I am experimenting with an idealized two-body system based on the Sun-Earth system. I want to calculate how much Earth's axis precesses and nutates given the simplified setup.

In my system, the orbit of the Earth about the Sun is a perfect circle. The sun is centered on the origin of the reference frame. At time $t = 0$, the Earth is located on the positive $x$-axis. Earth's axis of rotation is tilted 22.5° from vertical about the line of the equinoxes, which at $t = 0$ is parallel to and aligned with the $x$-axis. There are no other bodies in the system; the only thing gravitationally acting on the Earth is the Sun. The Sun is a perfect sphere, while the Earth is a spheroid with an equatorial radius of 6371 km and a polar radius of 6356 km.

Wikipedia gives this equation for the torque caused by a celestial body's gravity acting on the Earth:

$$\vec{T} = \frac{3GM}{r^3} (C − A) \sin(δ) \cos(δ) \begin{pmatrix} \sin(α)\\ −\cos(α)\\ 0\\ \end{pmatrix}$$


  • $GM$ is the standard gravitational parameter, the product of the gravitational constant $G$ and the mass $M$ of the perturbing body;
  • $r$ is the distance between the center of the Earth and the center of the perturbing body;
  • $C$ is the moment of inertia around Earth's axis of rotation;
  • $A$ is the moment of inertia around any equatorial diameter of Earth;
  • $(C − A)$ is the moment of inertia of Earth's equatorial bulge (C > A);
  • $δ$ is the declination of the perturbing body (positive for north of the equator, negative for south of the equator); and
  • $α$ is the right ascension of the perturbing body (east from vernal equinox)

I have values for all these variables, making the equation for the sun:

$$\vec{T_s}(t) = 6.916 × 10^{44} \frac{kg \: m^2}{s^2} \sin(δ_{s}(t)) \cos(δ_{s}(t)) \begin{pmatrix} \sin(α_{s}(t))\\ −\cos(α_{s}(t))\\ 0\\ \end{pmatrix}$$


  • $δ_{s}(t) = \arcsin(-\sin(t) \sin(π/8))$
  • $α_{s}(t) = \arctan(\tan(t) \cos(π/8))$, adjusted to cover the range [0, 2π)

Wikipedia says that the $y$ component of the vector averages to zero and can be neglected, leaving only the $x$ component. In my simplified system, the average value of the $x$ component of the vector should be

$$T_x = \left(\frac{3}{2}\right) \left(\frac{GM}{r^3}\right) (C-A) \sin(π/8) \cos(π/8)$$

Which works out to

$$T_x = 1.546 × 10^{22} \frac{kg \: m^2}{s^2} \sin(π/8) \cos(π/8)$$

Then Wikipedia says precession is

$$\frac{dψ}{dt} = \left(\frac{3}{2}\right) \left(\frac{GM}{r^3}\right) \left(\frac{C - A}{C}\right) \left(\frac{\cos(π/8)}{ω}\right)$$

Which works out to

$$\frac{dψ}{dt} = 2.46623 × 10^{-12} rad^{-1} s^{-1}$$

Which somehow converts to arcseconds per year.

That gives an average value, but is there a way to get the precession over time more exactly, rather than as an average?


4 Answers 4


As pointed out in the two already submitted answers, even when simplified to the minimum of circumstances that will give rise to precession the calculation will already be quite complicated.

It's not the kind of case you will find in an physics textbook. Physics textbooks will present how to set up calculations for cases such as a double pendulum.

It is of course the case that the motion of the Earth must be known exactly in order for the GPS system to achieve its optimal accuracy. The engineers that are optimizing the GPS technology are using very accurate models.

Those kind of models use the moment of Inertia of the Earth as input. Specifically: the difference between two moments of inertia:
Moment of inertia around Earth's axis of rotation
Moment of inertia around any equatorial diameter of Earth Wikipedia: the section equations of the wikipedia article about the Earth's gyroscopic precession.

The torque that the Sun exerts on the Earth is not constant, it depends on the angle of the Earth's equatorial plane relative to the Sun.

At the equinoxes the line of intersection of Earth-equatorial-plane and Earth-orbital-plane is oriented towards the Sun. Hence at equinox torque from the Sun is zero.

At the times of the year halfway between the equinoxes the torque from the Sun is maximal.

At all times that a torque from the Sun is acting it is acting to try and bring the Earth equatorial plane in alignment with the Sun.
That is: the effect of the torque does not reverse. During summer or during winter, during both times the torque acts to try and bring the Earth equatorial plane in alignment with the Sun.

As we know, the torque on the Earth from the Moon is stronger than that of the Sun, as the Moon is so much closer. The torque from the Moon acts to try and bring the Earth equatorial plane in alignment with the Moon.

As a first approximation you can treat the Moon as moving in the same plane as the Earth's orbit.

I'm guessing that as a first approximation you might try to use for the total torque just the average over time of the cyclicly changing torque

The period of the Earth's precession is about 26 thousand years, much longer than the year cycle of the torque exerted by the Sun.

I'm guessing that before electronic computers became available anybody who tried to actually calculate the Earth's rate of precession did so using the first approximation as described above. I very much doubt anybody went beyond that first approximation.

For the Earth precession the influences from celestial bodies other than the Sun and Moon are totally insignificant.

The reason why the Earth experiences a torque from other celestial bodies:

For a vivid picture we exaggerate the Earth's equatorial bulge, to the point of visualizing a celestial body in the form of a flat disk

The center of inertial mass of that disk is at the geometric center. The center of gravitational attraction to another celestial body is not at the geometric center. Gravity falls off with the square of distance. This means that the half of the disk that is closest to the other celestial body experiences more gravitational force than the half furthest away from the other celestial body. Hence the center of gravitational attraction to the other celestial body is somewhat away from the geometrical center, along the line that stretches between the two celestial bodies.


