Isn’t there some inverse square math rule about radiation like this? The further away you are the radiation reaching you is far less than it would seem? Not good at remembering this math so maybe someone can correct me.
Even if you could get the mirrors all focused accurately and tracking the object at speed it seems like it wouldn’t be any more of a concern than a really bright searchlight or something.
The power density square law is for an emitting light source that emits in all directions. Since the incoming light is basically parallel that doesn’t really apply. If you were able to accurately track a satellite (a feat I’m sure is pretty hard) you would definitely vaporize it pretty quickly I’m talking under a minute since space is a good insulator.
But the photons made it through the atmosphere in the first place to be collected by the reflectors. Is there just not enough energy left to make it back out before cooling off?
It’s not even coherent when the sun emits it. For one, it consists of a large range of wavelengths…
And I doubt there’s a way to make light coherent at that order of magnitude.
No that’s not true only about 30% of light energy scatters when traveling through the atmosphere to earth and certain wavelengths are almost completely absorbed in the way down. So on the way back up it should be a high portion make it to the satellite I would imagine 80%. Even worse case scenario 200 megawats shinning on a satellite would vaporize it almost instantly.
It holds if the light spreads wider than the target. So also for directed light sources at large enough distances. Even a perfect mirror must spread the light in the same angle as it is incomming. Hence the beam would at least 3 km wide at the satellite. Therefore the satellite can only recieve a Illumination of ~65W/m^2 which is a few percent of the normal sun brightness of 1300 W/m^2.
Another way to look at it, the mirrors cant make the sun seem brighter only larger. From the tower you see a large solid angle around you the mirror, therefore, it can seem like you are at the surface of the sun. However, fro. the position of a satellite, the power plant only takes a small solid angle, so it seems like a “smaller” sun.
Assuming 400 MW and 1 kW/m^2 (at surface) solar power, it has an area of 400000 m^2, so a solid angle of 4.5e-6 sr from 300km while the sun has 70e-6 sr. So ten times smaller, therefore weaker. Note however here i did not account for attenuation in the atmosphere
Light doesn’t have to spread wider than a target or else you wouldn’t be able to have telescopes or magnifine glasses. Each panel in unison can act like a giant magnifine glass. The difference in power density would be the ratio of the distance of sun to earth squared vs (sun to earth + earth to satellite) squared which is basically negligible.
Where do you get the 3km wide beam? Suns rays are almost parallel.
Yes, you are right, considering the rays emerge from a point. And yes, each panel or all panels in unison can act like a magnifying glass. However, if they focus the light on a point at the height of the satellite, they work like a magnifying glass, or telescope with a focal length of the satellite – power plant distance, so at least 300 km. Considering the angular size of the sun, this telescope would lead to an image of the sun, the size of 3 km.
No sun rays are not parallel. If you looked at the sun (don’t, it will burn your eyes), would you see it as a point or a disc? As a disc. Why? Because even looking in slightly different directions, you see the sun. So the rays from the sun are not almost parallel, the rays from other stars are, they look point like.
Two interesting images for you:
A solar eclipse viewed trough tree leaves: You can see the partial sun disc by using the small free points in the tree cover as pinhole cameras. Sure, the tree cover does not have lenses, but they only make the image sharper, not smaller. In this image the focal length is only the height of the trees and the image is already a few cm across. It also shows that the rays from the sun are not parallel. If they were,
all rays going through the small free spots in the tree cover would end up at the same spot on the ground.
International Space Station, ISS, flying in front of the Sun: As the sun and the satellite are far away, we can assume that the angular size of the original sun and the virtual sun image are approximately the same when viewed from the power plant. Hence, this image shows how the mirrors would form an image of the sun, where only a small part of it hits the sun.
As the sun is much larger than the ISS, the angle of rays which come from the sun is much larger than the angle of rays which hit the ISS.
