Not sure how much you’d care, but I came back to this and found (by hand) a function which closely approximates the solution - it’s not exact but it’s also not super far. Graph.
I think I could also solve the differential equation by hand at this point (getting the same solution as before) - I haven’t, but I’m pretty sure I could if I wanted to for whatever reason. I’m doubtful it’s possible to get an exact solution in terms of just x but if you ever manage I’d love to see how.
Kind-of solutionF = ma and m = 1, so a = F = 1/r^2. I used a = -1/r^2 so the force experienced by the particle at r = 5 would be negative. a = dv/dt. dv/dt = -1/r^2 v dv/dt = -1/r^2 dr/dt → v = dr/dt. Multiply both sides by v, but write it as dr/dt on the right v dv = -1/r^2 dr → Multiply both sides by dt 0.5v^2 = 1/r + C → Integrate v^2 = 2/r + C → Multiply both sides by 2, find C = -2/5 by plugging in r = 5, v = 0 v^2 = 2/r - 2/5 5arcsin(sqrt(r/5)) - sqrt(5r-r^2) = sqrt(2/5) t + C → Solve the differential equation y’^2 = 2/y - 2/5 with Wolfram Alpha, find C = 5π/2 by plugging in r = 5, t = 0 5arcsin(sqrt(r/5)) - sqrt(5r-r^2) = sqrt(2/5) t + 5π/2 → Can find that the particle reaches the origin at t = 5sqrt(5)π / 2sqrt(2) ≈ 12.418 seconds I made heavy use of internet resources on this one, not just for solving the differential equation but also for technique - I got a bunch of wrong answers before getting this one that I finally think is right, at least up until the moment of origin contact Graph (the hidden ellipse is because I thought the path may be elliptical, but it doesn’t appear to be)
![[Unsolved] [2024/05/17] 1 dimensional gravity](https://lemmy.world/pictrs/image/640321cf-f858-4f29-b57a-54742fed2637.png){kind=link}
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