0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 '18 at 15:42
  • $\begingroup$ @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"! $\endgroup$ – uhoh Aug 28 '18 at 15:49
  • 1
    $\begingroup$ 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 '18 at 1:18
  • 1
    $\begingroup$ 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 '18 at 2:44
  • 1
    $\begingroup$ 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 '18 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.