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I know that the aceleration vector in spherical coordinates is $$\vec{a}=(\ddot{r}-r\dot{\theta}^2-r\sin^2 \theta \dot{\phi}^2)\vec{u_r}+\dots$$ and since ever I have known the centripetal acceleration to be $a_r=\frac{v^2}{r}$ but it doesn't seem to appear on this expression of the aceleration at any moment. Where is it? Or how do we get there?

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the aceleration vector in spherical coordinates is $$\vec{a}=(\ddot{r}-r\dot{\theta}^2-r\sin^2 \theta \dot{\phi}^2)\vec{u}_r+\dots$$

The centripetal acceleration is right there. It is $$(-r\dot{\theta}^2-r\sin^2 \theta \dot{\phi}^2)\vec{u}_r$$

I have known the centripetal acceleration to be $a_r=\frac{v^2}{r}$

More excactly: the centripetal acceleration is $\frac{v_\text{perpendicular}^2}{r}$ where $v_\text{perpendicular}$ is the velocity component perpendicular to the radius.

The velocity vector written in spherical coordinates is (see e.g. Spherical coordinate system - Kinematics): $$\vec{v}=\underbrace{\dot{r}\vec{u}_r}_{\vec{v}_\text{radial}} +\underbrace{r\dot{\theta}\vec{u}_\theta+r\sin\theta\dot{\phi}\vec{u}_\phi}_{\vec{v}_\text{perpendicular}}$$

The first term is the radial velocity component. The second and third term together is the velocity component perpendicular to the radius.

Now let us focus the perpendicular velocity. Its square is $$v_\text{perpendicular}^2=r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2$$

Let us divide this by $r$. We get

$$\frac{v_\text{perpendicular}^2}{r}= r\dot{\theta}^2+r\sin^2\theta\dot{\phi}^2$$

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