[78666cdc]

The time derivative of that expression is indeed zero.

[toth]

It is not. $e_r$ is a rotating vector, not constant. Also $r$ is not really constant either.

Actually, if you think about it, the only plausible way you could have something in space with zero jerk and non-zero acceleration would be an interstellar ship maintaining constant acceleration, so the suggested rule is almost exactly wrong.

[78666cdc]

I'm afraid you'll have to explain your perspective to me as if I'm five years old. In your original comment, you say,

>a constant times the $-e_r/r^2$, and the time derivative of that is definitely not zero.

I don't see a term there that is time-dependent.