Sunday, May 16, 2010

Solar Sails I

Solar sails are in the news again, and this time not just for blowing up. The Japanese space agency is launching what they hope to be the first successful solar sail tomorrow. In honor of that, we will be discussing the physics of solar sails.

First of all, what the heck are solar sails? Solar sails are a means of propulsion based on the simple observation that "Hey, sails work on boats. Therefore, they should work on interplanetary spacecraft (in space)." Boat sails work when air molecules hit into the sail and bounce back. By conservation of momentum, this gives the boat sail an itty bitty boost in momentum. Summing over the large number of air molecules moving as wind, the boat gets pushed along in the water. A similar process works with solar sails, but instead of air molecules doing the hitting, it's photons. Since each photon of a given wavelength has some momentum, by reflecting that photon the solar sail can gain a tiny bit of momentum. Summing over the large number of photons coming from the sun over a long time frame we can get a considerable boost. So let's see how good solar sails are.

First we need to find the net force on our sail. We will certainly have to deal with gravitational forces (which will slow us down) :

\[ F_{g} = \frac{-GM_{\odot}m}{r^2} \]

where big M is the mass of the sun and little m is the mass of the sail. Now we need to find the radiation force on the sail. Since force is just rate of change of momentum, we can find the change of momentum of one photon per unit time, then find how many photons are hitting our sail. So for one elastic collision of a photon with the sail, the change in momentum will be

\[ \Delta p = 2 \frac{h\nu}{c} \]

and by conservation of momentum, this will also be the momentum gained by the sail. Now we want to find the number of photons incident on a given area in a given time. This will just be the energy flux output by the sun ( energy/ m^2 s ) divided by the energy per photon. In other words:

\[ f_n = \frac{L_{\odot}}{4\pi r^2}\frac{1}{h\nu} .\]

So now we can get a force by

\[ \text{Force} = \left(\frac{\Delta p}{\text{1 photon}} \right) \times \left(\frac{\text{number of photons}}{area \times time}\right) \times \left( Area\right) \]

which is just

\[ F_{rad} = 2 \frac{h\nu}{c} \times \frac{L_\odot}{4\pi r^2 h\nu} \times \pi R^2 = \frac{L_{\odot} R^2}{2cr^2} .\]

So combining the radiation force with the gravitation force, we have a net force on the sail of

\[ F = \left( \frac{L_{\odot} R^2}{2c} - GM_{\odot}m \right) \frac{1}{r^2} .\]

This can then be integrated over r to find an effective potential, giving:

\[ U = \left( \frac{L_{\odot} R^2}{2c} - GM_{\odot}m\right)\frac{1}{r} .\]

For simplicity, let's just write that

\[ \alpha = \frac{L_{\odot} R^2}{2c} - GM_{\odot}m \]


\[ U = \frac{\alpha}{r} .\]

Now we can start saying some things about this sail. The most straightforward quantity to find would be the maximum velocity. By conservation of energy (and starting from some r_0 at rest), we have that

\[ v_f = \left[\frac{2\alpha}{m} \left(\frac{1}{r_0} - \frac{1}{r_f} \right) \right]^{1/2} \]

So as r_f goes to very large values, the subtracted piece gets smaller and smaller. In the limit that r_f goes to infinity we have that

\[ v_{max} = \left(\frac{2\alpha}{mr_0}\right)^{1/2} .\]

Plugging back in our long term for alpha and plugging in some numbers we get:

\[ v_{max} = 42,000 m/s \left( \frac{1.5 \times 10^{-4}}{\sigma} - 1\right)^{1/2} \]

where sigma is just the surface mass density [g/cm^2] of the sail. Below is a plot of maximum velocity ( m/s) plotted against surface mass density (g/cm^2). For a sigma of 10^-4 g/cm^2, we get a max velocity of about 30,000 m/s. Not bad.

From this graph we see that there must be some maximum surface density, above which we don't get any (forward) motion at all. This makes sense since we want our radiation forces (which scale with area) to overcome our gravitational forces (which scale with mass). And below this maximal surface density we see a power law behavior. Cool.

We can also find the distance traveled as a function of time. Taking the final velocity equation above and writing v as dr/dt, we see that

\[ \frac{dr}{dt} = \left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r_f} \right)\right]^{1/2} \]

Rearranging and integrating, we can get time (in years) as a function of distance r (in AU):

\[ t = \frac{0.11 \left(\sqrt{(-1+r) r}+\text{Log}\left[1+\sqrt{\frac{-1+r}{r}}\right]+\frac{\text{Log}[r]}{2}\right)}{\sqrt{-1+\frac{1.5 \times 10^{-4}}{\sigma}}}\]

A plot of t vs. r is shown below for typical solar system distances and a sigma of 10^-4 g/cm^2. We assume that we are launching from earth (1 AU). Since Pluto is at a distance of about 40 AU, we see that our sail could get there in less than 7 years. For comparison, the New Horizons probe will use conventional propulsion to get to Pluto in 9.5 years (and it is the fastest spacecraft ever made).

Zooming in to our starting point around 1 AU, we see that there is a period of acceleration and then the maximum velocity is reached after a few months. Just eyeballing it, it looks like it takes at least a month to reach appreciable speed. That it takes so long is a result of the very small forces involved due to radiation pressure. But even a small acceleration amounts to a considerable speed if applied for long enough!

Now Pluto is fine I guess (it's the second largest dwarf-planet!), but how about some interstellar flight? Well, the nearest star is Proxima Centauri which is about a parsec away. A parsec is 3*10^16 m, or about 200,000 AU. From, the plot below (or plugging in to the equation above), we see that such a trip would take of order 10,000 years. That's a long time, but its not too shabby considering this craft uses no fuel of its own.

So solar sails can do some fairly impressive things simply by harnessing the free energy of the sun. Though this only provides a very small acceleration, it can be taken over a long enough time to be useful. However, since the radiation pressure of the sun falls off as 1/r^2, we start to observe diminishing returns and the sail reaches a max velocity. But overall the numbers seem fairly impressive. All that remains now is whether they are feasible to construct. Right now my only data point for feasibility was that it was in Star Wars, but as I recall that was a long time ago.


  1. Our class came across your post in a discussion of Star Wars Science. We're trying to work through your calculations but got stuck with the integral. Could you give us a clue to kick things off please?

  2. Which integral are you stuck on? I was glancing through Corky's post, and it looks like the two places he integrated are, first, integrating the force to get the potential energy, and, second, integrating dr/dt to get r(t). I'd guess you're having trouble with the last one since that looks like a pretty nasty result . . . .

  3. Sure thing. I would first suggest that you do a substitution so that your integration variable is a dimensionless parameter, then do a trig substitution. Then you'll need a bit of integration by parts magic, which should give the answer.

    Although, to do this post I just used mathematica (or equivently, one could use ).

    I will post an addendum which goes through the integration in a few moments.

  4. Well, I'm not sure if that counts as "a few moments," but you can find a full solution and a schematic solution (hints and such) in the two posts below this one.

    I had initially TEX-ed them up as pdfs, but apparently blogger doesn't let you post pdfs, so for now a post will have to do. If you would prefer pdfs, just send me an email through the virtuosi email address listed above and I will send them out.

    Good luck with your calculations and let me know if I've made any mistakes!

  5. Wow this is really amazing. This voile solaireseems to be well made and durable.