Why do people claim the inverse square law applies? The light has already travelled 147 million km, another 500km from the earth back to the satellite is mathematically insignificant, it’s a rounding error.
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 incoming. 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 in space.
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
No i am talking about all the mirrors as one surface, no matter they are really one or consist of small pieces
For the 65 W/m^2 i already used the size of the whole system, so all 10000 mirrors.
The sun has a angular diameter of 32 arcmin. (see here) Hence, the rays hitting one point of the one mirror, have come from different angles, namly filling a circle with this angular diameter. By reflection, the directions of the rays changes. But rays hitting the same spot on the mirror which were misaligned before by 32 arcmin are also misaglined by 32 arcmin after the mirror, independent of its shape. Therefore, the rays emerging from the power plant diverege by at least 32 arcmin. This is not a problem for operation, as this leads to a size of 4.6 m at an estimated maximum distance of 500 m between tower and mirrors. When the mirrors point at a satellite however, a distance of 300 km leads to a beam diameter of 2.8 km calculation
Even an ideal mirror can only project a point source onto a point. It is impossible to focus the rays of an extended source onto one point.
See https://en.wikipedia.org/wiki/Etendue if you want to know details. With conservation of etendue you can also calculate this in a similar way.
You seem to be neglecting that the lights already traveled 147 million km from the Sun, your math is wrong. You need to account for the distance from the sun to earth, plus to the satellite. Of course the math looks better on your end when you forget the most important detail. Come on lmfao.
Yes i am, because it is unimportant if the light comes from the sun or the moon or a 3km large satellite (assuming they would have the same radiance). It would be important if the power plant were ten times larger, the satellite would be closer or larger. However in this case the limit to the power is is the etendue of the light at the satellite. The maximum power is the etendue at the satellite times radiant flux of the sun.
If you want a fun and interesting read which does explain a related “problem”, there is a relevant xkcd
I could explain it to you in at least five different ways in detail, three of them i have already done in short here in the comments. However, you never argued directly against my point. You don’t talk to me seriosly but laugh about it.
This is not what a serious disussion looks like
like. If you want an explaination, i would be motiviated to take the time and explain it in detail.
Note that i listened to your point, considered it and argued why it plays no role. You have not considered my explainations.
.
Why do people claim the inverse square law applies? The light has already travelled 147 million km, another 500km from the earth back to the satellite is mathematically insignificant, it’s a rounding error.
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 incoming. 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 in space.
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
Sure that’s one mirror. But we are talking. Thousands and thousands upon thousands of them.
10,000x65= 65,000 which is now ~60x the Sun passing by it.
And why are you saying it would be 3km wide? I’d like to see the math please.
No i am talking about all the mirrors as one surface, no matter they are really one or consist of small pieces
For the 65 W/m^2 i already used the size of the whole system, so all 10000 mirrors.
The sun has a angular diameter of 32 arcmin. (see here) Hence, the rays hitting one point of the one mirror, have come from different angles, namly filling a circle with this angular diameter. By reflection, the directions of the rays changes. But rays hitting the same spot on the mirror which were misaligned before by 32 arcmin are also misaglined by 32 arcmin after the mirror, independent of its shape. Therefore, the rays emerging from the power plant diverege by at least 32 arcmin. This is not a problem for operation, as this leads to a size of 4.6 m at an estimated maximum distance of 500 m between tower and mirrors. When the mirrors point at a satellite however, a distance of 300 km leads to a beam diameter of 2.8 km calculation
Even an ideal mirror can only project a point source onto a point. It is impossible to focus the rays of an extended source onto one point. See https://en.wikipedia.org/wiki/Etendue if you want to know details. With conservation of etendue you can also calculate this in a similar way.
You seem to be neglecting that the lights already traveled 147 million km from the Sun, your math is wrong. You need to account for the distance from the sun to earth, plus to the satellite. Of course the math looks better on your end when you forget the most important detail. Come on lmfao.
Yes i am, because it is unimportant if the light comes from the sun or the moon or a 3km large satellite (assuming they would have the same radiance). It would be important if the power plant were ten times larger, the satellite would be closer or larger. However in this case the limit to the power is is the etendue of the light at the satellite. The maximum power is the etendue at the satellite times radiant flux of the sun.
If you want a fun and interesting read which does explain a related “problem”, there is a relevant xkcd
I could explain it to you in at least five different ways in detail, three of them i have already done in short here in the comments. However, you never argued directly against my point. You don’t talk to me seriosly but laugh about it.
This is not what a serious disussion looks like like. If you want an explaination, i would be motiviated to take the time and explain it in detail.
Note that i listened to your point, considered it and argued why it plays no role. You have not considered my explainations.
deleted by creator
Atmospheric attenuation of light is a real thing and the intensity of solar radiation is less on the surface.
https://pace.oceansciences.org/images/Solar_spectrum_en_v5.jpg