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The Coriolis effect is used to explain the formation of typhoons, hurricanes, and cyclones and other phenomena where we take the Earth as a rotating reference frame. How could we explain this phenomena in an inertial reference frame?

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Disclaimer: borrowed from wikipedia with minor modifications.

The acceleration is the second time derivative of position, $$ \mathbf{a}_{\mathrm{i}} \ \stackrel{\mathrm{ }}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}}\right)_{\mathrm{i}} = \left( \frac{d\mathbf{v}}{dt} \right)_{\mathrm{i}} = \left[ \left( \frac{d}{dt} \right)_{\mathrm{r}} + \boldsymbol\Omega \times \right] \left[ \left( \frac{d\mathbf{r}}{dt} \right)_{\mathrm{r}} + \boldsymbol\Omega \times \mathbf{r} \right] \ ,$$ where subscript $i$ means the inertial frame of reference. Carrying out the differentiations and re-arranging some terms yields the acceleration in the rotating reference frame $$ \mathbf{a}_{\mathrm{r}} = \mathbf{a}_{\mathrm{i}} - 2 \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}} - \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r}) - \frac{d\boldsymbol\Omega}{dt} \times \mathbf{r} $$ where $$\mathbf{a}_{\mathrm{r}} \ \stackrel{\mathrm{}}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}} \right)_{\mathrm{r}} $$ is the apparent acceleration in the rotating reference frame, the term $$-\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r})$$ represents centrifugal acceleration, and the term $$-2 \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}}$$ is the coriolis acceleration.

When the expression for acceleration is multiplied by the mass of the particle, the three extra terms on the right-hand side result in fictitious forces in the rotating reference frame, that is, apparent forces that result from being in a non-inertial reference frame, rather than from any physical interaction between bodies. Using Newton's second law of motion $\mathbf{F}=m\mathbf{a}$, we obtain:

the Coriolis force

$$\mathbf{F}_{\mathrm{Coriolis}} = -2m \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}}$$

the centrifugal force

$$\mathbf{F}_{\mathrm{centrifugal}} = -m\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r})$$ and the Euler force

$$\mathbf{F}_{\mathrm{Euler}} = -m\frac{d\boldsymbol\Omega}{dt} \times \mathbf{r}$$ where $m$ is the mass of the object being acted upon by these fictitious forces.

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  • $\begingroup$ I should've remembered that equation, I just read it yesterday. $\endgroup$ – Oscar Flores Nov 10 '14 at 23:27
  • $\begingroup$ Take it easy, I usually forget much more basic things :) $\endgroup$ – Wolphram jonny Nov 10 '14 at 23:32

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