What is the dot product of a vector and its derivative equal to the product of the magnitude of the vector and the magnitude of its derivative? On page 15 of "Fundamentals of Astrodynamics" by Bate et al., the energy constant of motion is derived. In the second step of the derivation, Bate asserts the following: > Since in general $\mathbf{a}\dot{}\mathbf{\dot{a}} = a\dot{a}$, $\mathbf{v}=\mathbf{\dot{r}}$ and $\mathbf{\dot{v}}=\mathbf{\ddot{r}}$, then... I am familiar with the latter two identities, which express kinematic relationships, but unfamiliar with the former: $\mathbf{a}\dot{}\mathbf{\dot{a}} = a\dot{a}$ In the preceding text, Bate does not precise a physical meaning of $\mathbf{a}$, so it would appear that he intends a general vector. (A similar, well-known identity is of course $\mathbf{a}\dot{}\mathbf{a} = a^2$, but this doesn't seem to suggest that the above identity should be valid.) Prove or refute Bate's identity.

Using the product rule for scalair multiplication $$2aa'=(a^2)' = (\mathbf{a}\cdot \mathbf{a})' = \mathbf{a}\cdot \mathbf{a}' + \mathbf{a}'\cdot \mathbf{a} = 2(\mathbf{a}\cdot \mathbf{a}')$$