6 ms·

We have; here's an example. A solar panel in geostationary orbit only has a few hours of shadow per year. The incoming power is so much greater than that at the

• dllthomas 13y ago
  "The incoming power is so much greater than that at the surface (about 4x the best desert)" But with 22k miles (3x earth diameter) of transmission losses.
  • comicjk 13y ago
    The losses when transmitting through vacuum are pretty negligible; it's only the last fifteen miles that really cost you.
    • dllthomas 13y ago
      That certainly might be the case, but I don't immediately see why.
      • csense 13y ago
        I'm not a physicist, but my understanding is that there are two ways a wireless photon-based energy transmission system loses energy in transit. First, each molecule has a specific set of frequencies where photons are "easy" to absorb because they correspond to transitions between quantum states. Any system whose goal is to deliver energy wirelessly through the atmosphere will obviously not use those frequencies. The other way for a photon to lose energy to the medium it's traveling through is, if the medium contains molecules like water that have different parts with positive and negative charges, the electromagnetic wave will move the differently charged parts in opposite directions. The energy to do that is lost [1]. [1] "Lost" is an application-specific notion. This mechanism is also how a microwave heats food, and in this case the energy imparted to the water molecules in the food can't be considered "lost" because it's supposed to go into the food!
        • dllthomas 13y ago
          Ah, you intend to keep them photons. I had been imagining converting to electrons and running down cabling attached to the elevator cable. The other approach might very well be better, but doesn't really require a space elevator does it?
          • adwf 13y ago
            I'd imagine they'd go with super-conducting cables personally. Space is extremely cold, so achieving a super-conducting state is simply a matter of shielding the cable from the sun. Shouldn't be too difficult... At least, not in comparison to building a space elevator in the first place!
            • dllthomas 13y ago
              Space is cold, but getting rid of heat is not easy in a vacuum. Not sure what that means with respect to this particular project, though.
              • lutusp 13y ago
                > Space is cold, but getting rid of heat is not easy in a vacuum. On the contrary, getting rid of heat in a vacuum is easier than getting rid of heat in an atmosphere. Here's a diagram of human heat loss at the surface: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.htm... According to this source, at the surface of earth, under the atmosphere, a person loses: Perspiration: 17 watts Conduction: 11 watts Radiation: 133 watts Without an atmosphere, the direct radiation of heat energy into space becomes more efficient (no greenhouse effect), and it's always the most efficient way to radiate heat. It is the efficiency of direct radiation of heat energy that explains why objects at the surface can fall below air temperature overnight under a clear sky, as they surely do.
                • dllthomas 13y ago
                  Hm, that's contrary to what I'd heard, but I'll certainly defer to you on the question.
  • bsder 13y ago
    You don't have to transmit the power. You can manufacture energy-intensive things right there and then transmit the end products. It's really easy to send stuff from orbit to Earth. The trick is keeping it from burning up on the way in.
    • mcv 13y ago
      And that's where the space elevator is very handy. It's good for getting satelites and factories up there, but it's just as good for getting the end products down again.
• sopooneo 13y ago
  How would geostationary orbit have only a few hours of shadow per year? My understanding is geostationary orbit requires you be pretty much over the equator, which would mean the satellite spent as much time on the far side of the earth as the near.
  • icegreentea 13y ago
    Geostationary orbit is 36,000 km above the earth surface. The earth is 12,000 km in diameter. The sun is 150 MILLION km away. The about of time the something in geo-stat stays in earth's shadow is tiny. Imagine a base runner in baseball after a home run. Spends half the time on the far side of the pitcher from the homeplate umpire, but he is only occluded from the homeplate umpire by the pitcher for a very short amount of time. That's (very!!) roughly the scales involved here.
    • gliese1337 13y ago
      Not only that, but thanks to the fact that the Earth's rotation (and thus the plane of the equator) is tilted off of the orbital plane, even equatorial satellites far below geosynchronous orbit will only pass through the Earth's shadow at all close to the equinoxes (twice a year). Most of the time, you have eternal sunshine throughout an orbit. It's as though the metaphorical baseball player takes a flying leap so that the umpire can see him over the pitcher's head.
    • sopooneo 13y ago
      Thank you. That answers it. I had no idea geostationary orbit was so high, but I probably should have.