Would mining huge amounts of resources on the Moon change its orbit?Could humans alter the moon's orbit significantly with current technology?What kind of event, if any, would knock the moon off its orbit, without destroying it?Could over-colonization throw our moon out of orbit?How would a nuclear blast on the moon affect its orbit?What kind of event, if any, would knock the moon off its orbit, without destroying it?Could a moon have its own satellites visible from the planet it orbits?How big can a moon be where you can physically jump out of its orbit, to its planet?Exoplanetary Review: Rock StormsHow would a nuclear blast on the moon affect its orbit?How plausible is this as a corporate motivation?Plausible way for the Sun to lose huge amounts of mass?Making a Planet Seem UninhabitableGiving a Planet SunburnWhat effect will mining on the Moon have long term?
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Would mining huge amounts of resources on the Moon change its orbit?
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Would mining huge amounts of resources on the Moon change its orbit?
Could humans alter the moon's orbit significantly with current technology?What kind of event, if any, would knock the moon off its orbit, without destroying it?Could over-colonization throw our moon out of orbit?How would a nuclear blast on the moon affect its orbit?What kind of event, if any, would knock the moon off its orbit, without destroying it?Could a moon have its own satellites visible from the planet it orbits?How big can a moon be where you can physically jump out of its orbit, to its planet?Exoplanetary Review: Rock StormsHow would a nuclear blast on the moon affect its orbit?How plausible is this as a corporate motivation?Plausible way for the Sun to lose huge amounts of mass?Making a Planet Seem UninhabitableGiving a Planet SunburnWhat effect will mining on the Moon have long term?
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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?
physics hard-science moons astronomy natural-resources
$endgroup$
This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.
|
show 5 more comments
$begingroup$
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?
physics hard-science moons astronomy natural-resources
$endgroup$
This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.
7
$begingroup$
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.
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– Erik
16 hours ago
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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.
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– DarthDonut
16 hours ago
2
$begingroup$
@Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul.
$endgroup$
– DarthDonut
16 hours ago
2
$begingroup$
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
15 hours ago
1
$begingroup$
@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.
$endgroup$
– Ville Niemi
15 hours ago
|
show 5 more comments
$begingroup$
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?
physics hard-science moons astronomy natural-resources
$endgroup$
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?
physics hard-science moons astronomy natural-resources
physics hard-science moons astronomy natural-resources
edited 6 hours ago
jdunlop
7,97811846
7,97811846
asked 16 hours ago
Mr.DMr.D
20016
20016
This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.
This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.
7
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
2
$begingroup$
@Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul.
$endgroup$
– DarthDonut
16 hours ago
2
$begingroup$
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
15 hours ago
1
$begingroup$
@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.
$endgroup$
– Ville Niemi
15 hours ago
|
show 5 more comments
7
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
2
$begingroup$
@Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul.
$endgroup$
– DarthDonut
16 hours ago
2
$begingroup$
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
15 hours ago
1
$begingroup$
@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.
$endgroup$
– Ville Niemi
15 hours ago
7
7
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
2
2
$begingroup$
@Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul.
$endgroup$
– DarthDonut
16 hours ago
$begingroup$
@Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul.
$endgroup$
– DarthDonut
16 hours ago
2
2
$begingroup$
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
15 hours ago
$begingroup$
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
15 hours ago
1
1
$begingroup$
@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.
$endgroup$
– Ville Niemi
15 hours ago
$begingroup$
@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.
$endgroup$
– Ville Niemi
15 hours ago
|
show 5 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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.
(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 $v=sqrtGM/r$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = sqrtM_0 / M_1$.
Let's plug in some numbers:
$v' = sqrtfrac7.342×10^227.342×10^22 - 100times26×10^6 = 1.0000000000000178$
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 $v=sqrtGM/r$ 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:
$frac7.342×10^22 - 100times26×10^67.342×10^22 = 0.99999999999996458$
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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10
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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.
