Question about axial precessions:

Earth's axial precession is rather well documented. I figured it's value is about -1.39 deg/cy. Does anybody know a source where I can find values of other planets and maybe even moons?

More generally:
Should the direction of (axial) precession not be determined by the rotation direction only (prograde, retrograde)? At least from conservation of angular momentum I would guess so. But if yes, rotation direction w.r.t. what? Ecliptic, Orbit, Invariable plane, ...?

Anybody any idea?

thanks,
Mike

Not sure if this will help, but here is a set of seven dialogues that lead up to a determination of the rate of the precession of the equinoxes.

Precession
In this set of dialogues I dispose of the idea that the precession of the equinoxes is caused by our Solar System being in orbit about Sirius and show that the Moon and the Sun account for more than 99 percent of it by actually calculating it. Accounting for 99 percent of the precession also disposes of precession caused by a companion brown dwarf as well.

Part One, Part Two, Part Three, Part Four, Part Five, Part Six, Part Seven.

Part Five contains a formula which I will rephrase here:

(3/2)*((C-A)/C)*G*(S/R_S³)*cos(eps)/w

Where G is the Newtonian gravitational constant, S is the mass of the Sun, R_S is the semimajor axis of the planet, eps is the obliquity of the planet relative to its orbital plane, w is the rotation rate of the planet (radians per second), and (C-A)/C is a ratio of moments of inertia for the planet. I believe this formula will work for the terrestrial planets but I'm not sure about the gas planets.

I am not sure about the direction; I'm going to have to give it some thought.

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Very nice essay indeed, CM! I was not doubting, that the precession was caused by a torque on the equatorial bulge, though.

One thing that puzzels me: for the torque you get an expression including sin(eps)*cos(eps), which makes sense to me. No torque for an axis "straight up" and also for "lying flat down". However you end up with only cos(eps) in your final precession rate. So you get a finite rate for a no-obliquity axis?

The cos(eps)/w part of your formula seconds my suspicion, that a retrograde rotating body (be it eps>90 deg or w<0) should have its axis "sweep the cone" counter-clockwise, right?
Say Venus, or does she have no precession at all due to the slow rotation?

Are there no values in the literature, at least for the major planets?

Mike

That looks good. Any elongated object will tend to align its long axis with the gravitational vector when there is a gravity gradient. Zero obliquity is the stable no-torque position. If we nudge it away from zero, it will tend to return. A coin lying flat on a table would be a good analogy.

At 90 degrees obliquity we have an unstable equilibrium, like that of a coin standing on edge. There is no torque at that position, but if we nudge it away a little bit, we get a small amount of torque that carries it away from 90 and toward zero.
The procession is with respect to a perpendicular coordinate axis. As I see it, the precession would be indeterminate at zero obliquity. I have not looked at the rest of the general formula.
Whenever the torque is tending to reduce the obliquity, the direction of precession will be opposite the spin direction. That is consistent with what I have observed with a toy gyro, and with what we learned in high school and college freshman physics.
For a given amount of gravitational torque, I would expect faster precession with a slower spin. The faster the spin, the more resistant the spinning body is to any attempt at changing its axial orientation. Once again, I have not checked out these calculations, but I know from observation how a toy gyro behaves.
I think computer simulations have been done for Mars, but I would have to search my collection of Sky and Telescope to find where I read it.

I would expect it to be difficult to observe it directly just by looking at the planets from afar.

Try Googling "Mars precession" and "Mars obliquity". That will turn up some interesting material on the precession rate and also the large variations in the amount of obliquity.

Fair enough. Faster spin -> axis more stable. That also agrees with the 1/w in CM's formula. But what about the transition to w=0? There should be no precession (perpendicular to the torque) whatsoever. Or is the formula in this case just not valid anymore?

Mike

At w=0, it becomes indeterminate. In the real world, no spin, no gyro action.

In theory, slow spin means rapid precession for a given amount of obliquity and torque. That is what a spinning top does. If the top is spinning too slowly when you let go of it, it simply will tumble instead of precessing. The equation as given does not easily handle the transient action at the start. It simply describes the steady state motion that continues once it is achieved by whatever means. If we spin it up to a high speed while holding it in an inclined position and then let go, it quickly settles into a stable slow precession.

