First, ignore air resistance. Always ignore air resistance.

Using kinematics for gravitational acceleration systems works within a specific scope, and when the system's scope widens too far, they become an oversimplification. Kinematics are great for systems subject to constant acceleration, and gravity fundamentally cannot be a constant acceleration (take a look at its formula).

I know how to accurately model velocity as a function of displacement in a gravitational system using potential energy, more specifically the changes in potential energy. Basically the difference between initial potential energy (using gravity calculated with the object's initial height) and final potential energy (with gravity calculated from the object's current position) tells us the change in the object's kinetic energy and thus velocity. I've checked these calculations against highly precise computer simulations and found them to be accurate.

This set up models velocity as a function of displacement. My question is how to derive a formula which models velocity as a function of time in a gravitational system. I've asked some colleagues about this and none can produce an accurate model. I tried to integrate the acceleration to yield velocity, but that approach can't convert acceleration vs displacement into velocity vs time.

Any insight would be much appreciated.

EDIT: To clarify, the model I'm looking for is not so much of an orbit as it is a simple (one dimensional) free fall straight down to the "earth"

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    $\begingroup$ There's not much information in here! Are you talking an mgh potential? 1/r potential? 1d? 2d? But yes, in general, Newton's equations can't be reduced to a single integral x (t)=..., you actually have to solve the ODEs. In lots of cases, however, it does reduce ("to quadrature"). "Kepler's problem" might be what you're looking for, or it might be overkill! But that's a keyword that might help you out. $\endgroup$
    – user12029
    Commented Dec 9, 2016 at 22:45
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    $\begingroup$ Related: physics.stackexchange.com/q/35878 and links therein (discourse on how big an error is made by assuming $g = \text{constant}$ near the surface of the Earth) $\endgroup$ Commented Dec 9, 2016 at 23:03
  • $\begingroup$ Related: physics.stackexchange.com/q/3534/2451 , physics.stackexchange.com/q/19388/2451 and links therein $\endgroup$
    – Qmechanic
    Commented Dec 10, 2016 at 6:34

1 Answer 1


I mean this depends strongly on what trajectories you're considering; it sounds like you're trying to use a $1/r^2$ force law on trajectories that are purely radial. Then we have:$$\ddot r = k/r^2,$$ which we can solve in a rather boring way by multiplying both sides by $\dot r$ and integrating to get:$$\dot r = \sqrt{C - \frac{2k}r},$$ which of course is your formula from the term for kinetic energy. To get a closed form for $r(t)$ you would now need to integrate:$$\int \frac{dr}{\sqrt{C_1 - 2k/r}} = t + C_2.$$The left hand side can be rewritten defining $\kappa = 2k/C_1$ and $v = \sqrt{C_1}$ as $$\int \frac{dr}{\sqrt{1 - \kappa/r}}=v t + C.$$Looking purely at the left-hand side we try substituting $u = 1 - \kappa/r,$ whence $r = \kappa/(1-u)$ and $dr = \kappa~du/(1-u)^2.$ So then we have$$\int \frac{du}{(1 - u)^2 \sqrt{u}}$$Now let $v = \sqrt{u}$ and this becomes $$\int \frac{dv}{(1 - v^2)^2},$$ which can be solved by the method of partial fractions.

The unfortunate thing is that even though you can get a $t(r)$ this way, inverting it to an $r(t)$ does not seem like an expression with a simple form.


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