Thrust for solar sail equation derivation

Question

In the new SMAD (Chapter 18 Space Mission Engineering and Analysis) they have the equation for thrust of a solar sail and a reduced version. Variables are described in this question.

I would like to reproduce this simplification. Can someone explain how they got from the first one to the simplified version? What numerical values (the actual values too, not only their descriptions) were likely to have been used to get the numerical value of $9.113 \times 10^{-6}$?

Answer 1

Unit

When dealing with a magic number, it's important to know what its unit should be.

$$F=\frac{2RSA}{c}\sin^2\theta=9.113\times10^{-6}\frac{RA}{D^2}\sin^2\theta$$

• $F$ is a thrust (=Force) in Newton.
• $R$, $2$ and $\sin^2$ are unitless.
• $A$ is an area in $m^2$.
• It means that $\frac{S}{c}$ is a pressure in Pascal.

By simplifying $R$, $A$ and $\sin$, we get:

$$\frac{2S}{c}=9.113\times10^{-6}\frac{1}{D^2}$$

• $D$ is a distance in $\mathrm{AU}$
• It means that $9.113\times10^{-6}$ is a force in a non-conventional unit : $\mathrm{Pascal} * \mathrm{AU}^2$

Value

This equation should be valid anywhere in the solar system. In particular, it should work on Earth ($D = 1 \mathrm{AU}$), where $S$ is the solar constant : $1361 \mathrm{W/m²}$.

Using qalculate with $2*S/c$:

> 2 * 1361W/m² / c to Pa

  ((2 * (1361 * watt)) / (meter^2)) / speed_of_light = approx. 9.0796147 uPa

Close enough! It seems that $1366 \mathrm{W/m²}$ has been used to get $9.113\times10^{-6} \mathrm{Pa}$.

Using inverse-square law, you get the desired equation.