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From the Wikipedia article on solar sails:

The force on a sail and the actual acceleration of the craft vary by the inverse square of distance from the Sun (unless extremely close to the Sun)

This is when you are accelerating straight away from the Sun, and "proportional to inverse square of distance" is $\frac{1}{r^2}$.

The problem is that this is only valid when you are far from the Sun, and can model it as a point source of light.

Of course the Sun is not a point, and its radius is important when close to it. After a little thinking, I came up with this modification to the proportionality law:

$$\frac{2}{3}\left(1-\cos^3\left(\sin^{-1}\left(\frac{r_S}{r}\right)\right)\right)$$

Or, if you prefer it non-trig:

$$\frac{2}{3}\left(1-\sqrt{\left(1-\left(\frac{r_S}{r}\right)^2\right)^3}\right)$$

Where $r$ is the distance from the centre of the Sun and $r_S$ is the radius of the Sun. This follows the $\frac{1}{r^2}$ law at a distance, but accounts for the fact that some of the light hits the sail at an angle when it is close to the Sun, following the squared cosine rule.

I have not been able to find a modified relation anywhere, but is this correct? Also, is there a way to include the diameter of the Sun in a formula taking the orientation of the sail into account? That seems difficult, as light is hitting both sides of the sail.

For a visualisation of the difference, this plots the simple proportionality in red, and the modified in grey. Distance from the middle of the Sun along the x-axis, and proportionality along the y-axis. enter image description here

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  • $\begingroup$ Interesting but... What are these equations? You say you came up with "this" but don't ever explain it! What are $r_S$ and $r$? Then you show a plot with no axes! Tsk tsk! I would assume this came from an integral of the solar incidence to a point -- if you want to include the orientation you could either just assume that your ratio of flux is locally uniform and simply apply the appropriate projection factor (sine or cosine of the orientation, depending on your convention). However, sounds like you prefer a detailed model so you might try a double integral along the sail and solar incidence. $\endgroup$ – Brian Lynch Dec 31 '15 at 10:21
  • $\begingroup$ @BrianLynch Better now? I cleaned up the question a bit. The formula is intended as a proportionality law, thereby the trouble of labelling the axes. I am, as you say, interested in a model not assuming uniform flux. Can you explain how exactly you calculate that integral? $\endgroup$ – Hohmannfan Dec 31 '15 at 11:17
  • $\begingroup$ Also see Can I derive a combined equation for velocity of solar sail? and Solar sail thrust calculation. And if you'll model this at closer distances, you'll also need visible fraction of a sphere (for idealized radiation pressure model of the Sun as a perfect sphere with no corona, uniform radiance at all latitudes, and no solar wind). $\endgroup$ – TildalWave Dec 31 '15 at 11:20
  • $\begingroup$ Yes, thanks! You didn't have to remove the figure, just state the axes in text (it looked like proportion vs distance from Sun). I don't think you will find much difference from a uniform model using your proportion law, but you would have to integrate along the sail and compute your proportion as it varies. Again, any spacecraft will be so small compared to the Sun so I don't think you need to do this. It would be important if you were interested in looking at torque on the sail due to the solar pressure gradient. $\endgroup$ – Brian Lynch Dec 31 '15 at 11:23
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    $\begingroup$ Interesting - and likely useful once someone tries any actual solar sail navigation. While my first instinct was "it's only useful at distances where the sail would melt anyway", after a consideration even very small errors early on in a space journey can accumulate into huge one upon arrival. Taking Sun as a point light source is just such a tiny error seemingly without significance. Still, probably it would need to be adjusted for solar wind emissivity - unlike light, which is emitted roughly equally in all directions only the central part of the solar disc emits solar wind towards the sail. $\endgroup$ – SF. Dec 31 '15 at 11:23
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In his text Solar sailing: Technology, Dynamics and Mission Applications (see here for some of his open access titles), Colin R. McInnes derives the pressure on a perfect sail face on to Sol as

$$P(r) = (L/(3 \pi c R^2)) * (1 -[1 - (R/r)^2]^{1.5})$$

$P$ is photon pressure
$c$ is speed of light
$r$ is distance from center of Sol
$L$ is luminosity of Sol
$R$ is radius of Sol