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– sp2danny
11 hours ago
4
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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.
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– Dave Sherohman
11 hours ago
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It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
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– jdunlop
7 hours ago
$begingroup$
@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.
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– Gimelist
7 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
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– Yakk
6 hours ago
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show 1 more comment
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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.
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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.
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– Pelinore
11 hours ago
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I'll try to come back later, when I have time to dig them up, and link the rocket equation, Lunar escape velocity, etc.
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– Zeiss Ikon
11 hours ago
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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.
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– Nuclear Wang
8 hours ago
$begingroup$
@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.
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– Zeiss Ikon
8 hours ago
$begingroup$
@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?
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– TheLuckless
4 hours ago
add a comment |
$begingroup$
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). The mass of the orbiting body 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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1
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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.
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– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
No
Mining on reasonable scales, in the sense you are used to thinking about, won't have any impact on planetary motion.
Yes
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.
TL;DR
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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add a comment |
$begingroup$
This is what one ton of iron looks like. "Thousands of tons" is maybe a couple of full trucks.
(This should really be a comment, but I just registered.)
New contributor
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add a comment |
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5 Answers
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5 Answers
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$begingroup$
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.
(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 $v=sqrtGM/r$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = sqrtM_0 / M_1$.
Let's plug in some numbers:
$v' = sqrtfrac7.342×10^227.342×10^22 - 100times26×10^6 = 1.0000000000000178$
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 $v=sqrtGM/r$ 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:
$frac7.342×10^22 - 100times26×10^67.342×10^22 = 0.99999999999996458$
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.
$endgroup$
10
$begingroup$
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
11 hours ago
4
$begingroup$
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.
$endgroup$
– Dave Sherohman
11 hours ago
$begingroup$
It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
$endgroup$
– jdunlop
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
$endgroup$
– Yakk
6 hours ago
|
show 1 more comment
$begingroup$
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.
(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 $v=sqrtGM/r$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = sqrtM_0 / M_1$.
Let's plug in some numbers:
$v' = sqrtfrac7.342×10^227.342×10^22 - 100times26×10^6 = 1.0000000000000178$
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 $v=sqrtGM/r$ 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:
$frac7.342×10^22 - 100times26×10^67.342×10^22 = 0.99999999999996458$
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.
$endgroup$
10
$begingroup$
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
11 hours ago
4
$begingroup$
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.
$endgroup$
– Dave Sherohman
11 hours ago
$begingroup$
It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
$endgroup$
– jdunlop
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
$endgroup$
– Yakk
6 hours ago
|
show 1 more comment
$begingroup$
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.
(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 $v=sqrtGM/r$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = sqrtM_0 / M_1$.
Let's plug in some numbers:
$v' = sqrtfrac7.342×10^227.342×10^22 - 100times26×10^6 = 1.0000000000000178$
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 $v=sqrtGM/r$ 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:
$frac7.342×10^22 - 100times26×10^67.342×10^22 = 0.99999999999996458$
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.
$endgroup$
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.
(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 $v=sqrtGM/r$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = sqrtM_0 / M_1$.
Let's plug in some numbers:
$v' = sqrtfrac7.342×10^227.342×10^22 - 100times26×10^6 = 1.0000000000000178$
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 $v=sqrtGM/r$ 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:
$frac7.342×10^22 - 100times26×10^67.342×10^22 = 0.99999999999996458$
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.
edited 7 hours ago
answered 15 hours ago
GimelistGimelist
2,549512
2,549512
10
$begingroup$
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
11 hours ago
4
$begingroup$
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.
$endgroup$
– Dave Sherohman
11 hours ago
$begingroup$
It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
$endgroup$
– jdunlop
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
$endgroup$
– Yakk
6 hours ago
|
show 1 more comment
10
$begingroup$
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
11 hours ago
4
$begingroup$
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.
