jaksa
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I got ##F=0.3849 \cdot F_{0}## and the approach from my first post gave me formula of:jbriggs444 said:I came up with a different formula following a different line of reasoning and used the angle of the sail rather than the angle of its normal. But I suspect that both yield the same result and that both optimize at the same angle.
Yes, after a little behind the scenes work, both come out the same.
[I used the sine of the angle of the reflected ray (twice my alpha) to get its tangential momentum component and the sine of the angle of the mirror to get the cross section. After some trig and double angle formula evaluation, that comes to ##2 \sin^2 \alpha \cos \alpha##. The factor of two comes out of the computation of ##F_{0}##. Mine is not doubled. The ##\sin## versus ##\cos## swap comes from the choice of using sail angle rather than sail normal angle].
You can make the substitution of ##1-\sin^2 \alpha## for ##\cos^2 \alpha##. That will give you a simple polynomial in ##\sin \alpha##. You can optimize that (solve for a zero first derivative) and then take the arc sine of the result. Then plug that angle into your formula for tangential force.
You will want to solve for ##F_0## either using the figures from the solar sail article or from first principles using ##E=pc##.
$$\dot{R} = \frac {2 \cdot 0.3849 \cdot F_{0} AR^2_{0} v} {GMm} \approx 3.5 \frac m s$$
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