Can someone who's better than me at math help me figure something out for a short story
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Can someone who's better than me at math help me figure something out for a short story
The solar constant is 1.361 kW/m² at 1 AU (the amount of solar energy that falls on a square meter). This figure obeys the inverse square law. How far away from the sun would 10 km² of 100% efficient photovoltaic panels need to be in order to collect a terawatt of energy? What equation can I use to play with these variables (distance from sun and surface area of solar panels)?
Edit: got an answer, thanks!
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@drewdevault so an area of 10 x 10 km, that is 10^8 sq meter. You want 10^12 W, so you need 10^4 W or 10000 W per square meter. You get 1361 per sq meter at 1AU, so your distance should be sqrt(1361/10000) AU.
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@dascandy this sounds about right, let me see...
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@dascandy yep that checks out and I can see how to play with the figures from your explanation, thanks very much!
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Though I have an answer, for anyone who wants to take a whack at this themselves, I double checked the work against a known solution, which is the solar flux at Mercury's orbit: 8,750 W/m² at 0.44 AU.
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@drewdevault spherical solar panel or flat one?
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@KasTasMykolas ooh good question
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@KasTasMykolas assume a spherical solar panel on a frictionless plane
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@drewdevault I see. Probably it won't make much of the difference at these "small" sizes of the panel. Only keeping the angles stable :}
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@KasTasMykolas indeed
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Follow up question, now for #optics people: what are the state of the art and near-future potential values for lasers which are optimized for a narrow beam divergence?
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Drew DeVaultreplied to Drew DeVault last edited by [email protected]
(Yes, the story involves a large photovoltaic installation stationed near the sun which shoots lasers at solar sails)
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@drewdevault Many collimating systems/lasers regularly reach the theoretical limit. You can achieve whatever divergence you want provided that the exit pupil of the collimating system is sufficiently large. This is limited purely by diffraction.
A rough equation for this, ASSUMING A SINGLE MODE LASER, is:
divergence (radians) = 1.22 lambda (meters)/diameter(meters)
where lambda is the wavelength of the laser and diameter is the exit pupil diameter. This is only correct when measured at a distance greater than about 10*diameter.
So choose your divergence and figure out how big the optics must be. Remember that with this configuration, your final spot will never be smaller than diameter.
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@uvbritelite thank you!
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@uvbritelite does it still hold that you can basically get whatever divergence you want if what you want is a laser that remains relatively narrow over astronomical distances? i.e. up to 500AU
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@drewdevault call me Kris De Decker but if an advanced civilisation is trying to shoot giant solar-powered lasers at solar sails, they should skip the conversion losses by just using a giant mirror array to focus the light directly.
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@PuddleOfKittens not sure that this works over extreme distances
