How is centrifugal force identified in equations of fluid motion in the rotating reference system of earth?

Question

I spent already days on the following: In one of my meteorology scripts the equations for conservation of momentum in the rotation reference system of earth are given by:

The apparent forces are in orange, the others are clear.

I wonder why there is no centrifugal force, however, the apparent force in a rotating system should be the sum of Coriolis and centrifugal force.

To learn more about the derivation of these equations I studied this video carefully: https://www.youtube.com/watch?v=nljud2UiWUk&list=PL_cuIb7hx5lg_zHfUVsUrw6I66U4jq8Dq

At it's end, the equations are written down:

First I'm surprised that there are "curvature" forces: In my opinion they are nothing else but the centrifugal forces, but usually the total acceleration a' seen in a rotating system is written down as

$$\vec a' = -\vec \omega \times \left(\vec \omega \times \vec r' \right) - 2 \cdot \vec \omega \times \vec v' \tag{1}$$

First term is called usually "centrifugal force" while the right term is the Coriolis force. But when I consider motion with (zonal) velocity u (along the east) on the equator, the total centrifugal force in radial direction z would be

$$a_z = \frac{(u+\omega R)^2}{R} = \omega^2R+u^2/R+2 \omega \cdot u \tag{2}$$

because the total tangential speed is the sum of earth's rotation and zonal speed.

Now I wonder, where the quadratic term would be derived from equation (1). I see only the first and third term arising from (1) but not the middle...

It seems a bit, that the u,v,w-driven part of what I call "centrifugal force" is absorbed into the Coriolis force and when we speak about centrifugal force only the static part of it (u=v=0) is regarded.

However, when this is the case, there is still one missing thing: when u=v=w=0 we have a point at rest in earth's reference frame. At a given northern latitude $\phi$ there is clearly a centrifugal force which doesn't point in pure z-direction but also has a y-component, pointing south:

$$F_y = \Omega^2 R \cos(\phi) \sin(\phi)$$

so I would expect this as a "static" part of the second equations for $Dv/dt$.

However, in neither in the first image (my script) nor in the second one (the video) such component is identified: everything is proportional to u,v,w without a static component.

Even worse, in the first image (taken from my lecture) the centrifugal forces are missing completely.

Final questions in particular:

1. How do quadratic terms arise from (1)?
2. Where have centrifugal components gone to in the first image? It drives me nuts...
3. Why "curvature force" when everywhere else (in my physics textbooks) total force is split just into centrifugal and Coriolis? Is there a new basic force "curvature" which I have overseen previously or is is just a matter of definition of whether we assign some mathematically arising components to "Coriolis" or "centrifugal"?
4. Could it be, that centrifugal force is just the "static" centrifugal force due to earth's rotation and the other components are distributed along linear and quadratic terms, where the quadratic terms are called "curvature forces" and the linear ones "Coriolis". Something here is not consistent with my textbook knowledge I learned in the past.

Answer 1

I wonder why there is no centrifugal force, however, the apparent force in a rotating system should be the sum of Coriolis and centrifugal force.

There is a centrifugal acceleration in those equations. What we call "gravity" in a lay sense is a combination of gravitational acceleration and centrifugal acceleration. In a technical sense, what we call "gravity" is the acceleration of a free-falling body at sea level as observed in a frame of reference fixed with respect to the rotating Earth. Centrifugal acceleration is baked into $g$.

This means centrifugal acceleration is present in your equations. In particular, it is in your equation (3) as $g$. If one takes $g$ as a constant, $g_0=9.80665\,\text{m}/\text{s}^2$ (which may be a bad idea), that's the acceleration due to gravitation and centrifugal forces at roughly the latitude of Paris. What we call $g$ varies with latitude and with height above the ellipsoid, plus minor local perturbations. (Gravity near the Himalaya can get quite complex if one wants to be very precise.)

A fairly simple approximation that accounts for latitude (but not height) is the Somigliana gravity formula,$$g = g_{\text{eq}}\frac{1+\kappa \sin^2\phi}{\sqrt{1-e^2\sin^2 \phi}}$$ where $g_{\text{eq}} = 9.7803267714\,\text{m}/\text{s}^2$ is the acceleration due to gravity (including centrifugal acceleration) at the equator, $\kappa = 0.00193185138639$, which reflects the observed difference between gravity at the equator versus the poles, $e^2=0.00669437999013$ is the square of the eccentricity of the figure of the Earth, and $\phi$ is the geodetic latitude.

The variation with height (the "free air correction") is approximately linear in the Earth's atmosphere, a decrease of about $3.086\,\text{μm}/\text{s}^2$ for every meter above sea level. If you are using $g$ as a constant, you are essentially ignoring both the latitude and altitude corrections. Centrifugal acceleration is baked in. If you are using the Somigliana formula (or some other approximation of gravitation at sea level as a function of latitude), centrifugal acceleration is once again baked in.