0.01c with a solar sail

[scienceguy]

If we had a square solar sail 10^5 m on one side (that’s right, 100 km on one side) and 1 nm thick, then the solar pressure on this sail summed over from 0.1 to 10 AU (see calculation below) and assuming that the radiation pressure is 4.6 x 10^-6 N/m^2 (from here: http://en.wikipedia.org/wiki/Radiation_pressure  ) would yield:

(4.6 x 10^-5 N/m^2)(10^10 m^2) = 4.6 x 10^5 N

Assuming payload is 10^5 kg:

F = ma,  a = F/m = 4.6 x 10^5 N /(10^5 kg) = 4.6 m/s^2

Mass of sail turns out to be almost negligible if it’s only 1 nm thick:

10^-9 m x 10^10 m^2 = 10 m^3

10 m^3 x (3 x 10^3 kg/m^3) = 3 x 10^4 kg (aluminum)

4.6 m/s^2 accelerated over 10 AU would take this much time:

t = sqrt(2x/a) = sqrt(2(1.5x10^12 m)/4.6 m/s^2) = 807 572s =~ 9 hours

which leads to a final velocity of v = at = (4.6 m/s^2)(807 572 s) = 3714831 m/s^2 = 0.01c

Thus we could accelerate 100 metric tons to 0.01c just by using a big and thin enough solar sail.

[orbitjunkie]

Cool idea! I don't think I can argue with you on your math but I think the underlying assumptions are what will get you.

I did a quick google search and found this:
http://www.niac.usra.edu/files/studies/final_report/333Christensen.pdf

For one thing, it claims a 4nm thick Al sail will only give you 0.47 m/s^2. The paper only alludes to the challenges of manufacturing a massive structure at that precision, much less launching, deploying and supporting it in flight. It takes some pretty exotic technologies to get to close to 4 m/s^2.

Still, there are some pretty interesting things you could do with it. I wonder how much benefit you could get if most of its payload was actually an RTG powering the highest ISP electric thruster you could find? I definitely wish NASA spent more money looking into technologies like this.

[alexterrell]

I'm not sure if integrating pressure is the right thing to do.

You want to start with DV/DT = Constant x Pressure.

Also F = K 1/x^2 is for a point source. I'm not sure if the sun is a point source at 0.1AU. It would also be interesting to consider how close you could start. When does the Al get to 400 degrees?

However, the result seems about right except the acceleration duration is 9 days.

The end velocity is about 2AU per day, which is pretty impressive.

Scale down the mission somewhat and you could do a mission to Kuiper belt with a J-130 launch.

 

[scienceguy]

Ah yes, 9 days instead of 9 hours.

I was integrating pressure because I figured it would just add up over the journey.

Note that, if you make the lower limit on the integral 0.01 AU instead of 0.1 AU, you end up with a final velocity of 0.03c. But that's of course assuming the sail doesn't rip apart being that close to the sun.

[Jim Davis]

I'm not sure if integrating pressure is the right thing to do.
You want to start with DV/DT = Constant x Pressure.

Alex is exactly right here.

Radiation pressure:

k = k0 * R0^2 / x^2

where

k0 = radiation pressure at earth orbit 4.6 x 10^-6 Pa
R0 = radius at earth orbit (1 AU) 1.5 x 10^11 m

For constant mass

F = m * dv/dt

where

m = 10^5 kg

but

F = k * S

where

S = 10^10 m^2

So

m * dv/dt = S * k0 * R0^2  / x^2

and since v = dx / dt

v * dv = S * k0 * R0^2 / m / x^2 * dx

Integrating the left side between 0 and Vfinal and the right side between .1 * R0 and 10 * R0 and solving for Vfinal gives

Vfinal = Sqrt(2 * S * k0 * R0 / m * 9.9) = 1,170,000 m/s

or Vfinal = .004 c

But you've neglected the sun's gravity entirely which is the biggest weakness of the scheme.

[khallow]

But you've neglected the sun's gravity entirely which is the biggest weakness of the scheme.

I'd have to disagree. The Oberth effect and no need for reaction mass guarantees the opposite. That is, your early delta v deep in the Sun's gravity well, is more effective when you climb out of the gravity well than if you weren't in a gravity well at all. And you're also guaranteed a bit more time near the Sun, which since you propulsion system is "always on", means you get more delta v from that effect as well.

[mlorrey]

Ah yes, 9 days instead of 9 hours.

I was integrating pressure because I figured it would just add up over the journey.

Note that, if you make the lower limit on the integral 0.01 AU instead of 0.1 AU, you end up with a final velocity of 0.03c. But that's of course assuming the sail doesn't rip apart being that close to the sun.

There's a paper out this month by a guy proposing reaching 0.03c with a lightsail, however it requires the sail be beryllium, not aluminum, due to mass issues. There is also a problem with getting too close to the sun, the temp the sun heats the sail to causes the beryllium to become transparent (at about 2900 K), so as I recall the closest approach it can make to the sun is about 0.14 AU. After perigee, the sail is undergoing 2 g of acceleration btw, but because the thrust is even over the entire surface of the sail, it doesn't cause structural issues.

