= Coriolis acceleration
{c}
{title2=$-2\boldsymbol\Omega\times\mathbf v$}
The Coriolis acceleration is the <Coriolis force> divided by the particle's <mass>:
$$
\mathbf a_{\rm cor}=-2\boldsymbol\Omega\times\mathbf v_{\rm rot}.
$$
It is therefore independent of <mass> for a specified relative <velocity> and <angular velocity>. This is one contribution to the <acceleration> measured in a <rotating reference frame>. For a fixed common origin with an <inertial frame> and constant $\boldsymbol\Omega$, the full <equation of motion in a rotating frame> is
$$
\underbrace{\ddot{\mathbf r}_{\rm rot}}_{\text{full rotating acceleration}}
=\frac{\mathbf F_{\rm real}}m
\underbrace{-2\boldsymbol\Omega\times\dot{\mathbf r}_{\rm rot}}_{\text{Coriolis}}
\underbrace{-\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r)}_{\text{centrifugal}}.
$$
All <vectors> here are expressed in the rotating axes; $\mathbf F_{\rm real}$ is the sum of actual <forces>. An accelerating origin adds a translational apparent <force>, and time-dependent <angular velocity> also adds the <Euler acceleration>.
The factor two follows directly from the <rotating-frame derivative formula>. First $\mathbf v_{\rm lab}=\dot{\mathbf r}_{\rm rot}+\boldsymbol\Omega\times\mathbf r$. Differentiate again in the <inertial frame>: differentiating $\dot{\mathbf r}_{\rm rot}$ contributes one $\boldsymbol\Omega\times\dot{\mathbf r}_{\rm rot}$, and differentiating $\boldsymbol\Omega\times\mathbf r$ contributes another. Move both terms to the rotating side of <Newton's second law> to obtain the minus sign and the factor two. The <centrifugal acceleration> comes from the remaining double <cross product>.
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