It is possible to estimate re-entry risk even if no detailed spacecraft/rocket stage breakup analysis is available. E.g., one may assume that any surviving fragments are dispersed in a rectangular or elliptical area based on the uncertainty in the predicted re-entry epoch [1/n]
This area could be, say, +/- 2000 km in the along-track direction and +/- 80 km in the cross-track direction. For an ellipse, these values would represent the 2-sigma values in a 2D Gaussian probability distribution [2/n]
We also need to know the equivalent casualty cross-section (i.e. the area occupied by a human being on the ground combined with the area of the surviving fragments). We can estimate this using a formula:
Here, Ai is the cross-section of individual fragments and Ah is the projected human risk-cross section (which can be assumed to be about 0.36 square metres, or rh = 0.33 m). [4/n]
We also need to know the population density along the projected trajectory. This can be obtained from a number of sources, including NASA& #39;s Socioeconomic Data and Applications Center (SEDAC): [5/n] https://sedac.ciesin.columbia.edu/data/collection/gpw-v4">https://sedac.ciesin.columbia.edu/data/coll...
Finally, we can estimate the casualty probability (again making a few assumptions) using another relatively simple formula: [6/n]
Here, Pi is the impact probability (our rectangular or elliptical probability distribution from earlier) in a grid cell (n, m) [7/n]
and pp (actually & #39;rho& #39; p) is the population density in the grid cell (n, m). As an example, I& #39;ve used a visualisation of the gridded population density for the Paris region in France: [8/n]
This explanation is from Heiner Klinkrad& #39;s excellent book, & #39;Space Debris: Models and Risk Analysis& #39;, if you are interested. [9/n]
In his book Heiner provided estimates based on this approach for the long-term risk of land impacts of fragments from an object in a near-circular orbit of inclination i, assuming an equivalent casualty cross-section of 10 square metres & world population in 2000: [10/n]
What& #39;s possibly interesting about the values in the table and the calculation of casualty probability I& #39;ve shown higher in this thread is that it is actually not dimensionless. It has units of & #39;people& #39;. In other words, the formula estimates the number of casualties [11/n]
So, even if the & #39;probabilities& #39; appear to be high (e.g. I had seen 1-in-100 mentioned in relation to the Long March re-entry) they can be interpreted as the probability of 1 casualty, not the probability of mass casualties. [12/n]
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