= Newtonian potential of a point mass
{c}
{title2=$\Phi(r)=-GM/r$}
= Gravitational potential
{synonym}
The <Newtonian potential of a point mass> $M$ at the origin is
$$
\Phi(r)=-\frac{GM}{r},\qquad r>0,\qquad \Phi(\infty)=0.
$$
It is the free-space <Green function of the Laplacian> with the normalization of the <Poisson equation for Newtonian gravity>. Differentiating gives inward <gravitational acceleration> $-GM/r^2$. With $z=r/r_0$, the normalized <Newtonian gravitational potential> is $\Phi/(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.
For a radially outward launch from $r_0$, <conservation of mechanical energy> in the fixed-source <test particle> approximation gives the minimum <escape velocity>:
$$
\frac{E}{m}=\frac{v_0^2}{2}-\frac{GM}{r_0}=0
\quad\Longrightarrow\quad v_{\mathrm{esc}}=\sqrt{\frac{2GM}{r_0}}.
$$
At this threshold the <speed> tends to zero only at infinite separation. The model neglects air resistance, external <tidal forces>, and <general relativity>. <University Physics Volume 1 by William Moebs and others/13.3 Gravitational Potential Energy and Total Energy>{ref}
A <circular orbit> at $r_0$ instead requires tangential <speed> $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt2\simeq1.41421$. A radial launch at <speed> $1$ reaches $r=2$ and returns; the same <speed> directed tangentially gives a <circular orbit>. <University Physics Volume 1 by William Moebs and others/13.4 Satellite Orbits and Energy>{ref}
\Image[/gravitational-escape-energy.png]
{title=Energy balance and outward radial speeds below, at, and above escape}
{description=Left: at the <escape velocity>, positive <kinetic energy> exactly balances negative <Newtonian gravitational potential energy>. Right: radial launches at normalized <speeds> $1$, $\sqrt2$, and $1.6$ respectively turn at $r/r_0=2$, approach zero <speed> at infinity, and retain a positive limiting <speed>. The plotted trajectories use a fixed source and the <test particle> approximation.}
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