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I am trying (unsuccessfully) to acquire an intuitive understanding of planetary precession.

Is Earth’s precession Torque-free Precession due to its asymmetry as per https://en.wikipedia.org/wiki/Precession ?

If an object is asymmetric about its principal axis of rotation, the moment of inertia with respect to each coordinate direction will change with time, while preserving angular momentum.

Accordingly, if Earth's precession is Torque-free, precession would not occur if the Earth were an ideal sphere.

The alternative is that Earth’s precession is Torque-Induced Precession . I’m not clear on how the Sun imposes a torque on a sphere, spinning or not. If it is imposing tidal forces due to Earth's oblate spheroid shape, these tidal forces would disappear if the Earth were an ideal sphere.

Apparently an ideal sphere would not process from either Torque-free or Torque-induced precession. Is this correct?


2 Answers 2


The component of Earth's precession attributed to the tidal torques of the Moon and Sun on the oblate Earth is called "precession of the equinoxes". Earth's much smaller and faster torque-free precession is called "Chandler wobble".

In theory, neither phenomenon would be observed if the Earth was a perfect sphere.

  • $\begingroup$ Do other deviations from an ideal sphere affect precession, except for “flattening” (bulges and dents on its surface, for example)? $\endgroup$
    – ayr
    Commented Dec 17, 2023 at 5:25
  • 1
    $\begingroup$ @dtn What matters for the precession behavior is the inertia tensor. Bodies with the same inertia tensor will precess in the same way. $\endgroup$
    – John Doty
    Commented Dec 17, 2023 at 13:24
  • $\begingroup$ Do I understand correctly, that if inertia tensors are not only equal in the same angular position, but also their change will be the same in different angular positions? $\endgroup$
    – ayr
    Commented Dec 18, 2023 at 6:00

Obviously, a spherical mass would not experience any torque as it behaves gravitationally exactly like a point mass (see Shell Theorem), but if the sphere is rotating, then its mass distribution would actually not appear as spherical anymore for an interacting body. This is due to the finite speed of gravitational interaction (assumed to be the speed of light $c$ in the theory of Relativity), which causes the mass density to appear higher in the receding half of the rotating sphere and lower in the approaching half, as illustrated in the below diagram

enter image description here

This figure shows the density of an initially (when at rest) uniform distribution when it rotates counter-clockwise with a speed of $0.5 c$ as it appears for an observer on the positive x-axis. The resultant apparent asymmetry for a rotating sphere results therefore in a torque which in general not only slows down the rotation but also causes a precession (depending on the orientation of the plane of rotation and the rotational speed).

I have published a paper a couple of years ago which discusses the retardation effect on the shape/density of moving masses in more detail ( https://iopscience.iop.org/article/10.1088/1361-6404/ab2578/pdf ; full text at https://www.physicsmyths.org.uk/retarded_positions.pdf ).


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