If building a space elevator is far-fetched for us, then using solar sail technology for space travel can be said to be “just around the corner.” Through articles in “Amateur Astronomer”, we can gain a considerable understanding of the technology and development of solar sails. However, this only tells us the “What” and the “How,” but not yet the “Why.” Now, let’s attempt to use the physics and astronomy knowledge we have encountered to perform a shallow-level analysis of solar sail technology.
The power of a solar sail primarily comes from light pressure (or radiation pressure), while the force from the solar wind itself is less than 1% of the light pressure. As is well known, pressure P = F/S, and force F = \frac{dp}{dt} (do not confuse P for pressure and p for momentum); in other words, the pressure of light is the momentum that light imparts to the receiver per unit time. For a single photon, its energy is E = mc^2 = h\nu, and its momentum is p = \frac{h}{\lambda} = mc = \frac{E}{c}. Thus, the force F is F = \frac{dE}{c dt}. For the Sun, its light pressure is the sum of the pressures of all photons; therefore, F = n \frac{dE}{c dt} = \frac{d(nE)}{c dt}. Here, (nE) is the total energy released by the Sun, and \frac{d(nE)}{dt} = L is the solar luminosity (radiation power). Thus, the solar radiation force is F = \frac{L}{c}, where L = 3.827 \times 10^{26} \text{ W}. This force acts on a spherical surface at a distance r from the center of the Sun, so the light pressure is P = \frac{L}{4\pi r^2 c}.
Assuming the solar sail is uniform with an area S exposed to light, the resulting thrust is F_{ray} = \frac{L \cdot S}{4\pi r^2 c}. If the solar sail is an ideal reflector, the thrust can reach F_{ray} = \frac{L \cdot S}{2\pi r^2 c} (this is a manifestation of the conservation of momentum; imagine that after you collide with a stationary object, if you stick to the object, you have clearly given it your momentum and it starts to move; if you also leave it at the same speed, you have to give it another kick, and this kick adds even more momentum to it). Since we are considering an ideal case, let’s assume F_{ray} = \frac{L \cdot S}{2\pi r^2 c}. If the mass of the solar sail is m = \rho Sh (where \rho is the density and h is the thickness), the acceleration obtained is:
a_{ray} = \frac{F_{ray}}{m} = \frac{L}{2\pi c \rho h} \cdot \frac{1}{r^2}
It can be seen that light pressure, like gravity, is also a force that is inversely proportional to the square of the distance. The above formula gives the maximum possible acceleration obtainable from solar radiation pressure, and it shows that we need to select lighter (\rho) and thinner (h) materials to manufacture solar sails.
Furthermore, if we hope to use light as power to fly into deep space, the light pressure must overcome the gravitational pull from the Sun (these two are always in opposite directions). Written mathematically: a = a_{ray} - a_G = \left( \frac{L}{2\pi c \rho h} - G M_{sun} \right) \frac{1}{r^2} > 0 Solving this gives \rho h < \frac{L}{2\pi G M_{sun} c} \approx 1.52 \times 10^{-3} \text{ kg/m}^2 = 1.52 \text{ g/m}^2.
\rho h is the so-called surface density. Japan’s “IKAROS” has clearly not yet met this requirement; its surface density is approximately 76 \text{ g/m}^2. Of course, its primary mission was to test whether it could generate power, rather than actual deep space exploration (its target was Venus).
Additionally, the above discussion is actually too idealized because it implicitly assumes that all the mass of the satellite comes from the sail. In reality, a significant portion of the satellite’s mass is “non-sail mass” (mass that does not see light or cannot provide power). Let this part of the mass be m', then:
\begin{aligned} a_{ray} &= \frac{L \cdot S}{2\pi c (m + m')} \cdot \frac{1}{r^2} \\ &= \frac{L}{2\pi c (m/S + m'/S)} \cdot \frac{1}{r^2} \\ &= \frac{L}{2\pi c (\rho h + m'/S)} \cdot \frac{1}{r^2} \end{aligned}
Similarly, to make a = a_{ray} - a_G > 0, we must have: \rho h + m'/S < \frac{L}{2\pi G M_{sun} c} \approx 1.52 \text{ g/m}^2 This places requirements on both the surface density and the area of the light sail, because at the very least we need: \rho h < \frac{L}{2\pi G M_{sun} c}, \quad m'/S < \frac{L}{2\pi G M_{sun} c}
This tells us that in addition to selecting lighter (\rho) and thinner (h) materials, we must also make the sail larger (S) while keeping the satellite’s payload as light as possible. From the above formula, it can be calculated that to use solar radiation pressure alone to send a 1kg object out of the solar system, a sail of approximately 660 square meters (a circle with a diameter of 29 meters) is required.
Above, BoJone has conducted a simple study of solar sail technology using basic physics knowledge, hoping to stimulate further discussion. If there are any errors, I hope readers will not hesitate to point them out!
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