Thank you. I didn't think you were one of the doubters, but periodically we get people who propose distant solar companions or very foolish ideas as the cause, so I now link to those dialogues as a resource whenever the question comes up.
Traditionally the rate of precession and the nutation in longitude are given in angular measurements of longitude. A degree of longitude at the equator subtends an angle of one degree, but a degree of longitude at co-latitude eps subtends an angle of 1*sin(eps) degree, hence the division by sin(eps).

It should be noted that as eps gets small it is possible for the loop from the largest periodic term in the nutation to be so big that it encircles the orbital pole. When that happens, the precessional rate of the rotational pole is then equal to the period of this term. Such a thing has happened to the Moon, so that its rotational pole precesses in the retrograde direction every 18.6 years, the same rate as the orbital pole. This causes the obliquity to have a particular value that is determined by the orbital inclination, the moments of inertia, and the precessional period.
I've had time to think about this a little more carefully, since (coincidentally) I've been looking over my earlier presentation about precession and I am contemplating another set of dialogues using a different method. Not only will I be able to re-derive the precessional rate, but I can also exhibit the main periodic terms of the nutation in longitude and nutation in obliquity, possibly say a few things about the "secular" parts of the precession and obliquity, and maybe even say something about the rotation of the Moon.

The most important thing to remember is the right-hand rule: the direction of the angular momentum vector is also the definition of "North" for a planet's rotational and orbital poles. I use n for a planet's angular rate of revolution and w for the angular rate of rotation, so that the planet's "revolution vector" is n*C where C is a unit vector pointing towards the (North) eCliptic pole, and the "rotation vector" is w*Q, where Q is a unit vector pointing toward the "North" pole (think eQuator). n and w are both positive, being the absolute values of their respective vectors.

Then C*Q = cos(eps), which is between 0 and 180 degrees, and is negative when one of them has a "north" pole in the "southern" hemisphere of the other.

Finally I define the unit vector X by QxC=sin(eps)*X. One then obtains the ecliptic coordinate system by choosing Z=C or the equatorial coordinate system by choosing Z=Q and completing the set of basis vectors with Y=ZxX.

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I have seen a few threads about that. Very funny, but what do they have to do with cash machines?
eps was the obliquity, so 90-latitude, right? Or is this what co-latitude means?
Is this when it locks into a Cassini state?
Fire away!
Although the IAU thinks otherwise.......

Correct. Co-latitude is 90-latitude in degrees. It runs from 0 to 180.
Yes.
I was afraid of that. I'll have to look it up.

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I have read through the above, and I follow most of it (particularly the worked example) except for the derivation of the constants C and A. On the third page it discusses the derivation of these constants. However, this derivation is unclear. One problem is the use of a constant m that doesn't seem to be defined anywhere. This is making their derivation hard to follow. In the worked example, these constants are obtained from a reference instead of being worked out.

On page 5, the figures given for C and A gave a result for the expression ((C-A)/C) roughly equal to 0.003293 (which for some reason is converted to 1/304). I couldn't help noticing that these figures are within 2% of the figure for the oblateness of the earth, 0.003353 [Wikipedia]. Given the nature of the figures for C and A, I'm sure these figures must be connected somehow.

What I would like to do is derive a general formula for use in a spreadsheet that gives rough estimates for the rate of precession for any terrestrial planet. The tricky part is these constants (C and A) which I lack the knowledge to derive from first principles. Ideally I would like to derive these from oblateness, equatorial radius and density because I have these figures on hand.

Thanks in advance for any advice.

Welcome to BAUTForum, Kingswood!

The constant m refers to the mass of the equatorial ring.
The 304 refers to the number of days in the free or Eulerian precession. The Earth is not rigid, however, so what is observed is a seemingly random wobble with a period of about 430 days called the Chandler Wobble.
That's OK; no one derives them from first principles because to do so would require detailed knowledge of the distribution of the Earth's mass. We have to content ourselves with measurements of these moments by other means.
I've gone looking for these quantities in two of my references, but I can only find J₂ and J₄ for the planets. Now J₂*M*R² is (C-A), where M is the mass and R is the equatorial radius. But I can't find C and/or A for anything other than the Earth and Moon. I would suggest Googling for +"moment of inertia" +Mars, etc. for each of the planets you're interested in.

Edited to add: I prefer the moments A, B, and C to be dimensionless constants, but others take them to have dimension kg*m², and I have followed that convention above.

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The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.