Two questions about the moon

[Farrago]

About the craters at the south pole, where the sun never shines: Have the moon's poles always been about where they are now? I understand the that the earth's north and south poles (relative to the ecliptic) move quite a distance through millions of years, and that they haven't always pointed to Polaris. Do the moon's poles also move relative to the sun's equator? Does the moon have a north star?

My other question is, how long has it been since a theoretical observer on earth could see the lunar far side? That is, how long has the moon been tidally locked? Could a dinosaur have seen the far side, or would you have to go further back than that? I understand that the moon was closer to the earth; did it also rotate faster?

I was wondering if either a wobble in the lunar poles, or the speed of the moon's rotation would matter, as to the idea of permanently shadowed craters.

Thanks for your input!

[grant hutchison]

The moon's axis precesses quite quickly, with an axial tilt of about one and a half degrees (IIRC) relative to the ecliptic plane. The axial precession takes place with the same period the lunar orbit precesses on the ecliptic plane: 18.6 years.
So an observer on the moon would see the moon's rotation pole move in a small circle around the ecliptic pole, once every 18.6 years.
The small tilt of the moon's axis means that polar craters don't need to be very deep before they're able to create permanent darkness at their bases.

The moon would have spun down to tidal lock quite quickly after it formed, in geological terms. But there are suggestions that it has flipped its nearside and farside more recently. We discussed this somewhere on BAUT, fairly recently: let me see if I can find the link ...

Edit: Here's the original paper: Did a large impact reorient the moon? The suggestion is that one of the early large impacts may have caused the moon to switch its near and far sides. But it seems to have been tidally locked in its current position for more than 3.5 billion years.

Grant Hutchison

[forrest noble]

Farrago,

About the craters at the south pole, where the sun never shines: Have the moon's poles always been about where they are now? I understand the that the earth's north and south poles (relative to the ecliptic) move quite a distance through millions of years, and that they haven't always pointed to Polaris. Do the moon's poles also move relative to the sun's equator? Does the moon have a north star?

On Earth we have plate tectonics where continents including the south pole have shifted around drastically over the eons, the Earth's crust floating around on a molten interior. The moon appears to have been solid and relatively dormant for a long time. Its surface features remain in the same relative position. The moons small axial tilt (relative to the sun) does not allow sunlight to enter the deepest craters at either pole. This is our good fortune since the moons surface in known to be very old, and meteorites and comets have been striking it since its beginning. Some of this cometary/ meteor residue seems to have left relatively large water-ice deposits in deep craters at the south pole according to recent discoveries.

Because the moon is believed to have been created from much of the same material as the Earth it is still possible that there are also underground water in aquifers, like on Earth, that did not boil off in the moons beginnings presently believed to have been created by a Mars sized planetary impact with the early Earth. We could also manufacture water from plentiful hydrogen and oxygen supplies chemically combined with surface materials.

Since most stars are relatively stationary compared to orbiting moons and planets, the moon as well as other planets with a small wobble (libration and/or precession), would have its own North star, but it may not be as bright or as easily identifiable as Polaris, nor as readily apparent by not moving in a background of "moving stars," since night-time lasts more than two weeks.

[grant hutchison]

The moon doesn't have particularly impressive pole stars.
Its north pole wanders in a circle three degrees across, tucked into the curve of the constellation Draco, centred on the patch of rather dark sky between ω Dra, 27 Dra, 36 Dra and 42 Dra.
Its south pole covers a similar circle in Dorado, centred between ε Dor and η¹ Dor.

Grant Hutchison

[Farrago]

Thanks for the information, Grant. This is very interesting, that the moon has been locked with us almost since its creation. Do you know, will it stay locked, as its orbit gets farther from earth over the next few billion years?

[grant hutchison]

Thanks for the information, Grant. This is very interesting, that the moon has been locked with us almost since its creation. Do you know, will it stay locked, as its orbit gets farther from earth over the next few billion years?

Yes, the moon should continue to be tidally locked during the future tidal evolution of the Earth-Moon system.

