11
$\begingroup$

In Space Mission Engineering: The new SMAD, page 555, section 18.7.2, the following thrust formula is given for a solar sail: $$F=\frac{2RSA}{c}\sin^2\theta=9.113\times10^{-6}\frac{RA}{D^2}\sin^2\theta$$ Where, $F$ is the thrust; $R$ is the fraction of incident light; $D$ is the distance from the Sun in astronomical units; $S$ the solar flux in $W/m^2$; $c$ the speed of light; $A$ the sail area in $m^2$ and $\theta$ the sail tilt angle.

Does this formula calculate instantaneous thrust? There is no time factor in this equation, so how does one go to calculate the thrust of a given solar sail over a period of time? Is there an integration friendly formula instead?

Improve this question
$\endgroup$
4
  • $\begingroup$ Can anyone explain how they simplified from first thrust equation to simplified version? Thank you $\endgroup$ – Ryan Dec 7 '17 at 22:34
  • $\begingroup$ θ is the angle between the plane of the sail and the radial from Sol, so that θ = 90 means the sail is face on to Sol. R is the efficiency factor that considers spectral reflection, diffuse reflection, absorption, and transparency. S must be the flux at one au, = Luminosity/(4*piau^2) = 3.83E26 W/(4*pi*(1.496E11^2)) = 1.36E3. Collecting R, S, and c, 2*1.36E3/3E8 = 9.1E-6 kgm/s^2. Divide the force F by mass of the sail (constant, unlike a rocket) to get the acceleration at any distance D. Except for θ = 0 or 90, or on a logarithmic spiral, the trajectory must be found numerically. $\endgroup$ – MBM Dec 8 '17 at 0:42
  • $\begingroup$ @MBM you can leave this as a complete answer to this question - it has now been asked separately. $\endgroup$ – uhoh Dec 8 '17 at 3:46
  • $\begingroup$ @Ryan after MBM has seen my comment and posts an answer to your new question, it would be best for your to delete this answer because of course it's not an answer. It's great that you've re-posted as a new question, that's the right way to do things! $\endgroup$ – uhoh Dec 8 '17 at 3:47
7
$\begingroup$

Thrust is a reaction force, so yes this formula provides instantaneous thrust. I think what you're asking is how does thrust (force) translate into velocity. Force is a rate of change of momentum (i.e. F=ma). To get velocity you need to integrate that over time.

If you think about it, you're given F and m, you solve for acceleration and then integrate over time with an initial condition to get velocity. I think by asking for thrust over time you are meaning how much work is performed. Use work formulas, such as this one:

enter image description here

for the amount of work over a dt of time or the full amount over a span of time here:

enter image description here

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for your answer. My physics courses have rusted a bit. This helps a lot. $\endgroup$ – ChrisR May 26 '14 at 15:54
4
$\begingroup$

Solar sails are unique when compared to other propulsion methods because in a solar sail, the acceleration is continuously available throughout the flight. Steady solar flux on the sail produces a steady force, albeit reducing with distance to the Sun.

Of course it depends on angle, and the occasional eclipse if passing through a body's shadow will cause a dip.

The equation you've provided calculates force at a given moment in the craft's flight. However as earlier mentioned, thrust in solar sails is not temporary, it is a steady process.

So to get the solution you want, you can integrate the equation using calculus and magic, but if you're like me, you want a simpler method for research purposes. The method I used was to simply solve the equation multiple times for differing values of distance from the Sun in AU, according to a hypothetical flight plan. Now I had a dataset of thrust values from Earth till Neptune, in steps of 1 AU.

It is simple now to calculate a rough estimate of the sail's final velocity using these 'steps of thrust'. Try this yourself using steps of any size, though I recommend writing a small computer program to automate the task like I did.

Improve this answer
$\endgroup$
2
  • $\begingroup$ If you're going to take a solar sail to Mars the D^2 term is going to more than double over the flight and the thrust drop correspondingly. It's nothing like constant. $\endgroup$ – Loren Pechtel May 22 '14 at 22:24
  • $\begingroup$ I didn't mean constant in value, I meant constantly present. You may edit my answers wording for clarity $\endgroup$ – Vedant Chandra May 23 '14 at 9:56

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.