description here

(source: Diego Delso)

Just as a comparison, one of the largest mines that ever operated on Earth, Chuquicamata in Chile, produced much more than "thousands of tons":

...it remains the mine with by far the largest total production of approximately 29 million tonnes of copper to the end of 2007...

And even though it's a big hole in the ground, it is completely negligible relative to the mass of the Earth (or the Moon for that matter).

To answer your question:

Does mining huge amounts of resources on Moon will change its orbit?

The answer is no.

EDIT - the below is wrong because I don't know physics as much as I thought

Let's assume you mined 100 Chuquicamatas on the Moon and removed this mass to build your colonisation fleet. Let's ignore effects of momentum and potential energies etc. The orbit velocity is defined by $v=\sqrt{GM/r}$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = \sqrt{M_0 / M_1}$.

Let's plug in some numbers:

$v' = \sqrt{\frac{7.342×10^{22}}{7.342×10^{22} - 100\times26×10^6}} = 1.0000000000000178$

The Moon's orbit is going to be 1.0000000000000178 times faster.

EDIT - so let's change it a bit.

Since $M$ in the equation $v=\sqrt{GM/r}$ is the mass of Earth, the mass of the Moon means nothing. Then the velocity to radius ratio is fixed regardless. It will not change a thing!

However, we can still calculate the mass change percentage. It will be:

$\frac{7.342×10^{22} - 100\times26×10^6}{7.342×10^{22}} = 0.99999999999996458$

The Moon's new mass will be 99.999999999996458% of the Moon's old mass. Completely meaningless.

Just as an aside, it is often talked about mining asteroids, moons, and all kinds of other extraterrestrial bodies. For example, as you said, the Moon has plenty of helium-3. Asteroids have plenty of precious metals like platinum or iridium. But this is pointless, because you know what place has even more helium, platinum, and iridium? Earth. Earth has much more. And it's easier to mine because you don't need to build spaceships and facilities in hostile environments to do it.

EDIT 2 - space mining

Some comments mentioned that getting things out of the Moon is easier than it is on Earth, and that you can dump things on the Moon without environmental impact. It doesn't work like this.

In films and video games you "mine a resource" (e.g. iridium) from a planetary body. In real life, you build the mining facilities, you build the refining and smelting facilities, you need people to do it, even if you have robots you need people to fix the robots, you need to feed the people, entertain the people, you need a constant supply of consumables to refine the stuff. And this is only the "resource". You don't build spaceships out of helium. You build them out of steel/aluminium/carbon-composites. So you need to mine that as well, and you need to smelt that as well. You also need to build everything on the Moon because otherwise you need to transport all your resources to another place, so you need factories. To do all of that, you are going to need quite a lot of population. And then the environmental factor becomes important. Your mining will generate huge amounts of dust (combination of dryness and low gravity). This just doesn't work.

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    $\begingroup$ Regarding your last paragraph; Mining the materials on earth, under the assumption that the final destination is not on earth, means that you trigger the tyranny of the rocket equation. Furthermore, most of the resources are on not earth. You can collect a billion times more solar power than on earth, for example. $\endgroup$
    – sp2danny
    Commented Mar 18, 2019 at 13:09
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    $\begingroup$ Your equation is wrong. Or, rather, you're applying it incorrectly. M is the mass of the primary (Earth), not the mass of the satellite (the moon). If you put a one-ton mass and a 10,000-ton mass into orbit beside the (419-ton) ISS, they'll all orbit at the same velocity. $\endgroup$ Commented Mar 18, 2019 at 13:54
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    $\begingroup$ @DaveSherohman this is what happens when you give a physics problem to a geologist, right? Then it’s even better. The orbit will not change at all. $\endgroup$
    – Gimelist
    Commented Mar 18, 2019 at 17:18
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    $\begingroup$ @Yakk the "Humans fixing robots" thing seems to fall under what I like to call the "New jobs fallacy" - The logical assumption that humans must work and have a direct and significant impact on a system in order for said system to operate. A weird human trait where we struggle to comprehend advances in automation potential and its impact on human employment. $\endgroup$ Commented Mar 18, 2019 at 20:14
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    $\begingroup$ @chrylis a million ton is 10^9 kg. $\endgroup$
    – Gimelist
    Commented Mar 19, 2019 at 4:54

In order to change the Moon's orbit, you'd have to change its velocity.

You could mine away half its mass without significantly changing its velocity, as long as all the mass was lifted off by something like rockets. If you use mass drivers, and they all throw the same direction, you'd start to see detectable changes in the Moon's orbit by the time you'd thrown a couple million tons (a tiny fraction of one percent of the Moon's mass, but thrown at Lunar escape velocity).

If you use mass drivers that throw in different directions (perhaps you're sending the product all over the Solar System, rather than all to Earth or LEO), once again, the average impulse could be adjusted to zero, keeping the Moon (the part you haven't thrown into space, at least) right where you found it.

