Suggested languages for you:

Americas

Europe

24P

Expert-verifiedFound in: Page 1002

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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.)**

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

Mass of ship and sail, m=1500kg .

Sail is perfectly reflected.

**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.

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,

$A=\frac{6.67\times {10}^{-11}\times 1500\times 1.99\times {10}^{30}\times 3\times {10}^{8}}{{\left(1.50\times {10}^{11}\right)}^{2}\times 2\times 1.40\times {10}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{6.67\times 15\times 1.99\times 3\times {10}^{29}}{1.50\times 2\times 1.40\times {10}^{25}}\phantom{\rule{0ex}{0ex}}A=9.5\times {10}^{5}{\text{m}}^{\text{2}}$

Therefore, the surface area of the sail, 9.5x10^{5} m^{2} .

94% of StudySmarter users get better grades.

Sign up for free