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.
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?
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.
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.
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.
Keep in mind that atmospheric interference would likely scatter the light enough to be ineffective
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.
So you’re saying we should weaponize the James Webb space telescope instead? :D
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?