Grant Hutchison

[neilzero]

Hi Farrago: Grant is correct. The tide lock will likely resume quickly even after a Ceres mass impact. The Earth will tide lock with the moon billions of years in the future, after which the moon will approach Earth for billions of years until the moon is well inside the Roche limit. The moon will likely then disintegrate into rings somewhat like Saturn's rings. Shortly after disintegration, few of the fragments will be tide locked. Neil

[grant hutchison]

The tide lock is likely stable enough to hold even with a Ceres mass impact.

Not so.
Wieczorek & le Feuvre actually did the calculations, and showed that impacts by objects an order of magnitude smaller than Ceres would be sufficient to disrupt the current synchrony. For the most likely impact velocities, and across a range of likely densities, an object between 50 and 100 km in diameter would do the trick. (See their Figure 1).

Grant Hutchison

[AndreasJ]

The Earth will tide lock with the moon billions of years in the future, after which the moon will approach Earth for billions of years until the moon is well inside the Roche limit. The moon will likely then disintegrate into rings somewhat like Saturn's rings. Shortly after disintegration, few of the fragments will be tide locked.

That is, unless the Sun destroyed the Earth and Moon during it's red giant phase first. The last word I read on that subject (which need not be the last one there is to read) is that that will happen, and the Earth thus will never get the time to ge tidally locked with the Moon.

[neilzero]

Hi Grant: Thanks for the interesting link. Do you think the tide lock is likely to resume in weeks following a large impact on the moon, perhaps with larger liberation? Neil

[grant hutchison]

Do you think the tide lock is likely to resume in weeks following a large impact on the moon, perhaps with larger liberation?

Suppose an impact today spun the moon up to a sidereal rotation period of 26 days. If we plug the parameters of the current Earth-Moon system into the standard tidal evolution formula, the spin-down time to synchrony is on the order of 10,000 years. Faster rotations give correspondingly longer spin-down times.
I've no idea how much of a change a single impact could plausibly make to the Moon's rotation period. Presumably we could ball-park that by comparing the necessary change in energy with the Moon's gravitational self-energy. Impacts approaching that limit would tend to disrupt the Moon rather than change its rotation.

Grant Hutchison

[neilzero]

Hi Andreas: My guess is lots of factors can determine the maximum radius of a red giant sun (a neutron star impact of the sun shortly before max radius which forces fresh hydrogen into the core, followed by quick departure of the neutron star?) If so, there is a slight possibility that a scorched Earth (not vaporized) will orbit about two AU from a cooled white dwarf, a trillion years from now. If so, Earth will have rings somewhat like Saturn? If so, will Earth's orbit be approximately circular?

[neilzero]

Hi Grant: How about the reverse? Slight spin down, instead of spin up, as a result of the impact? Do small impacts increase (or decrease) the liberation? Neil

[grant hutchison]

Hi Grant: How about the reverse? Slight spin down, instead of spin up, as a result of the impact?

The time to synchrony depends only on the difference between the current rotation rate and synchrony, not on the sense of the difference.

Do small impacts increase (or decrease) the liberation?

Libration?
Presumably impacts too slight to break the moon out of synchrony will increase physical libration, for a while.

Grant Hutchison

[George]

Yes, the moon should continue to be tidally locked during the future tidal evolution of the Earth-Moon system

Just speculating, but I wonder if this will be the case in the long run, assuming (in a big way) that much of the Earth's angular momentum is transfered to the Moon? Would the tidal locking strength be adequate to hold it all the way to this distant, and imaginative, orbit?

[grant hutchison]

Would the tidal locking strength be adequate to hold it all the way to this distant, and imaginative, orbit?

Not that distant: the Earth reaches synchrony with the Moon when the Moon has receded to just ~1.4 times its present distance. At that point the Earth will rotate, and the moon revolve, in 47 days; the Moon's tidal recession will then cease.
The tidal force on the Moon at that time will be reduce by a factor of 1.4³ = 2.7, so it seems it will still be reasonably firmly controlled.

Grant Hutchison

[George]

Not that distant: the Earth reaches synchrony with the Moon when the Moon has receded to just ~1.4 times its present distance. At that point the Earth will rotate, and the moon revolve, in 47 days; the Moon's tidal recession will then cease.
The tidal force on the Moon at that time will be reduce by a factor of 1.4³ = 2.7, so it seems it will still be reasonably firmly controlled.

