Maximum velocity achieved by solar sail

Question

What is the maximum (theoretical) velocity that can be achieved by solar-sail technology? Can we get close to the speed of light with that?

Answer 1

tl;dr:

Can we get close to the speed of light with that?

No, at least not very easily. The terminal velocity $v_{\infty}$ is only about 0.2% the speed of light if you start at 1 AU using a 10 nanometer thick sail, and scales only as the inverse square root of the distance to the Sun where you start accelerating (as well as sail thickness), so you'd melt without getting much benefit from starting really close to it. Passing by a second stars helps little because you'd be going past it so quickly that you wouldn't get much of a second kick.

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The relationship between the momentum and energy of a photon is $ p = E/c$, force is $dp/dt$, and acceleration is $F/m$. For a reflecting sail there is up to a factor of 2 for perfect reflection of normal incidence, and the total incident power (energy per unit time) would be the intensity $I$ of sunlight (energy per unit area per unit time) times the area $A$. So

$$a=\frac{F}{m} = \frac{2AI}{mc}.$$

To give this a test run, a one square kilometer sail made from a futuristic reflecting material with a thickness of 10 nanometers would weigh of order 10 kg. At 1 AU from the Sun, $I$ is about 1361 W/m^2 (the Solar Constant), giving an acceleration of 0.9 m/s^2 which is surprisingly big until you remember that you're redirecting a gigawatt of sunlight in the process.

If you balk at said futuristic 10nm thick mirror, a sub-wavelength spaced wire grid is discussed in the 1999 NASA Institute for Advanced Concepts report Ultra-Thin Solar Sails for Interstellar Travel: Phase I Final Report where values from 0.5 to 4 m/s^2 are plotted in Figure 7.

To get the asymptotic terminal velocity, we can make that work for all distances by normalizing to 1 AU :

$$a(r) = \frac{2AI}{mc} \frac{1AU}{r^2},$$

and then integrate it to infinity but unfortunately I can't remember how to solve differential equations any more, so I will cheat and use the result at the top of page 14 of that report:

$$v_{\infty} = 548,000 (m/s) \sqrt{a_{1AU} \frac{1AU}{r_{start}}},$$

which is only about 0.2% the speed of light starting at 1 AU, and scales only as the inverse square root of the distance to the Sun where you start accelerating.

Table 1 shows some the missions that can be performed for an ultra-light solar sail. An ultra-thin sheet of aluminum only a few nanometers thick can achieve ac of ~0.3 m/s2, could reach Pluto in ~100 days, and reach the Oort cloud at ~10,000 AU within a century. In contrast, current propulsive methods would take over a decade to reach Pluto and are totally impractical to reach interstellar space. A perforated light sail made of aluminum could reach the Oort cloud in a half century if we can achieve ac >0.5 m/s2, and could do so in 12 years if ac ~5 m/s2 can be achieved. In the far term, a sail made of doped carbon nanotubes could probably approach the sun within 4 solar radii, and if it had an ac of 10 m/s2 , could reach α Centauri in a century. A sail made of doped carbon nanostructures could reach our nearest star in a few decades if ac >100 m/s2 as suggested by extensions of microwave theory.

It can thus be seen that the ultrathin solar sail holds the potential to revolutionize the prospects for interstellar travel. Uniquely, with this technology, such missions could not only become feasible, but potentially cheap, since all the motive power is provided in the form of crude sunlight – no giant lasers or other power systems are required.

Answer 2

Assuming that photons are hitting the sail at an unchanging rate and removing variables such as gravity, obstacles, and anything else that would impede the crafts movement then theoretically, yes it is possible to approach the speed of light. However the faster you went the longer it would take to significantly accelerate the craft. This means it could take thousands of years if not more to actually reach the speed of light. Think of it this way. Two race car drivers are moving at different speeds. Car 1 is 20 feet in front of Car 2 yet, Car 1 is moving slower than Car 2, behind him . At this rate Car 2 will eventually catch up to Car 1 after a certain period of time. But if Car 1 speeds up it will take longer for Car 2 to catch up to Car 1. Car 1 is like the Solar Sail while, Car 2 is the Photon hitting the sail. In short, it is possible to reach the speed of light with a Solar Sail, however doing so in real life even outside of the solar system is very, very unlikely.

Answer 3

6.47% speed of light is the engineering limit.

Here are two key factors that determines the terminal velocity $v$.

1. $\sigma$: areal density of the solar sail.
2. $r_0$: perihelion distance.

For solar sail material, single layer graphene is the lightest material so far, and it has the largest areal density $\sigma =7.626 \times 10^{−7} kg/m^2$.

For perihelion distance, it is at least the radius of the sun, which is $r_0 = 6.9634\times 10^8 m$.

Assume the solar sail has surface area 1 square meter (then its mass is $\sigma$) and absorbs 100% light. Let's say the solar sail travels from distance $r_0$ to infinity. Theoretically, the total energy received from sun's radiational pressure is

$E = \int_{r_0}^{\infty} \frac{P_{au} r_{au}^2}{r^2} dr = \frac{P_{au} r_{au}^2}{r_0}$，

Where $r_{au} = AU = 1.49 \times 10^{11} m$ is the Astronomical Unit and $P_{au} = 4.5\times 10^{-6} N/m^2$ is the light pressure at distance 1 AU.

With Energy Conservation $E = \frac{1}{2} \sigma v^2$, we get terminal velocity

$v = \sqrt{ \frac{2 P_{au} r_{au}^2 }{\sigma r_0}} = 19,397,620 m/s = $ 6.47% c

"Too Optimistic:"

1. A single layer of graphene absorbs a mere 2.3% of incident light, allowing around 97.7% to pass through.

2. the single layer graphene sheet can hardly survive the high temperature near the sun's surface.

3. Most acceleration happens within 0.1 AU, which is around 20 times of the sun's radius.

In practice, the closest solar probe is around 10 times sun's radius. And if we also consider graphene transmittance, we get terminal velocity

$v' = \sqrt{0.023 / 10 } v = $ 0.31 % c.