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

• Gaba_p
In summary, when considering the Earth's precession, the tilt of its rotation axis and its asymmetry are important factors. The Sun's gravitational force gradient may not have a significant effect on precession compared to the luni-solar precession. The precession rate is not dependent on the tilt angle for small angles, but it becomes lower as the angle increases. When analysing precession, it is necessary to include the tilt angle in the model.
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!

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.

Last edited:
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!

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.

Filip Larsen said:
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?

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.

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:

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

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.

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

Filip Larsen said:
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!

## 1. What is Earth's precession?

Earth's precession is a slow and continuous change in the orientation of its rotational axis. This results in a gradual shift of the Earth's rotational axis in a circular pattern over a period of approximately 26,000 years.

## 2. How is the Sun's torque related to Earth's precession?

The Sun's torque is one of the main factors that contribute to Earth's precession. The gravitational pull of the Sun on the Earth's equatorial bulge causes a torque, or twisting force, that acts on the Earth's axis of rotation.

## 3. What are the causes of Earth's precession?

Aside from the Sun's torque, there are several other factors that contribute to Earth's precession. These include the gravitational pull of the Moon and other planets, the Earth's elliptical orbit around the Sun, and the distribution of mass within the Earth itself.

## 4. How does Earth's precession impact our planet?

Earth's precession has several effects on our planet, including changing the positions of the stars in the night sky over time, altering the length of the seasons, and affecting the Earth's climate patterns.

## 5. How is the study of Earth's precession relevant to our understanding of the universe?

The study of Earth's precession is important for understanding the movements and interactions of celestial bodies in our solar system and beyond. It also helps us to better understand the history and evolution of our planet and its place in the wider universe.

• Classical Physics
Replies
10
Views
1K
• Classical Physics
Replies
3
Views
669
• Classical Physics
Replies
4
Views
723
• Astronomy and Astrophysics
Replies
7
Views
956
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
33
Views
950
• Classical Physics
Replies
4
Views
2K
• Mechanics
Replies
10
Views
1K
• Classical Physics
Replies
4
Views
4K
• Special and General Relativity
Replies
5
Views
3K