A more accurate way of solar sail calculation?

Question

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.

Answer 1

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.

Answer 2

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:

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.

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.