  • $\begingroup$ Now that's nice observation, it has a hard science tag though so have you any links or math for us? I probably shouldn't (because, tags) but [+] anyway. $\endgroup$
    – Pelinore
    Commented Mar 18, 2019 at 13:23
  • $\begingroup$ I'll try to come back later, when I have time to dig them up, and link the rocket equation, Lunar escape velocity, etc. $\endgroup$
    – Zeiss Ikon
    Commented Mar 18, 2019 at 13:39
  • $\begingroup$ You are neglecting the fact that as mass is removed from the moon, the orbital radius increases because the barycenter moves toward the center of the earth. A massless moon has a ~1% higher orbital radius and a ~1% slower orbit. It may maintain the same speed, but will have a longer orbital path. $\endgroup$ Commented Mar 18, 2019 at 16:14
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    $\begingroup$ @NuclearWang While that's correct, unless you're going to remove enough mass to see the mines with the naked eye from Earth, and do it pretty asymmetrically, only physicists will be able to tell the orbit has changed (by bouncing lasers off those reflectors Apollo left behind). Don't forget, the Moon is moving away from Earth by something like an inch a year, just from tidal forces. $\endgroup$
    – Zeiss Ikon
    Commented Mar 18, 2019 at 16:41
  • $\begingroup$ @ZeissIkon wouldn't those tidal forces be important to factor in if doing a complete overview of end to end effects as larger portions of mass are removed? ie: What is the long term impact between the earth-moon system's tidal forces as with ship greater percentages of the moon out of the system, or spread it around a larger orbit? $\endgroup$ Commented Mar 18, 2019 at 20:18

To a first approximation, simply removing mass from an orbiting body doesn't change its orbit. The orbital radius is given by the velocity and the gravitational pull of the primary (in this case, Earth). When the mass of the satellite is small compared to the mass of the primary, the satellite mass cancels out in the equation and is not relevant.

To see a concrete example, what happens when an astronaut makes a spacewalk from the ISS? Nothing - the astronaut and the ISS continue to follow the same orbital path, even though the ISS is hundreds of times more massive than the astronaut.

As others have touched upon, the process of removing vast amounts of mass might have an effect, depending on how it's done:

  • If you use chemical rockets, for instance, the reaction to the rocket thrust at take-off would have some effect eventually.
  • An electric rail gun launcher (as imagined by Arthur C. Clarke) would transmit the acceleration force through its structure to the Moon and so deliver a torque.
  • a space elevator would have no direct effect.

But these are outwith the scope of the question.

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    $\begingroup$ The mass of the satellite is irrelevant only when it's a negligible fraction of the planet's mass, which is not true for the moon. The earth-ISS and earth-astronaut barycenters are in practically the same spot, the earth-moon and earth-"hollow moon" barycenters are thousands of miles apart. $\endgroup$ Commented Mar 18, 2019 at 16:11
  • $\begingroup$ @NuclearWang Fair point - the mass of the Moon is about 1.2% that of Earth, so yes, if you replaced the moon with a balloon that looks like the moon, but weighs 1kg, the semi-major axis of the system wouldn't change but the barycentre would now be at the centre of the earth, so the balloon Moon would come down about 1.2% closer to the surface of the Earth. $\endgroup$ Commented Mar 19, 2019 at 7:22


Mining on reasonable scales, in the sense you are used to thinking about, won't have any impact on planetary motion.


If we assume a civilization bewteen K1 and K2, you start having the energy budget to do things like dismantle moons.

The asteroid belt weighs 10^21 kg or so. The non-Titan moons of Saturn, and smaller Jupiter moons, have a few asteroid belt's worth of mass. After that comes Luna.

With a fully automated space mining economy (where robots build robots and mine and refine and recurse and space ships), dismantling low-gravity structures is far easier than ones where you have to go down a gravity well.

So a space faring civilization might start with asteroid tugs pulling asteroids to a central processing asteroid, build more solar panels for energy and smelt more metal for raw materials, and recurse.

As this can proceed exponentially, after a modest period of time you'll be running out of asteroids. The next targets would be building beanstalks on Saturns' (non-Titan) moons, and the other "small" moons of the solar system, and dismantling them for more raw materials. The material from this would then permit building beanstalks on the larger moons.

Before you reach the terrestrial planets you'll get through dismantling Luna, the Galilean moons, Titan, and the sub-planets like Pluto and other post-Neptune bodies.

After that, Mercury, Mars and Venus.

At this point you'll have a choice; if Earth is still important enough, you'll have to skip up to using the Gas Giants for raw material.

This kind of progression is plausible in hard science fiction, and it is a path to get from a K1 level civilization towards a K2 civilization (with Dyson Sphere's or Swarms to harvest the Sun's energy output efficiently).

The dismantling of the moons could alter their orbits; but the energy budgets involved in actually dismantling a planet is so large that the problem of "orbit is being altered" is a trivial one; the orbits alter if they care, and they move in the direction they want them to move.

A K2 civilization has 10^26 Watts of power. The Moon has a Binding Energy of 10^29 J. So a K2 civilization could dismantle the moon in 1000s of seconds; however, a K1 civilization would require 10^13 seconds, or 7 million years.

Hence this plan; the matter from smaller celestial bodies is used to harvest more and more of the sun's energy. At K1.3 taking apart the moon takes 10 thousand years. At K1.5 dismantling the moon takes less than a century. At K1.7 it is a year-scale project. At K1.9 it takes days. At K2.0 it takes hours.

The efficiency of turning matter into solar energy capture and the efficiency of mining the material (which includes lifting it from the gravity well) determine how fast you exponentially climb the Kardashev scale.


Climing the Kardashev by dismantling stellar bodies is "reasonable" path forward. And it could involve mining the moon for raw materials to the point where the moon isn't there anymore.

But barring that level of civilization, no, mining isn't going to mess with orbital mechanics.


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