What Is the Relation Between Gravitational Potential and Field?

Gravitational potential and gravitational field describe related features of the same gravitational influence. Potential tells you the gravitational potential energy per unit mass at a location; the field tells you the force per unit mass on a test object there.

The connection is a spatial derivative: the gravitational field points in the direction in which potential decreases most rapidly. This article gives the vector relation, its one-dimensional form, and a worked point-mass example.

The general relation

Let $\Phi(\mathbf r)$ denote gravitational potential and $\mathbf g(\mathbf r)$ the gravitational field. Their relation is

$$\mathbf g=-\nabla\Phi.$$

The minus sign matters: the field points downhill in potential. In Cartesian coordinates,

$$\mathbf g=-\left(\frac{\partial\Phi}{\partial x},\frac{\partial\Phi}{\partial y},\frac{\partial\Phi}{\partial z}\right).$$

Thus, a potential that changes rapidly with position corresponds to a strong field, while a region where potential is nearly constant has a weak field. The field’s magnitude is the rate of decrease of potential in the field’s direction.

In Newtonian gravity, potential is usually defined as potential energy per unit mass: $\Phi=U/m$. Its SI unit is joules per kilogram (J/kg), equivalent to square metres per second squared (m²/s²). The gravitational field has units of newtons per kilogram (N/kg), equivalent to metres per second squared (m/s²).

Potential differences and work

The gradient relation can also be written as a line-integral relation between two points $A$ and $B$:

$$\Phi(B)-\Phi(A)=-\int_A^B \mathbf g\cdot d\mathbf l.$$

This says that the change in potential is the negative of the work done by gravity per unit mass. If an object moves along the field, gravity does positive work and its potential decreases. Potential differences are physically meaningful; adding the same constant to $\Phi$ everywhere does not change the field.

For a mass $m$ in the field, its potential energy is $U=m\Phi$, and its gravitational force is

$$\mathbf F=-\nabla U=m\mathbf g,$$

provided the test mass does not significantly alter the source field.

Spherically symmetric sources

For a spherically symmetric source, potential depends only on distance $r$ from the centre. The gradient then has only a radial component:

$$\mathbf g=-\frac{d\Phi}{dr}\,\hat{\mathbf r}.$$

Here, $\hat{\mathbf r}$ points radially outward. For a point mass $M$, with the conventional choice $\Phi=0$ at infinity,

$$\Phi(r)=-\frac{GM}{r}, \qquad \mathbf g(r)=-\frac{GM}{r^2}\,\hat{\mathbf r}.$$

Indeed, $d\Phi/dr=GM/r^2$ is positive, so the minus sign makes the field point inward. For a spherical body, these expressions also apply outside the body as though its mass were concentrated at its centre. Inside an extended body, the potential and field depend on how mass is distributed.

Worked example: a 10 kg point mass

Take a point mass $M=10\,\text{kg}$ and calculate potential and field at $r=2\,\text{m}$. Use $G=6.6743\times10^{-11}\,\text{N}\,\text{m}^2/\text{kg}^2$ and set potential to zero at infinity.

The potential is

$$\Phi(2)=-\frac{GM}{2} =-\frac{(6.6743\times10^{-11})(10)}{2} =-3.33715\times10^{-10}\,\text{J/kg}.$$

The field is

$$\mathbf g(2)=-\frac{GM}{2^2}\,\hat{\mathbf r} =-1.668575\times10^{-10}\,\hat{\mathbf r}\,\text{m/s}^2.$$

The negative radial direction means inward. As a check, moving outward from $r=2\,\text{m}$ to $r=3\,\text{m}$ raises the potential:

$$\Phi(3)-\Phi(2)=GM\left(\frac{1}{2}-\frac{1}{3}\right) =1.1124\times10^{-10}\,\text{J/kg}.$$

That increase is consistent with the inward field: moving outward is against gravity, so potential rises.

Common mistakes

• Dropping the minus sign: $\nabla\Phi$ points toward increasing potential; gravity points toward decreasing potential.
• Confusing potential with potential energy: $\Phi$ is per unit mass. For a mass $m$, the potential energy is $U=m\Phi$.
• Treating potential as a vector: Potential is a scalar; the field is a vector. The gradient converts the spatial variation of the scalar into a vector.
• Assuming potential must be negative: Its value depends on the chosen zero. The point-mass potential is negative when zero is chosen at infinity, but adding a constant changes its value without changing the field.
• Using the radial formula without symmetry: $\mathbf g=-\frac{d\Phi}{dr}\hat{\mathbf r}$ applies when potential depends only on radius. In a general configuration, use $\mathbf g=-\nabla\Phi$.