4 ms·

Inverse square law. Every n AU you go out from the earth you will need n2 are of mirror to focus on a solar panel to get the same power. That assumes you can ev

by monob 9y ago

Inverse square law. Every n AU you go out from the earth you will need n2 are of mirror to focus on a solar panel to get the same power. That assumes you can even build a structure that can keep that shape. Say you have a mission to Uranus, you will need a mirror with 400 times the area of the solar cells to get as much power as they would in earth orbit.

hutzlibu 9y ago
Hm, ok the mirror is probably out of the question then. But if you use a laser? I imagine it will be easier to focus it more precisely?
- philjohn 9y ago
  Saw a documentary recently and they're doing just this.
- contact_fusion 9y ago
  Lasers are still subject to the diffraction limit, regardless of the quality of focus. We can examine this limit in the context of delivering energy to a remote object. This analysis will be simple - we assume that the optics are perfectly aligned (dubious - pointing is difficult); we assume nothing about the absorption properties of the object, which will necessarily need to be very high efficiency. The angular size T of a laser beam of wavelength L emitted by an aperture of diameter D_a is roughly L/D_a. Similarly, using the small angle approximation, this angular size T at the object itself will be the width of the beam, W, divided by the distance between the object and the aperture, D_o: T = W/D_o. Ultimately, this gives us the width: W = L*(D_o/D_a). What does this tell us about the practicality of such a system? Visible light, wavelength roughly 500 nm, is perhaps a solid guess for a real system. Realistically the aperture size is probably limited to about 10 meters, but we can go even further and assume a synthesized aperture of a realistic system being 100 meters. You would want to get all of your beam for power transmission, so lets assume an upper limit for the beam width at the object to be 100 meters as well - probably unrealistic, but maybe solar sail/ultralight absorbers could get there. Throwing these numbers in gives a maximum range of... 2x10^10 meters. This is roughly a tenth of an AU. In comparison, this is about 50 Earth-Moon distances... and only a quarter of the distance between Earth and Mars at their closest approach. Coincidentally, this is also about one light-minute. Any real power delivery system, using current tech and without assuming convenient fictions, will have a much more limited range. In short, lasers are not really perfect rays, even though they are approximately so over scales we typically encounter; at astronomical scale, diffraction always wins. This is why it is usually way better to bring the power with you - especially as you lose solar irradiance as you get further from the Sun. And for bringing power with you, nothing beats nuclear for energy density.
  - hutzlibu 9y ago
    Thanks! (also memorys of phsyik lectures are coming up again) And forgive a nonserious reply: "And for bringing power with you, nothing beats nuclear for energy density" I think fusion does ...
    - contact_fusion 9y ago
      Nuclear fusion is nuclear power.
    - hutzlibu 9y ago
      can't reply for some reasons ... anyway, yes sure fusion is nuclear power, I just read in your comment nuclear fission which you did not wrote ... so never mind I am just tired at some airport ...
imdsm 9y ago
Does this mean that if you go towards the sun, you would be able to generate more solar electricity with less surface area? I'm not thinking dyson sphere style, but perhaps some sort of electricity space-tanker, that sails toward (or around?) the sun, collecting solar energy, and then returns to earth with batteries packed to the brim with energy? Dibs on Solar Harvester
- TeMPOraL 9y ago
  Yes, it means exactly that - though soon you'll start hitting into issues with solar panel efficiency and thermal management. It's very hard to keep things cool in space. The space electricity tanker idea is theoretically possible, but might not make any economical sense. For it to be worth it, it would have to store a lot of energy, so that it could ship back more than the cost of moving it around, while also being competitive with simply building more bigger collectors further away (and closer to the industry). But maybe a highly eccentric orbit, with a very low perihelion, would work. Not totally sure if the math works out on this, but I could see moving the energy-intensive industry closer to the Sun, and having it use beamed energy to move resources and products to itself and back.
- InclinedPlane 9y ago
  Take a look at the solar panels on the Parker Solar Probe: http://parkersolarprobe.jhuapl.edu/index.php#spacecraft http://parkersolarprobe.jhuapl.edu/index.php#spacecraft
acidburnNSA 9y ago
I made a cool plot of this in the solar system the other day. https://whatisnuclear.com/img/solar-system-insolence.png https://whatisnuclear.com/img/solar-system-insolence.png

by monob 9y ago

hutzlibu 9y ago

Hm, ok the mirror is probably out of the question then. But if you use a laser? I imagine it will be easier to focus it more precisely?

