The deduction of the expression for $\vec{r} \cdot \dot{\vec{r}}$, where $\vec{r}$ is the displacement of the center of mass of the orbiting entity from the center of mass of the entity which it orbits and $\vec{v} := \dot{\vec{r}}$ is the velocity of the orbiting entity, is critical to the vector analysis which usually accompanies an orbital mechanics study. Denoting $r:=|\vec{r}|$ and $\hat{r} := \frac{\vec{r}}{|\vec{r}|}$, we can see that $\dot{|\hat{r}|} := \frac{d |\hat{r}|}{dt} = \dot{1} = 0 = \vec{r} \cdot \dot{\vec{r}}$$\dot{|\hat{r}|} := \frac{d |\hat{r}|}{dt} = \dot{1} = 0 = \frac{1}{2} \hat{r} \cdot \dot{\hat{r}}$, which implies that $\dot{\hat{r}} \perp \hat{r}$. Indeed, in general, the time derivative of a vector of constant magnitude is perpendicular to the vector. Therefore, $\vec{r} \cdot \dot{\vec{r}} = r \hat{r} \cdot (\dot{r} \hat{r} + r \dot{\hat{r}})$, so that $\vec{r} \cdot \vec{v} = r \dot{r}$.
Clearly, $\vec{r} \cdot \vec{v} = 0$ if and only if the motion is circular, that is $\dot{r} \equiv 0$. On the other hand, if the motion is not circular, the inner product of interest does not, in general, vanish. Notice that we did not have to use any property of the force field to arrive at this result, since the question asked by the OP is in fact a question on kinematics and does not involve the dynamics.