Unfortunately I can’t open the links on my phone so it’s hard to follow what you’re describing. I understand the rays aren’t perfectly parallel but they’re pretty close to parallel. Do you mind doing the math of where you are getting 3km from I’m not really following your logic. It doesn’t make sense to me that the light should suddenly spread out way more after bouncing off a mirror than if it has just continued traveling straight.
The links are actually only random images from an image search with the terms “solar eclipse through tree leaves” and “iss in front of sun”.
I think you have in mind, that the rays are not parallel because they have to arrive at different positions. As you say, this is negligible, and it can even be avoided by the tilting of the mirrors. However, the rays start from different parts of the sun, and as the sun is huge, this angle is not so small.
I’ll try to explain it in more detail, sorry for the wall of text, it got longer than expected. In this case, we can use simplified ray optics and ignore the wave nature of light. This means, the light of the different mirrors or even pieces of mirrors just adds to one another. An important point, even if obvious, is that each point on the mirror surface can only have one orientation. Now we “select” the orientation of this point, we orient it in a way, that it reflects the rays from the center of the sun1 directly on the sun.2 But until now, we have ignored the rays which come from the rim of the sun. These rays start at a different position, namely the sun radius (695700 km). Due to different starting position, the rays have a different angle to arrive at the power plant, arcsin(sun radius / sun-earth-distance), which is 0.27°. Now we already oriented the mirrors parts in a way, that the rays from the center of the sun are reflected onto the satellite, but the rays from the rim of the sun come at an 0.27° differing angle. If the incidence of the ray on a mirror is changed by an angle, the outbound ray is also changed by the same angle. This leads reflected rays leaving in a direction 0.27° offset from the direction to the satellite. Assuming the satellite is at a height of 300 km and directly above, it is 300 km away, the smallest realistic distance. With this angle, it leads to a miss of plant-satellite-distance * sin(angle) leading to 1.4 km. This thought is valid for all points on the rim. Similarly, the rays between the rim and the center land between the satellite and 1.4 km off target. Hence the plant projects an image of the sun onto the satellite with a radius of 1.4 km.
1 Well they actually do not come directly from the sun, they still come from very close to the surface, but they seem to come from the center of the sun and for rays it is not important how far they have already traveled. We can just assume the sun is a disc. 2 If we assume the mirror is optimally shaped, we can reflect every ray, which seems to come from the center of the sun, perfectly on the satellite. Such a mirror would be part of an ellipsoid, with focal points at the center of the sun and the center of the satellite. In practice, it would be practically indistinguishable from a paraboloid with the satellite (deviance of 1.5 µm with a guessed plant size of 1 km). This is possible as the rays through the center of the sun falling onto the plant are, as you say, almost parallel.
There is still a power density square law, but with focused energy you are only integrating power flux across a portion of the sphere’s surface instead of the whole thing.
It depends how the actual system was set up if they used flat reflectors then yeah it applies but the difference in power would be the ratio of the distance from earth to the sun vs the distance of E to S +mirror to satellite which would be negligible. If you had a parabolic mirror you could get no loss in power. The power density square law only apply because the area the light is being distributed over is growing at a square ratio to radius but if the beams are parallel the area doesn’t grow.
There is a cool easy-to-show fact that you can never make something hotter than the light source my focusing its light.
Since otherwise you could take heat and divide it into a hotter and colder region, decreasing entropy without using energy.
I’m not sure about the easy-to-show part, but take a look at the Brightness Theorem / Conservation of https://en.wikipedia.org/wiki/Etendue if you want to learn more.
Isn’t there some inverse square math rule about radiation like this? The further away you are the radiation reaching you is far less than it would seem? Not good at remembering this math so maybe someone can correct me.
Even if you could get the mirrors all focused accurately and tracking the object at speed it seems like it wouldn’t be any more of a concern than a really bright searchlight or something.
