I'm studying escape velocity and related orbital mechanics. Gravity of course follows the inverse square law against distance, and it's easy to thus compute velocity against distance for some object lifting out of a gravity well: v=sqrt(1/x). But how do I change that to velocity against time? Trying to graph this I know that, close to T=0 (launch) a given unit of time will include many units of distance, simply because we are traveling faster. There has to be a 'crossover' point where a unit of time is equal to a unit of distance. And, if the two curves are to meet at some specific height (radius above the launch point) then the time-curve will first be steeper than the distance-curve, and then become shallower such that the two lines meet. Past that, I'm stuck.
2026-05-15 11:31:16.1778844676
convert graph of velocity against distance to velocity against time.
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Subject to correction, it looks like the answer is
VelocityWRTtime=VelocityWRTdistance(r^(2/3))
... I was looking for something much more complicated.