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In this answer there's a geometrical derivation of the optimal angle of a solar sail to deorbit a spacecraft into the Sun.

The naive answer is 45° which would direct the reflected light directly prograde, but a shallower angle (reflecting slightly sunward (or nadir) to prograde) seems to substantially increases the area that the sail collects sunlight compared to the loss in the prograde component of the thrust (reflected sunlight). The value given there is about 35° rather than the naive 45°.

But Wait...There's More!

In this answer I show that for a modest, realistic scenario (the LightSail-2) with a cubesat mass of 5 kg and a solar sail of 32 m^2 at 45 degrees, the radial component reduces the the net radial acceleration by about 0.3% and tilting toward the sun from 45° to ~35° would make that larger, and of course for a larger area/mass ratio the reduction would be even larger.

What that means is that the central force is lower, and so the orbital velocity is lower as well and so the same delta-v will result in a large movement toward the Sun.

So for the fastest de-orbit towards the Sun (it could be to say Venus or Mercury), what is the new optimal angle when the radial thrust is not ignored?

The angle will depend on the area to mass ratio, so it would be interesting to do more cases, but at least do the current one; 5 kg, 32 m^2. I'm guessing it changes by only a quarter of a degree, but I don't know, and it it could be larger for a larger area/mass ratio.

You are welcome to start with the Python script or any other aspects of the linked answer. I was in a rush and so hard-wired it at 45°.

Assume an initial circular orbit, and that means that the initial velocity will be a bit slower than what I did in order to match the reduced net radial acceleration.

uhoh
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    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. – BowlOfRed Aug 28 '18 at 15:42
  • @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"! – uhoh Aug 28 '18 at 15:49
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    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. – MBM Aug 31 '18 at 01:18
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    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. – MBM Sep 01 '18 at 02:44
  • @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. – uhoh Sep 01 '18 at 03:35
  • 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 – MBM Sep 10 '18 at 01:19
  • @MBM that sounds really interesting! I'll try to locate a copy and take a look... https://www.springer.com/us/book/9783540210627 – uhoh Sep 10 '18 at 03:22
  • @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! – uhoh Sep 10 '18 at 12:44
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    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 http://solarsailingnotes.popelak.info/#orbit1. Someday a 2nd edition will add newer materials and designs, and the experience of IKAROS and LightSail, but not yet. http://www.u3p.net/u3p_fr/Accueil_U3P.html has simulator. – MBM Oct 07 '18 at 01:20
  • @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. – uhoh Oct 07 '18 at 01:23
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    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)^2sinA)/(1-B(cosA)^3).
    If for some R, speed^2 = (mu/R)(1-B    (cosA)^3+B(cosA)^2sinAtanG), it is true for all r. Substitute for TanG.
    Delta time = (R^1.5 – r^1.5)    (2/(BmusinAtanG))^0.5/(3cosA). For 0.05<B<0.15 and 30<A<37, Terra to Mars takes 300-900 days. Hohmann orbit 259.     – MBM Oct 14 '18 at 17:17
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    @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. – Roger Wood Dec 13 '20 at 03:56

1 Answers1

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Assuming I have understood the constraints correctly, you have a solar sail in a (very gentle) spiral trajectory inwards, and want to bleed orbital energy at the highest possible rate.

Considering edge cases, this is not always the optimal way of reducing transfer time. Imagine for instance falling straight down towards the sun, with no perpendicular velocity. Having the sail facing the Sun will clearly reduce orbital energy, but would be counterproductive in crashing into the Sun as fast as possible. The opposite scenario, escaping the solar system, have ideal solutions where apoapsis is increased and periapsis reduced, until a dive can be used to reach reach escape velocity (this is not time-reversible).

But for the scenario in question, we have the following conditions:

  1. Solar sails have a thrust proportional to $\cos^2(\theta)$, with a $\cos^3(\theta)$ radial component and $\cos^2(\theta) \sin(\theta)$ tangential component. (Where $\theta$ is the angle between the sail normal and the direction of the Sun. This comes from decomposing the thrust vector, which is of magnitude $\cos^2\theta$).
  2. Acceleration tangential to velocity does not affect orbital energy.

