Derivation of Earth's Precession: Sun's Torque & Causes

[Gaba_p]

Hi,

anybody knows of a book/article/source of any kind where I could check an actual rigorous derivation of the Sun's torque acting upon Earth?
There's a couple of things I don't fully comprehend, maybe someone here could help me.

1- Why is it that if Earth would be a perfect sphere there would be no precession? I mean, Earth's axis is still tilted, wouldn't Sun's gravitational force try to line it up with the ecliptic?

2- What are the causes of precession? I know that the 23º tilt and Earth's asymmetry are actual causes but, is the Sun's gravitational force gradient also a cause?

Any help much appreciated.

Cheers!

 

[Filip Larsen]

My edition of Goldstein [1] has reasonably thorough derivation in his chapter on general rigid body dynamics. I assume this also is included in later editions.

Goldstein gives the value of the solar precession, that is, the angular speed with which the Earth rotation axis is moving with respect to the the sun due to the gravitational effect from the sun, as

\frac{\dot{\phi}}{\omega_0} = -\frac{3}{2}\frac{\omega_0}{\omega_3}}\frac{I_3-I_1}{I_3}\cos \theta

where \omega_0 is the orbital angular speed of the Earth around the sun, \omega_3 is the angular speed of the Earth's rotation, \theta the obliquity of the ecliptic (the tilt of Earth's rotation axis), I_1 is the moment of inertia of the Earth around an axis in the equatorial plane, and I_3 the is the moment of inertia around the rotation axis. Note, that this is only for the solar precession; I assume a similar expression exist for the precession effect from the moon and planets.

Given that the Earth rotates in space and around the sun, the existence of the precession (i.e. the reason it is non-zero) can by seen as originating from the tilt of the rotation axis and that Earth has a bulge around equator. If either of these weren't there, that is, if the obliquity of ecliptic was zero or the Earth was an inertially perfect symmetric sphere (I_3=I_1) then there would be no precession. Note, that if the obliquity is zero the precession is zero not because the rate, as given by above equation, is zero but because the rotation axis precess around an axis normal to the ecliptic and at zero obliquity the rotation axis is then already normal to the ecliptic.

Regarding the gravity gradient (tidal force) from the suns gravity, I do not believe it has any significant effect on precession, at least not compared to the luni-solar precession. Gravity gradients are usually mostly significant in the coupling between orbital and rotational energy of a moon circling close to a planet (like with our own moon), often ending up with the moon being in "tidal lock" with its orbital rotation.

[1] Classical Mechanics 2nd, Herbert Goldstein, Addison-Wesley 1980.

 

[Gaba_p]

Of course! Goldstein!

Filip you've been a tremendous help. It really bugged me that many of the internet sources I found about the topic claimed that one of the causes of precession was the stronger gravitational pull from the Sun on the side of the bulge closer to it. Here's an extract of what's in spanish written Wikipedia:

But the Earth is flattened at the poles and the attraction of gravity varies as the square of the distance between the masses. The consequence is that the attraction of the sun on the equatorial swelling is a little stronger on the side nearest the sun than on the far side.

From what I read from Goldstein's derivation, this effect is not relevant regarding the precession phenomenon (please correct me if I'm wrong)

One thing that still bothers me though: if Earth's axial tilt was 0 (\theta = 0), shouldn't the precession rate (\dot{\phi}) also be zero?

Thanks!

 

[Filip Larsen]

The angle between the rotation axis and the axis it precess around is \theta, so when \theta = 0 there is no precession because the two axis coincide. As you can see from the equation I quoted from Goldstein, the rate does not go to zero when \theta = 0; in fact, the rate has a maximum for \theta = 0.

But even if \theta = 0, then this would only cancel the solar precession. The lunar precession would still be present since the moons orbital plane around Earth is not coplanar with the ecliptic.

 

[Gaba_p]

As you can see from the equation I quoted from Goldstein, the rate does not go to zero when \theta = 0; in fact, the rate has a maximum for \theta = 0.

Well, that's exactly what I don't get. Shouldn't that equation reduce to \dot{\phi}=0 for \theta = 0? I mean, what does it mean to have a maximum of precession when there's no tilt to precess around?

 

[Filip Larsen]

You should probably consider the precession rate as a "natural" rate that is independent of the tilt angle for small angles, and only when the angle becomes larger it begins to have the effect of lowering the precession rate.

 

[Gaba_p]

Ok, that's one way to see it.
But what if the tilt was zero from the beginning? Would this equation make any sense?
Should I (making the same assumptions and calculations but taking \theta = 0 from the start) arrive at a similar expression but without the angle? Ie:

<br /> \frac{\dot{\phi}}{\omega_0} = -\frac{3}{2}\frac{\omega_0}{\omega_3}}\frac{I_3-I_1}{I_3}

Thanks for your patience.

 

[Filip Larsen]

That expression would describe the solar precession rate as the angle goes toward zero, yes.

When analysing precession you would always include the tilt angle in your model (however small its value then is in practice). Modelling precession with a fixed zero angle would not make much sense, I guess. It would be like analysing how a ball would roll down a slope with the constrain that the slope is horizontal.

 

[Gaba_p]

I've re-read what you wrote earlier and found the part you edited in:

Note, that if the obliquity is zero the precession is zero not because the rate, as given by above equation, is zero but because the rotation axis precess around an axis normal to the ecliptic and at zero obliquity the rotation axis is then already normal to the ecliptic.

That's what I was aiming at. Thank you very much!

Cheers!