How did the Earth capture the Moon ?

Thanks for this. A couple of questions, though.

- How do we know the CG is offset from geometrical centre by 2km?

- What would be the process by which the moon would "spin down" to become tidally locked and what would happen to the excess angular momentum?

 

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The Moon's rotational axis passes through its CoG. This axis is tilted by about 6 degrees with respect to the axis of the Moon's orbit. In addition, the Moon's rotational speed is steady, while its orbital speed is not. This means that the Moon exhibits both latitudinal and longitudinal librations, which are centered on the CoG. By its shape, we can locate the geometrical center. By careful examination of the the libration you could determine if the two coincide. we have also been able to map the Moon's gravitational field with orbiters ( this is how we located the mascons)

Some of it is converted to orbital angular momentum, and the rest carried off as waste heat. (since photons have momentum they can also carry off angular momentum. Consider the image below. The circle is a body rotating in the clockwise direction. The arrows are photons emitted from the surface in opposite directions, and of the same wavelength as measured from the emitting surface. The blue arrow is emitted in the same direction as the surface is moving relative to the axis of rotation. Thus as measured from the axis, which is the reference against which angular momentum is measured, the right moving photon is blue shifted and the left moving photon is red-shifted) Thus the right moving photon has more momentum relative to the axis, and more angular momentum. Thus you end up with a net clockwise angular momentum for the pair.

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Thanks, yes, libration, of course.

Interesting about emission of photons. That's a process that would never have occurred to me. For a moment I struggled to see how these photons would carry off angular momentum from the system (as opposed to their own intrinsic spin), but their motion relative to the axis of rotation of the moon does include angular momentum I suppose.

But is this really though to be a significant process, over astronomical periods of time?

 

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Photon drag as described would seem a tiny effect compared to tidal friction. The Earth distorts the moon from the spherical shape it would otherwise tend to form, and this continuous distortion as the moon rotates generates very significant heat, which radiates away the angular kinetic energy. Similarly Earth used to rotate about once every 10 hours and it is close to 24 hours now and slowing. Most of our heat is generated via friction of ocean water on the ocean floor, but there is also similar bending of the shape of the planet which generates heat as well.

Mercury similarly has a tidal force that resists its rotation, but it also has a positive force that exactly cancels that due to its oblate shape, so Mercury is currently locked into its 3-2 ratio of spin to orbital rate.

 

Yes, tidal friction would be a factor, I imagine and yes energy would be radiated. But regarding angular momentum, you don't radiate that away (except via the mechanism Janus mentions). Surely what happens instead is there is angular momentum transfer. I can see two possibilities: either to the other body (the Earth) or from spin of the moon to its orbital angular momentum. How would you see that working?

 

I wouldn't consider it a "drag" effect. As far as the surface emitting the radiation there is no differential force felt between emitting a photon in the direct vs. retrograde direction We are mainly considering the conservation of the total angular momentum. It is just that the extra radiation created by the heating carries away some of the angular momentum rather than it being transferred to the Moon.
Granted, it would be a very minor part of the angular momentum lost.

 

In the image below, the Moon orbits the Earth( not shown but at bottom of image) in the counterclockwise direction, While also rotating in the same direction.

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The thicker green arrow represents the majority of Earth's gravity acting on the Moon. The gravity differential produces tidal bulges ( shown extended from the Moon). While the Earth's gravity tries to align these bulges with the thick green arrow, friction with the rotating Moon tries to drag them along with it's rotation, and you end up with an equilibrium with the bulges not in perfect alignment with the Earth.
This results in the further bulge leading the Moon a bit in its orbit and the closer bulge trailing a bit. The net gravitational pull from the Earth on these bulges are shown as the thinner green lines. The red arrows are the components of these pulls that act along the orbital motion of the Moon. The component on the near bulge tends to aid the Moon in its orbit, and the component for the far bulge tend to retard the orbit. However, since the Near bulge is closer to the Earth, the Earth's gravitational pull on it will be slightly greater as a whole, which results its orbital acting component being greater than the far bulge's. The Moon ends up with a net force pushing it forward in its orbit, which results in an increase of orbital angular momentum.
So loss of counterclockwise angular momentum from a slowing rotation results in a gain of counter clockwise orbital angular momentum.
Note that if the Moon had been rotating clockwise and orbiting counterclockwise, it would be the near bulge that leads the Moon's center of gravity in the orbit, the net pull would work against the Moon's orbital motion, and thus a loss of clockwise rotation of the Moon would result is a loss of counterclockwise orbital angular momentum. The Moon would still end up in eventual tidal lock, but the Moon would have spiraled in towards the Earth rather than away as a result.

