[charli (OP)]

Listener Amber wants to know:

"If we eventually start mining the moon for minerals etc., would most of it be removed from the moon, would the lightening of the moon cause it to move away from earth more quickly than now?  If we continue to put rockets, infrastructure and people on the moon, would that make it heavier, and cause it to fall into the Earth (over time of course)?

In other words, would any change to the weight of the moon cause divergences in its orbit?"

[Halc]

First of all, the moon, like anything in orbit, is in free-fall, and thus does not weigh anything.  So I think you're asking if we changed the moon's mass, would that change its orbit.

That all depends on how the mass is removed (or added). If it is all accelerated in one direction, it would serve as a sort of rocket exhaust which would put a net thrust on the moon with the corresponding change in orbit. You could also get it spinning this way.  I'm picturing what is probably the most efficient way to get ore off the surface which is by rail gun, shooting a steady stream of material to something that catches it in space, possibly a Lagrange point. Another likely method is by space elevator.

If the mass were to be removed in some crazy way that doesn't exert a force on the remainder, then no orbital change will occur, even if they get the size of the moon down to a pebble.  I cannot think of how one might go about doing that, but what I can imagine (even with the above two methods) is simply putting the ore into space in equal and opposite directions.  Shoot a kg in one direction, and at the same time shoot a different kg in the opposite direction. Then there's no net force on the part of the moon that's left, and no orbital change will occur.

Adding mass is the same thing. Each addition may or may not exert a net momentum to the moon. If that momenum is all in random directions, it will cancel out and no net change occurs, but if it is all in one consistent direction then a change could very much occur relative to the path the moon would have otherwise taken. Keep in mind that forward momentum is continuously being added to the moon via tidal forces, which results in an increase of its orbit by several cm per year. Any reasonable amount of landings or mining activity is going to be dwarfed by these natural forces already accelerating the moon.

[alancalverd]

Contentious and confusing definition of weight, Halc.

Weight is the gravitational force exerted on one body by another, and the fact that the moon is in orbit radius R with tangential speed |v| is due to its weight (mg) being equal to the centripetal force  mv²/R required to keep it there

The pedantic conclusion is correct: if no other force is involved, a change in mass will not alter R because m appears on both sides of the equation.

But whilst the preamble talked about mining etc., the final question asked about weight, not mass, and g only appears on one side of the equation. A change in g will induce a corresponding change in R.

[Bored chemist]

Contentious and confusing definition of weight, Halc.

Which is consistent with everyone's notion of "weightlessness" when in free fall.

[alancalverd]

So weight = mg but only sometimes?

The problem with ascribing weightlessness to free vertical fall, orbit, or any parabolic locus between them, is it doesn't consist with the fact that the falling body is accelerating with F = mg. The fact is that if other objects in the vicinity are following the same trajectory, you can't measure mg as a force between them, but that doesn't mean it zero.

The original question gets more interesting however if we look at the second order effects.

We know F = mg = GmM/R². What happens if we transfer bits of mass from M to m? At present the earth-moon barycenter is some 3000 miles moonwards of the center of the earth. If we terraform the moon by exporting 48.9% of the mass of the earth, the barycenter will move to a point halfway between them. It is left as an exercise to the reader to calculate the final diameter and speed of rotation of the redistributed system   

[Halc]

Contentious and confusing definition of weight, Halc.

It is indeed contentious. Wiki lists several conflicting definitions. You seem to use the gravitational one where g is the coordinate acceeleration due to gravity. With such a definition, the guys on the ISS have weight, albeit reduced by several percent of their weight at sea level.
The ISO definition is close to the gravitational definition, but with adjustments for centripetal and buoyant effects.
BC seems to use the operational definition, which is closer to the one I use.
I work frequently in relativity where gravity is not a force at all. Weight under relativity is still mg where g is the proper acceleration of the object being weighed. That's pretty close to the operational definition, and it gives the moon zero weight.
Doesn't mean any particular definition is the right or wrong one.

A change in g will induce a corresponding change in R.

We're mining the moon, not mining the Earth. g (the gravitational acceleration at lunar radius) is unaffected by the mining of the moon.

The original question gets more interesting however if we look at the second order effects.

We know F = mg = GmM/R2.

Newtonian approximation, but absolutely fine for this exercise.

What happens if we transfer bits of mass from M to m?

Their product changes, hence so does F. It will have little effect on R if the mass moved is significantly less than the larger of the two masses M.

If we terraform the moon by exporting 48.9% of the mass of the earth, the barycenter will move to a point halfway between them. It is left as an exercise to the reader to calculate the final diameter and speed of rotation of the redistributed system 

Interesting exercise. The new dynamic is completely a function of the forces used to do said transfer. Where did all this new angular momentum come from? You can't just pull that out of your arse. If this is a new system, the new separation would be massively larger to preserve said angular momentum, but I don't think it's physically possible to do what you suggest in a closed system.

[alancalverd]

Where did all this new angular momentum come from?

No new angular momentum. That's part of the discussion, like the skater's arms or a coalsescing galaxy: change the mass distribution and the rotational speed changes.

