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2.6.3.2. Gravitational Potential

Table of Contents

Meaning of Gravitational Potential

Gravitational potential is the gravitational potential energy per unit mass at a point in space. It tells us how much potential energy a small test mass would have at that location, divided by the value of that mass.

If a mass $m$ is placed at a point where the gravitational potential is $V$, then its gravitational potential energy is

$$
U = mV
$$

This idea is useful because the potential depends only on the source mass and the position in space, not on the test mass we use to probe it.

Important relation:
$$
V = \frac{U}{m}
$$
Gravitational potential is potential energy per unit mass.

Potential Around a Point Mass

Consider a mass $M$ producing a gravitational field. The gravitational potential at a distance $r$ from that mass is

$$
V(r) = -\frac{GM}{r}
$$

where $G$ is the gravitational constant.

The negative sign is very important. It means that the potential is chosen to be zero infinitely far away, and everywhere closer to the attracting mass the potential is negative. A mass must gain energy to escape from that region and go to infinity.

For a point mass or a spherically symmetric mass distribution, with zero potential at infinity,
$$
V(r) = -\frac{GM}{r}
$$
The potential is always negative for finite $r$.

Why the Potential Is Negative

Gravity is an attractive force. If an object is very far from a planet or star, we define its gravitational potential to be zero. As the object moves closer, gravity pulls it inward, and the system loses potential energy. That is why the potential becomes negative.

A more negative value means the object is deeper in the gravitational influence of the source mass.

For example, if one point has potential $-2.0 \times 10^7 \,\text{J/kg}$ and another has potential $-6.0 \times 10^7 \,\text{J/kg}$, the second point is at a lower potential.

Difference in Potential

In many problems, what matters is not the absolute value of potential, but the change in potential between two points. If the distances from the mass $M$ are $r_1$ and $r_2$, then

$$
\Delta V = V_2 - V_1 = -\frac{GM}{r_2} + \frac{GM}{r_1}
$$

If a test mass $m$ moves between these points, the change in gravitational potential energy is

$$
\Delta U = m \Delta V
$$

So gravitational potential helps us calculate energy changes quickly.

Potential difference between two distances from a mass $M$:
$$
\Delta V = -\frac{GM}{r_2} + \frac{GM}{r_1}
$$
And for a mass $m$,
$$
\Delta U = m \Delta V
$$

Units of Gravitational Potential

Since gravitational potential is energy per unit mass, its SI unit is

$$
\text{J/kg}
$$

Using base SI units,

$$
1\ \text{J/kg} = 1\ \text{m}^2/\text{s}^2
$$

This is not the same as gravitational field strength, even though the units are closely related in meaning through force and energy ideas.

Potential of a Spherical Body

For any spherically symmetric body, such as an ideal planet or star, the gravitational potential outside the body is the same as if all its mass were concentrated at the center. So for $r \ge R$,

$$
V(r) = -\frac{GM}{r}
$$

where $R$ is the radius of the body.

At the surface,

$$
V_{\text{surface}} = -\frac{GM}{R}
$$

This is often used for planets.

Example with Earth

For Earth, using mass $M_E$ and radius $R_E$, the gravitational potential at the surface is

$$
V_E = -\frac{GM_E}{R_E}
$$

Numerically, this is about

$$
V_E \approx -6.3 \times 10^7\ \text{J/kg}
$$

This means that each kilogram at Earth's surface has gravitational potential energy about $6.3 \times 10^7\ \text{J}$ below the zero level at infinity.

Visualizing Potential

A graph of $V$ versus $r$ is a negative curve that approaches zero from below as $r$ becomes very large. Close to the mass, the potential becomes more negative.

Gravitational potential around a point mass

This picture shows that the potential never becomes positive if we choose zero at infinity.

Superposition of Potentials

If several masses are present, the total gravitational potential at a point is the sum of the individual potentials:

$$
V_{\text{total}} = V_1 + V_2 + V_3 + \cdots
$$

For point masses,

$$
V_{\text{total}} = -G\left(\frac{M_1}{r_1} + \frac{M_2}{r_2} + \frac{M_3}{r_3} + \cdots \right)
$$

This scalar addition is often simpler than adding gravitational fields, because potential is a scalar quantity.

Gravitational potentials add algebraically:
$$
V_{\text{total}} = \sum_i V_i = -G \sum_i \frac{M_i}{r_i}
$$

Comparison with Gravitational Potential Energy

It is easy to confuse gravitational potential with gravitational potential energy, so it helps to compare them directly.

QuantityMeaningSymbolDepends on test mass?SI unit
Gravitational potentialPotential energy per unit mass$V$No$\text{J/kg}$
Gravitational potential energyEnergy due to position in gravitational field$U$Yes$\text{J}$

If two objects are placed at the same point, they have the same gravitational potential $V$, but their potential energies $U = mV$ are different if their masses are different.

Near Earth's Surface

Very close to Earth's surface, where height changes are small compared with Earth's radius, the change in gravitational potential can be approximated using

$$
\Delta V \approx g \Delta h
$$

if the object moves upward by height $\Delta h$. Since the object gains potential energy when it rises, the potential increases, becoming less negative.

This approximation comes from the more exact expression and works well only for small height changes.

Near Earth's surface, for a small upward change in height $\Delta h$,
$$
\Delta V \approx g\Delta h
$$
This is an approximation, not the exact formula for all distances.

Physical Interpretation

Gravitational potential tells us how much work per unit mass must be done by an external agent to bring an object slowly from infinity to a point in the gravitational field. Because gravity pulls inward naturally, the required external work is negative when moving from infinity to a closer point.

It can also be understood as a measure of how tightly bound a mass is by gravity at that location. The more negative the potential, the more energy per kilogram would be required to remove the object to infinity.

Summary Relations

The central formulas of this chapter are these:

$$
V = \frac{U}{m}
$$

$$
V(r) = -\frac{GM}{r}
$$

$$
\Delta V = -\frac{GM}{r_2} + \frac{GM}{r_1}
$$

$$
U = mV
$$

$$
\Delta U = m\Delta V
$$

These formulas connect position in a gravitational field to energy in a very direct way.

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