The power density square law is for an emitting light source that emits in all directions. Since the incoming light is basically parallel that doesn’t really apply. If you were able to accurately track a satellite (a feat I’m sure is pretty hard) you would definitely vaporize it pretty quickly I’m talking under a minute since space is a good insulator.
Keep in mind that atmospheric interference would likely scatter the light enough to be ineffective
So you’re saying we should weaponize the James Webb space telescope instead? :D
I like the cut of your jib
But the photons made it through the atmosphere in the first place to be collected by the reflectors. Is there just not enough energy left to make it back out before cooling off?
That’s the assumption, yes. But if the beams are coherent (like a laser) atmospheric interference would be a lot smaller.
The real question is whether the light would be coherent, which I lean towards no on.
It’s not even coherent when the sun emits it. For one, it consists of a large range of wavelengths… And I doubt there’s a way to make light coherent at that order of magnitude.
No that’s not true only about 30% of light energy scatters when traveling through the atmosphere to earth and certain wavelengths are almost completely absorbed in the way down. So on the way back up it should be a high portion make it to the satellite I would imagine 80%. Even worse case scenario 200 megawats shinning on a satellite would vaporize it almost instantly.
It holds if the light spreads wider than the target. So also for directed light sources at large enough distances. Even a perfect mirror must spread the light in the same angle as it is incomming. Hence the beam would at least 3 km wide at the satellite. Therefore the satellite can only recieve a Illumination of ~65W/m^2 which is a few percent of the normal sun brightness of 1300 W/m^2.
Another way to look at it, the mirrors cant make the sun seem brighter only larger. From the tower you see a large solid angle around you the mirror, therefore, it can seem like you are at the surface of the sun. However, fro. the position of a satellite, the power plant only takes a small solid angle, so it seems like a “smaller” sun. Assuming 400 MW and 1 kW/m^2 (at surface) solar power, it has an area of 400000 m^2, so a solid angle of 4.5e-6 sr from 300km while the sun has 70e-6 sr. So ten times smaller, therefore weaker. Note however here i did not account for attenuation in the atmosphere
Light doesn’t have to spread wider than a target or else you wouldn’t be able to have telescopes or magnifine glasses. Each panel in unison can act like a giant magnifine glass. The difference in power density would be the ratio of the distance of sun to earth squared vs (sun to earth + earth to satellite) squared which is basically negligible. Where do you get the 3km wide beam? Suns rays are almost parallel.
Yes, you are right, considering the rays emerge from a point. And yes, each panel or all panels in unison can act like a magnifying glass. However, if they focus the light on a point at the height of the satellite, they work like a magnifying glass, or telescope with a focal length of the satellite – power plant distance, so at least 300 km. Considering the angular size of the sun, this telescope would lead to an image of the sun, the size of 3 km.
No sun rays are not parallel. If you looked at the sun (don’t, it will burn your eyes), would you see it as a point or a disc? As a disc. Why? Because even looking in slightly different directions, you see the sun. So the rays from the sun are not almost parallel, the rays from other stars are, they look point like.
Two interesting images for you: A solar eclipse viewed trough tree leaves: You can see the partial sun disc by using the small free points in the tree cover as pinhole cameras. Sure, the tree cover does not have lenses, but they only make the image sharper, not smaller. In this image the focal length is only the height of the trees and the image is already a few cm across. It also shows that the rays from the sun are not parallel. If they were, all rays going through the small free spots in the tree cover would end up at the same spot on the ground.
International Space Station, ISS, flying in front of the Sun: As the sun and the satellite are far away, we can assume that the angular size of the original sun and the virtual sun image are approximately the same when viewed from the power plant. Hence, this image shows how the mirrors would form an image of the sun, where only a small part of it hits the sun.
As the sun is much larger than the ISS, the angle of rays which come from the sun is much larger than the angle of rays which hit the ISS.