We're mining the moon, not mining the Earth. g (the gravitational acceleration at lunar radius) is unaffected by the mining of the moon.

True, but it's another way of changing the weight of the moon, and the one (the OP said "any change...") that would alter R.

[Janus]

Where did all this new angular momentum come from?

No new angular momentum. That's part of the discussion, like the skater's arms or a coalsescing galaxy: change the mass distribution and the rotational speed changes.

We're mining the moon, not mining the Earth. g (the gravitational acceleration at lunar radius) is unaffected by the mining of the moon.

True, but it's another way of changing the weight of the moon, and the one (the OP said "any change...") that would alter R.

The equation for orbital period is T = 2pi sqrt(a^3/G(m1+m2)
(In many practical cases m2 is small enough compared to m2, that you really don't need to include it)
Thus, if we imagine magically transferring some percentage of the Moon's mass to the Earth, The sum m1+m2 doesn't change, so at first glance neither does the orbital period.
However, The barycenter will shift closer to the Earth, which means the Moon would have to travel around a longer path in the same amount of time if it were to maintain a circular orbit.  Ergo, it would need a higher orbital velocity.
But since we magically transferred just the mass, and changed nothing else, our new less massive Moon, will have too low an orbital speed to hold a circular orbit, and will transition to an elliptical one, with a perigee lower than its present distance. And since this in turn decreases "a", the semi-major axis of the orbit, you'll see a decrease in the length of the orbital period.

That being said, there is no way to magically transfer mass from the Moon to Earth, and how it is done would have a large effect.    An efficient means would be by magnetic rail gun mounted on the surface of the Moon. (With one at each pole they will always be pointing in pretty much the right direction.)
You fire your mass opposite to the Moon's orbital velocity (~1 km/sec) to drop it into an Earth atmosphere grazing orbit (it would likely need to be equipped with some type of retro-rocket package)
This would also act like a rocket engine adding orbital velocity to the Moon with each launch.
But, we are talking about pretty insignificant changes. 
The other thing that the OP should be made aware of is that orbits aren't that fragile.  People sometimes get the idea that the slightest deviation will cause an orbiting object to either fall to the planet or fly off into space.  This is far from the case.  In order to get the Moon to fly off you would need to increase its orbital velocity by a good fraction of its present orbital speed of 1 km/sec, and decrease it by even more to put it into an Earth intersecting trajectory.
In reality, anything we do on the Moon would not have any significant effect on it. ( for example, even if you were to transfer the entire population to of the Earth to the Moon, you'd only increase the mass of the Moon by roughly 1/100,000,000,000.)

[CliffordK]

There are other things we could do... For example a favorite energy boost for rockets is the "slingshot maneuver".  Generally adding energy to a rocket by taking a tiny bit of energy away from a moon or planet. 

In theory, one could also do slingshot braking to reduce speed or change trajectory before doing a reentry, adding energy to a moon.

Again, the amounts are so small it is insignificant to a moon or planet's orbital energy...  at least at this time.

[evan_au]

If we eventually start mining the moon for minerals etc

We have already added a few tons to the Moon since the 1960s, with various landers, rovers and crash-landers.

But the mass of space dust and bigger bits of space debris hitting the Moon every day must be far higher.
- It will be many years before anything humans can add to the Moon (or remove from the Moon) comes anywhere near this daily influx of "meteoric" material.

The moon is moving away from the Earth at a rate of 3.8 cm per year.
- That is a huge amount of energy which is gradually eroding the seashores and ocean basins of the Earth, through tides.
- Humans are nowhere near harnessing the tidal energy of all the world's oceans, so we are nowhere near affecting the rate at which the Moon is receding from the Earth.

[charles1948]

If we eventually start mining the moon for minerals etc

The moon is moving away from the Earth at a rate of 3.8 cm per year.
- That is a huge amount of energy which is gradually eroding the seashores and ocean basins of the Earth, through tides.

Why is the Moon moving away from the Earth.  Shouldn't it be getting closer, pulled towards the Earth by gravity?

[Eternal Student]

Why is the Moon moving away from the Earth.  Shouldn't it be getting closer, pulled towards the Earth by gravity?

  Earth isn't a perfect sphere, it's oblate  (more like an egg than a good round ball).   The moon is actually causing most of this deformity in the earth (by pulling on it with the force of gravity).   The thing is the earth is spinning on it's own axis quite fast, so these bulges get pulled slightly forward of the line between the earth and the moon.   The moon goes in a prograde orbit around the earth (it moves around the earth in the same direction as the earth spins).  The bulges on the earths surface are then ever so slightly pulling the moon forward a bit,  while at the same time slowing the rotation of the earth.   By giving the moon a bit more of a pull in the direction tangential to it's orbit the moon's speed is increasing.  As that speed increases, the moon drifts out a bit further to an orbit with a greater radius.
    There's some info here:
https://en.wikipedia.org/wiki/Tidal_acceleration

[charles1948]

Thanks Eternal, for your post, and the link you kindly provided.  I've read this, the math contained in it is very formidable!  But I get the general idea, I think - thanks again!

[evan_au]

Why is the Moon moving away from the Earth.  Shouldn't it be getting closer, pulled towards the Earth by gravity?

