What is the physics or math behind that? Light from the sun is essentially aligned by the time it reaches earth. If the mirror is perfectly reflective, a 10 m^2 mirror should light up a patch of Earth roughly 10 m^2 times the cosine of the angle of the mirror. So unless the angle is close to 90°, most of the losses would be from poor reflectivity.
I totally agree it’s a stupid idea. But maybe it’s even worse than I am thinking of?
The Sun has an angular diameter of about half a degree viewed from Earth. 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. You can’t get away with less because a mirror can’t appear brighter than what it’s reflecting; this is a fundamental property of optical systems.
A mirror 600km away and 5km in diameter has an angular diameter of arctan(5/600) = 0.48°, close enough to half a degree. It has an area of π(5km/2)² = 19.6km² which is pretty much 20km².
You can’t get away with less because a mirror can’t appear brighter than what it’s reflecting; this is a fundamental property of optical systems.
I can understand that a single flat mirror cannot ever appear brighter than whatever is being reflected. But why can’t multiple mirrors pointed at one spot have a total intensity greater than that of any one of the mirrors (or a curved dish that focuses the light)?
Multiple mirrors reflecting sunlight can indeed result in an irradiance greater than that of the Sun (this is how solar power towers work). But to do that, you need to cover more than a 0.5° circle of the sky with mirrors. The total irradiance can be as high as you want (up to covering the whole sky with mirrors, which would be equivalent to standing on the surface of the Sun) but the irradiance per angular area can never be higher than the Sun’s.
You can think of it as dividing a mirror of any shape into a bunch of flat mirrors. Each mirror can only show you a reflection of part of the Sun (which is exactly as bright as that part of the Sun) but you can have as many mirrors as you want showing the same part of the Sun, multiplying the light you receive from it (up to filling the entire sky once again).
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).
To get daylight illumination on even a small area from a 600km orbit you’d need about 20 km² of reflectors. Which is obviously absurd.
What is the physics or math behind that? Light from the sun is essentially aligned by the time it reaches earth. If the mirror is perfectly reflective, a 10 m^2 mirror should light up a patch of Earth roughly 10 m^2 times the cosine of the angle of the mirror. So unless the angle is close to 90°, most of the losses would be from poor reflectivity.
I totally agree it’s a stupid idea. But maybe it’s even worse than I am thinking of?
The Sun has an angular diameter of about half a degree viewed from Earth. 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. You can’t get away with less because a mirror can’t appear brighter than what it’s reflecting; this is a fundamental property of optical systems.
A mirror 600km away and 5km in diameter has an angular diameter of arctan(5/600) = 0.48°, close enough to half a degree. It has an area of π(5km/2)² = 19.6km² which is pretty much 20km².
I can understand that a single flat mirror cannot ever appear brighter than whatever is being reflected. But why can’t multiple mirrors pointed at one spot have a total intensity greater than that of any one of the mirrors (or a curved dish that focuses the light)?
Multiple mirrors reflecting sunlight can indeed result in an irradiance greater than that of the Sun (this is how solar power towers work). But to do that, you need to cover more than a 0.5° circle of the sky with mirrors. The total irradiance can be as high as you want (up to covering the whole sky with mirrors, which would be equivalent to standing on the surface of the Sun) but the irradiance per angular area can never be higher than the Sun’s.
You can think of it as dividing a mirror of any shape into a bunch of flat mirrors. Each mirror can only show you a reflection of part of the Sun (which is exactly as bright as that part of the Sun) but you can have as many mirrors as you want showing the same part of the Sun, multiplying the light you receive from it (up to filling the entire sky once again).
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).