<p>The component of Earth's precession attributed to the tidal torques of the Moon and Sun on the oblate Earth is called &quot;precession of the equinoxes&quot;. Earth's much smaller and faster torque-free precession is called &quot;Chandler wobble&quot;.</p>
<p>In theory, neither phenomenon would be observed if the Earth was a perfect sphere.</p>

CC BY-SA 4.0
2023-12-17T05:25:53.973Z
Do other deviations from an ideal sphere affect precession, except for “flattening” (bulges and dents on its surface, for example)?
ayr
CC BY-SA 4.0
2023-12-17T13:24:54.973Z
@dtn What matters for the precession behavior is the inertia tensor. Bodies with the same inertia tensor will precess in the same way.
John Doty
CC BY-SA 4.0
2023-12-18T06:00:05.683Z
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?
ayr
CC BY-SA 4.0
2023-12-16T15:16:12.570Z
2023-12-16T15:16:12.570Z
John Doty
<p>Obviously, a spherical mass would not experience any torque as it behaves gravitationally exactly like a point mass (see <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">Shell Theorem</a>), 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 <span class="math-container">$c$</span> 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</p>
<p><a href="https://i.sstatic.net/Tlk6F.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Tlk6F.gif" alt="enter image description here" /></a></p>
<p>This figure shows the density of an initially (when at rest) uniform distribution when it rotates counter-clockwise with a speed of <span class="math-container">$0.5 c$</span> 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).</p>
<p>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 ( <a href="https://iopscience.iop.org/article/10.1088/1361-6404/ab2578/pdf" rel="nofollow noreferrer">https://iopscience.iop.org/article/10.1088/1361-6404/ab2578/pdf</a> ; full text at <a href="https://www.physicsmyths.org.uk/retarded_positions.pdf" rel="nofollow noreferrer">https://www.physicsmyths.org.uk/retarded_positions.pdf</a> ).</p>

CC BY-SA 4.0
2023-12-23T18:08:52.290Z
2023-12-24T13:15:18.323Z
2023-12-24T13:15:18.323Z
Thomas
<p>Migrated from Space Exploration</p>
<p>I am trying (unsuccessfully) to acquire an intuitive understanding of planetary precession.</p>
<p>Is Earth’s precession <strong>Torque-free Precession</strong> due to its asymmetry as per <a href="https://en.wikipedia.org/wiki/Precession" rel="noreferrer">https://en.wikipedia.org/wiki/Precession</a> ?</p>
<blockquote>
<p>If an object is <em>asymmetric about its principal axis of rotation</em>, the
moment of inertia with respect to each coordinate direction will
change with time, while preserving angular momentum.</p>
</blockquote>
<p>Accordingly, if Earth's precession is <strong>Torque-free</strong>,  precession would not occur if the Earth were an ideal sphere.</p>
<p>The alternative is that Earth’s precession is <strong>Torque-Induced Precession</strong> . 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.</p>
<p>Apparently an ideal sphere would not process from either Torque-free or Torque-induced precession. Is this correct?</p>

<orbital-mechanics><precession>
Would the Earth still precess if it were an ideal sphere?
Woody
# Would the Earth still precess if it were an ideal sphere?

(Asked by: username_0 on 2023-12-16)

Migrated from Space Exploration

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?

**Tags:** orbital-mechanics, precession


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## 2 answers

### Answer by username_1 on 2023-12-16. Score: 3

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.

> **Comments:**
> **username_2**: Do other deviations from an ideal sphere affect precession, except for “flattening” (bulges and dents on its surface, for example)?  
> **username_1**: @dtn What matters for the precession behavior is the inertia tensor. Bodies with the same inertia tensor will precess in the same way.  
> **username_2**: 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?  

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### Answer by username_3 on 2023-12-23. Score: 1

Obviously, a spherical mass would not experience any torque as it behaves gravitationally exactly like a point mass (see [Shell Theorem](https://en.wikipedia.org/wiki/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](https://i.sstatic.net/Tlk6F.gif)](https://i.sstatic.net/Tlk6F.gif)

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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