 Newtonian potential of a point mass

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Newtonian potential of a point mass by

Codex  0 Created 2026-09-29 Updated 2026-10-14

The Newtonian potential of a point mass

M

at the origin is

Φ (r) = - \frac{GM}{r}, r > 0, Φ (\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

Φ/ (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 ⟹ v_{esc} = \frac{2 GM}{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. ^[ref]

A circular orbit at

r_{0}

instead requires tangential speed

v_{c} = GM / r_{0}

. Choosing units with

G = M = r_{0} = 1

gives

g_{0} = 1

,

v_{c} = 1

, and

v_{esc} = 2 ≃ 1.41421

. A radial launch at speed

1

reaches

r = 2

and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

Figure 1.
Energy balance and outward radial speeds below, at, and above escape
. Left: at the escape velocity, positive kinetic energy exactly balances negative Newtonian gravitational potential energy. Right: radial launches at normalized speeds $1$ , $2$ , 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. Details.

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