Gravitational Potential

Mathematical description of gravitational potential energy

Understanding the energy contained in the gravitational field provides us with a lot of useful physical information regarding the nature of gravity and the role it plays in nature.

Equations of gravitational potential energy

We are going to work with Newton’s law of gravitation because it offers a simple description, even though it is not the theory that best describes gravitation. The main feature of this description of gravity is the spherical symmetry, signalling that there are no special spatial directions. We start by writing down Newton’s law for the gravitational field strength.

\[\vec{Z} = G \cdot \frac{M}{r^2} \cdot \vec{e}_r\]

Here, the vector Z is the gravitational force, M is the mass of the body, r is the radial distance to the source body, the vector e_r is the unit radial vector pointing towards the source body, and G is the universal constant of gravitation whose value is approximately \(G = 6.67 \cdot 10 ^ {-11} m^3/kg \cdot s^2\).

We now present the expression for the gravitational potential energy V:

\[V= -G \cdot \frac{M}{r}\]

We see that spherical symmetry is preserved: all points distributed in a sphere of constant radius \(r = r_0\) have the same gravitational potential energy. For instance, a body in the North Pole has the same potential energy as a body in the South Pole. In fact, all bodies on the surface of the earth have the same gravitational potential energy because they are all at a constant radial distance from the centre of the earth.

It can be readily seen that, by differentiating with respect to the radial distance r, we obtain the expression for the gravitational field strength (given our knowledge, we have to believe in the appearance of the vector e_r ):

\[\vec{Z} = \frac{dV}{dr} \cdot \vec{e}_r\]

These are expressions for the fields produced by a mass M. If we want to consider the force experienced by a mass m under the influence of the gravitational field and its gravitational potential energy, we simply have to multiply the previous expressions by a factor m:

\(\vec{F} = m \cdot \vec{Z}\)(force on mass m under the influence of field Z )

\(U = m \cdot V\)(gravitational potential energy of mass m under the influence of field Z )

Gravitational potential: energetic considerations

We start by analysing the units of the gravitational potential energy of a mass m. Knowing the units of all the quantities implied, we find that

\(m \cdot V = m \cdot G \cdot \frac{M}{r}\) corresponds to \((kg) \cdot (\frac{m^3} {kg \cdot s^2}) \cdot (\frac{kg}{m}) = \frac {kg \cdot m^2}{s^2} = Joules\), which are the units of energy. This is one of the reasons that led to the name of gravitational potential energy.

With this notion in mind, we now recover the principle of stability, which states that bodies tend to minimise their energy. If we take the expression of the gravitational potential energy, we see, due to the negative sign, that lower energies are achieved by reducing the radial distance with the source. The following graph should help us to understand this.

Figure 1. Gravitational potential energy vs radial distance. The gravitational force is computed as the derivative at a certain point along with a distance r (radial distance).

How to solve gravitational potential energy examples

In general, if we are asked about spatial distributions, we usually have to draw or describe graphs that involve the field equations (without including the test mass m). If we are asked about specific values for a certain mass m, we have to use the other two, remembering that U is the potential energy of an object that enters the total energy formula with a minus sign (in case we were considering other forms of energy like the kinetic energy).

Imagine we have a body sourcing the gravitational field with a mass M = 1.50 · 10¹⁰ kg and place a body with a mass m = 1.00 kg to see its behaviour under the gravitational field created by the body with mass M.

Since the body with mass m has the unit mass, the numerical value of the gravitational field and the gravitational force, on the one hand, and the numerical value of the gravitational potential field and the gravitational potential energy, on the other, will be the same. However, they are different concepts with different units.

We can compute all the functions and study their behaviour:

\[vec{Z} = G \cdot \frac{M}{e^2} \cdot \vec{e}_r = \frac{1}{r^2} \cdot \vec{e}_r [N/kg]\]

\[\vec{F} = m \cdot \vec{Z} = \frac{1}{r^2} \cdot \vec{e}_r [N]\]

\[V = -G \cdot \frac{M}{r} = -\frac{1}{r} [m^2/s^2]\]

\[U = m \cdot V = -\frac{1}{r} [J]\]

For the gravitational potential energy, we get a graph similar to the one in figure 1. For the gravitational force, we are going to ignore the radial space orientation, which would require a complex graph, and represent its magnitude as a function of the radial distance r:

Figure 2. Magnitude of gravitational force vs radial distance.

Both the strength of the interaction and the energy asymptotically approach zero at an infinite radial distance. This makes sense because there should be no gravity present between them if the objects are infinitely separated and, therefore, the energy associated with the field should be zero.

We also observe that both quantities diverge when approaching a null radial distance, which signals that the objects cannot be infinitely close. This restricts the validity of Newton’s law in certain circumstances.

When considering spheres centred on the gravitational source, neither the strength of the force nor the energy change (although the direction of the force changes as it is directed towards the centre/source). This is why by moving around on the earth, which is an almost spherical surface, we do not experience changes in the strength of gravity. Furthermore, we do not need to put energy into doing so (we get tired because of friction) unless we go upwards, like when climbing a mountain because we are gaining energy. On the other hand, when we go down an inclined surface, we lose energy, which is why it does not cost us any effort. Many extreme sports are based on this principle.

Gravitational potential approximation on the surface of a planet

For the earth, there is an approximation for objects close to the surface, which simplifies the formula considerably. It shifts the zero energy (which is located at an infinite radial distance) to the surface of the earth so that the final expression is:

\[E = m \cdot g \cdot h\]

Here, E is the gravitational potential energy for objects close to the surface of the earth and the origin of energy at the surface, m is the mass of the body whose gravitational potential energy we are computing, g is the gravitational surface acceleration for the earth (equal to G·M / R² = 9.81 m/s² ), and h is the height over the surface of the earth.