3
$\begingroup$

Is there a point where the larger solar sail would not add more thrust? What is the most thrust a solar sail can produce?

Can a solar sail be added to an ion engine and work better?

$\endgroup$
  • 2
    $\begingroup$ A bigger solar sail always adds thrust. The thrust/weight ratio may become a problem though, so the question is, 'at what size does a bigger solar sail stop increasing your acceleration'. $\endgroup$ – Hobbes Aug 4 '18 at 17:11
10
$\begingroup$

As long as other variables are held constant (such as the orientation with respect to the direction to the sun, reflectivity of the sail surface, etc.) the force produced by a solar sail is always proportional to its area—at least, until the sail size gets rediculously large, a significant fraction of the distance to the sun!—so bigger always produces more thrust. The area referenced is the area of the sail projected onto a plane perpendicular to the direction to the sun; if you turn a solar sail edge-on to the sun, you won't get much thrust!

But, as @Hobbes says, even though more area produces more force, it doesn't always produce more acceleration, which is what's important.

Solar sails have a figure of merit—a performance spec—of the amount of mass it takes to make a unit area of the sail: the areal density. Given the areal density $\rho$, the mass of a sail is $\rho \times A$, where A is the total area of the sail.

The maximum thrust a sail can produce (oriented face-on to the sun) is a function of its area, its reflectivity, and the intensity of sunlight falling on it: $$F = \frac {Io A} {r^{2}}\frac{1+R}{2}$$ where F is the thrust; Io is the force produced on a perfectly reflecting surface of unit area oriented face-on to the sun, at a distance of 1 AU; A is the total sail area (without any projection effect); r is the heliocentric distance (easiest if you use AU); and R is the sail's reflectivity, 0 for perfectly absorbing and 1 for perfectly specularly reflecting—for diffusely reflecting surfaces R is smaller than 1.

Holding everything constant except area, this reduces to $$F = k A$$ where k is a constant determined by all the other factors: $$k = \frac {Io} {r^{2}}\frac{1+R}{2}$$

The acceleration produced by a solar sail is just the force produced divided by the total mass of the spacecraft, which is the sail's mass plus the mass of everything else, call it Msc:$$Acc = \frac {k A} {\rho A + Msc}$$When the mass of the sail ($\rho$A) is roughly the same as, or smaller than, Msc, increasing the sail area increases the acceleration. But if you make the sail huge, many times Msc, then increases in the sail area barely affect the resultant acceleration. As the sail mass gets much, much larger than Msc, Msc becomes negligible, and the acceleration equation approaches $$Acc = \frac {k A} {\rho A} = \frac{k}{\rho}$$ or the force per unit area divided by the mass per unit area. This is called the characteristic acceleration of the sail, and is the quantity that sets the limits of a solar sail's performance.

This is the net result: as a sail attached to a fixed spacecraft mass is made larger, the craft's acceleration potential asymptotically approaches the characteristic acceleration, but can never meet or exceed it.

Note: solar sails oriented this way generally don't gain much energy—or any, in the case of a circular orbit. Solar sails are most effective at adding net energy when the force vector aligns with the spacecraft's velocity vector, which is almost never radially away from the sun.

$\endgroup$
4
$\begingroup$

As Tom Spilker showed, there comes a point where more sail adds almost nothing. However, there's another factor: you don't just have a solar sail and a spacecraft. In reality you need something to connect them together. Yes, the forces are low but the lengths are long.

While the sail weight scales linearly with it's area (and thus thrust) the bigger the sail the heavier the cables become in comparison to the sail. There will come a point where making your sail bigger lowers your acceleration.

$\endgroup$
  • $\begingroup$ Indeed, in my effort to keep it simple I oversimplified - thanks for the correction. $\endgroup$ – Tom Spilker Aug 6 '18 at 18:21
1
$\begingroup$

I am agree with @Loren Pechtel.

The math in @Tom Spilker answer doesn't tell about stiffness of construction materiel for solar sail. But when we scale the size of sail's structure we encounter square-cube law. (Scale Eiffel tower 10 times with all proportions - it will fall)

I am not a specialist in engineering and strength of materials, but I suppose as the sail size reaches several hundreds of meters (or, optimistically, several kilometers) - we'll have a problem that supporting constructions become too heavy.

$\endgroup$
  • 1
    $\begingroup$ Fortunately, almost all solar sail designs work entirely on tension, not compression, which makes things a lot more favorable. $\endgroup$ – Nathan Tuggy Mar 25 '19 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.