$$ \begin{aligned} x&=\text{East} \\ y&=\text{North} \\ z&=\text{Up} \\ \lambda&=\text{Latitude} \end{aligned} $$

• We can deduce

$$ \vec \omega = \omega \begin{pmatrix} 0 \\ \cos \lambda \\ \sin \lambda \end{pmatrix} $$

The centrifugal force

• $\alpha$ is the angle from the vertical where the pendulum is at rest due to earths rotation

$$ \frac{\vec F_\text{cen}}{m}=\vec a_\text{cen}=\omega^2\vec r'-(\vec r' \cdot \vec \omega) \, \vec \omega =\omega^2 R_E \begin{pmatrix} 0 \\ -\sin(\lambda)\cos(\lambda) \\ \cos^2\lambda \end{pmatrix} \; \text{where: }\vec r'=\begin{pmatrix} 0 \\ 0 \\ R_E \end{pmatrix} $$

• Thus the effective gravitational acceleration is given by

$$ \vec g_\text{eff}=\vec g+\vec a_\text{cen}=\begin{pmatrix} 0 \\ -\omega^2R_E\sin(\lambda)\cos(\lambda) \\ -g+\omega^2R_E\cos^2(\lambda) \end{pmatrix} $$

• We can calculate the amplitude

$$ |\vec g_\text{eff}|=g\left [ 1-\frac{2\omega^2R_E}{g} \cos^2\lambda+\left(\frac{\omega^2R_E}{g} \right)^2\cos^2\lambda \right]^\frac 12 \approx g\left( 1-\frac{\omega^2R_E}{g}\cos^2\lambda\right) $$

• Furthermore we can calculate $\alpha$

$$ \tan \alpha = \frac{R_E\omega^2}{g}\sin\lambda\cos(\lambda)\left( 1-\frac{\omega^2R_E}{g}\cos^2\lambda\right)^{-1} $$

• so:

$$ \alpha\approx\frac{R_E\omega^2}{g}\sin\lambda\cos\lambda $$

Particle falling to the earth

• Initial conditions

$$ \vec r(0)=\begin{pmatrix} 0 \\ 0 \\ R_E+h \end{pmatrix} \; ; \; \dot{ \vec r }=0 \; ; \; \ddot{ \vec r }=\vec g-2\vec \omega \times \dot{ \vec r } + \omega^2 \vec r-(\vec r \cdot \vec \omega)\, \vec \omega $$

we take: $\vec r = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

• then the Coriolis force can be defined as:

$$ \frac 1m \vec F_\text{cor}=-2(\vec \omega \times \dot{ \vec r })=-2\omega\begin{pmatrix} \dot z \cos\lambda-\dot y \sin \lambda \\ \dot x \sin \lambda \\

• \dot x \cos \lambda \end{pmatrix} $$