Ah, I muffed it. I was wrapped-up in estimating a simple coefficent for the moment of inertia for the Earth, that I failed to convert rotational velocity to angular velocity and it put me off by about 6 orders of mag. Nuts! [I suppose I should have been a little suspicious of my math when I noticed the Moon was floating out there at 100,000,000,000 km orbital radius after absorbing the Earth's ang. momentum. I also failed to check my units. ]

Recrunching with ang. vel., I see that I am close to your 1.4x orbital distance figure, with the difference likely being in my down-n-dirty coefficient.

[Hornblower]

Just speculating, but I wonder if this will be the case in the long run, assuming (in a big way) that much of the Earth's angular momentum is transfered to the Moon? Would the tidal locking strength be adequate to hold it all the way to this distant, and imaginative, orbit?

We can take the step from speculating to estimating. See this Wiki article.
http://en.wikipedia.org/wiki/Orbit_of_the_Moon
Scroll down to Tidal Evolution of the Lunar Orbit.

As usual, don't take Wiki as gospel, but this one is in good agreement with what I read in an astronomy textbook decades ago, and with my own rough extrapolation from the present-day values for the rates of change of the Earth's rotation rate and the Moon's orbital radius.

A lunar orbit period of 47 days means a radius only about 50% greater than now. That should be close enough to keep the Moon tidally locked.

Suppose the Earth and Moon have survived the red giant flareup and are orbiting the remaining white dwarf. The Sun, even with its reduced mass, will continue slowing the Earth's spin down, with the result that the Moon will overrun the rotation and start spiralling in. In enough tens of billions of years, it should come inside the Roche radius and create stupendous rings.

[grant hutchison]

See this Wiki article.
http://en.wikipedia.org/wiki/Orbit_of_the_Moon
Scroll down to Tidal Evolution of the Lunar Orbit.

As usual, don't take Wiki as gospel ...

Indeed. The author of that section has mixed sidereal and synodic months. The Moon currently "circles the Earth" in about 27 days, not 29. It's the former figure which should be compared with the 47-day period for the double synchronous state.

Grant Hutchison

[George]

We can take the step from speculating to estimating. See this Wiki article.
http://en.wikipedia.org/wiki/Orbit_of_the_Moon
Scroll down to Tidal Evolution of the Lunar Orbit.

That's a nice page.

As usual, don't take Wiki as gospel,.. .

Yes, and here's a good example....

Earth's day thus lengthens by about 17 microseconds every year. (This would make each Earth day one second longer every 60,000 years or so, and one minute longer every four million years). Looking back, the day was a mere 23 hours in length when the dinosaurs roamed the Earth 65 million years ago.)

65 million/4 milllion would yield a 16 minute change, not near an hour. [I'm startin' to feel a little better. Thanks ]

A lunar orbit period of 47 days means a radius only about 50% greater than now. That should be close enough to keep the Moon tidally locked.

Yes, that is about right, though I get about 50 days using the dual mass equation for orbital periods. [I could be wrong; this is not my day on many levels.]

Suppose the Earth and Moon have survived the red giant flareup and are orbiting the remaining white dwarf. The Sun, even with its reduced mass, will continue slowing the Earth's spin down, with the result that the Moon will overrun the rotation and start spiralling in.

Hmmm. I see that if we have the Sun swell out to about a 120,000,000 km (bye Venus), that it will be rotating at this point at about 10 meters per sec. (11.6 mps using only today's equitorial conditions), which is a snail's pace and would tend to slowly draw Earth toward it (tidal interaction), ignoring all the drag from the Solar blow-off.

But when the Sun is a white dwarf....[but it's already....ok, nevermind ], a degenerate white dwarf, it would have much greater rotational speed and would cause the Earth to move out, right? What am I missing?

[Added: Ok, I read it too quick. Yes, assuming the Earth is close enough to the white dwarf, what you stated makes sense.]

[AndreasJ]

Hi Andreas: My guess is lots of factors can determine the maximum radius of a red giant sun (a neutron star impact of the sun shortly before max radius which forces fresh hydrogen into the core, followed by quick departure of the neutron star?)

