I'm currently studying the coriolis effect in my undergrad degree and I can't find anywhere on the internet the solution to this question. Please tell me if I'm right or not:
Suppose we have an inertial frame of reference $S$ and another frame of reference $S'$ which its origin rotates with angular velocity $\mathbf{\Omega}$ with respect to the $S$ frame origin, and its base rotates with angular velocity $\mathbf{\omega}$. This is, it orbits with angular velocity $\mathbf{\Omega}$ and spins with angular velocity $\mathbf{\omega}$. Let $\mathbf{r}$ be the position vector of a point $P$ as viewed from $S$, $\mathbf{R}$ the vector that goes from the origin of $S$ to the origin of $S'$, and $\mathbf{r'}$ the position vector of $P$ as viewed from $S'$. We can write $\mathbf{r}=\mathbf{R}+\mathbf{r'}$. We know that $\frac{d\mathbf{R}}{dt}=\mathbf{\Omega}\times\mathbf{R}$, and we may say that $\frac{d\mathbf{r'}}{dt}=\mathbf{\omega}\times\mathbf{r'}+\mathbf{v'}$, where $\mathbf{v'}$ is the velocity vector of $P$ as seen from $S'$. Therefore $$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{d\mathbf{R}}{dt}+\frac{d\mathbf{r'}}{dt}=\underbrace{\mathbf{\Omega}\times\mathbf{R}}_{\text{"orbit velocity"}}+\underbrace{\mathbf{\omega}\times\mathbf{r'}}_{\text{"spin velocity"}}+\underbrace{\mathbf{v'}}_{\text{velocity w.r.t $S'$}}$$
If we suppose $\frac{d\mathbf{\Omega}}{dt}=\frac{d\mathbf{\omega}}{dt}=0$, then $$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\mathbf{\Omega}\times\frac{d\mathbf{R}}{dt}+\mathbf{\omega}\times\frac{d\mathbf{r'}}{dt}+\frac{d\mathbf{v'}}{dt}=\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{R})+\mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{r'}+\mathbf{v'})+\frac{d\mathbf{v'}}{dt}$$
We can work out a little bit $\frac{d\mathbf{v'}}{dt}$. Suppose we write the base vectors of the $S'$ coordinate system as the set $\{\mathbf{e_1'},\mathbf{e_2'},\mathbf{e_3'}\}$. Then, $$\frac{d\mathbf{v'}}{dt}=\frac{d}{dt}\sum_{i}v_i'\mathbf{e_i'}=\sum_i\frac{dv_i'}{dt}\mathbf{e_i'}+\sum_iv_i'\frac{d\mathbf{e_i'}}{dt}=\mathbf{a'}+\sum_iv_i'\frac{d\mathbf{e_i'}}{dt}$$
Where $\mathbf{a'}$ is the acceleration of $P$ as seen from $S'$. We can notice $\frac{d\mathbf{e_i'}}{dt}=\mathbf{\omega}\times\mathbf{e_i'}$. Then, $$\frac{d\mathbf{v'}}{dt}=\mathbf{a'}+\sum_iv_i'\frac{d\mathbf{e_i'}}{dt}=\mathbf{a'}+\sum_iv_i'(\mathbf{\omega}\times\mathbf{e_i'})=\mathbf{a'}+\mathbf{\omega}\times\sum_iv_i'\mathbf{e_i'}=\mathbf{a'}+\mathbf{\omega}\times\mathbf{v'}$$
Which leads us to the following: \begin{align} \mathbf{a}&=\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{R})+\mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{r'}+\mathbf{v'})+\frac{d\mathbf{v'}}{dt} \\ &=\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{R})+\mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{r'}+\mathbf{v'})+\mathbf{a'}+\mathbf{\omega}\times\mathbf{v'} \\ &=\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{R})+\mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{r'})+\mathbf{a'}+2(\mathbf{\omega}\times\mathbf{v'}) \end{align}
If we rearrange the equation to get the acceleration in the $S'$ frame, we get: $$\mathbf{a'}=\mathbf{a}-\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{R})-\mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{r'})-2(\mathbf{\omega}\times\mathbf{v'})$$
And we kind of could call the term (maybe?) $-\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{R})$ "orbital centrifugal acceleration", the term $-\mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{R})$ "spin centrifugal acceleration", and finally the term $-2(\mathbf{\omega}\times\mathbf{v'})$ Coriolis acceleration.
Is this a correct approach or it misses something? Thank you.