Built on Facts

More Sailing

There’s an interesting question in the comments of the last solar sail post:

I have a question that’s been bugging me about solar sails for ages: what about the fact that light pressure falls off over distance? Every time I see the idea discussed, this is never mentioned…

He’s right. As the sail gets farther from the sun, the intensity of the light reaching the sail diminishes. By the same token, the sun’s gravitational force is diminishing as well. To keep going, the radiation pressure has to be greater than the force from the sun’s gravity.

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Where I0 is the power output of the sun (3.07 x 1025 W, by my rough calculation), and k is some proportionality factor depending on the sail, used to write that intensity as a force. And the distances d cancel. So if the sail is ever good enough to resist the sun’s gravity, it will never completely stop working due to diminished light because gravity is falling off at the same rate.

The total force (and thus the acceleration) is the difference of those two quantities, and will be decreasing since d is getting larger, reducing the forces in both directions. So the sail will become less and less effective.

You can find the final kinetic energy by integrating that force from the starting point at the Earth’s orbit to infinity. You can use that to find the final velocity.

Reader Challenge: Find that speed! Don’t worry about orbits, just assume the sail is going in a straight line path directly out of the solar system. Bonus points for also plugging in some reasonable values for k and finding an order-of-magnitude estimate for the actual speed.

Update: Here’s the equation I get for final velocity:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

R is the sail’s starting distance (earth’s orbital distance), and sigma is the mass per area of the sail. Assuming no payload other than the mass of the sail itself, the size of the sail cancels out of the final equation.

Comments

  1. #1 Zifnab
    August 6, 2008

    “– and k is some proportionality factor depending on the sail –”

    This would be based on the size and shape of the sail, right? Like, a sail the size of an office building would give you a bigger “k” than a sail the size of a piece of writing paper?

    If that’s the case, what is the suitable size for a space sail? I mean, you’re in space. You can basically build the damn thing as big as you want, can’t you?

  2. #2 grad
    August 6, 2008

    I’d guess as big as you can go until drag from the few molecules out there starts to kick in. Frankly that’s probably going to be many times larger than the biggest sail you could want.

  3. #3 Nentuaby
    August 6, 2008

    I’d say the actual ability of human space industry to fabricate and launch a sail ought to be taken into account, Zifnab. We probably won’t have the ability to truly build in space anytime soon, so at most we could launch about a 71 tonne vehicle (that’s the estimated capacity of an Ares V to the moon; I don’t know what it could truly escape with, but we’re working on orders of magnitude.)

    The current lightest sail material we can actually make is a carbon fibre fabric at about 3 grams a square metre. (I’m sourcing all this from That Famous Wiki, btw, so pinch of NaCl and all that).

    So… Ignoring structure… We get about 23,666,666 square meters of fabric up… Or about 4.8 km on a side. No significant digits were harmed in this egregiously back-of-the-napkin calculation.

  4. #4 Antagonist
    August 6, 2008

    Hey Matt,

    I found it curious that you didn’t mention that the amount of force on the sail also depends on the portion of the light reflected. The difference between complete absorption and complete reflection is a factor of 2, which is a big deal when we are talking about such weak forces. Also, I hadn’t thought about it before, but the way the 1/r^2 properties of gravity and an isotropic power source cancel out is rather exquisite.

  5. #5 Matt Springer
    August 6, 2008

    Thanks Nentuaby for the numbers, and Antagonist for making another good point. The equation I’ve written gives the radiation pressure of light being absorbed my a material. if the light actually gets turned around by reflection there’s been twice the momentum transfer.

    Using Nentuaby’s numbers and that factor of 2, I get a final velocity of about 72 km/s. This is several times faster than Voyager 2′s current speed asuuming I haven’t screwed up the calculation somewhere. I’ve added the equation to the original post.

  6. #6 Dunc
    August 7, 2008

    Thank you very much for answering my question. Now I have another: what about the solar wind? Can you “sail” on that, and how would it stack up against pure light pressure?

  7. #7 Winter Toad
    August 7, 2008

    One comment about solar sails. Currently feasible designs using aluminized mylar do not feel a radiation pressure higher than the sun’s gravity. The effect of deploying such a sail in normal incidence to the sunlight is as if you had reduced the mass of the sun. The sail moves in a conic section orbit, and unless you gave it a good initial velocity, it traces an elliptical path that returns it to its starting point after one orbit.

    If a single sheet solar sail were to be made using the currently available materials, it would be deployed angled to the sun, in such a way as to maximize the tangential thrust. The radial thrust does no work. The sail should be angled to maximize sin(theta)sin(theta)cos(theta), where theta is the angle between the radial vector and the normal vector of the sail plane. That’s about 0.96 radians.

    Note that this means the solar sail can move inwards toward the sun just as easily as it can move out, because it is applying tangential thrust, and the radial thrust component has no effect to first order.

    Of course, with all that big sail material angled, you sacrifice a lot of light-capturing ability, so a more sophisticated system would have the sail set up as a curved concentrator that reflected the light back to a secondary mirror. That way the primary could stay face-on to the sun, while the much smaller secondary bounced the light in almost any direction, making it easy to adjust the direction of thrust. With a spin-stiffened sail this would be very handy, since it takes a lot of torque to alter the angle of a rotating planar sail.

  8. #8 Rick Pikul
    August 7, 2008

    Regarding solar wind v. light pressure: The solar wind imparts 3-4 orders of magnitude less force than the light from the sun. If you want to use the solar wind you have to build a magsail, which has its own set of advantages and disadvantages.

    Of course, there is one other issue that large solar sails start bumping into: Jon’s Law, because the larger solar sails enter into the territory of being interesting space drives, and any interesting space drive is also a WMD. (Solar sails are, however, rather hard to aim.)

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