3 ms·
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
by toth 11y ago
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 11y agoI'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.