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A spacecraft is powered through a solar sail, which reflects sunlight. The area of the sail is 1 km^2, oriented perpendicular to the direction to the Sun. The spacecraft is at 1 AU from the Sun. The sunlight falling on the sail has a power of 1380W/m^2.

(a) Calculate the momentum in this sunlight, and from this derive the pressure exerted on the solar sail by the sunlight;

(b) What is the maximum mass of the sail for which the force from light exceeds the gravitational force from the Sun?

For (a), which forumulae should be used here? I didn't find anything useful online and have no idea of a systematic textbook that I can read.

For a single photon, the momentum p=E/c which depend on frequency or wavelength. But here only enegry flux and surface area are given.

For (b), I presume I should equate the gravitational force and the light pressure to solve for the mass. Or do I have to consider the escape speed of the solar system?

PS: This question is at the level of first year uni physics.

Thank you.

lkjhgtf
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  • I am not completely familiar with this, but you know that the sunlight falling on the sale has a power of 1380W/m^2 (so not what the solar panel actually generates). I think this means that 1380J worth of photon energy is falling on the panel every second. From this you can just fill in p = E/c where E=1380J – Thomas Wagenaar Oct 08 '17 at 18:14
  • @ThomasW Although the dimensions are correct, I'm still a bit afraid to use it. Actually I'm not even sure how p=E/c for a photon is derived. – lkjhgtf Oct 08 '17 at 18:24
  • For the last part of your comment see this question https://physics.stackexchange.com/questions/2229/if-photons-have-no-mass-how-can-they-have-momentum – Thomas Wagenaar Oct 08 '17 at 18:25
  • Given the enegry flux and surface area, I can get the power. And as P=Fv=Fc, I can get the F, namely the light pressure. It is correct? But the question asks me to derive it from the momentum. Confused now. – lkjhgtf Oct 08 '17 at 18:27
  • For part (a) of your question, you might like to read parts of this Feynman lecture, particularly below Equation (27.15) and around Equation (27.21). – diracula Oct 08 '17 at 18:28
  • @diracula Unluckily I haven't learned enough maths for that... – lkjhgtf Oct 08 '17 at 18:32

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The force exerted by radiation on a surface has a magnitude of $$ F = \frac{1+R}{c}\int \vec{N}\cdot d\vec{A},$$ where $\vec{N}$ is the time-averaged Poynting vector (power per unit area) of the incoming radiation in Wm$^{-2}$, with a direction given by the direction of the electromagnetic waves (or photons). This is integrated over the area $A$ of the solar sail, but the scalar product takes account of the angle of the sail to the radiation. The term $R$ is the reflectivity of the sail, where $R=1$ doubles the force applied to the sail, because the momentum change of the light is doubled.

An important point though, is that this is only the magnitude of the force. The direction of the force will be opposite to the net momentum change of the radiation.

ProfRob
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