Radial free fall from rest ($v(r)=\sqrt{2\mu (1/r-1/r_0)}$)

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 givesThe 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 $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

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.