 

Thank you, yes, that explains it nicely.

 

A further thought about this....

The result is that the spinning down of the moon adds to its orbital angular momentum. The effect of the tides on earth also spins down the earth, again transferring angular momentum to the moon's orbit. Eventually the earth would in principle become tidally locked to the moon, I suppose.

So it looks as if one can conclude that tidal effects will always tend to convert any spin of the participating bodies into orbital angular momentum of the system. That is something I had not appreciated. Is that right?

Also, in the case of the moon, both effects will cause the moon to move gradually away from earth.

 

Well not always since spin is not always greater than the orbital motion. So sure, Phobos has long since transferred its spin to its orbit about Mars, but Mars is currently speeding up its spin due to Phobos, not slowing like Earth is. Eventually (not long from now), the entire angular momentum of the Mars/Phobos system will be converted to Mars spin, not to orbital energy. Phobos will fall to the planet, breaking up as it does so, giving Mars rings for a while. That's the fate of our moon as well if Earth lives long enough to see it. Jury is out on that.
So long term, tidal effects will always eventually tend to convert orbital energy back to spin. It just takes a long time. The day on Earth will eventually be considerably shorter than 10 hours if it survives long enough for that to happen.

One effect in that case. Only tidal drag of the moon on Earth transfers some Earth spin Energy to moon orbital energy. OK, the spin of the moon is technically still slowing, but most of the tidal forces of Earth on the moon goes into circularizing its orbit, not increasing it. The tidal drag of the sun on the system decreases the moon's orbit faster than the drag on the moon's spin rate increases it.

 

OK I'm not sure I follow all of this.

Yes, I can certainly see that if the orbital frequency> spin frequency then the orbital angular momentum will get converted to spin rather than the other way round, as it is, I suppose, essentially a drag effect between the bodies.

But why do you say that long term tidal effects always convert orbital angular momentum to spin? Surely the moon/earth system shows the opposite, doesn't it? I thought the earth day is lengthening, not shortening, due to transfer of angular momentum to the moon's orbit.

 

I took it as including both positive and negative contribution via "conversion", despite the ambiguity of the subsequent references in the posting.

 

Well yes I follow that the spin can - in principle though rarely in practice* - be aligned with or against the orbit.

But in the case of the moon/earth system my understanding is that the earth will lose spin and the moon will gain orbital angular momentum, until the earth becomes tidally locked to the moon's orbit. So orbital angular momentum is not ultimately converted to spin in that case.

*Because most of these system result from condensation of a rotating dust cloud, which is where they got their angular momentum from originally.

 

Hence Phobos, yes. It's orbital frequency > Mars spin, so orbital energy is transferred to Mars and Phobos is falling as a result, which makes it orbit even faster, not slower.

Eventually everything gets tide locked, so there's no more spin to add energy to orbits. Any further loss of energy to the system is thus going to take from orbital energy, and a lower orbit must increase spin. So assuming Earth isn't destroyed, the moon will orbit once every ~1500 hour day, and after that it will start falling, which shortens the day, until it's back to a single digit of hours. We're talking a long time for this to happen.

Yes, because the moon/Earth are not yet tide locked, so there's still spin energy being put into orbit energy. That is a finite resource, so it has to eventually end, after which the moon falls. That's why I said it must eventually come to that.

Yes, but what happens after that? There's still energy being removed from the system due to solar tides, and orbital energy is the only thing to take it from, so the orbit drops, which increases the spin of both objects.

 

Oh I see, so now you are invoking the tidal effects from the sun as well. I had not thought about that.