= Radial free fall from rest
{title2=$v(r)=\sqrt{2\mu(1/r-1/r_0)}$}
Let a <test particle> start at rest at radius $r_0$ outside a fixed spherically symmetric source. The <spherical shell theorem> gives the exterior <Newtonian gravitational potential> $\Phi(r)=-\mu/r$, where $\mu=GM$ is the source's <standard gravitational parameter>. Zero initial transverse <velocity> makes the motion radial. <Conservation of mechanical energy> per unit <mass> gives
$$
\frac{\dot r^2}{2}-\frac{\mu}{r}=-\frac{\mu}{r_0},\qquad
\dot r=-\sqrt{2\mu\left(\frac1r-\frac1{r_0}\right)}.
$$
The minus sign selects inward motion; the positive square root is its <speed>. For a fall from $r_0=2R$ to the surface $r=R$, that <speed> is $\sqrt{\mu/R}$. In the limit of release from rest infinitely far away, the familiar <free-fall speed> is $\sqrt{2\mu/r}$, equal in magnitude to the local <escape velocity>. A finite starting radius gives a smaller speed.
The arrival time follows from the same first integral. Set $r=r_0\cos^2\eta$. Then
$$
t(\eta)=\sqrt{\frac{r_0^3}{2\mu}}\left(\eta+\sin\eta\cos\eta\right),\qquad
\eta=\arccos\sqrt{\frac r{r_0}},\qquad 0\leq\eta<\frac\pi2.
$$
For arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta=\pi/4$ gives
$$
t_{\mathrm{arrival}}=\left(1+\frac\pi2\right)\sqrt{\frac{R^3}{\mu}}\approx2.570796\sqrt{\frac{R^3}{\mu}}.
$$
Keeping the smaller initial gravitational magnitude $g(r_0)=\mu/(4R^2)$ constant would predict $\sqrt8\,\sqrt{R^3/\mu}$, about $10.0\%$ longer. The actual <acceleration> increases during the fall.
\Image[/radial-free-fall-gravity-variation.png]
{title=Radial free-fall height and arrival time compared with constant gravity at the release point}
The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of <nondimensionalization>: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu}$ supplies the dimensional time scale.
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