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Hi Jonathan and list, I wrote in part, > Many people have been erroneously applying > > illumination fraction to satellite observing (and the > > prediction of satellite brightness), and it has no > > place there. > Jonathan asked: > So ... these are only theoretical equations, and don't truly reflect > reality? Illumination fraction is a "real thing" -- it just doesn't have any (direct) value in computing the visual magnitude of an object. Take the Moon as an extreme example. At 90-degree phase (1st Quarter or 3rd Quarter Moon), the illumination fraction is 50%. How bright is the Full Moon (100% illumination) by comparison? If you blindly assumed a linear relationship between illumination fraction and brightness, you'd confidently say that the Full Moon is twice as bright. Fact is, the Full Moon is over TEN times as bright because its surface is highly non-Lambertian. If the moon's surface was a perfect Lambertian reflector, the Full Moon would be only pi times brighter than 1st or 3rd Quarter -- about 1 1/4 visual magnitudes. So illumination fraction is not very helpful in determining brightness. > Jonathan went on to ask about the example of the satellite at 1600-km range, 90-degree phase: > So, just confirming - that is just the application of the inverse square > law? Yes. >> Mag = Std. mag + 2.5*LOG((1600/1000)^2) -or- >> Mag = Std. mag + 5*LOG(1600/1000) >> = 5.5 + 1.02 = 6.52 > I'm sorry, but the above equations seem to be ignoring the phase angle. Yes -- the equations are specific to the example you gave: an object at 90-degree phase. > > It must be possible for an object to be at 1600km distance in more than > one spot. Or perhaps the phase angle is irrelevant? Of course phase is VERY important. I deliberately excluded the phase-dependence from the equation because your example did not require it. I posted a full equation w/phase-dependence on Seesat many months ago, but I'll include one here: Mag = Std. Mag - 15 + 5*LOG(Range) - 2.5*LOG(SIN(B) + (pi-B)*COS(B)) where Range is in km, and B is in radians and measures the angle from the sun to the satellite to the observer. At full phase, B is 0; at new phase, B is pi (i.e. satellite transiting the sun). I want to emphasize that this is the correct equation for a spherical satellite with a perfectly Lambertian surface. Of course, few satellites are spherical, and none have perfectly Lambertian surfaces. Still, it is a good approximation when you don't know the orientation of a satellite. > Perhaps a better approach would be to assume that all satellites are cylinders (since most of the brightest satellites are rocket bodies) and compute the mean reflected radiance (as a function of B) for all orientations. Question for the list -- what do most rocket bodies look like when viewing their ends? Cheers, Rob ----------------------------------------------------------------- Unsubscribe from SeeSat-L by sending a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org http://www2.satellite.eu.org/seesat/seesatindex.html

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