If there were no gravity, the Moon would depart from the Earth on a straight-line path
- As Newton observed, objects will continue traveling in a straight line, unless acted on by an external force (gravity, in this case)

The definition of an orbit is that the tendency of an object to travel away in a straight line is balanced by the gravitational attraction that bends its path into an ellipse or a circle.
- The Moon has an elliptical orbit, ranging from 363,000km to 405,000km (and back) over the course of each month
- But its average distance is not changing over millennia.

...Unless you look extremely closely - 3.8cm per year means 38 meters in 1,000 years, which is almost nothing compared to 363,000km.

[Bored chemist]

Earth isn't a perfect sphere, it's oblate  (more like an egg than a good round ball).   

I feel sorry for hens in your universe.

[Janus]

Why is the Moon moving away from the Earth.  Shouldn't it be getting closer, pulled towards the Earth by gravity?

  Earth isn't a perfect sphere, it's oblate  (more like an egg than a good round ball).   The moon is actually causing most of this deformity in the earth (by pulling on it with the force of gravity).   The thing is the earth is spinning on it's own axis quite fast, so these bulges get pulled slightly forward of the line between the earth and the moon.   The moon goes in a prograde orbit around the earth (it moves around the earth in the same direction as the earth spins).  The bulges on the earths surface are then ever so slightly pulling the moon forward a bit,  while at the same time slowing the rotation of the earth.   By giving the moon a bit more of a pull in the direction tangential to it's orbit the moon's speed is increasing.  As that speed increases, the moon drifts out a bit further to an orbit with a greater radius.
    There's some info here:
https://en.wikipedia.org/wiki/Tidal_acceleration

A few corrections here:
While the Earth is oblate (a greater radius at the equator), an egg is prolate( a greater polar radius).
The Earth is oblate due to its spin, giving the equator a radius 21 km greater than the polar radius.
The tidal effect of the Moon, while it does distort this shape some, is far from the dominant effect, as it is in the range of 1 meter.
A complicating factor with the Earth-Moon system is that there are two tides: The Earth tide, and the ocean tide. And these are not in phase with each other.  The ocean tides are effected by the way the water interacts between basins and coastlines.
So ocean tidal bulges tend to be drug along by the Earth's rotation more than Earth tides are.   

[CliffordK]

I think the issue with the question...

The moon has has X amount of kinetic energy.

If one could remove mass without removing kinetic energy, then the moon would speed up.

However, if you remove mass, you also remove kinetic energy.  Even if you had a nuclear reactor slowly reducing mass in situ, you would also lose the kinetic energy associated with that mass.  Thus, no change in motion.

Your changes in energy then come with how you are adding/removing matter from the moon. 

A couple of people have mentioned some kind of an electromagnetic rail gun that would push off of the moon to go either towards Earth or away from Earth, either accelerating or decelerating the moon. 

I'm not sure a chemical rocket would follow the same.  While the exhaust bombards the moon it would push off from the moon.  But, say one uses a tangential acceleration, the moon wouldn't necessarily be impacted with the exhaust in the same manner.  So not being pushed off of like the electromagnetic accelerator.

[Eternal Student]

While the Earth is oblate (a greater radius at the equator), an egg is prolate( a greater polar radius)......(etc.)...

   Hi Janus.    Thanks, I didn't know that. I thought oblate meant the 3-D version of oval.

[Halc]

While the Earth is oblate (a greater radius at the equator), an egg is prolate( a greater polar radius)......(etc.)...

   Hi Janus.    Thanks, I didn't know that. I thought oblate meant the 3-D version of oval.

A prolate the shape you get from an ellipse (not an oval) spun on its long axis. An oblate shape is an ellipse spun on its short axis.
Tides do indeed induce a very mild prolate shape, (a one-meter deformation), but the oblateness  is from the spin, a deformation of many km, kind of the same way a pizza crust is stretched by spinning it. It gets fat and wide (oblate), not skinny and long (prolate) like a french bread.

[Eternal Student]

Thanks Halc and Janus.
    I wonder if we are losing sight of the wood for the trees?  I haven't studied planetary systems much and would just like to check the following:    The tidal distortion (bulges on both sides of the earth nearest and furthest away from the moon) is what's important for the recession of the moon.   The oblate-ness of the earth due to it's rotation is a more permanent bulge around the equator, which is interesting but nothing to do with why the moon is receding?

     Returning to the main topic,  the tidal effects from the moon would seem to generally increase if the moon did come closer to the earth (with the moon having much the same mass).  Which would then tend to oppose progress toward the earth and encourage recession.  Unless the mining operations on the moon are astonishingly rapid, this may help to prevent the moon spiraling into the earth.
   On the other hand, increased tidal effects generally means kinetic and gravitational potential energy is being lost by the system (and heat increases), which is a bit of a worry.  However, maybe most of the energy lost would be noticed in the reduction of earth's rate of rotation.  It's often said that we need an extra hour in the day.  Anyway, returning to the original question again,  if the earth quickly moves to a state of being tidally locked to the moon's orbit then I suppose there is an "apparent" change in the moon's path as the original listener, Amber, would view things from the earth.