Attempt no landings.
January 27, 2012 4:27 PM Subscribe
Reading about Gingrich's plan to mine the moon, I starting wondering: could the orbital stability of the moon be changed if enough mass was transferred to the Earth?
I would expect a heavier Earth, already spinning faster than the moon, would cause an even larger leading force which would accelerate the moon along its trajectory faster than it currently is, causing it to drift away. But I'm definitely no physicist. On what scale would the mass have to change to to have a catastrophic effect, if any?
For the sake of argument, let's define catastrophe as doubling the current speed at which the moon is already moving away from the Earth.
I would expect a heavier Earth, already spinning faster than the moon, would cause an even larger leading force which would accelerate the moon along its trajectory faster than it currently is, causing it to drift away. But I'm definitely no physicist. On what scale would the mass have to change to to have a catastrophic effect, if any?
For the sake of argument, let's define catastrophe as doubling the current speed at which the moon is already moving away from the Earth.
I'm having trouble finding a readable equation (ie: for the non-astronomer) that governs the moon's recession. And what I did find suggests that calculating that recession historically is very, very difficult.
But as a thought experiment, I'd guess that the two major terms in the equation are the mass of the earth and the mass of the moon. But in this case, we're not just shooting moon mass into space, we're transferring it to earth. So the mass of the earth will increase at the same rate that the mass of the moon decreases. I suspect that the gravitational force between the earth and the moon will remain the same.
My buddy who's a physics PhD (and studied astrophysics for awhile) says that the period of the moon will double in 50 billion years. But the sun will have gone nova long before that. Then again, my buddy was unaware that the moon was even receding so I'm not sure what help he is :)
posted by sbutler at 4:59 PM on January 27, 2012
But as a thought experiment, I'd guess that the two major terms in the equation are the mass of the earth and the mass of the moon. But in this case, we're not just shooting moon mass into space, we're transferring it to earth. So the mass of the earth will increase at the same rate that the mass of the moon decreases. I suspect that the gravitational force between the earth and the moon will remain the same.
My buddy who's a physics PhD (and studied astrophysics for awhile) says that the period of the moon will double in 50 billion years. But the sun will have gone nova long before that. Then again, my buddy was unaware that the moon was even receding so I'm not sure what help he is :)
posted by sbutler at 4:59 PM on January 27, 2012
Best answer: The moon weighs something like 10^23 kg.
The largest mine on earth is, according to wikipedia, 1.2 km deep and 7.7 km^2 in area. It's absolutely huge. Assuming all that is one big rectangle (it'd be considerably less due to surface shape, taper, etc., but we're doing rough figuring here): 9.2 km^3 of earth, or 9.2e9 m^3
Rock density is somewhere around 3000 kg/m^3
So that mined earth is ~3e13 kg. If you removed a similar mass from the moon, that would represent about 0.000 000 01% of the moon's total mass.
posted by introp at 4:59 PM on January 27, 2012 [5 favorites]
The largest mine on earth is, according to wikipedia, 1.2 km deep and 7.7 km^2 in area. It's absolutely huge. Assuming all that is one big rectangle (it'd be considerably less due to surface shape, taper, etc., but we're doing rough figuring here): 9.2 km^3 of earth, or 9.2e9 m^3
Rock density is somewhere around 3000 kg/m^3
So that mined earth is ~3e13 kg. If you removed a similar mass from the moon, that would represent about 0.000 000 01% of the moon's total mass.
posted by introp at 4:59 PM on January 27, 2012 [5 favorites]
While I don't know the real answer, this gravity simulator is a fun way to try out the question (I had to cut out half the moon's mass before I saw anything exciting and apocalyptic, but then, I was getting impatient waiting for catastrophe).
I wasted a heap of time in the past playing with that gravity simulator; it's great fun. You can speed it up to see what happens many many orbits into the future. My (incredibly limited) understanding, though, is that when you have more than two bodies involved (as we, in fact, do in the solar system) the system becomes inherently somewhat unstable. I imagine that a full-scale simulator might make the moon's orbit more sensitively dependent on initial conditions than in a simple two-body simulation.
posted by yoink at 5:00 PM on January 27, 2012
I wasted a heap of time in the past playing with that gravity simulator; it's great fun. You can speed it up to see what happens many many orbits into the future. My (incredibly limited) understanding, though, is that when you have more than two bodies involved (as we, in fact, do in the solar system) the system becomes inherently somewhat unstable. I imagine that a full-scale simulator might make the moon's orbit more sensitively dependent on initial conditions than in a simple two-body simulation.
posted by yoink at 5:00 PM on January 27, 2012
Best answer: AIUI, the Moon is currently receding because angular momentum from Earth's rotation is slowly being transferred to the Moon through interaction with the tidal bulge; the Earth is (very very slowly) attempting to become tide-locked to the Moon. I think it's unlikely any Moon-mining efforts would have an effect even comparable to that.
