Is the vector product of a vector with its derivative equal to the linear product of their magnitudes?

Question

I’m reading “Fundamentals of Astrodynamics” by Roger Bate and in the first chapter he states “in general

$ \vec{a }\cdot\dot{\vec{a}}= a\dot{a}. $

My intuition tells me that the dot product of the position and velocity of something in a circular orbit for example should in fact be zero.

Just to check it for myself I tried to prove it and got the following:

$\frac{d}{dt} (\vec{r}\cdot\vec{r})=\dot{\vec{r}}\cdot\vec{r}+\vec{r}\cdot\dot{\vec{r}}\\ \frac{d}{dt}r^{2}= 2(\vec{r}\cdot\dot{\vec{r}})\\ \vec{r }\cdot\dot{\vec{r}}= r\dot{r}$

This is the result that I don’t intuitively believe since the two vectors aren’t always parallel. Where did I go wrong with my logic?

Answer 1

Note that $\dot a$ is not the length of $\dot{\vec a}$, it is the time derivative of $|\vec a|$. So for instance, for an object moving at constant speed in a circular orbit, where $\vec a$ and $\dot{\vec a}$ are perpendicular, as you say, then we have $\dot a=0$.

Answer 2

Generally, $\frac{d}{dt} \! \lVert \vec{a}\rVert \ne \lVert \frac{d}{dt}\! \vec{a} \rVert$ for vector derivatives. The notation used in the book is potentially ambiguous if you're not careful.

Put another way, the rate of change of a vector's length is not the same thing as the length of the vector's rate of change.

You're using the latter while the book is using the former.