I think it is useful to have a simplified calculation to understand how the period of precession of the earth depends on the tidal forces of Moon and Sun.

I made the steps:

  1. Subtract the mass of the earth from the mass of a sphere with the smaller radius (from pole to pole), and same density. This mass is supposed to be distributed along a ring over the equator.($m = 4.5\times 10^{22}$kg)

  2. Calculate the sum of tidal force of Moon and Sun as distributed along the ring $\frac{GM\delta}{r^3}$, where $r$ is the average distance from Sun or Moon to the earth, and $\delta$ is the distance between 2 opposing points of the ring.

  3. Calculate the average lever arm for the torque on each point of the ring, considering that each quarter of a year the angle of inclination ($\phi$) varies from zero to $23.4^\circ$.

  4. Integrating $F(\theta).b(\theta,\phi)$ over the ring, where $\theta$ is the angle from the center to the ring, for Sun and Moon and add them. $\tau = 1.89 \times 10^{22}$Nm

  5. From the moment of inertia of the earth $(8.034 \times 10^{37})$ calculate its angular momentum (supposing only the daily rotation and no precession).

  6. Divide total torque by the radial component (projection on the ecliptic plane) of the earth angular momentum of (5). That is the velocity of precession $\omega_p = 8.16\times 10^{-12}$ rad/s.

  7. From the numbers of seconds per year, that gives $24412$ years for a complete turn, which is reasonable (due to the approximations, mainly the step 1) of the standard value of $26000$ years


Reason for Precession

Its important to understand why earth precesses the way it does and then start of by deciding whether the earth in your model will precise at all.

Here is a paragraph that I decided to viciously copy from here.

The cause of the precession is the equatorial bulge of the Earth, caused by the centrifugal force of the Earth's rotation (the centrifugal force is discussed in a later section). That rotation changes the Earth from a perfect sphere to a slightly flattened one, thicker across the equator. The attraction of the Moon and Sun on the bulge is then the "nudge" which makes the Earth precess.

To explain it a bit more, owing to the fact that -

  1. Earth is not perfectly spherical and
  2. The sun and moon are not always in the same plane

hence at some point in time (or maybe always) there will exist a torque perpendicular to the earth's axis of rotation, as the moon is out of plane and so attracts the bulge that's closer to it more than the bulge away.

Here is a picture, to help visualize the out of plane trajectories. Image made using Geogebra. ![Geogebra Image: Trajectory

To Summarize

So to have precision you have a
i. a distorted sphere (either in terms of non uniform density or in terms of the actual shape)
ii. The moon has to go out of plane of the trajectory of earth (assuming the distortion is equatorial). If the distortion is along the axis of rotation of earth itself (bit unphysical I guess), then a moon need not be out of plane.

  • $\begingroup$ It is called precession, not precision. $\endgroup$ Commented Jul 7, 2021 at 18:51
  • $\begingroup$ Ohh sorry, I have corrected it. My voice typing gets very bad at times. $\endgroup$ Commented Jul 7, 2021 at 18:55
  • $\begingroup$ Afaik iframe does not work on the SE. Probably on security reasons (could be used for fishing attacks). $\endgroup$
    – peterh
    Commented Jul 7, 2021 at 19:00
  • $\begingroup$ In my model the Earth is a distorted sphere with a tilted axis, as I say in my question, so I expect there will be some precession. How do I calculate how much precession there will be? $\endgroup$
    – Lawton
    Commented Jul 7, 2021 at 19:03
  • $\begingroup$ If the bulge is equatorial then you have to have both the trajectory out of plane for precession to happen, otherwise the pull on upper and lower hemisphere is exactly equal nullifying the torque. Given that, Actually calculating the values is a whole another task. I have some ways of going about it. I will update the answer when I figure out which ones work. $\endgroup$ Commented Jul 7, 2021 at 19:12

I'm thinking that even with your simplifications this is a very complex problem. It is my understanding that the torque responsible for precession results primarily from the moon pulling more strongly on the nearside (tipped) equatorial bulge. This is a variable torque which goes to zero as the moon crosses the equatorial plane. The sun would have a similar effect with a smaller amplitude and longer period. Variations in distances would also be a (small) factor. Your first problem would be finding a center of gravity (and effective mass) for each side of the bulge.

  • $\begingroup$ Do you have any ideas for how to go about finding the centers of gravity and mass for the bulge, as you put it? $\endgroup$
    – Lawton
    Commented Jul 7, 2021 at 19:06
  • $\begingroup$ You might see what the oblateness tells you about the shape of the bulge. $\endgroup$
    – R.W. Bird
    Commented Jul 8, 2021 at 14:30
  • $\begingroup$ What do you mean by "see what the oblateness tells you about the shape of the bulge"? $\endgroup$
    – Lawton
    Commented Jul 8, 2021 at 14:50
  • $\begingroup$ How thick is the bulge? Is there an equation for the non-spherical shape? What about variations in density? $\endgroup$
    – R.W. Bird
    Commented Jul 8, 2021 at 17:29
  • $\begingroup$ The equation I'm using for the Earth-spheroid is $x² / 6372² + y² / 6372² + z² / 6349.875² = 1$, making the equator 22.125 km farther from the center of the Earth than the poles. For density I'm using the Preliminary Reference Earth Model (PREM): en.wikipedia.org/wiki/File:RadialDensityPREM.jpg $\endgroup$
    – Lawton
    Commented Jul 8, 2021 at 18:23

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