Solar Sail Physics - Do they work on a large scale?

  • #1
Hello physicists! I'm hoping you can help me with a question I'm pondering for a book.

At what acceleration could a 4 km^2 solar sail move a 1 metric megaton starship? Assume we're using the solar radiation from the Sol system as our force and that the solar sail in questions is as low mass and highly reflective as theoretically possible.

Basically, I'm trying to understand if solar sail is an economical method of intrasystem travel in terms of time, not energy. How large would the solar sail have to be to move something that big at an acceleration of greater than 100 m/s^2? (Assume g-forces on the crew are not an issue.)

How far from the sun is too far for a solar sail to be useful? At what point does solar radiation become too weak to maintain a reasonable acceleration?

Are there any simple rules of thumb for this? (Can we invent them?) Such as 1 km^2 sail per X number of tons? X% decrease in acceleration per 1 AU from the sun? X% change in acceleration from solar radiation per X% change in the luminosity of the star compared to the Sun?

Any wisdom you can add would be helpful. Thanks!

-Monica Mostly
 

Answers and Replies

  • #2
K^2
Science Advisor
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A metric megaton is a lot. That's 10^9 kg. That's the mass of a cube of water 1km across. What could you possibly need to move that is that massive? Trajectory of that mass would remain virtually unchanged with or without a 4km² sail.

Effect on solar sail is dominated by radiation pressure, which is easy enough to compute. At 100% reflectivity, you can get 2x the momentum carried by the light. So the maximum force on the sail is given by following expression.

[tex]F=\frac{A L}{2 \pi R^2 c}[/tex]

Where A is area of the sail, L is total luminosity of the Sun, R is the distance from the Sun's center, and c is the speed of light. Sun's luminosity is 3.846×1026W. So at Earth's orbit, R≈150Gm, the force comes out to 9N per square kilometer.

So to accelerate even a tiniest ship at 1G, you will need thousands of square kilometers of sail, made of material that adds up to considerably less than 1kg per square kilometer.

Hopefully, that answers all of your questions on time-efficiency of solar sails.
 
  • #3
Filip Larsen
Gold Member
1,296
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Welcome to PF!

In short, the acceleration for any realistic solar sail will be minuscule and not appropriate for short transit times.

Just to give you a rough idea what we are talking about the upper limit for the force obtainable. Since solar light has a power P = 1360 W/m2 at the Earths distance from the sun, you can expect a reaction force of F = 2*A*P/c for fully reflected light over an area of A (with c being the speed of light) which means that to give the mass m an acceleration a from this force you need an area of A = 2*m*a*c/P. To give m = 1 kg an acceleration of one G, a = 9.8 m/s2 you therefore need no less than 4.3 km2 of sail, and that is even with the very unrealistic assumption that the sail itself has no mass. Add solar sail mass and maneuvering that gives realistic interplanetary trajectories and you quickly realize that solar sails are not going to accelerate fast or "go" anywhere quick, at least not compared with other existing or feasible propulsion technologies.

You may want to read some more background material about solar sails on [1].


[1] http://en.wikipedia.org/wiki/Solar_sail
 
  • #4
K^2
Science Advisor
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Filip, A=m*a*c/(2P) from your equations above, giving a little over 1km², which is the same result that I give in a post right above yours.
 
  • #5
Filip Larsen
Gold Member
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Filip, A=m*a*c/(2P) from your equations above, giving a little over 1km², which is the same result that I give in a post right above yours.
You are right. I managed to mess up moving a factor of 2 to the other side :rolleyes:
 
  • #6
That's what I was afraid of. I was hoping I was doing the math wrong. Thanks everyone!
 

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