= Coriolis acceleration in local Earth coordinates
{c}
{title2=$a_E=2\Omega\sin\varphi\,v_N-2\Omega\cos\varphi\,v_U$}
Use local east, north, and up <unit vectors> $(\mathbf e_E,\mathbf e_N,\mathbf e_U)$, with $\mathbf e_E\times\mathbf e_N=\mathbf e_U$. In a spherical-<Earth> approximation at <geographical latitude> $\varphi$, the <angular velocity> has components $\boldsymbol\Omega=(0,\Omega\cos\varphi,\Omega\sin\varphi)$. Expanding the <cross product> for relative <velocity> $(v_E,v_N,v_U)$ gives
$$
\begin{aligned}
a_E&=2\Omega\sin\varphi\,v_N-2\Omega\cos\varphi\,v_U,\\
a_N&=-2\Omega\sin\varphi\,v_E,\\
a_U&=2\Omega\cos\varphi\,v_E.
\end{aligned}
$$
These are <Coriolis acceleration> components, not the total <acceleration>. For horizontal motion, the horizontal part is $(fv_N,-fv_E)$, where the <Coriolis parameter> is $f=2\Omega\sin\varphi$. Thus northward motion in the northern hemisphere has an eastward <Coriolis acceleration>, to the right of its <velocity>; in the southern hemisphere the horizontal turning is to the left. At the equator $f=0$, but the full <Coriolis acceleration> need not vanish: vertical or eastward motion still couples through $\Omega\cos\varphi$.
For a numerical example, take the <Earth> rotation rate $\Omega=7.292115\times10^{-5}\ \mathrm{s^{-1}}$, <geographical latitude> $\varphi=45^\circ$, and an initially northward relative <velocity> $\mathbf v_{\rm rot}=20\mathbf e_N\ \mathrm{m\,s^{-1}}$. Then
$$
f\simeq1.03126\times10^{-4}\ \mathrm{s^{-1}},
\qquad
a_E=fu\simeq0.0020625\ \mathrm{m\,s^{-2}}.
$$
\Image[/coriolis-earth-latitude-and-eastward-acceleration.png]
{title=Latitude, local rotation geometry, and the scale of eastward Coriolis acceleration for northward motion}
For a local, short-time estimate with no other horizontal <force> and zero initial eastward <velocity>, hold the initial northward <velocity> in the leading term: $\Delta x_E\simeq\tfrac12fu t^2$. At $t=10\ \mathrm s$ this gives about $0.1031\ \mathrm m$ eastward. It assumes nearly fixed local axes and <geographical latitude>, negligible changes of height, and $|ft|\ll1$. It is not a complete wind <trajectory>: <pressure gradients>, <friction>, and evolving <velocity> must enter an atmospheric <equation of motion>. The <effective gravity in a rotating frame> already combines ordinary <gravity> with the steady terrestrial <centrifugal acceleration>.
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