Hello all! Kerry Kirkland asked about the visibilitiy of a 19-inch sphere studded with 1-inch circular mirrors as seen at an altitude of 300 km. It may come as a surprise that such an object would be quite visible, and very interesting to observe! Even more surprising is that only one one-inch mirror would be contributing to the observed brightness of the satellite at any given time (more on this below). Also, if the satellite isn't spinning, then it will flash at a variable rate that increases as it approaches culmination, and decreases again as it descends. If the satellite is spinning, then it will "scintillate" -- flashing approximately 60 times for every revolution. The reason that only one mirror contributes to the observed brightness at any given time is that adjacent mirrors are angled 6-degrees from one another. The solar image only subtends about a half-degree. What this also means is that more than 90% of the time the satellite would not be directly reflecting the sun. So how bright are the flashes? That depends on several factors, including the satellite range, phase angle and mirror reflectivity, but I'll give you a ballpark estimate based on the following assumptions: 1. range of 400 km 2. phase of 90 degrees 3. mirror reflectivity of 95% 4. sun visual magnitude of -26.7 Since the phase is 90 degrees, the individual mirror causing the specular reflection would be tilted 45 degrees from normal to the observer. This reduces the projected area of the mirror from pi*(0.5*0.5 inch^2) to COS(45) * pi * (0.5*0.5 inch^2) = 0.555 inch^2 = 3.58 cm^2. The solid angle subtended by the mirror (as seen by the observer) is the projected area divided by the range squared: 3.58 cm^2 / (400 km)^2 = 2.24 x 10^-15 steradians By comparison, the sun subtends roughly 6.0 x 10^-5 steradians. This means that just based on projected area alone, the sun would be brighter than the mirror reflection by a factor of: (6.0 x 10^-5) / (2.24 x 10^-15) = 2.68 x 10^10 Including the reflectivity factor of 0.95, the sun is brighter by a factor of 2.82 x 10^10. In visual magnitudes, the difference is: 2.5 * LOG10(2.82 x 10^10) = 26.1 Since the sun is magnitude -26.7, this means the flashes would be magnitude (-26.7 + 26.1) = -0.6. Most definitely naked-eye visible! Now if the satellite isn't spinning, or is spinning extremely slowly, the flashes may be slightly brighter than this. The reason is that the sun is not uniformly bright across its disk -- the center is more than 3 times brighter than the edge. If the entire sun had the same intensity as the center, it would perhaps be as much as one magnitude brighter. So the flashes in the above example could actually rival the star Sirius. On the other hand, if this "disco ball" is spinning too fast, then the shortness of the individual flashes will begin to affect their perceived brightness. But I can compute a lower limit on the brightness by assuming an 8.33% duty cycle (0.5 degree sun / 6-degrees between mirrors) -- i.e. a factor of 12. That's 2.7 visual magnitudes, so the -0.6 becomes +2.1. Still an easy object. Others out there, please check my math! It's possible that I've goofed somewhere along the way, but the result seems reasonable when you compare with the -8 magnitude flashes of Iridium at twice the range, but with mirrors 3200 times larger. --Rob