In another form,

$$F(r) = (2/3) * (r/R)^2 * (1 -[1 - (R/r)^2]^{1.5})$$

meaning at $R$ the force experienced is 2/3 that expected from the inverse square law. After 10 $R$, $F(r)$ becomes very close to one. McInnes also covers limb darkening. This is a good text, and I hope an updated version will be out soon. If using the original version, watch out for printer's errors: missing parentheses, incorrect or misplaced exponents, dropped parameters, and so on.

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  • $\begingroup$ The first one works as a charm, but are you sure about the second one? $\endgroup$ – Hohmannfan Jan 11 '16 at 21:39
  • $\begingroup$ Do you mean $F(r) = (2/3) * (1 -[1 - (R/r)^2]^{1.5})$? $\endgroup$ – Hohmannfan Jan 11 '16 at 21:41
  • $\begingroup$ It does need the (r/R)^2. Altho it looks like force, McInnes uses F(r) as a dimensionless parameter that multiplies the usual 1/r^2 dependence that he calls P*(r). His equation. 2.44a has a misprint, should be P*(r) F(r) instead of P(r)*F(r). $\endgroup$ – MBM Jan 11 '16 at 22:28
  • $\begingroup$ Tidal wave, thank you for the edits. I could not find pi, and using too many * caused some of them to disappear. $\endgroup$ – MBM Jan 11 '16 at 22:31
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The usual proportionality law for solar sails is

$$\frac{\cos^2\left(\alpha\right)}{r^2}$$

where $\alpha$ is the sails angle from zenith, and $r$ is the distance from the Sun. This assumes that the Sun a point source of light, and that the size of the sail is negligible compared to the Sun.

This can simply be decomposed into

$$\frac{\cos^3\left(\alpha\right)}{r^2}$$

for the vertical component, and

$$\frac{\cos^2\left(\alpha\right)\sin\left(\alpha\right)}{r^2}$$

for the horizontal.

A naive solution for this is simply combining the $\cos^2\left(\alpha\right)$,$\cos^3\left(\alpha\right)$ and $\cos^2\left(\alpha\right)\sin\left(\alpha\right)$ with the modified law, however this is not completely accurate, but it is a little bit better than the point source assumption.

For a per unit of angle solution, we must make a decision about if we assume that the backside of the sail is reflective or not. Why this is important can be illustrated in the following way:

light hitting both sides of the sail

This can be decomposed multiplying with either $\cos\left(\alpha\right)$ or $\sin\left(\alpha\right)$ as usual.

Important this is a formula per unit of angle, so to get a law similar to $\frac{\cos^3\left(\alpha\right)}{r^2}$, you must integrate your decided function per degree angle for the whole range. This is possible, as the function obviously has an enclosed area. Total acceleration in green, vertical in blue and horizontal in orange. Y-axis is proportionality, X-axis is the different angles of illumination. In this case, the angle of the sail is 20 degrees, and the distance from the centre of the Sun is 2 solar radii. Note the part where the horizontal acceleration is negative, due to light hitting the backside of the sail.

graph showing decomposed acceleration

I gave up calculating the integral after about an hour, as did my CAS.

For any practical calculation, the usual solar sail equations are accurate enough, as you have to be extremely close to the Sun for any error by the point light source to be detectable.

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  • $\begingroup$ What about limb darkening? :-) $\endgroup$ – LocalFluff Jan 2 '16 at 16:27
  • $\begingroup$ @LocalFluff Yeah, ideally you must compensate for that too, but that makes the problem a lot harder. Feel for posting an answer? :-) $\endgroup$ – Hohmannfan Jan 2 '16 at 16:35

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