Gravitational Field and Gravitational Potential

TL;DR

• Newton’s law of universal gravitation: $\mathbf{F} = -G\cfrac{mM}{r^2}\mathbf{e}_r$
• For continuous mass distributions and objects with size: $\mathbf{F} = -Gm\int_V \cfrac{dM}{r^2}\mathbf{e}_r = -Gm\int_V \cfrac{\rho(\mathbf{r^\prime})\mathbf{e}_r}{r^2} dv^{\prime}$
  • $\rho(\mathbf{r^{\prime}})$: Mass density at a point located at position vector $\mathbf{r^{\prime}}$ from an arbitrary origin
  • $dv^{\prime}$: Volume element at a point located at position vector $\mathbf{r^{\prime}}$ from an arbitrary origin
• Gravitational field vector:
  • A vector representing the force per unit mass experienced by a particle in a field created by an object of mass $M$
  • $\mathbf{g} = \cfrac{\mathbf{F}}{m} = - G \cfrac{M}{r^2}\mathbf{e}_r = - G \int_V \cfrac{\rho(\mathbf{r^\prime})\mathbf{e}_r}{r^2}dv^\prime$
  • Has dimensions of force per unit mass or acceleration
• Gravitational potential:
  • $\mathbf{g} \equiv -\nabla \Phi$
  • Has dimensions of (force per unit mass) × (distance) or energy per unit mass
  • $\Phi = -G\cfrac{M}{r}$
  • Only the relative differences in gravitational potential have meaning, not the specific values themselves
  • The ambiguity is typically removed by arbitrarily setting $\Phi \to 0$ as $r \to \infty$
  • $U = m\Phi, \quad \mathbf{F} = -\nabla U$
• Gravitational potential inside and outside a spherical shell (Shell Theorem)
  • When $R>a$:
    • $\Phi(R>a) = -\cfrac{GM}{R}$
    • When calculating the gravitational potential at a point outside a spherically symmetric distribution of matter, the object can be treated as a point mass
  • When $R<b$:
    • $\Phi(R<b) = -2\pi\rho G(a^2 - b^2)$
    • Inside a spherically symmetric mass shell, the gravitational potential is constant regardless of position, and the gravitational force is $0$
  • When $b<R<a$: $\Phi(b<R<a) = -4\pi\rho G \left( \cfrac{a^2}{2} - \cfrac{b^3}{3R} - \cfrac{R^2}{6} \right)$

Gravitational Field

Newton’s Law of Universal Gravitation

Newton had already formulated and numerically verified the law of universal gravitation before 11666 HE. Nevertheless, it took him 20 more years to publish his results in his work Principia in 11687 HE, because he could not justify his calculation method that treated the Earth and Moon as point masses without size. Fortunately, using calculus that Newton himself later invented, we can prove this problem much more easily than Newton could in the 11600s.

According to Newton’s law of universal gravitation, each particle of matter attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, this is expressed as:

\[\mathbf{F} = -G\frac{mM}{r^2}\mathbf{e}_r \label{eqn:law_of_gravitation}\tag{1}\]

Image source

• Author: Wikimedia user Dennis Nilsson
• License: CC BY 3.0

The unit vector $\mathbf{e}_r$ points from $M$ toward $m$, and the negative sign indicates that the force is attractive. In other words, $m$ is pulled toward $M$.

Cavendish’s Experiment

The experimental verification of this law and the determination of the value of $G$ was accomplished in 11798 HE by the British physicist Henry Cavendish. Cavendish’s experiment used a torsion balance consisting of two small spheres fixed at the ends of a light rod. The two spheres are attracted toward two other larger spheres placed nearby. The official value of $G$ obtained so far is $6.673 \pm 0.010 \times 10^{-11} \mathrm{N\cdot m^2/kg^2}$.

Despite being one of the earliest known fundamental constants, $G$ is known with lower precision than most other fundamental constants such as $e$, $c$, and $\hbar$. Even today, much research is being conducted to determine the value of $G$ with higher precision.

