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I have been looking through my old dynamics books and derived the acceleration for a particle on a circular orbit. To begin with, I assume the particle is on a circular orbit around some center $x_0$. With a vector to the moving particle $\overline{e}_r$ and a tangential vector $\overline{e}_{\phi}$. The simple location of the particle can be written as:

$$\overline{x}(t)=r(t)\cdot e_r$$

Differentiating this gives:

$$\dfrac{d\overline{x}(t)}{dt}=\dfrac{dr(t)}{dt}\cdot \overline{e}_r + r(t) \cdot \dfrac{d\overline{e}_r}{dr}$$

Using $\frac{d\overline{e}_r}{dt}=\dot{\phi}\cdot \overline{e}_{\phi}$ and $\frac{d\overline{e}_{\phi}}{dt}=-\dot{\phi}\cdot \overline{e}_{r}$ which comes from a simple geometrical idea, I can simplify to (ignoring all the $(t)$ which should be obvious for all the variables):

$$\dot{\overline{x}}=\dot{r}\cdot \overline{e}_{r} + r\cdot \dot{\phi} \cdot \overline{e}_{\phi}$$

Differentiating this once more yields the acceleration: $$ \ddot{\overline{x}} = \left(\ddot{r}-r\dot{\phi}^2 \right)\overline{e}_{r} + \left(\ddot{\phi} r+ 2\dot{\phi}\dot{r} \right) \overline{e}_{\phi} $$

The part I was wondering about is the Coriolis-acceleration:

$$ a_c =2\dot{\phi}\dot{r} \overline{e}_{\phi} $$


I was trying to apply this to the passat winds on the earth.

enter image description here

So regarding the "Nordost Passat", we have wind particles flowing from North to South towards the equator. Since the formula above is only for a disc, yet the acceleration into the 3rd dimension (in this case the rotational axis of the earth) can be disregarded, I came up with this set of assumptions:


  1. Since we are moving away from the rotational axis (towards the equator, I assume that $\dot{r} > 0$

  2. Since the derivative of the radius with respect to the latitude at the equator is 0, the second derivative of the radius has to be smaller than $0$: $\ddot{r} < 0$

  3. The wind is moving together with the earth's surface (roughly) which means a positive rotation around the rotational axis if the rotation axis is supposed to be "upwards" which corresponds to: $\dot{\phi} > 0$

  4. Since the earth does not accelerate: $\ddot{\phi} = 0$

Also it is important to note that $\overline{e}_{\phi}$ shows to the "right" in the image which equals the direction the earth is rotating. Furthermore $\overline{e}_{r}$ shows outwards from the rotational axis.

Applying all of this to the Coriolis-acceleration, I end up with $a_c=(\text{something positive}) \overline{e}_{\phi}$. Since, according to the image, the wind is accelerated towards the negative $\overline{e}_{\phi}$, I am curious where I have messed up.

I am happy for any hints or advice.

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SHORT ANSWER

The Coriolis Force is not equal to mass times the Coriolis Acceleration. (Meanings used as is depicted in the underbraces below)

LONG ANSWER

The formula you wrote down is for an inertial frame of reference $(2D)$ and the Earth's frame is non-inertial and $3D$, instead you should use: $$m\vec{a}_{\text{rot}}=m\vec{a}_{\text{inertial}}\ \underbrace{-\ 2m\ \vec{\Omega}\times\vec{v}_{\text{rot}}}_{\text{Coriolis Force}}-m\vec{\Omega}\times(\vec{\Omega}\times\vec{r})\tag{1}$$

The above is valid for $3$ dimensions too.

A Slight digression not particularly germane to OP:

If I use equation $(1)$ for $2D$ and try to reconcile it with $$\vec{a}_{\text{inertial}}=(\ddot \rho - \rho \dot\phi^2)\hat \rho + (\rho\ddot \phi + \underbrace{2\dot \rho \dot \phi)\hat \phi}_{\text{Coriolis Acceleration}}\tag{2}$$ , the first difficulty I face is that equation $(1)$ has a radial Coriolis Force term too due to the component of $\textbf{v}_{\text{rot}}$ in the tangential direction. It can be made out by considering the below example.

Suppose you're on a merry-go-round rotating with constant angular velocity $\vec\Omega=\Omega\hat k\ $ along with your friend who is standing exactly opposite to you. You throw a ball in their direction but slightly missing so that $\vec{v}_{\text{rot}}=\dot\rho\hat\rho+\rho\dot\theta\hat\phi$. Seeing the moment you throw from an inertial frame of reference, the equation $(1)$ says \begin{align} \vec{a}_{\text{inertial}}&=\vec{a}_{\text{rot}}+2\ \vec{\Omega}\times\vec{v}_{\text{rot}}+\vec{\Omega}\times(\vec{\Omega}\times\vec{r})\tag{3} \\ &=\vec{a}_{\text{rot}}+2\ \vec{\Omega}\times(\dot\rho\hat\rho+\rho\dot\theta\hat\phi)+\vec{\Omega}\times(\vec{\Omega}\times\vec{r})\tag{4} \\ &=\vec{a}_{\text{rot}}+(2\dot\rho\Omega\hat\phi-2\rho\Omega\dot\theta\hat\rho)-\rho\Omega^2\hat\rho\tag{5} \end{align} and the equation $(2)$ says \begin{align} \vec{a}_{\text{inertial}}&=(\ddot \rho - \rho \dot\phi^2)\hat \rho + (\rho\ddot \phi + 2\dot \rho \dot \phi)\hat \phi\tag{6} \\ &=(\ddot \rho - \rho (\Omega + \dot\theta)^2)\hat \rho + 2\dot \rho(\Omega + \dot\theta)\hat \phi\tag{7} \\ &=(\ddot \rho - \rho\Omega^2 - \rho\dot\theta^2 - 2\rho\Omega\dot\theta)\hat \rho + 2\dot \rho(\Omega + \dot\theta)\hat \phi\tag{8} \end{align}

As a final step of this reconciliation, we see that the equations $(5)$ and $(8)$ are the same and $\vec{a}_{\text{rot}}$ contains the unmatched terms. Just like the centripetal force maintains (part of) the radial acceleration of the particle, the Coriolis Force maintains the Coriolis Acceleration and (part of) radial acceleration.

PS: Feel free to critique after reviewing my answer. I haven't given it the time it deserves due to my other engagements.

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Reference: Kleppner, Kolenkow- Introduction to Mechanics.

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  • $\begingroup$ Since the earth rotating around the sun has no impact on the winds, I did assume an inertial frame of reference. Do you think this assumption is wrong? Also in your example, the coriolis force would also go into the opposite direction as depicted in the image? $\endgroup$ Jun 13 at 11:35
  • $\begingroup$ 1/2 Earth is a rotating frame of reference like a merry-go-round. 2/2 How can it be in the opposite direction as depicted for both of us? I have used an expression exactly opposite to yours. If you are getting an inconsistency, I shouldn't. The Coriolis Force on wind is to its right in the Northern Hemisphere and to its left in the Southern. Example: As depicted, $\vec v\times\vec\Omega$ is to the left(but wind's right) in the Northern Hemisphere. $\endgroup$ Jun 13 at 14:06
  • $\begingroup$ ah i see. thanks! $\endgroup$ Jun 14 at 13:11

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