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?

  • $\begingroup$ Can anyone explain how they simplified from first thrust equation to simplified version? Thank you $\endgroup$
    – Ryan
    Commented Dec 7, 2017 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
    Commented Dec 8, 2017 at 0:42
  • $\begingroup$ @MBM you can leave this as a complete answer to this question - it has now been asked separately. $\endgroup$
    – uhoh
    Commented Dec 8, 2017 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
    Commented Dec 8, 2017 at 3:47

3 Answers 3


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

  • $\begingroup$ Thanks for your answer. My physics courses have rusted a bit. This helps a lot. $\endgroup$
    – ChrisR
    Commented May 26, 2014 at 15:54

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.

  • $\begingroup$ 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. $\endgroup$ Commented May 22, 2014 at 22:24
  • $\begingroup$ I didn't mean constant in value, I meant constantly present. You may edit my answers wording for clarity $\endgroup$ Commented May 23, 2014 at 9:56

Just a little calculations on the acceleration for a solar Sail:

The impulse p of a photon of frequency ν, thus of energy hν , thus of wavelength λ = c/ν on the sail is p = hν/c = h/λ where h is Planck's constant (6.6 10^-34 Joules). To simplify the calculations, assume that the sun does not emit a whole rainbow of colors, but only the yellow-green color (λ = 500 nm) . We have then p = 1.32 10^-27 kg.m.s-1 and the energy Ep = hν of a photon will be 3.957 10^-19 J

Our sun has a total power Es of 3.9x10^26 W. It emits np = Es/Ep = 9.85 10^44 photons per second. But the Earth is 149 million kilometers from the sun. All these photons are distributed on a sphere of the same radius R, whose surface is S = 4πR2 or 2.79 10^23 m2. This gives us a density d = np/S = 3.53 10^21 photons per square meter per second.

Quite a lot! But each of these photons carries a tiny pulse : (the p above). The impulse received by a square meter of solar sail will therefore be i = p.d = 4.66 10-6 kg.m-1.s-2 In other words, a sail of 1m2 weighing 4.66 milligrams will accelerate at 1 m.s-2, that is to say about one tenth of a g. If we want to accelerate a mass of 1 kg with this same acceleration, the surface of the sail will have to be 1/i = 214 456 m2, a square of 463 m of side (and thus weighing 1 kg!). This requires extremely light materials, but it is not utopian.

Note : if the craft is a a distance D (in astronomical units) from the sun, just multiply the above acceleration with 1/D^2

However there is an easier way.

Radiation pressure from reflection is just $$P = 2\frac{I}{c}$$ Newtons per square meter where $I$ is the intensity of the light in watts/m^2 and $c$ is the speed of light. At 1 AU $I$ is the solar constant of about 1361 W/m^2. If we do it this way we don't need to worry about approximating with a single wavelength.

To conclude and answer the question, if you have the radiation Pressure (in Newtons), you divide it by the craft mass (in kg) to get acceleration in m.s-2(Newton's F=mA). Once you have the craft acceleration, just integrate over time to get the velocity increase. Obviously, as the craft distance from the sun may change, you will need to do it in (likely thousands of) steps with a computer.

  • $\begingroup$ Welcome to the site! It's not clear where in your answer you specifically address the central question in the OP's post. Can you edit to clarify? $\endgroup$
    – called2voyage
    Commented Jun 10, 2021 at 15:15

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