To light up a location as brightly as the Sun would, you need to cover a half-degree circle in the sky (viewed from that location) with mirrors that reflect the Sun directly at the location.
That’s the best, simplest example I’ve seen for why this doesn’t work. But…I wanted to look at it from the perspective of irradiance losses from the beam spreading. It’s been a long time since I did any optics, so I could be way off-base with my approach. Feel free to correct anything I screw up.
Here are my assumptions:
Near space irradiance from the sun is 1,367 W/m^2 [0]. Let’s round up and assume the mirror gets 1400 W/m^2 from the sun.
We want 1000 W/m^2 on the ground to qualify as daylight [1]
Collimated light
No attenuation or scatter from the atmosphere, but we will assume the beam diameter spreads 0.5 degrees [2]
Perfectly reflective mirror
Mirror 600 km away from the earth
Beam spreading loss is a function of distance. So however large the beam width (mirror diameter) starts, it’ll be this much bigger when it reaches the ground:
600km * tan (0.5 degree) = 5.24km
That means if we have a 1m diameter mirror, we get a beam 5.24km + 1m on the ground. If we have a 5km diameter mirror, we get a 10.24km beam on the ground.
To get our target of 1000 W/m^2, we need at least 1000/1400=0.71 of what hits the mirror to hit our target.
mirror/(mirror+spread) >= 0.71
mirror >= 12.83km
[0] https://en.wikipedia.org/wiki/Sunlight#Measurement
[1] Wikipedia says that we actually get more like 1100 W/m^2 when the sun is at its zenith.
[2] https://en.wikipedia.org/wiki/Collimated_beam#Distant_sources
I don’t think that’s right. You’re assuming the intensity of the beam is constant over its entire area, which isn’t true. With a 5.24km flat circular mirror*, someone at the center of the beam would see the entire Sun reflected in the mirror and thus get the full solar irradiance. Someone near the edge would see only a small sliver of Sun and get much less.
*Technically a circular mirror would require the Sun’s light to arrive perpendicular to the mirror surface and reflect directly back, which would mean the Sun is behind the Earth, but that’s beyond the scope of this hypothetical (and can be solved with a suitable ellipse).
That’s the best, simplest example I’ve seen for why this doesn’t work. But…I wanted to look at it from the perspective of irradiance losses from the beam spreading. It’s been a long time since I did any optics, so I could be way off-base with my approach. Feel free to correct anything I screw up.
Here are my assumptions:
Beam spreading loss is a function of distance. So however large the beam width (mirror diameter) starts, it’ll be this much bigger when it reaches the ground:
600km * tan (0.5 degree) = 5.24km
That means if we have a 1m diameter mirror, we get a beam 5.24km + 1m on the ground. If we have a 5km diameter mirror, we get a 10.24km beam on the ground.
To get our target of 1000 W/m^2, we need at least
1000/1400 = 0.71of what hits the mirror to hit our target.mirror/(mirror+spread) >= 0.71 mirror >= 12.83km
I don’t think that’s right. You’re assuming the intensity of the beam is constant over its entire area, which isn’t true. With a 5.24km flat circular mirror*, someone at the center of the beam would see the entire Sun reflected in the mirror and thus get the full solar irradiance. Someone near the edge would see only a small sliver of Sun and get much less.
*Technically a circular mirror would require the Sun’s light to arrive perpendicular to the mirror surface and reflect directly back, which would mean the Sun is behind the Earth, but that’s beyond the scope of this hypothetical (and can be solved with a suitable ellipse).