For Objects with Size

The law in equation ($\ref{eqn:law_of_gravitation}$) can strictly be applied only to point particles. If one or both objects have size, an additional assumption that the gravitational force field is a linear field must be made to calculate the force. That is, it is assumed that the total gravitational force on a particle of mass $m$ from multiple other particles can be found by adding the vector forces from each. For objects with continuous mass distribution, the sum is replaced by an integral as follows:

\[\mathbf{F} = -Gm\int_V \frac{dM}{r^2}\mathbf{e}_r = -Gm\int_V \frac{\rho(\mathbf{r^\prime})\mathbf{e}_r}{r^2} dv^{\prime} \label{eqn:integral_form}\tag{2}\]

• $\rho(\mathbf{r^{\prime}})$: Mass density at a point located at position vector $\mathbf{r^{\prime}}$ from an arbitrary origin
• $dv^{\prime}$: Volume element at a point located at position vector $\mathbf{r^{\prime}}$ from an arbitrary origin

If both objects of masses $M$ and $m$ have size and we want to calculate the total gravitational force, a second volume integral over $m$ is also needed.

Gravitational Field Vector

The gravitational field vector $\mathbf{g}$ is defined as the force per unit mass experienced by a particle in a field created by an object of mass $M$:

\[\mathbf{g} = \frac{\mathbf{F}}{m} = - G \frac{M}{r^2}\mathbf{e}_r \label{eqn:g_vector}\tag{3}\]

or

\[\boxed{\mathbf{g} = - G \int_V \frac{\rho(\mathbf{r^\prime})\mathbf{e}_r}{r^2}dv^\prime} \tag{4}\]

Here, the direction of $\mathbf{e}_r$ varies with $\mathbf{r^\prime}$.

This quantity $\mathbf{g}$ has dimensions of force per unit mass or acceleration. The magnitude of the gravitational field vector $\mathbf{g}$ near the Earth’s surface is what we call the gravitational acceleration constant, with $|\mathbf{g}| \approx 9.80\mathrm{m/s^2}$.

Gravitational Potential

Definition

The gravitational field vector $\mathbf{g}$ varies as $1/r^2$, and therefore satisfies the condition ($\nabla \times \mathbf{g} \equiv 0$) for being expressed as the gradient of a scalar function (potential). Thus, we can write:

\[\mathbf{g} \equiv -\nabla \Phi \label{eqn:gradient_phi}\tag{5}\]

Here, $\Phi$ is called the gravitational potential, and it has dimensions of (force per unit mass) × (distance) or energy per unit mass.

Since $\mathbf{g}$ depends only on the radius, $\Phi$ also varies with $r$. From equations ($\ref{eqn:g_vector}$) and ($\ref{eqn:gradient_phi}$):

\[\nabla\Phi = \frac{d\Phi}{dr}\mathbf{e}_r = G\frac{M}{r^2}\mathbf{e}_r\]

Integrating this, we get:

\[\boxed{\Phi = -G\frac{M}{r}} \label{eqn:g_potential}\tag{6}\]

Since only the relative differences in gravitational potential have meaning, not the specific values themselves, the integration constant can be omitted. The ambiguity is typically removed by arbitrarily setting $\Phi \to 0$ as $r \to \infty$, and equation ($\ref{eqn:g_potential}$) satisfies this condition.

For a continuous mass distribution, the gravitational potential is:

\[\Phi = -G\int_V \frac{\rho(\mathbf{r\prime})}{r}dv^\prime \label{eqn:g_potential_v}\tag{7}\]

For a mass distributed on a thin shell with surface density $\rho_s$:

\[\Phi = -G\int_S \frac{\rho_s}{r}da^\prime. \label{eqn:g_potential_s}\tag{8}\]

And for a linear mass source with linear density $\rho_l$:

\[\Phi = -G\int_\Gamma \frac{\rho_l}{r}ds^\prime. \label{eqn:g_potential_l}\tag{9}\]

