Angle Binocular
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Can an artificial satellite reflect "Moonlight" and be seen by an observer on earth?
Just curious If angles are always right, and the satellites have a reflectivity sufficient, you can see in our night sky in the moonlight (reflected sunlight and by) or satellite must be illuminated only directly by the sun to be visible to an observer on earth. I accept the responses of the eyes and binoculars and telescope response, if it does indeed.
Let me see if I can get technical about it. My first guess is that the moon does not provide sufficient reflected light to be seen by reflection on an artificial satellite. First order and the Moon the Sun is about the same apparent size and its light reaches the satellite in nearly parallel rays, it which means that the reflection takes place in the same way and that is the brightness of the sun or the moon are important in comparison. Okay? 2) Now, you know that the apparent magnitude is: http://en.wikipedia.org/wiki/Apparent_magnitude. My argument would be followed in the assertion that the scale is a logarithmic scale, which means a point lower on the scale means 2512 times more light. For he has done the calculations for me, and Table shows the apparent magnitude Sol - 26.73; application magn Luna - 12.6; Sol 449 brighter than the moon 000 times. QED) Satellite should be greater than 449 000 times reflect moonlight as much as its normal size reflects sunlight. Let see if an expert in this section finds no reason to refine this proposal my side. In fact, taking a blow to the unknown, I would say that if the satellite resembles a star of magnitude -1, full sun, as it should be the size 12 on the moon. Remember, at first glance, we see that up to 6, and 9.5 with binoculars. Is that correct? Any mathematician at home? ... ********************************* For a complete review of the response I would indicate how I felt the greatness of a satellite reflection. A detailed calculation included angle, albedo, shape / Reflecting field and brightness the source. If the optimal viewing conditions, and all things are equal, the more practical factors seem to be driving, therefore, as I said in discussing the optimal conditions of sunshine is even brighter 449 000 times more than a reflection of the moon. Some of the respondents examined in detail the probability to observe a planet that reflects in a practical way. I am interested in setting an upper limit to the brightness at best. However, if this best 449 000 times less light is reflected by the Moon, the magnitude difference between the thought of sun and moon would think of the difference in the apparent magnitude of sources, or 13, maybe 14 points. Note that factors such as surface albedo, reflecting an increase detailed calculation and affect the intensity of reflected light in the same way. Moreover, the brightness of the source that is also multiplied in a logarithmic scale corresponds to a difference as between the Sun and the Moon. Finally, I had my share of seeing the satellites artificial cruise back in the day, and I agree that some hours after sunset, we see what the stars are apparently the average, unless they move straight. These orbits correspond to reasonably high, since the satellite you want to be always in the line of sight with the sun. Its brightness, it could correspond to a star with a 2 or 3 or higher, the apparent magnitude. However, I had the opportunity to observe a satellite in a race for north-south at sunset with the sun still partially in the distance. It was very bright, perhaps brighter than Sirius (this was long before the ISS). Prudente can assign a luminosity in terms of magnitude of -1. So all things being equal, the same satellite to perform a service when the full moon has risen only in the eyes to the horizon like a star of magnitude say 12 as my argument. The size 12, I suggested then the best of scenarios and practical observation to maintain its light clearly more in the actual size. No hope of ever seeing one with binoculars.
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