[mackger (OP)]

The further the Earth/Moon barycenter from the center of the Earth, the greater the Earth’s ‘wobble’ due to the moon.
Given r=a/(1+(m1/m2)) then, apparently, the Earth’s barycentric wobble should increase as the Moon moves away from the Earth.
But surely- as the Moon moves further away from the Earth- the force of gravity between them should reduce with the square of the distance- so, shouldn’t this reduce the Earth’s wobble?
Should the equation for the barycenter not be expressed in terms of the force between the bodies?
As the Moon continues to move away from the Earth, at some point (Hill Sphere?) shouldn’t the barycenter start to move back towards the center of gravity of the Earth?

[Halc]

The further the Earth/Moon barycenter from the center of the Earth, the greater the Earth’s ‘wobble’ due to the moon.
Given r=a/(1+(m1/m2)) then, apparently, the Earth’s barycentric wobble should increase as the Moon moves away from the Earth.

That it does.

But surely- as the Moon moves further away from the Earth- the force of gravity between them should reduce with the square of the distance- so, shouldn’t this reduce the Earth’s wobble?

The gravity between the two does decrease with distance, but that only reduces the acceleration of each object. It also increases the orbital period, so it undergoes that acceleration for a longer time, which accounts for the greater total distance traveled (the wobble). So if the moon doubles the distance it has to travel, so must Earth, even though the force drops to a 4th that of the closer orbit.

Should the equation for the barycenter not be expressed in terms of the force between the bodies?

No. The force between the two is equal an opposite, but the barycenter is not halfway.

As the Moon continues to move away from the Earth, at some point (Hill Sphere?) shouldn’t the barycenter start to move back towards the center of gravity of the Earth?

Given just a two body system, there is no hill radius, and the barycenter is always a fixed ratio of the total distance divided by the respective masses, or a little over 1% of the total distance. That means if the moon was a light year away (but still orbiting), the barycenter would be just over 1% of a lightyear from the center of Earth.

Of course Earth does have a finite hill radius because there are more than two bodies in the universe, so if the moon gets far enough away, it will just simply cease to orbit Earth and will make its independent way around the solar system, and there won't be a barycenter between the two at all.

[evan_au]

Why is the Barycenter Equation expressed in terms of mass?

Because it literally means "the center of mass"...
- I guess if you wanted to be pedantic, it might be debatable whether the "bary-" part is talking about mass or weight...
- but since the ancient Greeks breathed air and lived within a couple of kilometers of sea level, it doesn't make too much of a difference
- unless you were trying to detect an adulterated gold crown!

[mackger (OP)]

Thank you Halc- here's my issue- as you mention, the Earth does actually have a Hill radius- but surely the 'Earth wobble radius' due to the moon moving away cannot keep increasing linearly right to the point where the moon crosses the hill radius, and then suddenly the barycentric Earth Wobble just disappears? Surely, when the moon is about half-way to the Earth's Hill radius, the Earth Wobble radius must start to reduce linearly with increased distance, until it disappears as the moon crosses the Earth's Hill radius? Sorry I didn't make this point clearer in my initial post.

[Halc]

Thank you Halc- here's my issue- as you mention, the Earth does actually have a Hill radius- but surely the 'Earth wobble radius' due to the moon moving away cannot keep increasing linearly right to the point where the moon crosses the hill radius

There is more than one hill radius to Earth. For a satellite, an orbit can only be so high before the moon interferes and the orbit becomes unstable.
The more usual hill radius (about 1.5 m km) is the maximum the moon could be before other objects pull it away from Earth.

and then suddenly the barycentric Earth Wobble just disappears?

If the orbit disappears, then so does a meaningful barycenter. But the center of gravity between the two would always remain, and they're mathematically the same thing. So for instance, there's a center of gravity of the Earth/Neptune pair, but no barycenter since neither orbits the other.

Surely, when the moon is about half-way to the Earth's Hill radius, the Earth Wobble radius must start to reduce linearly with increased distance, until it disappears as the moon crosses the Earth's Hill radius?

As explained in the posts above, this cannot be.

[mackger (OP)]

Thank you again Halc- please persist and be patient with me as I try and get my head around this- so as the moon heads out towards the 1.5m km threshold, the radius of Earths barycentric wobble increases towards a maximum of about 15k km (1%) around a barycenter 15k km out from the Earths own center of gravity. Then as the moon crosses the 1.5m km threshold the Earths barycentric wobble completely collapses from the maximum 15k radius to totally gone? There's no extended transition zone where the weakening gravity between the Earth and moon allows the barycenter to recede back towards the Earths own center of gravity? Allowing the Earths radius of barycentric wobble to spin down and decay away naturally? 

[evan_au]

There's no extended transition zone where the weakening gravity between the Earth and moon...

One possible result of the Moon drifting away from Earth into its own orbit is called a "Horseshoe orbit".
- Both Earth and Moon remain at the same average distance from the Sun
- But the Moon drifts ahead of the earth in its orbit, until it catches up on the other side, then reverses course, forming the horseshoe shape.
- When the Moon is "ahead" of the Earth in its orbit,  the Earth will accelerate towards the barycenter, and accelerate slightly in its orbit
- When the Moon is "behind" the Earth in its orbit,  the Earth will accelerate towards the barycenter, and decelerate slightly in its orbit
- When the Moon is on the far side of the Sun, the Earth will still accelerate towards the barycenter, but this acceleration is miniscule compared to its acceleration towards the Earth-Sun barycenter
See https://en.wikipedia.org/wiki/Horseshoe_orbit

[Halc]

so as the moon heads out towards the 1.5m km threshold, the radius of Earths barycentric wobble increases towards a maximum of about 15k km (1%) around a barycenter 15k km out from the Earths own center of gravity.

Right.  The moon isn't going to get that far out since the system lacks the angular momentum for it, but if new angular momentum was somehow introduced pushing the moon that far out, then yes. a 15k km separation of Earth CoM from the barycenter.

Then as the moon crosses the 1.5m km threshold the Earths barycentric wobble completely collapses from the maximum 15k radius to totally gone?

As a regular near-sine-wave sort of wobble, yes,l it would abruptly disappear. You'd probably end up with the horseshoe orbit that evan describes if only a little bit too much push was given to the moon. If enough push was given, the moon would become an independent body, perhaps capable of hitting any planet.

If the horseshoe orbit occurs, the Earth would orbit for a long time (years) at about 15k km closer to the sun than its mean, and then when it does the other half,l it would orbit for years at 15k km further from the sun than its mean. There are already several known natural objects in such an orbit with Earth.

There's no extended transition zone where the weakening gravity between the Earth and moon allows the barycenter to recede back towards the Earths own center of gravity?

No, that would violate conservation of momentum and Newton's first law.

[mackger (OP)]

OK- so the barycentric radius in terms of a two body relative mass system holds good within the Hill radius- and it’s NOT the case that as the Moon approaches the Hill radius the acceleration towards the Earth-Moon barycenter just becomes increasingly miniscule/negligible relative to the acceleration towards the Earth-Sun barycenter. But it is the case that the Earth-Moon barycentric radius suddenly/abruptly collapses as the Moon crosses the Hill threshold. That is not intuitive. That is pretty mind-bending. Thank you both for your time and effort in taking my question- very much appreciated. I’ve self-published advertising removed about the Eratosthenes method and very basic star and moon observation, and I was thinking about expanding the notes on barycentric wobble- but now I think I’ll leave well alone! Thank you.