Physical Meaning

Consider the work per unit mass $dW^\prime$ done when an object moves by $d\mathbf{r}$ in a gravitational field:

\[\begin{align*} dW^\prime &= -\mathbf{g}\cdot d\mathbf{r} = (\nabla \Phi)\cdot d\mathbf{r} \\ &= \sum_i \frac{\partial \Phi}{\partial x_i}dx_i = d\Phi \label{eqn:work}\tag{10} \end{align*}\]

In this equation, $\Phi$ is a function of position coordinates only, expressed as $\Phi=\Phi(x_1, x_2, x_3) = \Phi(x_i)$. Therefore, the work per unit mass done when moving an object from one point to another in a gravitational field equals the difference in potential between those two points.

If we define the gravitational potential at infinity to be $0$, then $\Phi$ at any point can be interpreted as the work per unit mass required to move the object from infinity to that point. The potential energy of an object is the product of its mass and the gravitational potential $\Phi$, so if $U$ is the potential energy:

\[U = m\Phi. \label{eqn:potential_e}\tag{11}\]

Therefore, the gravitational force on an object is obtained by taking the negative gradient of its potential energy:

\[\mathbf{F} = -\nabla U \label{eqn:force_and_potential}\tag{12}\]

When an object is placed in a gravitational field created by some mass, it always has a potential energy. Strictly speaking, this potential energy resides in the field itself, but conventionally, we refer to it as the potential energy of the object.

Example: Gravitational Potential Inside and Outside a Spherical Shell (Shell Theorem)

Coordinate Setup & Expressing Gravitational Potential in Integral Form

Let’s find the gravitational potential inside and outside a uniform spherical shell with inner radius $b$ and outer radius $a$. The gravitational force due to the shell can be obtained by directly calculating the force components acting on a unit mass in the field, but using the potential method is simpler.

Let’s calculate the potential at point $P$ at distance $R$ from the center. Assuming a uniform mass distribution in the shell, $\rho(r^\prime)=\rho$, and due to symmetry about the azimuthal angle $\phi$ along the line connecting the center of the sphere and point $P$:

\[\begin{align*} \Phi &= -G\int_V \frac{\rho(r^\prime)}{r}dv^\prime \\ &= -\rho G \int_0^{2\pi} \int_0^\pi \int_b^a \frac{1}{r}(dr^\prime)(r^\prime d\theta)(r^\prime \sin\theta\, d\phi) \\ &= -\rho G \int_0^{2\pi} d\phi \int_b^a {r^\prime}^2 dr^\prime \int_0^\pi \frac{\sin\theta}{r}d\theta \\ &= -2\pi\rho G \int_b^a {r^\prime}^2 dr^\prime \int_0^\pi \frac{\sin\theta}{r}d\theta. \label{eqn:spherical_shell_1}\tag{13} \end{align*}\]

According to the law of cosines:

\[r^2 = {r^\prime}^2 + R^2 - 2r^\prime R \cos\theta \label{eqn:law_of_cosines}\tag{14}\]

Since $R$ is constant, differentiating this equation with respect to $r^\prime$:

\[2rdr = 2r^\prime R \sin\theta d\theta\] \[\frac{\sin\theta}{r}d\theta = \frac{dr}{r^\prime R} \tag{15}\]

Substituting this into equation ($\ref{eqn:spherical_shell_1}$):

\[\Phi = -\frac{2\pi\rho G}{R} \int_b^a r^\prime dr^\prime \int_{r_\mathrm{min}}^{r_\mathrm{max}} dr. \label{eqn:spherical_shell_2}\tag{16}\]

Here, $r_\mathrm{max}$ and $r_\mathrm{min}$ are determined by the position of point $P$.

