Timeline for What is the optimal angle for a solar-sail deorbit towards the Sun when radial thrust is included?
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22 events
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| S Dec 19, 2020 at 23:57 | history | bounty ended | uhoh | ||
| S Dec 19, 2020 at 23:57 | history | notice removed | uhoh | ||
| Dec 13, 2020 at 3:56 | comment | added | Roger Wood | @uhoh Interesting problem. From the various comments, it does seem like this is already well studied and that the fastest way of changing orbital energy results in a logarithmic (constant angle) spiral. That being said, you can gradually get to infinity at zero velocity but you can never exceed escape velocity unless the maximum radial solar-sail acceleration exceeds the gravitational acceleration. Then the spiral becomes a straight radial line outward. Spiraling inward is different and sooner or later you will hit the radius of the sun. Calculating the time for that seems a bit tricky. | |
| Dec 12, 2020 at 17:49 | answer | added | SE - stop firing the good guys | timeline score: 2 | |
| Dec 12, 2020 at 12:00 | history | tweeted | twitter.com/StackSpaceExp/status/1337728942212599809 | ||
| S Dec 12, 2020 at 3:34 | history | bounty started | uhoh | ||
| S Dec 12, 2020 at 3:34 | history | notice added | uhoh | Draw attention | |
| Oct 14, 2018 at 17:17 | comment | added | MBM | I know three simple sailcraft orbits. Edge-on to Sol (Alpha = 90), regular Kepler conic. Face-on to Sol (A = 0), Kepler with (mu)(1-Beta), modified gravitational parameter. You may want logarithmic spiral with flight angle Gamma (R. H. Bacon (1959). It requires (tanG/(2+(tanG)^2) = (B*(cosA)^2*sinA)/(1-B*(cosA)^3). If for some R, speed^2 = (mu/R)(1-B*(cosA)^3+B*(cosA)^2*sinAtanG), it is true for all r. Substitute for TanG. Delta time = (R^1.5 – r^1.5)*(2/(BmusinAtanG))^0.5/(3*cosA). For 0.05<B<0.15 and 30<A<37, Terra to Mars takes 300-900 days. Hohmann orbit 259. | |
| Oct 7, 2018 at 1:23 | comment | added | uhoh | @MBM what, you are far from the internet because you are out solar sailing? Very nice! Okay I will try to hunt that down and have a look, thank you. I can imagine that a solar still could still work in space, but collecting sea water is hard to do except for a few moons of Jupiter and Saturn. But for this question I have a hunch the solution may have to be numerical, though there's always a chance there is an analytical approximation as well for gradual spirals. I may take a crack at it myself as nobody has bitten. | |
| Oct 7, 2018 at 1:20 | comment | added | MBM | Uhoh, sorry for delayed reply, I am often far from the internet. Currently my copy of Solar Sailing is loaned. The text by McInnes ist much superior to that by Friedman, Vulpetti or Wright, and is worth the high price, but the printing mistakes can confuse the mathematics. I list known errata at solarsailingnotes.popelak.info/#orbit1. Someday a 2nd edition will add newer materials and designs, and the experience of IKAROS and LightSail, but not yet. u3p.net/u3p_fr/Accueil_U3P.html has simulator. | |
| Sep 10, 2018 at 12:44 | comment | added | uhoh | @MBM believe it or not I can't find a copy in any library close by, nor not-so-close by, and so far when I search google books I don't find page numbers anywhere near there. If you can post an answer and add a screen shot or block quotes to the relevant parts, that would be great! | |
| Sep 10, 2018 at 3:22 | comment | added | uhoh | @MBM that sounds really interesting! I'll try to locate a copy and take a look... springer.com/us/book/9783540210627 | |
| Sep 10, 2018 at 1:19 | comment | added | MBM | McInnes (page 265) graphs Vulpetti’s H-reversal orbit for alpha – 40 degrees, beta = 1.0. This is a very lightly loaded sailcraft, but after about 160 days the path is pointed directly at Sol. With beta = 0.1, it may take several orbits of Sol, but eventually the perihelion will be smaller than the solar radius. For a logarithmic spiral starting at 1 au with alpha = -40 degrees and beta = 0.1, gamma = 4.53 degrees, and the orbit will just graze Sol after 500 days | |
| Sep 1, 2018 at 3:35 | comment | added | uhoh | @MBM that would be an extraordinarily large & light-weight solar sail to be able to simply stop and fall radially in practice, but it's interesting to think about theoretically. | |
| Sep 1, 2018 at 2:44 | comment | added | MBM | Sorry, was interrupted and timed out. The total force goes as cos(alpha)^2, with alpha the tilt from face on to Sol. The component perpendicular to the Sol-line that changes angular momentum is sin(alpha)cos(alpha)^2, with maximum at tan(alpha)^2 = 0.5. After your angular momentum drops to zero, jettison the sail and fall directly to Sol. | |
| Aug 31, 2018 at 1:18 | comment | added | MBM | Roughly 37 degrees. As you tilt the sail, the light intercepted diminishes by cos(alpha), but the component of net thrust perpendicular to the radius from Sol increases. See the text by Colin. R. McInnes. | |
| Aug 28, 2018 at 16:48 | history | edited | uhoh | CC BY-SA 4.0 |
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| Aug 28, 2018 at 15:53 | history | edited | uhoh | CC BY-SA 4.0 |
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| Aug 28, 2018 at 15:49 | comment | added | uhoh | @BowlOfRed It looks like I misread the edit history and thought someone else had made what I'd called an "impromptu edit" to your question. I've deleted the comment there and adjusted the wording here. I of course think it's fantastic when people "do the math"! | |
| Aug 28, 2018 at 15:42 | comment | added | BowlOfRed | The calculations I did in the other answer simply assume the radial velocity is close enough to zero to ignore. If that's true, then almost anything else (mass, velocity, etc.) doesn't matter. | |
| Aug 28, 2018 at 12:33 | history | edited | uhoh | CC BY-SA 4.0 |
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| Aug 28, 2018 at 12:25 | history | asked | uhoh | CC BY-SA 4.0 |