3.7.2.3 - Gravitational Potential

Potential, the reference zero and work signs

Gravitational potential $V$ at a point is the gravitational potential energy per unit mass there. Equivalently, it is the work done per unit mass by an external agent in bringing a small test mass from infinity to the point without changing its kinetic energy. Its unit is $\mathrm{J\,kg^{-1}}$.

For an isolated finite source, the conventional zero is $V=0$ at infinity. The zero is a reference choice, not a rule that zero field always means zero potential. Gravity does positive work when a mass moves inward. To bring it inward slowly at constant kinetic energy, an external agent must remove energy: the external work is negative. Therefore finite-distance potential is negative relative to infinity.

Between positions A and B define $\Delta V=V_B-V_A$. For a constant test mass $m$:

The external work against gravity is $W_{\text{ext}}=m\Delta V$ when there is no change in kinetic energy and no other energy transfer. More generally, $W_{\text{ext}}=\Delta K+\Delta U$ under those same no-loss assumptions. Do not call every launch-energy requirement $m\Delta V$: a launch may also change kinetic energy.

Moving outward makes $V$ less negative, so $\Delta U>0$ and gravity does negative work. Moving inward makes $\Delta U<0$ and gravity does positive work. With gravity alone, $\Delta K=-\Delta U$; the loss in potential energy becomes kinetic energy.

For a $200\,\mathrm{kg}$ probe moved slowly from $-30$ to $-24\,\mathrm{MJ\,kg^{-1}}$, $\Delta V=+6\,\mathrm{MJ\,kg^{-1}}$ and $W_{\text{ext}}=+1.2\times10^9\,\mathrm J$. Gravity does $-1.2\times10^9\,\mathrm J$ of work on that same move.

Potential outside a spherical mass

For a point mass, or at/outside a spherically symmetric source of mass $M$, the potential with zero at infinity is

Use the centre distance $r$, not altitude. The expression does not describe the interior of a planet using its entire mass. It is a negative inverse-distance relationship, distinct from the inverse-square field magnitude $g=GM/r^2$.

Doubling $r$ halves the magnitude of $V$, making its signed value higher: $-40\,\mathrm{MJ\,kg^{-1}}$ becomes $-20\,\mathrm{MJ\,kg^{-1}}$. At the same time $g$ becomes one quarter as large. Both tend towards zero far away, but neither vanishes at a finite exterior radius around a single nonzero source.

For Earth, $G=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$, $M=5.97\times10^{24}\,\mathrm{kg}$ and $R=6.37\times10^6\,\mathrm m$ give $V_s=-6.25\times10^7\,\mathrm{J\,kg^{-1}}$. Moving one kilogram slowly from the surface to infinity requires $+6.25\times10^7\,\mathrm J$ against gravity in this isolated-Earth model.

Potential is a scalar: add signed potentials from separate sources. Two positive masses both contribute negative potential even at a point where their field vectors cancel. A zero resultant field therefore need not mean zero potential.

Equipotentials and zero gravitational work

An equipotential surface has the same potential at every point. A displacement along it has $\Delta V=0$, so gravity does no work and gravitational potential energy does not change. A force can be present while its work is zero: the gravitational force is perpendicular to the instantaneous displacement.

Around an isolated spherical source, equipotentials are concentric spheres; circles in a cross-section represent those spheres. Their normals are radial, so gravitational field lines meet them at right angles. In a small approximately uniform field they are approximately parallel planes perpendicular to the field.

[DIAGRAM: asset_name: 7.2.3 - Gravitational Potential - Diagram 1; asset_slug: 7.2.3 - Gravitational Potential - Diagram 1; recommended_method: retained_png; caption: Exact equal-step equipotentials outside a spherical source. R is its radius. Field arrows point inward.; description: Exact equipotential cross-section for V = −60R/r in MJ kg⁻¹. The labelled surfaces differ by 10 MJ kg⁻¹, so their spacing grows outward. Radial arrows point inward and meet each circle normally.]

For equal potential steps, closer surfaces indicate a larger field magnitude: the same energy change per kilogram occurs over a shorter distance. This comparison requires equal steps; arbitrary unlabelled contour spacing is insufficient.