A collision between the red giant Sun and a neutron star is extremely unlikely, and I don't see how it would force hydrogen into the solar core, or how that would stop the Sun swallowing the Earth. One'd think the associated pyrotechnics of such a collision would increase the chance of the Earth being fried. Also, something as massive as a neutron star (more massive than the Sun) passing through the system is going to have an affect on the Earth's orbit.

If so, there is a slight possibility that a scorched Earth (not vaporized) will orbit about two AU from a cooled white dwarf, a trillion years from now.

Something massive passing near the Sun and lifting the Earth into a higher orbit and thus saving it from engulfment is possible. It particularly being a neutron star that actually hits the Sun and then zooms of to infinity doing it is vanishingly unlikely.

It's also possible the calculations I refered to are somewhat off and the Earth isn't on schedule to be swallowed in the first place.

If so, Earth will have rings somewhat like Saturn? If so, will Earth's orbit be approximately circular?

Once the Moon is broken up, Earth will have rings for a while. The material will eventually be dragged down by continued interactions.

Whether Earth's orbit will be approx. circular depends how we got there.

[chornedsnorkack]

What are the periods of free libration for small amplitudes? What are the current amplitudes of free libration of Moon, and has any dissipation been observed?

Also, Moon would presumably have imaginary period of libration when turned at a right angle. How long period?

[grant hutchison]

Yes, that is about right, though I get about 50 days using the dual mass equation for orbital periods. [I could be wrong; this is not my day on many levels.]

What are you using for the moment of inertia of the Earth? Fifty days seems pretty close to what you'll get if you treat the Earth as a homogeneous sphere.
Plugging in appropriate values from Allen's Astrophysical Quantities (4th Ed.) I get 46.918 days, at a separation 1.435 times the present separation.

Grant Hutchison

[Farrago]

Getting away from the maths, which are way beyond me, are there a lot of other tidally locked moons in the solar system? Around Jupiter or Saturn?

[ShinAce]

There certainly are other cases of tidal lock. It's not unique to the earth-moon system. Pluto and Charon have a 'double lock' on each other.

A list is at the bottom of the wiki page:
http://en.wikipedia.org/wiki/Tidal_locking

You might also want to look at orbital resonances(mercury for example).

[grant hutchison]

A list is at the bottom of the wiki page:
http://en.wikipedia.org/wiki/Tidal_locking

In the "likely to be locked" table there are three Saturnian objects which were probably just transient ring clumps rather than moons: S/2004 S 3, S/2004 S 4 and S/2004 S 6 have not been recovered since the discovery images.
But we could replace them with three recently discovered inner moons of Saturn which aren't listed: Anthe, Aegaeon and S/2009 S 1.

Grant Hutchison

[George]

What are you using for the moment of inertia of the Earth?

I very roughly extrapolated from the fact the coefficient for a thin shell sphere is 2/3rds. and for a solid sphere it's 2/5ths, so I went to 1/5th for a heavy core and a lighter outer layer for the Earth -- crude, but a ball-park value is fine for me. This gives an angular momentum for the Earth of 3.54 x 10²⁷kg-km²-sec^-1. If this transfers to the Moon, it moves the lunar orbit out by 26% of its current orbit, 55% if I use the 2/5ths coefficient (solid body).

Fifty days seems pretty close to what you'll get if you treat the Earth as a homogeneous sphere.

I was using Hornblower's 50% orbital increase that produces the 50 day result, which is independent of the moment of Inertia, but in close agreement with what you just said [when I work backwards to get the matching angular momentums].

Plugging in appropriate values from Allen's Astrophysical Quantities (4th Ed.) I get 46.918 days, at a separation 1.435 times the present separation.

Yes, solving for period given both masses and orbital distance, that's very close to what I get and I'm using Wiki values with some rounding to boot. Yet if that's correct, then I get a moment of inertia coefficient of 0.32 which is between a solid body and a thin shell body, which doesn't make sense to me, assuming the Earth has greater density at the core. [Quickly added: But then again, 0.32 is less than 2/3rds and 2/5ths. Ug. ]