Usually the total acceleration a' seen in a rotating reference frame (like earth) 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...

The answer must be simple, because it is just kinematics, not even physics, but I cannot find it.


1 Answer 1


Your equation $(1)$ is valid only for a stationary point in the inertial frame. In general, you have: $$a’+\omega’\times (\omega’\times r’)+2\omega’\times v’=a$$ for uniform rotation ($\dot\omega=0$) and no translation.

The more general formula is: $$a’+\omega’\times (\omega’\times r’)+\dot \omega’\times r’+a_t’+2\omega’\times v’=a$$ with $a_t’$ the translation acceleration. In fact you can identify the terms: $$ \omega’\times (\omega’\times r’)+\dot \omega’\times r’+a_t’ $$ as the acceleration in the inertial frame of a point fixed in the non inertial frame coinciding at the given time with the point of interest. This usually makes the calculations more intuitive.

In your example, you have $a’=\frac{u^2}{R}$ since even in the Earth’s reference frame, the point is accelerating from following the curved path of the Equator. The formulas are therefore consistent.

Hope this helps.

  • $\begingroup$ Hmmm, for decades I took equation (1) as fully true. What do you mean by "stationary point". Stationary means for me at rest? But we are talking about points not at rest in the earth's frame (Coriolis force is for a point not at rest). Where is this formula for the force in a rotating reference frame fishy in particular, because I miss what you refer as at? en.wikipedia.org/wiki/Coriolis_force#Formula I must have a closer look as I did all the time before. I'll review your answer soon. $\endgroup$
    – MichaelW
    Dec 8, 2022 at 12:40
  • $\begingroup$ As I mentioned, stationary (not moving) is from the perspective of the inertial frame. Check out Arnold’s Mathematical Methods of Classical Mechanics for example, where the formula is proven cleanly. $\endgroup$
    – LPZ
    Dec 8, 2022 at 13:33
  • $\begingroup$ your last line above (a'=u²/R) opened my eyes - thanks for the hint. $\endgroup$
    – MichaelW
    Dec 8, 2022 at 22:06

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