Timeline for answer to Doubt on conservation of angular momentum for Kepler's laws by GiorgioP-DoomsdayClockIsAt-85
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Jul 12, 2024 at 7:34 | comment | added | Davide Masi | @GiorgioP-DoomsdayClockIsAt-90 now I see. Probably, as the other answer suggested, in this case it is $\vec{L} \cdot \vec{r}=0$, right? | |
| Jul 12, 2024 at 7:31 | comment | added | Davide Masi | @Polaris5744 I know that the cross product of $\vec{r}$ and $\vec{v}$ is perpendicular to both, but this fact alone didn't help me to understand why the motion must take place in a plane. The last answer I received clarifies this point but I still have a doubt on the last part | |
| Jul 12, 2024 at 3:38 | comment | added | GiorgioP-DoomsdayClockIsAt-85 | @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. | |
| Jul 12, 2024 at 0:31 | comment | added | Polaris5744 | @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. | |
| Jul 11, 2024 at 22:19 | comment | added | John Doty | Note that if $L\neq 0$, the plane of motion is simply the plane perpendicular to $\vec{L}$. | |
| Jul 11, 2024 at 21:38 | comment | added | Davide Masi | 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 | |
| Jul 11, 2024 at 21:33 | comment | added | Davide Masi | 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. | |
| Jul 11, 2024 at 21:26 | history | answered | GiorgioP-DoomsdayClockIsAt-85 | CC BY-SA 4.0 |