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:

enter image description here

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:

enter image description here

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.
  • 1
    $\begingroup$ I am pretty sure your equation (2) is the correct way to actually introduce all those terms. But two things are frequently confused: 1. Transformation from cartesian to spherical coordinates. Doing that, you already pick up a number of those terms for free (I think I may have this written down somewhere on SE, I'll check later...) and I am pretty sure th centrifugal terms is amongst them. ($\vec r = r \vec e _r$-> $\dot{\vec r}$ etc. ) 2. Inertial to non-inertial frame transformation: That's when you pick up $\vec v_{inertial} = \vec v_{noninertial} + \vec \Omega$, and the mix/curvature terms. $\endgroup$ Dec 8, 2022 at 16:58

1 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.

  • $\begingroup$ Yes, but when I'm not at equator, then centrifugal force is not radial with respect to center of earth and has an "off zenith" component. The formulas have g only in z-direction (equ. 3) - what about y? $\endgroup$
    – MichaelW
    Dec 8, 2022 at 12:43
  • 2
    $\begingroup$ @MichaelW If you use geodetic coordinates (e.g., geodetic east-north-up) as opposed to geocentric coordinates (e.g., geocentric east-north-up) you will find that gravitation plus centrifugal acceleration is only in the geodetic up coordinate. The acceleration due to gravity (gravitational plus centrifugal acceleration) is normal to the Earth's non-spherical surface. That is essentially the definition of sea level, or more specifically, the geoid. (continued) $\endgroup$ Dec 8, 2022 at 13:04
  • 2
    $\begingroup$ Gravitational acceleration is not quite directed toward the center of the Earth thanks to the Earth's equatorial bulge, and centrifugal acceleration is directed away from the rotation axis rather than away from the center of the Earth. The vectorial sum of the two is however normal to the geoid. Mathematically, this has to be the case as the geoid is defined as an equipotential surface of the potentially energy due to gravitational energy plus centrifugal energy. What we call gravity is the gradient of this potential at the geoid. $\endgroup$ Dec 8, 2022 at 13:08
  • $\begingroup$ That would mean that earths surface has adapted such that the sum of centrifugal force and gravitational force points always normal to the surface. Considering this, makes above equations even more complicated. Is the deviation from ideal sphere considered at all when we model atmosphere or is this just ignored? Then off-zenith centrifugal forces could also be ignored, because they must be of same order of magnitude. $\endgroup$
    – MichaelW
    Dec 9, 2022 at 18:05
  • 1
    $\begingroup$ @MichaelW With regard to That would mean that earths surface has adapted such that the sum of centrifugal force and gravitational force points always normal to the surface -- that is correct, more or less. This condition is called hydrostatic equilibrium. Objects smaller than about 600 km across tend to look like lumpy potatoes while larger objects have enough self-gravitation to pull themselves into hydrostatic equilibrium. This breaking point is the potato radius. $\endgroup$ Dec 10, 2022 at 5:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.