The Wobbling of the Earth and Other Curiosities

Ted Bunn's Comments

December 26, 1999

John Baez wrote:

>Jim Heckman wrote:

>> (John Baez) wrote:
>>> Are you also guessing that the wobble period doesn't depend on
>>> the wobble tilt, or something like that?  And is there some reason
>>> you're guessing this?

>>Hmm... Wobble tilt would indeed seem to be an initial condition that I
>>overlooked, wouldn't it? :-) 

>Right, but it's okay!  The period of the wobble is almost independent
>of the angle of tilt, as long as the angle is fairly small.  I just 
>wanted to see if you realized you were making this assumption, and 
>whether you knew how to justify it.  
>So: why doesn't the period of the wobble depend much on the angle of
Without actually working out the details again -- I have solved this problem before, but not for years, so I don't remember all the details -- I'd guess that one possible answer is based on the wonderfully useful fact that everything in nature is linear, to first order. When you work out the equation of motion for this system, you'll be able to do perturbation theory in some small quantity. The first-order equation will be the harmonic oscillator equation, since it always is for small departures from equilibrium, and we all know that the period of a harmonic oscillator doesn't depend on the amplitude.

Of course, this doesn't *prove* anything: I'm just guessing that the first-order equation of motion will be the harmonic oscillator equation, because that's what it always seems to be in these situations.

>I defined this to be the effect that occurs when you have a rigid 
>rotating body which has been set spinning around some axis other 
>than its principal axes of inertia.  And perhaps I should emphasize
>that this effect occurs even in the absence of external forces!
>In the theory of rigid rotating bodies, this effect is called
I always think of "precession" as being what a top does -- i.e., change in the axis of rotation due to an external torque. But I guess you're right, it's also the change in the direction of omega in the absence of torque. Goldstein, for example, uses the term for both.
>But in astronomy, the "precession" of the earth 
>refers to a *different* effect!  
>So: who knows what this other effect is?  
It's more like the precession of a top: it's a slow change in the direction of the Earth's rotation due to an external torque on it.
>Also: who knows what "nutation" means in the theory of spinning tops?
Say you have a spinning top. One point (the point that's touching the ground) is fixed. For a given amount of angular momentum, there's a natural angle for the axis of the top to be tipped at. If the top is at that angle, it'll just precess around the vertical axis at a steady rate. If on the other hand it starts off tipped at some other angle, it'll wobble up and down as it precesses. That wobble is called nutation.
>And who knows what it means in astrononmy, when applied to the motion 
>of the earth?
Pretty much the same thing. There's no fixed point now, but there's still a big spinning thing subjected to torques, and the motion turns out to be similar. That is, the direction of the Earth's axis slowly circles around the z axis (the line perpendicular to the ecliptic plane) -- that's precession -- and also wobbles up and down a bit as it goes -- that's nutation.
>And finally, just for the sake of a well-rounded education: what's 
>the period of the precession of the earth?  What's the period of 
>the nutation of the earth?
I can look those up if you want. Or did you want me to estimate them from first principles? That sounds hard. If I recall correctly, the precession period is 26,000 years. The nutation period is probably the same order of magnitude, but I don't really know. There's probably some relatively simple relation between the two, at least when the nutation amplitude is small, but I don't know what it is.
>Okay, now for another criticism of your proposed solution.  You wrote:

>>Well... On dimensional grounds alone, it seems like it would almost
>>*have* to be ~( I1 / (I1-I2) ) times the period of the earth's rotation...
>>(I'm *guessing* that the wobble period -> infinity rather than -> 0 as
>>earth -> spherical.)

>First it sounds like you're claiming that since the answer has units of
>time and it depends only on the earth's period of rotation, I1, and 
>I2, it must some dimensionless function of I1 and I2 times the
>earth's period of rotation.  And I think that's right - at least after 
>you give an argument explaining why the answer doesn't depend much
>on the earth's tilt.  
>But then it sounds like you're claiming that the only dimensionless
>functions of I1 and I2 are I1/(I1-I2) and its reciprocal.  But that's
>not right.  I can think of all sorts of other possibilities
Oh, I think he was just making the simplest guess that had the right asymptotic behavior, not asserting that it was really the only possibility. Guesses like that are right a surprisingly large percentage of the time!
>So.... what's the right dimensionless function of I1 and I2? 
I'm not sure I can justify Heckman's guess without doing at least a little bit of work. I get the impression from the way you stated this puzzle, and from your later comments, that you're looking for some sort of intermediate level of analysis, more rigorous than Heckman's guess but less messy than a full calculation. I don't immediately see how to justify the guess that the relevant small quantity is (I1-I2)/I1 without doing a slightly messy calculation. Do you have such a way? I'd be interested in seeing it.

In any case, the calculation's not that messy, and it beats grading final exams, so here it is. Let's write the angular velocity as

w = w0 + w1,

where w0 points in the direction of the Earth's axis and w1 is a small perturbation to it. w0 is fixed; w1 varies with time.

(Digression: I can't convince my intro physics students to pronounce the symbol for angular velocity "omega" rather than "double-you"; now here I go exacerbating the problem. But I find lots of Greek letters in Usenet posts distracting. Even if I did call it "omega," I wouldn't put a backslash in front of it. I like TeX as much as the next person, but Usenet posts are for human readers, and I have confidence that a human sci.physics.research reader can tell that the character string "omega" is supposed to represent a Greek letter without the help of a backslash.)

We want to write down an equation for the rate of change of w. Remember that, for any vector V,

(dV / dt)_body = (dV / dt)_space + w x V,

where the two derivatives refer to the rates of change of the components of the vector in coordinate systems fixed in the body and fixed in space. I may have the sign of the w x V term wrong; I don't care. Let V = L, the angular momentum. Then the first term on the right is zero, since angular momentum is conserved. So, in the body's coordinate system,

dL / dt = w x L.

Say w0 points in the z direction and w1 points instantaneously in the x direction. Then the dominant terms on the right (that is, the ones with the fewest powers of the small quantity w1) are have magnitudes w1 (I1 w0) and w0 (I2 w1) and point in the y direction. (I1 is the large moment of inertia, about the z axis, I2 is the other moment. I don't remember if that was the original notation or not, but I've hijacked the notation, and that's what it is now.) There must be a relative minus sign between those two terms. Again, I don't care to work it out. So to first order, the right-hand side is (I1-I2) w0 w1 and points in the y direction.

The dominant term on the left is I1 dw1/dt, since w0 is constant in time. So dw1 / dt has magnitude w0 w1 (I1-I2)/I1 and points perpendicular to w1. The solutions to that equation have w1 precessing about the z axis with angular frequency w0 (I1-I2)/I1, just as Heckman guessed.


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© 1999 John Baez