Hi Bjoern and List, > The area of a band around a latitude is R*R*width * 2 * pi * cos(lat), > not considering oblateness. So between two latitudes, it is the > integral, or 2*pi*(sin(lat1)-sin(lat2)), or for your example > 4*pi*0.3987. For the whole sphere it is 4*pi, so 40% is your > answer. The more interesting (and far more difficult) calculation is the land-to-water ratio as a function of satellite inclination. Because of the irregular boundaries of continents and islands, there will be two extreme cases: the inclination with maximum land-to-water ratio, and the inclination with the corresponding minimum ratio. Since the computation encompasses more and more area as the inclination increases, both extremes are likely to be found at fairly low inclinations. The 90-degree-inclination case simply reduces to the global land-to-water ratio, which is known (though I don't recall the figure). Of course, for this list ~observation~ of the reentry is of more interest than actual recovery of debris; the two are obviously related, but the probability of observation is much improved over probability of being within the debris field boundary (somewhat akin to the difference between seeing a partial solar eclipse and a total one). Best, Rob ----------------------------------------------------------------- Unsubscribe from SeeSat-L by sending a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org http://www.satellite.eu.org/seesat/seesatindex.html
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