Question

It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. How large must the surface area of the sail be if the radiation force is to be equal in magnitude to the Sun’s gravitational attraction? Assume that the mass of the ship + sail is 1500 kg, that the sail is perfectly reflecting, and that the sail is oriented perpendicular to the Sun’s rays. See Appendix C for needed data. (With a larger sail, the ship is continuously driven away from the Sun.)

Short answer

The surface area of the sail, $\mathrm{A}=9.5\times {10}^5{\text{m}}^{\text{2}}$.

Step-by-step solution

Step 1: Given data

Mass of ship and sail, m=1500kg .

Sail is perfectly reflected.

Determining the concept

Here, we need to use the concept of force developed due to the radiation pressure. The force acting on a perfectly reflective surface of the area as a result of incident radiation of intensity I is given by$\mathbf{F}\backslash \mathrm{user}1=2\mathbf{IAc}$, where$\mathbf{c}$is the speed of light.

Formulae are as follows:

For purely reflecting surface, the radiation pressure, $p=\frac{2I}{c}$.

Where c is the speed of light, and I is the intensity of radiation.

Attractive force due to Sun on mass m is,$F=\frac{Gm{M}_s}{d^2}$

Where M_S is the mass of the sun, F is the force, d is the distance.

(a) Determining the surface area of the sail

The spaceship is propelled in the solar system by the radiation pressure, and at the same time, it’s attracted by the sun. So write force equations as follows:

$F_s=\frac{Gm{M}_s}{d^2}=pA=f_r$

Where A is the surface area of the sail, and p is the radiation pressure, and where F_s, F_r are forces due to sun and radiation.

For purely reflecting surface, radiation pressure,$p=\frac{2I}{c}$

$\frac{Gm{M}_s}{d^2}=\frac{2IA}{c}$

Hence, the Surface area can be calculated as,

$A=\frac{Gm{M}_s}{d^2}\times \frac{c}{2I}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} A=\frac{6.67\times {10}^{-11}\times 1500\times 1.99\times {10}^{30}\times 3\times {10}^8}{{\left(1.50\times {10}^{11}\right)}^2\times 2\times 1.40\times {10}^3} \\ =\frac{6.67\times 15\times 1.99\times 3\times {10}^{29}}{1.50\times 2\times 1.40\times {10}^{25}} \\ A=9.5\times {10}^5{\text{m}}^{\text{2}} \end{gathered}$$

Therefore, the surface area of the sail, 9.5x10⁵ m² .