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.

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    $\begingroup$ 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. $\endgroup$
    – BowlOfRed
    Aug 28, 2018 at 15:42
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    $\begingroup$ 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. $\endgroup$
    – MBM
    Aug 31, 2018 at 1:18
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    $\begingroup$ 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. $\endgroup$
    – MBM
    Sep 1, 2018 at 2:44
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    $\begingroup$ 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. $\endgroup$
    – MBM
    Oct 7, 2018 at 1:20
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    $\begingroup$ 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. $\endgroup$
    – MBM
    Oct 14, 2018 at 17:17

1 Answer 1


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".

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    $\begingroup$ @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), $\endgroup$ Dec 12, 2020 at 21:51
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    $\begingroup$ @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. $\endgroup$
    – Roger Wood
    Dec 13, 2020 at 1:42
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    $\begingroup$ @RogerWood Nice to hear that beta stays constant under realistic conditions! $\endgroup$ Dec 13, 2020 at 1:48
  • $\begingroup$ @RogerWood I've just asked Are solar sail spirals logarithmic? Can this be shown analytically or by dimensional analysis alone? $\endgroup$
    – uhoh
    Dec 13, 2020 at 22:58
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    $\begingroup$ @uhoh - that's an enticing rabbit hole if ever I saw one ... $\endgroup$
    – Roger Wood
    Dec 14, 2020 at 1:53

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