What is the optimal angle for a solar-sail deorbit towards the Sun when radial thrust is included?

Question

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.

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".