I'm pretty sure that the effect depends a lot on exactly how the mass is transferred. Lunar mass-drivers will have one result; rockets will have a different one. (Transferring by rocket doesn't conserve angular momentum, because the exhaust escapes the system.)
posted by hattifattener at 6:00 PM on January 27, 2012
I'm pretty sure that the effect depends a lot on exactly how the mass is transferred. Lunar mass-drivers will have one result; rockets will have a different one. (Transferring by rocket doesn't conserve angular momentum, because the exhaust escapes the system.)
posted by hattifattener at 6:00 PM on January 27, 2012
If you're just talking moon mining, then the scales involved would mean nothing even vaguely significant would change. Mines can be big compared to people, but are absolutely miniscule in terms of the total mass of a planet (or moon).
posted by twirlypen at 6:03 PM on January 27, 2012
posted by twirlypen at 6:03 PM on January 27, 2012
The real point of mining the moon is not transfer to earth, it is to process materials for use in space for infrastructure there. The gravity well (a measure of how far you have to lift something up to escape the pull of a planet or moon) of the earth is so deep once you have something in orbit you are halfway to anywhere in the solar system in how much energy you need to get there. Mining the moon lets gets your raw materials much easier and cheaper than lifting them from earths surface. What you would return to earth is energy (solar power satellites and helium 3) and satellites of all kinds and the infrastructure to get to mars and the rest of the planets. It might be even better to establish a colony on a near earth asteroid for resource extraction since the moon appears to be defecient in heavier elements like iron and such.
posted by bartonlong at 6:14 PM on January 27, 2012 [1 favorite]
posted by bartonlong at 6:14 PM on January 27, 2012 [1 favorite]
Best answer: The moon isn't exactly stable anyway. It's already moving away form the earth a around 4cm a year. Which doesn't seem like much, but that adds up to 40,000 kilometers over a billion years. So yes, if we moved a bunch of mass around we would, over the long term (geologically) have an observable effect.
However what it will be we can't yet calculate because there's too much stuff going on in the solar system to know where the planets will be in a billion years. At least given today's ability to calculate.
posted by Ookseer at 9:55 PM on January 27, 2012
However what it will be we can't yet calculate because there's too much stuff going on in the solar system to know where the planets will be in a billion years. At least given today's ability to calculate.
posted by Ookseer at 9:55 PM on January 27, 2012
1- We could just send our garbage up there to account for the difference. (As seen in THGTTG where a planet required visitors to leave the same mass that they took, and payment that wasn't rendered in waste was rendered via amputation, if memory serves.)
2- But yeah, the moon is huge. It's the 13th most massive thing in the solar system (not counting the sun), and the Earth is the 5th.
posted by gjc at 10:45 PM on January 27, 2012
2- But yeah, the moon is huge. It's the 13th most massive thing in the solar system (not counting the sun), and the Earth is the 5th.
posted by gjc at 10:45 PM on January 27, 2012
Note that some asteroid-deflection schemes involve shining a really big laser at them for a long time such that material is ablated away (I'm not saying this is feasible now). But luckily for us, very few asteroids are remotely close to the moon in size. (Unluckily for us, it doesn't take a very big asteroid to do a lot of civilization-impacting damage.) For example, the K-T extinction event may have been caused by an asteroid some 10km in diameter.
But anyway, bartonlong outlines the more usual scenario for mining the moon. It's a key to, for example, Gerard K. O'Neill's L5 Society orbital colonies, and some Mars exploration/settlement scenarios, as well as more fanciful ones for interstellar travel.
One of the most useful things to mine from the moon, by the way, might be water -- if we can find enough of it.
posted by dhartung at 1:08 AM on January 28, 2012
But anyway, bartonlong outlines the more usual scenario for mining the moon. It's a key to, for example, Gerard K. O'Neill's L5 Society orbital colonies, and some Mars exploration/settlement scenarios, as well as more fanciful ones for interstellar travel.
One of the most useful things to mine from the moon, by the way, might be water -- if we can find enough of it.
posted by dhartung at 1:08 AM on January 28, 2012
No.
Even if we could take a huge amount of stuff off the moon, like half of it, the total gravitational force between the two bodies would be the same, because earth's gravity would be strong.
The formula is fg = G(M + m)/r2. Transferring mass from M to m has no impact on the value fg
posted by delmoi at 6:07 PM on January 28, 2012
Even if we could take a huge amount of stuff off the moon, like half of it, the total gravitational force between the two bodies would be the same, because earth's gravity would be strong.
The formula is fg = G(M + m)/r2. Transferring mass from M to m has no impact on the value fg
posted by delmoi at 6:07 PM on January 28, 2012
(er, that should "earths gravity would be stronger.")
posted by delmoi at 6:08 PM on January 28, 2012
posted by delmoi at 6:08 PM on January 28, 2012
That's f=GMm/r^2. Yeah, if you could move half the mass of the moon on to the earth, you'd change things. It sure does make a difference.
(But as above, even a very big mine would be a very small percentage of the mass of either earth or moon).
posted by nat at 5:27 PM on January 29, 2012
(But as above, even a very big mine would be a very small percentage of the mass of either earth or moon).
posted by nat at 5:27 PM on January 29, 2012
This thread is closed to new comments.
posted by mittens at 4:53 PM on January 27, 2012 [2 favorites]