Radial and transverse components of acceleration

Polar coordinates $(r,\theta)$ locate a particle by its distance $r$ from the pole O and the angle $\theta$ of the radius vector from a fixed line. Two unit vectors are used: $\mathbf{e}_r$ along the radius vector and $\mathbf{e}_\theta$ perpendicular to it (in the direction of increasing $\theta$).

          y
          |      P
          |    / |
          |  /   | e_theta
          | / r  |-> e_r
          |/ th
     O----+------------ x

Rate of change of unit vectors

Because the directions of $\mathbf{e}_r$ and $\mathbf{e}_\theta$ rotate with $\theta$:

$$\dot{\mathbf{e}}_r = \dot\theta\,\mathbf{e}_\theta \qquad \dot{\mathbf{e}}_\theta = -\dot\theta\,\mathbf{e}_r$$

Velocity

The position vector is $\mathbf{r} = r\,\mathbf{e}_r$. Differentiating:

$$\mathbf{v} = \dot r\,\mathbf{e}_r + r\dot{\mathbf{e}}_r = \dot r\,\mathbf{e}_r + r\dot\theta\,\mathbf{e}_\theta$$

So $v_r = \dot r$ (radial) and $v_\theta = r\dot\theta$ (transverse), with $v = \sqrt{v_r^2 + v_\theta^2}$.

Acceleration

Differentiating again:

$$\begin{aligned} \mathbf{a} &= \ddot r\,\mathbf{e}_r + \dot r\dot\theta\,\mathbf{e}_\theta + (\dot r\dot\theta + r\ddot\theta)\,\mathbf{e}_\theta - r\dot\theta^2\,\mathbf{e}_r \\ &= (\ddot r - r\dot\theta^2)\,\mathbf{e}_r + (r\ddot\theta + 2\dot r\dot\theta)\,\mathbf{e}_\theta \end{aligned}$$ $$a_r = \ddot r - r\dot\theta^2 \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta$$

The term $2\dot r\dot\theta$ is the Coriolis component; $r\dot\theta^2$ is the centripetal term.

Equations of motion

Applying Newton's second law $\Sigma \mathbf{F} = m\mathbf{a}$ along the two directions:

$$\Sigma F_r = m(\ddot r - r\dot\theta^2) \qquad \Sigma F_\theta = m(r\ddot\theta + 2\dot r\dot\theta)$$

For a central force ($\Sigma F_\theta = 0$) the second equation gives $\dfrac{d}{dt}(r^2\dot\theta) = 0$, i.e. constant angular momentum $h = r^2\dot\theta$ (Kepler's law of areas).