Free-fall time for two objects on a radial path (starting from rest) is given by:

$t(y)=\sqrt\frac{y_0^3}{2u}\biggl(\sqrt{\frac{y} {y_0} \bigl({1} -\frac{y}{y_0}}\bigr) + \arccos\sqrt\frac{y}{y_0}\biggr)$


$t$ is the time after the start of the fall

$y$ is the distance between the centers of the bodies

$y_0$ is the initial value of $y$

$u$ = $G(m_1 + m_2$) is the standard gravitational parameter.

The solution of this equation of motion yields time as a function of separation. See also this Phys.SE and links therein.

Question: Now let's say these objects start with some initial velocity. How would you incorporate some initial $v$ into these calculations?

  • $\begingroup$ Try starting with the derivation given here and add a $\dot r = ...$ at the appropriate point. $\endgroup$ Commented Mar 21, 2014 at 14:25

1 Answer 1


This is only a partial answer but for gravitationally bound objects, i.e. the kinetic + potential energy < 0, you can simply use the equation you already have.

If you start at a distance $y$ with velocity $u$ simply find the distance $y'$ where the velocity would be zero:

$$ \frac{GM}{y_0^2} - \frac{GM}{y'^2} = \tfrac{1}{2}m u^2 $$

Then use this value of $y_0$ in the equation you already have.


You'll need to substitute this expression in the equation you already have and replace $t$ by $t + t(y_0)$ so that $t = 0$ at your desired starting position. I haven't attempted this because while straightforward it produces a messy and unilluminating equation.

For gravitationally unbound objects I guess you could use the equation for a hyperbolic orbit and take the limit of the angular momentum equal to zero. I did a quick Google but couldn't find a closed form expression for a hyperbolic orbit, so I suspect it's not pretty.

I suspect the reason for the lack of response to your question is that it's not a terribly exciting one. The differential equation that controls the equation of motion is a very simple one:

$$ \frac{d^2y}{dt^2} = \frac{GM}{y^2} $$

but as you've already found even simple differential equations can have cumbersome analytical solutions. Solving the equation for the initial conditions $v(0) <> 0$ isn't physically very enlightning and is algebraically tedious.


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