# Does polar motion cause the Earth ellipsoid to change?

The Earth is oblate because it spins and is not a rigid body. Since its spin axis changes relative to its crust, then the meridian of maximum diameter changes too, right?

I understand that polar motion is very small (just a few meters), but the tidal bulge is larger (over 20 kilometers), so I wonder if the side effects of polar motion are more dramatic than I had initially thought.

To be clear: I am talking here about polar motion, not precession or nutation.

• Reading the wiki article, I wonder whether the definition of North (usually averaged over a year) doesn't follow the polar wander. en.m.wikipedia.org/wiki/Polar_motion. As tides play a large role, it matters whether it is spring tide or not (this sun-moon relative alignment). Hard to guesstimate ðŸ˜€ Commented Sep 25, 2022 at 9:45
• 20km is more than I have read before, I've heard of crustal tides in the order of 30cm. Commented Sep 25, 2022 at 10:43
• The way they are modeled in the Explanatory Supplement to the Astronomical Almanac, the answer is no. The algorithm for computing an observer's geocentric position does not account for polar motion. The observer's lat, lon and altitude may change, but this is modeled as the observer moving, but the reference ellipsoid doesn't change because of it. Put another way: of course it does, but to simplify things, different effects are classified into different models and accounted for separately. Commented Sep 25, 2022 at 15:29
• You can check Earth SE. See if you find anything related. Commented Oct 20, 2023 at 1:18

The answer to this question is obviously affirmative, though the change of shape will be minuscule.

Let $$\boldsymbol{\omega}$$ be the angular velocity vector, and $$\bf r$$ be a point on the Earth surface. Employing the vector formula $$\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times{\bf r}\right) = \boldsymbol{\omega}~(\boldsymbol{\omega}\cdot{\bf r}) - {\bf r}~\boldsymbol{\omega}^{\,2} = \nabla\left[\,\frac{1}{2}\,\left(\boldsymbol{\omega}\cdot{\bf r}\right)^2-\frac{1}{2}\,\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}\,\right]\;,$$ and introducing the colatitude $$\varphi$$ through $$\cos\varphi=\frac{\boldsymbol{\omega}}{\,|\boldsymbol{\omega}|\,}\cdot\frac{\bf r}{\,|{\bf r}|\,}\;,$$ we split the centrifugal force into a second-harmonic part and a purely radial part: $$-\rho~\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times{\bf r}\right) = -\nabla\left[\frac{\rho}{3}~\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2} \,P_2\left(\cos\varphi\right)\right] +\nabla\;\left[\frac{\rho}{3}~\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}\right]~,\qquad(*)$$ where we assumed the body homogeneous, and used the expression $$P_2(\cos\varphi)=\frac{\textstyle 1}{\textstyle 2}\left(3\,\cos^2\varphi-1\right)$$ for the degree-2 Legendre polynomial.

Interested in oscillating deformation solely, we now must change, in equation (*), $$\boldsymbol{\omega}^{\,2}~$$ to $$~\boldsymbol{\omega}^{\,2}-\langle \boldsymbol{\omega}^{\,2}\rangle$$, where $$\langle ... \rangle$$ denotes time averaging.

From expression (*), we observe that the forcing splits into two parts -- quadrupole and purely radial.

Spherical asymmetry aside, the radial forcing causes a radial deformation. For terrestrial bodies, this is a teeny-tiny effect, because under weak forcing these bodies are virtually incompressible.

The quadrupole term mimics the quadrupole component of tides, wherefore the resulting deformation will be easy to calculate using the standard tidal machinery. I am not expert on ocean tides, but can say that the deformation of the solid part of the Earth will be expressed via the Love number $$h_2$$.

• It's possible, but a bit painful. See math.meta.stackexchange.com/q/29967/207316 I suppose you could make a macro using \newcommand Commented Apr 9 at 6:36
• @PM2Ring Thank you. I didn't realise that here we can make macros. But I also see that `planetmaker' has kindly fixed my answer using $\backslash$boldsymbol Commented Apr 9 at 12:25