11
$\begingroup$

In this Wikipedia article about solar sails, I read the following:

The force on a sail and the actual acceleration of the craft vary by the inverse square of distance from the sun (unless close to the sun), and by the square of the cosine of the angle between the sail force vector and the radial from the sun

This was followed by the equations:

$F = \frac{F_0 \cos^2 \theta} {R^2}$ (perfect sail)

$F = \frac{F_0 (0.349 + 0.662 \cos{2\theta} − 0.011 \cos{4\theta}) } {R^2}$ (realistic sail)

I understand that the second equation is an expansion of the first, taking into account realistic factors that affect the performance of the sail. However I have two doubts here.

  1. What does "angle between the sail force vector and the radial from the sun" mean?
  2. What exactly are the realism factors the second equation adds to the first?
  3. BONUS: What's the best way to graph these equations in order to better understand them?

UPDATE: Could I also get clarification on the variables in the equations and their sources?

$\endgroup$
  • 3
    $\begingroup$ A perfect sail reflects all the incident light, a realistic sail transforms part of incident light into heat which is radiated into the environment. $\endgroup$ – Deer Hunter Dec 2 '13 at 13:03
  • $\begingroup$ I am still looking for a valid answer. $\endgroup$ – Vedant Chandra Dec 7 '13 at 11:15
  • $\begingroup$ @VedantChandra, did you find an answer somewhere? If so, could you post it below? I am interested in the answer. Thank you. $\endgroup$ – ChrisR May 21 '14 at 8:13
  • $\begingroup$ Thank you for your interest @ChrisR, but I have still not received a valid answer on this or it's sister question here: space.stackexchange.com/q/2997/1028 $\endgroup$ – Vedant Chandra May 21 '14 at 13:55
  • $\begingroup$ @VedantChandra, okay thanks for the heads up and for letting me know about that question. Since you're interested in solar sails, maybe could you provide some insight on my question? space.stackexchange.com/questions/4587/… Thanks. $\endgroup$ – ChrisR May 21 '14 at 16:24
3
$\begingroup$

This is a partial answer.

Given a flat reflector, as shown below, the force resulting from the incident photons will always be normal (90°) to the reflector plane. This is because the angle of reflection is always equal to the angle of incidence.

The angle between the incident photons and the force vector is labelled θ. The magnitude of the force will vary according to cos θ. When sail is at right angles to the incident photons, as when accelerating away from the Sun, θ will be zero, and the sail will deliver the maximum force (cos 0 is 1).

Solar Sail (after Jerry Wright, Wikipedia)

θ increases as the sail is oriented away from the Sun, and the thrust obtained from every photon is reduced. For example, with the sail oriented at 45°: cos 45° is $\sqrt{2}$, about 0.71. But the area of the sail available to collect photons is reduced also; it also varies according to cos θ. Hence the $\cos^2 {\theta}$ term in the ideal formula.

The Wikipedia Solar sail article you mention does not clearly explain how the actual square sail model is derived, nor does the associated Radiation pressure article.

All I can add at this point is that the performance of a real solar sail will be affected by

  • the sail reflectivity; the linked articles assume 90% is feasible.
  • the sail re-radiation of absorbed energy.
  • the sail mass per unit area; the linked articles say

    At a thickness of 20 nm, lithium has an areal density of 0.011 g/m2.

  • the sail's surface accuracy (how flat it is).
  • the efficiency with which the force on the sail is translated to the spacecraft (eg, is there a turning moment that must be balanced).
$\endgroup$
1
$\begingroup$
  1. "Angle between the sail force vector and the radial from the sun": the sail is moved by the incident photons from the Sun, and by the reflected photons (reaction).

enter image description here

(source: spaceplan2020.com)

Whatever the angle of the Sun, the resultant force will be close to the normal to the sail surface (normal only if 100% of the incident photons are reflected). This means there is an angle between the radial from the Sun and the sail force (direction of thrust) which depends on the reflectivity and also, more important, on the sail orientation.

  1. Coefficients 0.349, 0.662 and 0.011 are for a square sail. More on this.

  2. Some clues:

    • $\frac{1}{r^2}$ is the effect of the inverse square law (because the light from the Sun is emitted on a sphere, which total surface area is proportional to the square of distance.
    • $\cos \theta$ reflects the ratio between the actual surface area of the sail, and its apparent surface area as seen from the Sun (1 when the angle is null, then decrease down to 0 when the angle is 90°).
  3. Variables are the solar pressure exerted on the sail when the sail is oriented in the Sun direction ($F_0$), the angle between the Sun and the normal to the sail ($\theta$) and the distance to the Sun ($R$).

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.