Solar Sail. NASA is giving serious consideration to the concept of solar sailing. A solar sail craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is $\mathbf{3}\mathbf{.}\mathbf{9}\mathbf{x}{\mathbf{10}}^{\mathbf{26}}\mathbf{W}$. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometres.(c) Explain why your answer to part (b) is independent of the distance from the sun.

The wave exerts an average force F per unit area and this is the radiation pressure Prad and it is the average value of dp/dt divided by the area. So, the radiation pressure of the wave that totally absorbed is given by equation in the form ${\mathbf{p}}_{\mathbf{abs}}\mathbf{=}\frac{\mathbf{l}}{\mathbf{c}}$While the radiation pressure of the wave that totally reflected is given by equation${\mathbf{p}}_{\mathbf{abs}}\mathbf{=}\frac{\mathbf{2}\mathbf{l}}{\mathbf{c}}$

If we compare the above two equations, we conclude that, the pressure due to the radiation of reflection larger than for absorbing. Therefore, the sail should be reflective

The force of gravity of the Sun on the sail is given by Newton's general law $\mathrm{F}=\mathrm{G}\frac{\mathrm{mM}}{{\mathrm{r}}^2}$Where m is the mass of the spacecraft while M is the mass of the Sun. The intensity I of the output light is the power per unit area. The average power output of the light P is related to the intensity and the area of the spread spot by $\mathrm{l}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{\mathrm{P}}{4{\mathrm{\tau \tau r}}^2}$

The wave exerts an average force ${\mathrm{F}}_{\mathrm{rad}}$per unit area and this is the radiation pressure Prad and it is the average value of dp/dt divided by the area. So, the radiation pressure of the wave that totally reflected is given by equation (in the form

$$\begin{gathered} {\mathrm{p}}_{\mathrm{abs}}=\frac{2\mathrm{l}}{\mathrm{c}} \\ =\frac{2\mathrm{P}/4{\mathrm{\pi r}}^2}{\mathrm{c}} \\ =\frac{\mathrm{P}}{2{\mathrm{\pi r}}^2\mathrm{c}} \end{gathered}$$

For the area of the sphere, we can get the force that exerted on that spacecraft by the radiation by the next equation

The force of radiation is the same for the gravitational force to propel the spacecraft So, we equal both equations to get A

$$\begin{gathered} \mathrm{F}={\mathrm{F}}_{\mathrm{rad}} \\ \mathrm{G}\frac{\mathrm{mM}}{{\mathrm{r}}^2}=\left(\frac{\mathrm{P}}{2{\mathrm{\pi r}}^2\mathrm{c}}\right)\mathrm{A} \\ \mathrm{A}=\left(\frac{2\mathrm{\pi GmM}}{\mathrm{P}}\right) \\ =\frac{2\mathrm{\pi }\left(3\times {10}^8\right)\left(6.67\times {10}^{-11}\right)\left({10}^4\right)\left(2\times {10}^{30}\right)}{3.9\times {10}^{26}} \\ =6.5\times {10}^6{\mathrm{m}}^2 \\ =6.5{\mathrm{km}}^2 \end{gathered}$$

As shown by equation $\mathrm{A}=\left(\frac{2\mathrm{\pi GmM}}{\mathrm{P}}\right)$the distance between the sun r and the spacecraft is not included. This occurs because both forces the gravitational force and the radiation force are inversely proportional to the square of this distance, and when both forces have the same magnitude, the distance r is canceled.