= Inverse-square potential
{title2=$V(r)=-k/r$}
Here <inverse-square potential> means the <potential energy> associated with an attractive <inverse-square law> for a <central force>: the <potential energy> itself varies as inverse distance. For $k>0$ and <position> $\mathbf r$ with $r=|\mathbf r|>0$,
$$
V(r)=-\frac{k}{r},\qquad V(\infty)=0,\qquad \mathbf F=-\nabla V=-\frac{k}{r^2}\widehat{\mathbf r}=-\frac{k\mathbf r}{r^3}.
$$
The <gradient> makes the distinction explicit: doubling $r$ halves the magnitude of <potential energy> but quarters the magnitude of <force>. The neighboring <repulsive inverse-square potential> instead has <potential energy> proportional to $r^{-2}$ and an inverse-cube <force>; the two naming conventions must be distinguished.
For <Newton's law of universal gravitation>, $k=GMm$, where $G$ is the <gravitational constant>. Treat the attracting <mass> $M$ as an isolated fixed <point mass>, or as a body with <spherical symmetry> and exterior <test particle> of constant <mass> $m\ll M$. The <spherical shell theorem> gives the same exterior <Newtonian gravitational potential energy> $U=-GMm/r$ without requiring uniform <mass density>. Here $r$ is measured from the centre, and the formula applies outside the body; it does not describe its interior. The <Newtonian gravitational potential> per unit <mass> is $\Phi=U/m=-GM/r$.<The Feynman Lectures on Physics, Volume I by Richard Feynman and others/Chapter 13: Work and Potential Energy (A)>[§13–2, equations 13.11–13.12: radial work and conserved energy; §13–4: gravitational field outside a sphere]{ref}
The negative sign records binding relative to infinite separation, rather than a negative <kinetic energy>. Moving from $r_1$ to $r_2>r_1$ increases <potential energy> by
$$
\Delta V=k\left(\frac1{r_1}-\frac1{r_2}\right)>0,\qquad W_{\mathrm{gravity}}=-\Delta V.
$$
Because the <force> is a <conservative force>, this <work> depends on the endpoints, including for a curved path. Integrating $dV/dr=k/r^2$ with zero at infinity gives $V=-k/r$; adding a constant leaves the <force> unchanged but shifts the numerical escape <energy> threshold by that same constant. This is why the reference at infinity must accompany an <escape velocity> calculation.
For an exact synthetic comparison, stipulate the <standard gravitational parameter> $\mu=GM=8\,\mathrm{m^3/s^2}$ and a $1\,\mathrm{kg}$ <test particle>. At $r=2\,\mathrm m$, $U=-4\,\mathrm J$ and the inward <force> has magnitude $2\,\mathrm N$; at $r=4\,\mathrm m$, $U=-2\,\mathrm J$ and the inward <force> has magnitude $0.5\,\mathrm N$. The increase is $2\,\mathrm J$ of <potential energy>. These stipulated parameters are an arithmetic example, not observations of a particular body. The <escape-energy balance diagram> uses the same parameters.
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