Read on Twitter
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]
![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]](https://pbs.twimg.com/media/E08MW1IWUAIZgVL.jpg "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's Socioeconomic Data and Applications Center (SEDAC): [5/n] https://sedac.ciesin.columbia.edu/data/collection/gpw-v4
![We also need to know the population density along the projected trajectory. This can be obtained from a number of sources, including NASA's Socioeconomic Data and Applications Center (SEDAC): [5/n] https://sedac.ciesin.columbia.edu/data/collection/gpw-v4](https://pbs.twimg.com/media/E08NHiTWEAAVSNr.jpg "We also need to know the population density along the projected trajectory. This can be obtained from a number of sources, including NASA's Socioeconomic Data and Applications Center (SEDAC): [5/n] https://sedac.ciesin.columbia.edu/data/collection/gpw-v4")
Finally, we can estimate the casualty probability (again making a few assumptions) using another relatively simple formula: [6/n]
![Finally, we can estimate the casualty probability (again making a few assumptions) using another relatively simple formula: [6/n]](https://pbs.twimg.com/media/E08N2AKWYAAZbZN.png "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]
![Here, Pi is the impact probability (our rectangular or elliptical probability distribution from earlier) in a grid cell (n, m) [7/n]](https://pbs.twimg.com/media/E08ORQ5WYAEPR4t.jpg "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 'rho' p) is the population density in the grid cell (n, m). As an example, I've used a visualisation of the gridded population density for the Paris region in France: [8/n]
![and pp (actually 'rho' p) is the population density in the grid cell (n, m). As an example, I've used a visualisation of the gridded population density for the Paris region in France: [8/n]](https://pbs.twimg.com/media/E08OsEgWUAEdZyY.png "and pp (actually 'rho' p) is the population density in the grid cell (n, m). As an example, I've used a visualisation of the gridded population density for the Paris region in France: [8/n]")
This explanation is from Heiner Klinkrad's excellent book, 'Space Debris: Models and Risk Analysis', if you are interested. [9/n]
![This explanation is from Heiner Klinkrad's excellent book, 'Space Debris: Models and Risk Analysis', if you are interested. [9/n]](https://pbs.twimg.com/media/E08O6YFXsAQMFvp.jpg "This explanation is from Heiner Klinkrad's excellent book, 'Space Debris: Models and Risk Analysis', 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]
![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]](https://pbs.twimg.com/media/E08PaJ-X0AE2RcE.png "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's possibly interesting about the values in the table and the calculation of casualty probability I've shown higher in this thread is that it is actually not dimensionless. It has units of 'people'. In other words, the formula estimates the number of casualties [11/n]
So, even if the 'probabilities' 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]