Unfortunately I can’t open the links on my phone so it’s hard to follow what you’re describing. I understand the rays aren’t perfectly parallel but they’re pretty close to parallel. Do you mind doing the math of where you are getting 3km from I’m not really following your logic. It doesn’t make sense to me that the light should suddenly spread out way more after bouncing off a mirror than if it has just continued traveling straight.
The links are actually only random images from an image search with the terms “solar eclipse through tree leaves” and “iss in front of sun”.
I think you have in mind, that the rays are not parallel because they have to arrive at different positions. As you say, this is negligible, and it can even be avoided by the tilting of the mirrors. However, the rays start from different parts of the sun, and as the sun is huge, this angle is not so small.
I’ll try to explain it in more detail, sorry for the wall of text, it got longer than expected. In this case, we can use simplified ray optics and ignore the wave nature of light. This means, the light of the different mirrors or even pieces of mirrors just adds to one another. An important point, even if obvious, is that each point on the mirror surface can only have one orientation. Now we “select” the orientation of this point, we orient it in a way, that it reflects the rays from the center of the sun1 directly on the sun.2 But until now, we have ignored the rays which come from the rim of the sun. These rays start at a different position, namely the sun radius (695700 km). Due to different starting position, the rays have a different angle to arrive at the power plant, arcsin(sun radius / sun-earth-distance), which is 0.27°. Now we already oriented the mirrors parts in a way, that the rays from the center of the sun are reflected onto the satellite, but the rays from the rim of the sun come at an 0.27° differing angle. If the incidence of the ray on a mirror is changed by an angle, the outbound ray is also changed by the same angle. This leads reflected rays leaving in a direction 0.27° offset from the direction to the satellite. Assuming the satellite is at a height of 300 km and directly above, it is 300 km away, the smallest realistic distance. With this angle, it leads to a miss of plant-satellite-distance * sin(angle) leading to 1.4 km. This thought is valid for all points on the rim. Similarly, the rays between the rim and the center land between the satellite and 1.4 km off target. Hence the plant projects an image of the sun onto the satellite with a radius of 1.4 km.
1 Well they actually do not come directly from the sun, they still come from very close to the surface, but they seem to come from the center of the sun and for rays it is not important how far they have already traveled. We can just assume the sun is a disc.
2 If we assume the mirror is optimally shaped, we can reflect every ray, which seems to come from the center of the sun, perfectly on the satellite. Such a mirror would be part of an ellipsoid, with focal points at the center of the sun and the center of the satellite. In practice, it would be practically indistinguishable from a paraboloid with the satellite (deviance of 1.5 µm with a guessed plant size of 1 km). This is possible as the rays through the center of the sun falling onto the plant are, as you say, almost parallel.
There is still a power density square law, but with focused energy you are only integrating power flux across a portion of the sphere’s surface instead of the whole thing.
It depends how the actual system was set up if they used flat reflectors then yeah it applies but the difference in power would be the ratio of the distance from earth to the sun vs the distance of E to S +mirror to satellite which would be negligible. If you had a parabolic mirror you could get no loss in power. The power density square law only apply because the area the light is being distributed over is growing at a square ratio to radius but if the beams are parallel the area doesn’t grow.
There is a cool easy-to-show fact that you can never make something hotter than the light source my focusing its light.
Since otherwise you could take heat and divide it into a hotter and colder region, decreasing entropy without using energy.
I’m not sure about the easy-to-show part, but take a look at the Brightness Theorem / Conservation of https://en.wikipedia.org/wiki/Etendue if you want to learn more.
The easy to show part was the second sentence of my comment.
This is really useful physics trivia, because the basic truth is easy to show from a simple law, but the detailed explanations go quite in-depth.
With lenses, you trade bewteen angular accuracy and light density.
For a challenge, try it with LEDs. Where do you find the source “temperature”, you can get from focusing an LEDs light?
That’s an intriguing question. My first guess would be it corresponds to the diode’s band gap?
Thats definitely hugely relevant, yes.