Photons, their power output and force they exert on objects.

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rakkabal

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I have been trying to find some info on the subject matter but its seems more elusive than I thought, wondering if anyone here has some info.<br /><br /><blockquote><font class="small">In reply to:</font><hr /><p>NASA researchers have found that at 1 astronomical unit (AU), which is the distance from the sun to Earth, equal to 93 million miles (150 million km), sunlight can produce about 1.4 kilowatts (kw) of power. If you take 1.4 kw and divide it by the speed of light, you would find that the force exerted by the sun is about 9 newtons (N)/square mile (i.e., 2 lb/km2 or .78 lb/mi2).<p><hr /></p></p></blockquote><br /><br />I would expect that the 1.4 kw is over an area but my source failed to mention how big that area is. I assumed that this info was attained from a satellite in orbit while the solar cells faced the sun.<br /><br />Also if anyone does know how many kw are generated on the dark side of a planet (out of our suns direct light) please let me know.
 
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doubletruncation

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Hi and welcome to SDC!<br /><br />You're right, it is per unit area - square meter in this case. See for example: Wikipedia article on the Solar Constant. Note that this is the *total* energy emitted at all wavelengths, a solar-cell would not be 100% efficient in converting all of this energy into electricity.<br /><br />As to how this number is measured, it's a bit complicated, but basically yes the best measurements come from satellites though the results have to be calibrated to a model somehow since you can't actually measure every wavelength of light with perfect efficiency. You can see for example the ACRIM experiment which is an experiment on a satellite that measured the "total solar irradiance" (which is directly related to the 1.4 kw/m^2 number).<br /><br />Regarding the amount of power per unit area output by the dark side of the Earth - that's a very good question. You can estimate what it would be by figuring out the effective temperature of the Earth and using the black-body radiation law to convert this into light output. You find that the effective temperature is about 255 K (a class notes discussing this is at: http://eesc.columbia.edu/courses/ees/climate/lectures/radiation/ ) which corresponds to about 239 W/m^2 just above the Earth's atmosphere (most of the energy is in the infrared - you can't see it). I imagine that it isn't evenly distributed from day-side to night-side (as that value assumes) but I don't know the scale of the variation. <div class="Discussion_UserSignature"> </div>
 
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