Aluminum melts at too low a temp to be useful in this application.

[scienceguy]

Thanks for the responses. Doesn't hurt to toss some calculations out there!

[hop]

Mass of sail turns out to be almost negligible if it’s only 1 nm thick:

nm ? as in one nanometer ? This won't work, even if you can handwave away the practical difficulties of manufacturing and deploying it. Hint: figure out how much of the light from the sun will pass straight through 1nm aluminum, and revise your thrust estimate accordingly.

http://arxiv.org/ftp/arxiv/papers/0901/0901.0047.pdf

Thus we could accelerate 100 metric tons to 0.01c just by using a big and thin enough solar sail.

You can do all kinds of cool things on paper if you make enough absurd assumptions.

[alexterrell]

Thanks for the responses. Doesn't hurt to toss some calculations out there!

So you going to come back with some more numbers?

Assume total mass = 1 SDHLV, say 70 tons. If you can get a cargo to 10 AU in 100 days that's transformational.

By the way, have you read Flight of the Dragonfly (seems to be renamed to rocheworld http://www.amazon.com/Rocheworld-Robert-L-Forward/dp/0671698699/ref=pd_sim_b_1)

[scienceguy]

OK so here’s some more calculations:

The Ikaros sail, currently deployed, has a thickness of 0.0075 mm, or 7.5x10^-6 m.

(http://www.jaxa.jp/projects/sat/ikaros/index_e.html)

I couldn’t find the density of the sail, so I just went with Beryllium (1.85 kg/m^3). (On the Ikaros web page they say that they used polyimide, which has a density of 1.43 x 10^3 kg/m^3, but I was pretty sure they coated it with some metal)

The area of the Ikaros sail is just under 200 m^2, being 14 m on one side.

If we scale up the sail to a side of 1 km, keeping the same thickness, the area is then 10^6 m^2, so the volume of the sail would be:

10^6 m^2 * 7.5x10^-6 m = 7.5 m^3

And the mass would be:

7.5 m^3 * 1.85x10^3 kg = 1.39 x 10^4 kg

Assuming the payload is 8.61x10^4 kg, for a total mass of 10^5 kg, we have (based on Jim Davis’ formula for final velocity):

Vfinal = sqrt(2 * S * k0 * R0 / m *9.9)
= sqrt (2 * 10^6 m^2 * 4.6x10^-6 Pa * 1.5x10^11 m)/10^5 kg * 9.9) = 1.17 x 10^4 m/s, which is nowhere near the speed of light, but it is 2.46 AU/year.

[hop]

If we scale up the sail to a side of 1 km, keeping the same thickness

You need to learn about scaling laws.

[aero]

OK so here’s some more calculations:

The Ikaros sail, currently deployed, has a thickness of 0.0075 mm, or 7.5x10^-6 m.

(http://www.jaxa.jp/projects/sat/ikaros/index_e.html)

I couldn’t find the density of the sail, so I just went with Beryllium (1.85 kg/m^3). (On the Ikaros web page they say that they used polyimide, which has a density of 1.43 x 10^3 kg/m^3, but I was pretty sure they coated it with some metal)

The area of the Ikaros sail is just under 200 m^2, being 14 m on one side.

If we scale up the sail to a side of 1 km, keeping the same thickness, the area is then 10^6 m^2, so the volume of the sail would be:

10^6 m^2 * 7.5x10^-6 m = 7.5 m^3

And the mass would be:

7.5 m^3 * 1.85x10^3 kg = 1.39 x 10^4 kg

Assuming the payload is 8.61x10^4 kg, for a total mass of 10^5 kg, we have (based on Jim Davis’ formula for final velocity):

Vfinal = sqrt(2 * S * k0 * R0 / m *9.9)
= sqrt (2 * 10^6 m^2 * 4.6x10^-6 Pa * 1.5x10^11 m)/10^5 kg * 9.9) = 1.17 x 10^4 m/s, which is nowhere near the speed of light, but it is 2.46 AU/year.

Unfortunately, with the reduced area of the sail your solar force is now about two orders of magnitude less than solar gravity whereas in your previous example your solar force was about two orders of magnitude greater than the solar gravitational force. In the first example it was acceptable to ignore gravity for a first approximation. In the quoted example you can't.

[mlorrey]

you also need to account for the structural mass, which averages about 15% of total mass. Furthermore, you need to figure in a carbon fiber grid as a form of rip-stopping.

The sail that calculated reaching 0.03c was something like 500 sq km in area. This is without any sort of laser augmentation. I spoke to the author of that paper about that possibility and he hasn't yet done any calculations involving thrust augmentation from lasers.

Note also that in order to decelerate from that speed once your sail reaches its target star system, the star needs to have at least the same level of output as our sun or the sail will overshoot its target with excess velocity.

[TrueBlueWitt]

In the longer term..  would a graphene sheets be a good candidate for a solar sail?