This would require adding the force of drag to the dynamics timestep, since the purpose of the belly-flop (after re-entry) is to minimize terminal velocity by m

This would require adding the force of drag to the dynamics timestep, since the purpose of the belly-flop (after re-entry) is to minimize terminal velocity by maximizing drag. Currently his solver assumes no drag. Calculating the drag on an ideal cylinder is non-trivial to begin with. It gets vastly more complicated once you start adding the flaps, which are what Starship uses to get itself into the belly-flop position in the first place. Simulating the whole belly-flop maneuver isn't going to be feasible on the scale of a blog post.