Timeline for How much delta-v would Hayabusa-2 need per day to remain 20 km sunward of Ryugu?
Current License: CC BY-SA 4.0
11 events
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| Oct 3, 2018 at 13:11 | comment | added | asdfex | What I wanted to say is: It's orbiting the Sun, but it doesn't follow any of the usual formulas, like semi-major axis defines period - at least not to the precision we require to talk about correction maneuvers. | |
| Oct 3, 2018 at 13:05 | comment | added | uhoh | @asdfex the spacecraft is absolutely in a heliocentric orbit without question. That it may make parts-per-billion adjustments to that orbit does not make it a non-orbit. Thanks for the edit btw. | |
| Oct 3, 2018 at 12:56 | comment | added | asdfex | @uhoh I rewrote that part. | |
| Oct 3, 2018 at 12:55 | history | edited | asdfex | CC BY-SA 4.0 |
added 896 characters in body
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| Oct 3, 2018 at 12:37 | comment | added | asdfex | You can't actually call it an orbit if it requires constant station keeping. | |
| Oct 3, 2018 at 12:29 | comment | added | uhoh | I was just trying to understand the shape, parameters, and purpose of the elliptical orbit to which you refer, right now I don't understand the nature of this elliptical orbit. The reason I thought you might be talking about one is that a circular orbit 20km closer will move faster, so in 30 days it will advance in its orbit by about 15 km from its "home position". I had thought you'd matched the semi-major axes as a way to reduce that sizable walk-off rate. Right now I'm still not clear what your proposed orbit is, is it possible to add a summary separate from the paragraphs of text? Thanks! | |
| Oct 3, 2018 at 11:29 | comment | added | asdfex | @uhoh If your current state is "closer to the Sun" and "same speed (actually lower as you specified 'between Sun and asteroid') as the asteroid", you're not on an orbit with the same semi-major axis. For that you would need to be faster than the asteroid. But that's all minuscule contributions because the difference is so small. | |
| Oct 3, 2018 at 2:25 | comment | added | uhoh | Two orbits w/ same period (return to the same configuration after one year) will have the same semi-major axis. In the sentence "...a slightly elliptical trajectory that takes us further away from the asteroid ever so slightly over the course of a quarter of a year, then brings us back." do you mean further away from the Sun instead? A slightly elliptical orbit would intersect the circular orbit of the same period twice a year, so if the spacecraft is closer to the Sun than the asteroid now, it intersects in 3 months and is farther from Sun in three more. | |
| Oct 2, 2018 at 22:12 | history | edited | asdfex | CC BY-SA 4.0 |
fix units
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| Oct 2, 2018 at 21:32 | comment | added | DrSheldon | "or 2.4 m per year". Should this read m/s? | |
| Oct 2, 2018 at 17:45 | history | answered | asdfex | CC BY-SA 4.0 |