If we then introduce another angle, $\beta$, which is the angle between perfectly perpendicular velocity and actual velocity (positive towards the Sun), the ideal angle would come from maximising the effect of the two components:

$$\cos^2(\theta) \sin(\theta) \cdot \cos(\beta) + \cos^3(\theta) \cdot \sin(\beta)$$

In the case of perpendicular velocity ($\beta \approx 0$), this is $\theta = 2\tan^{-1}\left( \sqrt {5 - 2\sqrt{6}} \right)$

But the general case actually has an analytic solution!

$$\theta = \frac{1}{2}\left(\cos^{-1}\left(\frac{\cos(\beta)}{3}\right) - \beta\right)$$

This does not stay constant as $\beta$ changes, which a simple scaling argument should indicate: At half distance from the sun, the sail provides 4x the acceleration, but the circular orbital velocity is only $\sqrt{2}$ times larger, meaning the spiral does not have constant "angle of attack".

SE - stop firing the good guys
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    @uhoh $\theta$ is the angle between the normal vector and the direciton towards the Sun. The radial and tangential components come from decomposing a vector of magnitude $\cos{^2}\theta$. As for sourcing that, pretty much any solar sail paper, like this one should do. (look at equations 3 and 4), – SE - stop firing the good guys Dec 12 '20 at 21:51
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    @SE - stop firing the good guys Nice analysis. I might be wrong, but for a weak solar sail and small beta, I'm concluding that the result will be a very shallow logarithmic spiral with constant beta. I've got thrust going as 1/r^2, velocity as 1/sqrt(r), power as 1/r^(5/2), period as r^(3/2), and hence energy loss per rev as 1/r. Since dEnergy/dr goes as 1/r^2, this means that the change in radius per rev is proportional to r, which gives a logarithmic spiral with constant beta. Anyway, I think this has to be the answer since there's no absolute scale parameter in the problem. – Roger Wood Dec 13 '20 at 01:42
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    @RogerWood Nice to hear that beta stays constant under realistic conditions! – SE - stop firing the good guys Dec 13 '20 at 01:48
  • @RogerWood I've just asked Are solar sail spirals logarithmic? Can this be shown analytically or by dimensional analysis alone? – uhoh Dec 13 '20 at 22:58
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    @uhoh - that's an enticing rabbit hole if ever I saw one ... – Roger Wood Dec 14 '20 at 01:53
  • I appreciate the work that's gone into this answer so +100 but I don't understand it yet, mostly because I keep delaying setting aside time to go through it carefully. I'm confident that thrust being proportional to $\cos^2(\theta)$ can be shown, but less confident that any effect of the radial component on the final result can be dismissed via hand-waving without being shown rigorously. I see that with $\beta=0$ the expression does have a maximum at $\theta = 2\tan^{-1}\left( \sqrt {5 - 2\sqrt{6}} \right)$ but $\beta=0$ in the analytical general case expression gives a different $\theta$. – uhoh Dec 20 '20 at 00:06
  • Eventually I'll write a script and just test a bunch of cases to confirm once the possible disagreement between the two theta's is resolved, but that will take time because it's easy to do it incorrectly given that I appear to misunderstand solar sail force vectors. – uhoh Dec 20 '20 at 00:11
  • @uhoh For $\beta = 0$, I have the equality $2\tan^{-1}\left( \sqrt {5 - 2\sqrt{6}} \right) = \frac{1}{2}\left(\cos^{-1}\left(\frac{1}{3}\right)\right)$ – SE - stop firing the good guys Dec 20 '20 at 11:10
  • Ya I get that too now, strange. I tried it a few times and kept getting something like 53° for the general expression with $\beta=0$ so I must have been making the same strange mistake each time. I'm going to chalk that up to pre-coffee brain. Thanks for the feedback! – uhoh Dec 20 '20 at 12:23