Skip to main content
Physics

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

7
  • $\begingroup$ I'm still not sure why the trajectory should be on the plane of $\vec{r}$ and $\vec{v}$. I know it seems very intuitive, but I don't see how it should be proved formally. $\endgroup$ Commented Jul 11, 2024 at 21:33
  • $\begingroup$ Probably my problem is that I don't have a clear idea of what the definition or plane/rectilinear motion is and how this is related to the direction of $\vec{r}$ and $\vec{v}$. In one exercise we solved in class, which required to prove that a trajectory was an ellipse, we were able to find the equation of the trajectory and observe it was the equation of an ellipse, but here I don't have any equations so I don't get how it is possible to say that the trajectory is plane or rectilinear. Hope I was clear enough $\endgroup$ Commented Jul 11, 2024 at 21:38
  • $\begingroup$ Note that if $L\neq 0$, the plane of motion is simply the plane perpendicular to $\vec{L}$. $\endgroup$ Commented Jul 11, 2024 at 22:19
  • $\begingroup$ @DavideMasi, as the answer stated, if the motion is not fixed to a plane, that implies that angular momentum is changing (cross product produce a vector that is perpendicular to the plane two vectors form), which violates the conservation of angular momentum, meaning the motion must take place in a fix plane that is not changing, so angular momentum does not change direction. $\endgroup$ Commented Jul 12, 2024 at 0:31
  • $\begingroup$ @DavideMasi Equations are not a substitute for concepts. However, if you feel more comfortable with equations, a "constructive" formal proof goes as follows: in the case of ${\bf L} \neq 0$, the orthogonality of angular momentum and position (easily proved algebraically) implies the constraint on the trajectory ${\bf L}\cdot {\bf r}= constant$, which is the equation of a plane. $\endgroup$ Commented Jul 12, 2024 at 3:38