tell my why this thing should not be able to melt satelites that cross over during the day
My scientific research of squinting at the poster says a spy satellite is probably about as long as a pickup truck which is probably about 20 feet long.
xkcd says space is 100 km away and I’m sure there’s nothing else I need to understand about that.
At 100 km away, the change of angle that will move your beam by 20 feet (enough to make the difference between hitting or not, if the thing and the flat mirror are both about 20 feet long I guess) is (20 feet / 100 km / pi) radians or 0.0000194 radians, meaning you raised or lowered one edge of the mirror by 0.004 inches or around the width of pretty-thick hair. I would be a little surprised if the mirrors even stayed within that tolerance just from flexing around in the wind for as big as they are.
On the other hand, you wouldn’t have to hit the spy satellite with every mirror; you could probably heat it up significantly just by hitting it with a bunch of the beams as they were swinging wildly around and mostly missing it. And if it was specifically a spy satellite, you could probably fry its optics with not really a lot of mirrors for not a long time actually managing to hit it.
On the other other hand the thing would be flying along at around 8 km/s, so you’d have to get your mirrors positioned accurately enough, and then start moving them at a relatively insane speed while still keeping their absolute positioning dead accurate when their motors and overall construction clearly weren’t designed for either of those tasks at the required level of precision.
TL;DR Let’s try it
and then start moving them at a relatively insane speed while still keeping their absolute positioning dead accurate when their motors and overall construction clearly weren’t designed for either of those tasks at the required level of precision.
That’s what they want you to think.
Props on your Internet math and research. It was a fun read.
You still have a crap-ton of atmosphere you have to get through, and the beams being reflected aren’t coherent. So the light reflected is subject to the inverse square law, which means that the energy diminishes as the inverse square of the distance. So the actually energy reaching the satellite would be minuscule. If you want to effectively use light to punch all the way through the atmosphere, you’ll need beam coherence.
The difference in the angles of the beams is the angle difference of a beam that came from an object 149,597,871 km away at a separation of 20 feet i.e. basically fuck-all. For this purpose I think they’re effectively (edit: coherent) parallel. And I think the atmospheric reduction would be significant but not defeating-to-the-purpose; I mean the sunbeam on its way in still had plenty of effectiveness after getting through the same atmosphere. If you did it on a cloudy day or something then yeah it wouldn’t work at all.
(Edit: Wait, I don’t understand optics; I mean parallel, not coherent. I don’t think coherence enters into it?)
Fine fine fine fine fine fine OH GOD WHY
For some reason it’s really funny to me. It would be in the beam for a vanishingly small time: 762 microseconds, which if every mirror in the 392-megawatt array were properly focused, is still enough to receive a burst of 300,000 joules of radiant energy. I have not enough physics to tell you if that’s a big deal or not, but I feel like it would be and I don’t think the cameras would work after.
I realize we’ve had disagreements in other regards but this is excellent
I think solar-powered lasers would be a better bet. That would eliminate any surface irregularities of the mirrors and reduce the effective focus area . This would also reduce the number of moving parts required for focusing.
On the other hand, the amount of particulate diffusion within the atmosphere would complicate both the accuracy of the beam and the effective beam area, so who knows.
Let’s try it.
The mirrors are flat, and the sun has an apparent diameter of about half a degree, so at 100 km, the spot diameter would be 900 meters.
You could use concave mirrors, but since you’re moving them independently, you’d also have to consider the diffraction limit for each one.
The suns angular diameter is about 0.01 radian, so at a distance of 100km, the suns reflection will spread out to a disc about 1km across.
392MW over a disc that size is 500w/m2, which is weaker than direct sunlight.
Yes; it is well known that if you look at yourself in a flat mirror, and then back up, your reflection will spread out bigger and bigger and get dimmer and dimmer, the further away you get.
Wait
“You” are the sun in this scenario.
As you back up, you fill a smaller and smaller fraction of the mirror. The reflection becomes less sun and more space.
Uh… losses from transmitting through the atmosphere a second time?
Damn. I wonder what its operational range would be.
this thing is big enough to alter the average reflective index of a whole state if it swings around its mirrors
the focus spot in theorie could be set to any range, just as u go more far the precision of each mirror angle will be the limiting factor amongst atmospheric losses distortions.
Even if the actuators had enough precision, which they almost certainly do not, there’s no way the mirrors are flat enough to keep the light collimated that far out. The angular spread would make the intensity much lower at orbital altitudes.
That is 300 ft, not 600-1,200 miles.
The Sun puts more energy on a spy satellite than the array could do.
I suspect in order to stay focused on such distances you’d need extremely flat mirrors. Like, telescope grade stuff.
I doubt the mirrors they have is even within an order of magnitude flat enough.
You might even need adaptive mirrors to deal with atmospheric distortion. Also, they would have to move relatively quickly and very precisely (read: an impossible combination) to track satellites in low orbit. Plus, you could only hit satellites that crossed overhead at a relatively high angle.
But yeah, one solar tower plant did a stunt where they reflected an image made of sunlight at the ISS and an astronaut took a picture. They didn’t melt.
where they reflected an image made of sunlight at the ISS and an astronaut took a picture
got a link to said picture? it may make for a good meme template. I saw that the chinese did that kind of ‘pixel art’ with there own near identical solar thermal plant
They’re misremembering Italian astronaut Samantha Cristoforetti photographing the light dot where the mirrors shine at in the power plant. They were NOT shining it at the ISS.
https://www.jpost.com/omg/article-715675
https://twitter.com/AstroSamantha/status/1562833775293186049
I’m no optical physicist, but based on empirical evidence of not melting due to light arriving from a huge ball of thermonuclear fire 8 light minutes away (and sure it’s not exactly focused), I propose a hypotesis that light-based energy transfer in atmosphere is very lossy and not feasible as a weapon.
Which is perfect for this community, of course.
So, the light from the sun is very spread out. This is why it doesn’t hurt you. When you take almost all of the light from a large area and focus it to a point, it starts to add up fast.
You can test this with a leaf and a magnifying glass. The mirrors in this scenario are acting as a giant lens.
The biggest question is whether the mirror array is large enough to redirect enough energy to make it though the atmosphere.
It would likely be probable if the facility was specifically designed to do this.
Those are designed to focus on a large, stationary, object not far away, not a small hypersonic object very very far away.
Satellites be zoomin, it would be hard to hit one for more than a split second. But I’m definitely down to try!
Because of how far it is I’m guessing they could move the focal point very fast too with just the slightest of movements in the mirrors.
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
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?
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.
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?
