$begingroup$
Imagine that humanity builds mining facilities on moon. We know that 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 various metals will be mined from moon in order to sustain giant fleet of military, civilian and scientific ships for further colonization of our Solar System.
Because all of these actions, moon will start to lose its own mass. Is it going to have some kind of impact to its orbit? If it is, how severe will be its changes?
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 4 more comments
$begingroup$
Imagine that humanity builds mining facilities on moon. We know that 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 various metals will be mined from moon in order to sustain giant fleet of military, civilian and scientific ships for further colonization of our Solar System.
Because all of these actions, moon will start to lose its own mass. Is it going to have some kind of impact to its orbit? If it is, how severe will be its changes?
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.
3
$begingroup$
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
4 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
4 hours ago
$begingroup$
@Erik. I think I need to clarify that in my question :), We know that mining volume will increase after some time because of rising demand. I think after several decades it will rise to considerable level.
$endgroup$
– Mr.D
4 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
4 hours ago
1
$begingroup$
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
3 hours ago
|
show 4 more comments
$begingroup$
Imagine that humanity builds mining facilities on moon. We know that 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 various metals will be mined from moon in order to sustain giant fleet of military, civilian and scientific ships for further colonization of our Solar System.
Because all of these actions, moon will start to lose its own mass. Is it going to have some kind of impact to its orbit? If it is, how severe will be its changes?
physics hard-science moons astronomy natural-resources
$endgroup$
Imagine that humanity builds mining facilities on moon. We know that 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 various metals will be mined from moon in order to sustain giant fleet of military, civilian and scientific ships for further colonization of our Solar System.
Because all of these actions, moon will start to lose its own mass. Is it going to have some kind of impact to its orbit? If it is, how severe will be its changes?
physics hard-science moons astronomy natural-resources
physics hard-science moons astronomy natural-resources
edited 35 mins ago


Pelinore
2,646725
2,646725
asked 4 hours ago
Mr.DMr.D
1704
1704
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.
3
$begingroup$
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
4 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
4 hours ago
$begingroup$
@Erik. I think I need to clarify that in my question :), We know that mining volume will increase after some time because of rising demand. I think after several decades it will rise to considerable level.
$endgroup$
– Mr.D
4 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
4 hours ago
1
$begingroup$
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
3 hours ago
|
show 4 more comments
3
$begingroup$
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
4 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
4 hours ago
$begingroup$
@Erik. I think I need to clarify that in my question :), We know that mining volume will increase after some time because of rising demand. I think after several decades it will rise to considerable level.
$endgroup$
– Mr.D
4 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
4 hours ago
1
$begingroup$
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
3 hours ago
3
3
$begingroup$
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
4 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
4 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
4 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
4 hours ago
$begingroup$
@Erik. I think I need to clarify that in my question :), We know that mining volume will increase after some time because of rising demand. I think after several decades it will rise to considerable level.
$endgroup$
– Mr.D
4 hours ago
$begingroup$
@Erik. I think I need to clarify that in my question :), We know that mining volume will increase after some time because of rising demand. I think after several decades it will rise to considerable level.
$endgroup$
– Mr.D
4 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
4 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
4 hours ago
1
1
$begingroup$
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
3 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
3 hours ago
|
show 4 more comments
1 Answer
1
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.
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:
$v' = sqrt{frac{7.342×10^{22}}{7.342×10^{22} - 100times26×10^6}} = 1.0000000000000178$
The Moon's orbit is going to be 1.0000000000000178 times faster.
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.
$endgroup$
add a comment |
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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.
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:
$v' = sqrt{frac{7.342×10^{22}}{7.342×10^{22} - 100times26×10^6}} = 1.0000000000000178$
The Moon's orbit is going to be 1.0000000000000178 times faster.
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.
$endgroup$
add a 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.
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:
$v' = sqrt{frac{7.342×10^{22}}{7.342×10^{22} - 100times26×10^6}} = 1.0000000000000178$
The Moon's orbit is going to be 1.0000000000000178 times faster.
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.
$endgroup$
add a 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.
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:
$v' = sqrt{frac{7.342×10^{22}}{7.342×10^{22} - 100times26×10^6}} = 1.0000000000000178$
The Moon's orbit is going to be 1.0000000000000178 times faster.
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.
$endgroup$
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.
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:
$v' = sqrt{frac{7.342×10^{22}}{7.342×10^{22} - 100times26×10^6}} = 1.0000000000000178$
The Moon's orbit is going to be 1.0000000000000178 times faster.
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.
edited 11 mins ago
answered 3 hours ago


GimelistGimelist
2,419411
2,419411
add a comment |
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3
$begingroup$
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
4 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
4 hours ago
$begingroup$
@Erik. I think I need to clarify that in my question :), We know that mining volume will increase after some time because of rising demand. I think after several decades it will rise to considerable level.
$endgroup$
– Mr.D
4 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
4 hours ago
1
$begingroup$
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
3 hours ago