$endgroup$
– Dave Sherohman
11 hours ago
$begingroup$
It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
$endgroup$
– jdunlop
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
$endgroup$
– Yakk
6 hours ago
10
10
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
4
4
$begingroup$
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.
$endgroup$
– Dave Sherohman
11 hours ago
$begingroup$
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.
$endgroup$
– Dave Sherohman
11 hours ago
$begingroup$
It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
$endgroup$
– jdunlop
7 hours ago
$begingroup$
It's also worth noting that you can mine a megaton of iridium (or whatever other mineral suits your fancy) from the moon, or an asteroid, and just dump the spoilage wherever you want, effectively. No environmental protection laws, no rivers to poison, no species to squash. Assuming getting to space is made cheap by some means, there's a lot to recommend resource extraction up there.
$endgroup$
– jdunlop
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
$endgroup$
– Yakk
6 hours ago
$begingroup$
I'm unclear why robots aren't fixing the robots. Can you clarify?
$endgroup$
– Yakk
6 hours ago
|
show 1 more comment
$begingroup$
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.
$endgroup$
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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?
$endgroup$
– TheLuckless
4 hours ago
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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?
$endgroup$
– TheLuckless
4 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 11 hours ago
Zeiss IkonZeiss Ikon
1,811115
1,811115
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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?
$endgroup$
– TheLuckless
4 hours ago
add a comment |
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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?
$endgroup$
– TheLuckless
4 hours ago
$begingroup$
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.
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– Pelinore
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
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.
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– Nuclear Wang
8 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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?
$endgroup$
– TheLuckless
4 hours ago
$begingroup$
@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?
$endgroup$
– TheLuckless
4 hours ago
add a comment |
$begingroup$
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). The mass of the orbiting body 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.
$endgroup$
1
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
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). The mass of the orbiting body 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.
$endgroup$
1
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
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). The mass of the orbiting body 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.
$endgroup$
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). The mass of the orbiting body 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.
answered 11 hours ago
Oscar BravoOscar Bravo
36816
36816
1
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
1
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
1
1
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
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.
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
No
Mining on reasonable scales, in the sense you are used to thinking about, won't have any impact on planetary motion.
Yes
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.
TL;DR
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.
$endgroup$
add a comment |
$begingroup$
No
Mining on reasonable scales, in the sense you are used to thinking about, won't have any impact on planetary motion.
Yes
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.
TL;DR
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.
$endgroup$
add a comment |
$begingroup$
No
Mining on reasonable scales, in the sense you are used to thinking about, won't have any impact on planetary motion.
Yes
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.
TL;DR
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.
$endgroup$
No
Mining on reasonable scales, in the sense you are used to thinking about, won't have any impact on planetary motion.
Yes
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.
TL;DR
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.
answered 6 hours ago
YakkYakk
8,90911238
8,90911238
add a comment |
add a comment |
$begingroup$
This is what one ton of iron looks like. "Thousands of tons" is maybe a couple of full trucks.
(This should really be a comment, but I just registered.)
New contributor
$endgroup$
add a comment |
$begingroup$
This is what one ton of iron looks like. "Thousands of tons" is maybe a couple of full trucks.
(This should really be a comment, but I just registered.)
New contributor
$endgroup$
add a comment |
$begingroup$
This is what one ton of iron looks like. "Thousands of tons" is maybe a couple of full trucks.
(This should really be a comment, but I just registered.)
New contributor
$endgroup$
This is what one ton of iron looks like. "Thousands of tons" is maybe a couple of full trucks.
(This should really be a comment, but I just registered.)
New contributor
New contributor
answered 2 hours ago
markemusmarkemus
1
1
New contributor
New contributor
add a comment |
add a comment |
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7
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
2
$begingroup$
@Mr.D That sounds like whoever wants to calculate the effects will have to integrate over time... That poor soul.
$endgroup$
– DarthDonut
16 hours ago
2
$begingroup$
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
15 hours ago
1
$begingroup$
@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.
$endgroup$
– Ville Niemi
15 hours ago