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What does the math look like to calculate the size of an object in LEO that when angled off the sun, would illuminate a given portion of the Earth's surface?

For example, if I wanted to "light up" a square mile of Earth's surface, how big would the reflecting object need to be to achieve this?

I'd like to see the math that calculates this.

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    $\begingroup$ How much light do you want? $\endgroup$ Aug 27 '18 at 16:05
  • $\begingroup$ Also, what is the shape of the area to be lit? Square? Circular? And is the light level to be constant over the entire area, or can it vary a little from center to edges, as long as it stays above a minimum value? $\endgroup$ Aug 27 '18 at 16:49
  • $\begingroup$ You might want to look into "iridium flares", where something like this happens accidentally. With a normal flat mirror, you're collecting a small amount of sun and spreading it out over a large portion of the Earth. You might be able to accomplish what you want with a parabolic mirror. Under unachievably perfect conditions, you could probably fry ants (or people) from space :) $\endgroup$
    – user7073
    Aug 27 '18 at 17:05
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    $\begingroup$ Keep in mind that LEO isn't far away, it is fast. If you are trying to constantly illuminate a fixed point you will need a new satellite every few minutes as the previous one falls toward the horizon. Unrelated, but worth looking into is Rjukan Norway that already does what you are asking on a smaller scale with the mirrors mounted on a hillside. $\endgroup$
    – Lex
    Aug 27 '18 at 20:55
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    $\begingroup$ This has been done as an experiment: smithsonianmag.com/smart-news/… $\endgroup$
    – Hobbes
    Aug 28 '18 at 9:24
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A rough calculation based on a small (less than km) mirror:

Light from a small mirror will have the sam divergence as the incident light. The Sun’s angular diameter is about 10mr. So a mirror in a 300 km orbit will create a 3km diameter spot on the earth

So the intensity of the illumination, compared to Sunlight, is just the ratio of the mirror area to that of the spot:

$$ I_{\rm{delivered}} = (r/1500m)^2 I_{\rm{sun}}$$

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  • $\begingroup$ Diffraction will increase the divergence especially for small mirrors. $\endgroup$
    – Uwe
    Mar 28 '20 at 10:20
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    $\begingroup$ True, but it would have to be a very small mirror to have a divergence larger than the spread of light from the Sun. Your eye, a small optic, can resolve features on the Sun and moon unaided (well, except for attenuation of the Sun’s light) $\endgroup$ Mar 28 '20 at 17:21
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Any size of object will reflect enough light to reach the ground. The question is how much light do you want?

As a tangible near-baseline example, a Starlink satellite is visible from the ground. enter image description here

They are about 2.4 m by 1.4 m in base area. (Source)

enter image description here

So at least an area of ~3.3 m^2 is sufficient to illuminate the ground.

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    $\begingroup$ Due to the high sensitivity of the human eye a satellite with a reflecting are of some square meters may be visible. But to get a visible ilumination of the surface of Earth a much larger reflecting area will be neccessary $\endgroup$
    – Uwe
    Feb 27 '20 at 21:38
  • $\begingroup$ Though not really an answer to the question as asked, +1 for introducing a new and more intuitive unit of measure for payload size! $\endgroup$
    – uhoh
    Mar 28 '20 at 4:45

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