philjohn 9y ago
Saw a documentary recently and they're doing just this.
contact_fusion 9y ago
Lasers are still subject to the diffraction limit, regardless of the quality of focus. We can examine this limit in the context of delivering energy to a remote object. This analysis will be simple - we assume that the optics are perfectly aligned (dubious - pointing is difficult); we assume nothing about the absorption properties of the object, which will necessarily need to be very high efficiency. The angular size T of a laser beam of wavelength L emitted by an aperture of diameter D_a is roughly L/D_a. Similarly, using the small angle approximation, this angular size T at the object itself will be the width of the beam, W, divided by the distance between the object and the aperture, D_o: T = W/D_o. Ultimately, this gives us the width: W = L*(D_o/D_a). What does this tell us about the practicality of such a system? Visible light, wavelength roughly 500 nm, is perhaps a solid guess for a real system. Realistically the aperture size is probably limited to about 10 meters, but we can go even further and assume a synthesized aperture of a realistic system being 100 meters. You would want to get all of your beam for power transmission, so lets assume an upper limit for the beam width at the object to be 100 meters as well - probably unrealistic, but maybe solar sail/ultralight absorbers could get there. Throwing these numbers in gives a maximum range of... 2x10^10 meters. This is roughly a tenth of an AU. In comparison, this is about 50 Earth-Moon distances... and only a quarter of the distance between Earth and Mars at their closest approach. Coincidentally, this is also about one light-minute. Any real power delivery system, using current tech and without assuming convenient fictions, will have a much more limited range. In short, lasers are not really perfect rays, even though they are approximately so over scales we typically encounter; at astronomical scale, diffraction always wins. This is why it is usually way better to bring the power with you - especially as you lose solar irradiance as you get further from the Sun. And for bringing power with you, nothing beats nuclear for energy density.
- hutzlibu 9y ago
  Thanks! (also memorys of phsyik lectures are coming up again) And forgive a nonserious reply: "And for bringing power with you, nothing beats nuclear for energy density" I think fusion does ...
  - contact_fusion 9y ago
    Nuclear fusion is nuclear power.
  - hutzlibu 9y ago
    can't reply for some reasons ... anyway, yes sure fusion is nuclear power, I just read in your comment nuclear fission which you did not wrote ... so never mind I am just tired at some airport ...

imdsm 9y ago

Does this mean that if you go towards the sun, you would be able to generate more solar electricity with less surface area? I'm not thinking dyson sphere style, but perhaps some sort of electricity space-tanker, that sails toward (or around?) the sun, collecting solar energy, and then returns to earth with batteries packed to the brim with energy? Dibs on Solar Harvester

TeMPOraL 9y ago
Yes, it means exactly that - though soon you'll start hitting into issues with solar panel efficiency and thermal management. It's very hard to keep things cool in space. The space electricity tanker idea is theoretically possible, but might not make any economical sense. For it to be worth it, it would have to store a lot of energy, so that it could ship back more than the cost of moving it around, while also being competitive with simply building more bigger collectors further away (and closer to the industry). But maybe a highly eccentric orbit, with a very low perihelion, would work. Not totally sure if the math works out on this, but I could see moving the energy-intensive industry closer to the Sun, and having it use beamed energy to move resources and products to itself and back.
InclinedPlane 9y ago
Take a look at the solar panels on the Parker Solar Probe: http://parkersolarprobe.jhuapl.edu/index.php#spacecraft http://parkersolarprobe.jhuapl.edu/index.php#spacecraft

acidburnNSA 9y ago

I made a cool plot of this in the solar system the other day. https://whatisnuclear.com/img/solar-system-insolence.png https://whatisnuclear.com/img/solar-system-insolence.png