Timeline for Why does the eccentricity vector always point towards the periapsis of an orbit?
Current License: CC BY-SA 4.0
12 events
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| Jun 4 at 13:13 | comment | added | Puffin | Please excuse my naivety: I had previously understood that the answer to the question as written in the title would be that the eccentricity vector always points to the periapsis because that is how it is defined, thus no need for a proof. I'm happy to move on from that idea though looking through this answer leads me to want clarification for how else is the eccentricity vector defined, i.e. such that it is worthwhile finding a proof? | |
| Feb 17, 2022 at 20:34 | comment | added | uhoh | got it, thanks! I can only speak three languages with even a minimal degree of utility; English Math and Python. My MathJax is barely 7-11 quality (i.e. enough to get through trying to buy something at 7-11 and to answer coherently to "Do you want a bag?") Kudos to your MathJax fluency! | |
| Feb 17, 2022 at 16:10 | comment | added | Ryan C | @uhoh That part is supposed to not display. It's setting up commands that I can reuse later, so I can just type "\ e" over and over, rather than "\ mathbf { \ vec { e } }" every time. If MathJax allowed $\LaTeX$ commands that take arguments, I could have done what I really wanted, which was "\ newcommand { \ v } [ 1 ] { \ mathbf { \ vec { # 1 } } }", and then "\ v { e }", "\ v { L }", and \ v everything else. | |
| Feb 17, 2022 at 2:36 | comment | added | uhoh | @RyanC I noticed here that your post as a big chunk of MathJax that doesn't display on my screen. Can you see it? If so, I've got a problem somewhere. i.sstatic.net/wqyEQ.png | |
| Jan 4, 2022 at 15:46 | vote | accept | John | ||
| Dec 31, 2021 at 23:52 | comment | added | David Hammen | What you didn't do was to use the fact that $$\frac{d}{dt} \left(\frac{\dot{\mathbf r} \times \mathbf h}{\mu}\right) = -\,\frac{\mathbf r \times \left(\mathbf r \times \dot{\mathbf r}\right)}{||r||^3}$$ | |
| Dec 31, 2021 at 22:46 | comment | added | David Hammen | Nicely done, Ryan. | |
| Dec 31, 2021 at 22:03 | comment | added | Ryan C | @DavidHammen I tried, but I'm not sure what I produced is comprehensible enough to be valuable. Any additional criticism you could offer at this time would be most welcome. | |
| Dec 31, 2021 at 22:02 | history | edited | Ryan C | CC BY-SA 4.0 |
added 5451 characters in body
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| Dec 31, 2021 at 14:20 | comment | added | David Hammen | You stated that "The eccentricity vector is a conserved quantity". It would be worthwhile to add to this answer a proof that in the two body problem the eccentricity vector is constant. In other words, its time derivative is zero (or since its time derivative is zero, it is constant). | |
| Dec 31, 2021 at 13:34 | comment | added | David Hammen | In addition to the eccentricity vector being a conserved quantity in the two body problem, so is the angular momentum vector. | |
| Dec 31, 2021 at 0:52 | history | answered | Ryan C | CC BY-SA 4.0 |
