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Selected Articles from the
May 2001 Odyssey

Editor: Terry Hancock

Equations don't look right? You can also view this article with the equations rendered as GIF files.

Theory of Light Sails

Terry Hancock

Why Does a Photon Have Momentum?

Photons have no mass, and momentum is proportional to mass and velocity, right?

p = mv

So why does a photon have momentum?

The answer is that, in relativity, a photon does have mass - it just has no ``rest mass,'' but then again, photons are never at rest! The mass we're interested in here is the equivalence mass from the photon's kinetic energy (which is simply the energy of the photon)

E = hν
where h is ``Planck's Constant'' and ν is the frequency of the photon. So far, so good.

But now along comes Einstein, who tells us that this energy can be rewritten as a mass:

E = mc2
(I know you remember this one!), where c is the speed of light. For the photon, this energy, E , is the same as the one in the previous equation, so we can stick them together and solve for the equivalent mass of a photon:
mc2 = hν
m = hν 


Then we just convert this to momentum (Newton and Einstein agree on this part):

p = mv =hν 


We could also have figured this out by relating the energy to the momentum first, which is what physicists usually prefer, but I think this approach is more intuitive, since we're used to thinking of momentum in terms of mass and velocity.

Pushing a Sail

Okay, so now we have a momentum in theory - but how much is this momentum, and how do we capture it?

Well, capturing it is easy - all you have to do is reflect the photon with a mirror. When you do this, the momentum of the photon will be reversed, and by the conservation of momentum and Newton's third law, the mirror will get a tiny little push. I say tiny because it is, and that hardly does it justice! In fact, even with hordes of photons from a very bright light source hitting it, this force is almost impossible to measure, much less use as a propulsion system.

But we can calculate exactly what this force should be for any incoming flux of light. Each photon imparts a tiny quantum of momentum that we computed in the last section:

p =hν 

of which we get twice the effect by reversing the direction of travel. If we didn't reflect it straight back, but rather at some angle (by turning the mirror), we'll get a slightly different amount and direction of momentum - an important point that we'll come back to. Meanwhile, we have to sum up the impact of all those N photons hitting our mirror every second, multiply in that factor of two, and finally do a little algebra to make the result easier to interpret:
p = 2  


= 2  




Let's say our mirror is exactly one square meter in size. Why those specs? Because, if we pull the speed of light out of the sum like this, then we've reduced this to just the amount of light energy hitting each square meter of the mirror per second, and that's called the flux, F. Why is that good? Because now we can drop the equations and go look up the flux for our light source. Assuming that's the Sun, and that we're somewhere in the vicinity of Earth, this is about 1.4kW/m2 which gives us a momentum per square meter of sail per second, which can also be called a pressure (force per unit area) of:
P = 2

F =2·1.4kW/m2

= 9.3×10-5N/m2

Not a lot, is it? In space, though, where there's no friction to slow us down, we can still make this work. We have to use extremely light-weight sails made of highly-reflective Mylar (or a similar material). If we can keep the mass down to only a few grams per square meter of sail (including the payload!), we can get useful accelerations out of this. Keeping that mass down is the main engineering challenge for solar sails.

Now we can figure out what kind of acceleration we'll get for a 5g/m2 sail, by just using Newton's second law ( F = mα ):

α = F


= 0.0186m/s2 = 0.0019gees
Whew! This is the acceleration we can expect for a sail with that much weight per area, regardless of its actual size. Larger size sails can carry more cargo, of course, as the payload weight is included in the mass figure above.

Just to see how useful that amount of acceleration is (it may not seem like much now), let's see how long this sail would take to get up to 10,000m/s , a fairly typical ΔV for interplanetary flight:


= 5.38×105s = 6.2days

Wow, less than a week! That's a very practical acceleration for travelling to Mars, for example, where the transit time is normally measured in months (in other words, this extra time to accelerate won't hurt us much compared to chemical rockets which accelerate much faster). And we don't use a gram of fuel to do it! So we can come back and do it all over again, which is something you definitely can't do with chemical rockets.

The Cosmos-1 Solar Sail

Given the published specifications on the Cosmos-1, we can compute the numbers above for that specific sail:

Mass per unit area (for a round sail):

μ =m


3,1416·( 30m) 2
= 14g/m2


α =F

= 9.3×10-5N/m2

= 0.0066m/s2

Time to accelerate to 10,000m/s :


= 1.51×105s = 17.5days

Of course, these are only rough ``ballpark'' figures - there are a lot of details missing from this calculation. But this should give you a good idea of what is possible with this technology.

File translated from TEX by TTH, version 2.25.
On 22 Jul 2001, 21:34.