When $R>a$

\(\begin{align*} \Phi(R>a) &= -\frac{2\pi\rho G}{R} \int_b^a r^\prime dr^\prime \int_{R-r^\prime}^{R+r^\prime} dr \\ &= - \frac{4\pi\rho G}{R} \int_b^a {r^\prime}^2 dr^\prime \\ &= - \frac{4}{3}\frac{\pi\rho G}{R}(a^3 - b^3). \label{eqn:spherical_shell_outside_1}\tag{17} \end{align*}\)

The mass $M$ of the spherical shell is given by:

\[M = \frac{4}{3}\pi\rho(a^3 - b^3) \label{eqn:mass_of_shell}\tag{18}\]

Therefore, the potential is:

\[\boxed{\Phi(R>a) = -\frac{GM}{R}} \label{eqn:spherical_shell_outside_2}\tag{19}\]

Comparing the gravitational potential due to a point mass $M$ in equation ($\ref{eqn:g_potential}$) with the result we just obtained in ($\ref{eqn:spherical_shell_outside_2}$), we see they are identical. This means that when calculating the gravitational potential at a point outside a spherically symmetric distribution of matter, we can consider all the mass to be concentrated at the center. Most spherical celestial bodies of significant size, such as the Earth or Moon, fall into this category, as they can be considered as countless overlapping spherical shells with the same center but different diameters, like Matryoshka dolls. This provides the justification for treating celestial bodies like the Earth or Moon as point masses without size mentioned at the beginning of this article.

When $R<b$

\(\begin{align*} \Phi(R<b) &= -\frac{2\pi\rho G}{R} \int_b^a r^\prime dr^\prime \int_{r^\prime - R}^{r^\prime + R}dr \\ &= -4\pi\rho G \int_b^a r^\prime dr^\prime \\ &= -2\pi\rho G(a^2 - b^2). \label{eqn:spherical_shell_inside}\tag{20} \end{align*}\)

Inside a spherically symmetric mass shell, the gravitational potential is constant regardless of position, and the gravitational force is $0$.

This is also a key reason why the “Hollow Earth Theory,” a well-known pseudoscience, is nonsense. If the Earth were a hollow shell as claimed by this theory, gravity would not act on any object inside that cavity. Considering the mass and volume of the Earth, not only is it impossible for such a hollow to exist, but even if it did, life forms there would not be walking on the inner surface of the shell but rather floating in a weightless state like on a space station.
While microorganisms can live several kilometers deep in the Earth’s crust, it is certainly not possible in the form claimed by the Hollow Earth Theory. I also really enjoy Jules Verne’s novel “Journey to the Center of the Earth” and the movie “Journey to the Center of the Earth,” but creative works should be enjoyed as fiction, not taken seriously as fact.

When $b<R<a$

\(\begin{align*} \Phi(b<R<a) &= -\frac{4\pi\rho G}{3R}(R^3 - b^3) - 2\pi\rho G(a^2 - R^2) \\ &= -4\pi\rho G \left( \frac{a^2}{2} - \frac{b^3}{3R} - \frac{R^2}{6} \right) \label{eqn:within_spherical_shell}\tag{21} \end{align*}\)

Results

The gravitational potential $\Phi$ and the magnitude of the gravitational field vector $|\mathbf{g}|$ in the three regions, plotted as functions of distance $R$, are shown below:

• Python visualization code: yunseo-kim/physics-visualization repository
• License: See here

We can see that both the gravitational potential and the magnitude of the gravitational field vector are continuous. If the gravitational potential were discontinuous at any point, the gradient of the potential (i.e., the magnitude of gravity) would be infinite at that point, which is physically implausible, so the potential function must be continuous at all points. However, the derivative of the gravitational field vector is discontinuous at the inner and outer surfaces of the shell.

Example: Galactic Rotation Curves

According to astronomical observations, the observable mass in many spiral galaxies that rotate around their centers, such as our Milky Way or the Andromeda galaxy, is mostly concentrated near the central region. However, as can be seen in the following graph, the orbital velocities of masses in these spiral galaxies significantly disagree with the theoretical predictions based on the observable mass distribution and remain nearly constant beyond a certain distance.

Image source

• Author: Wikipedia user PhilHibbs
• License: Public Domain

Rotation Curve of Spiral Galaxy M33 (Triangulum Galaxy)

• Author: Wikimedia user Mario De Leo
• License: CC BY-SA 4.0

Let’s predict the orbital velocity as a function of distance when the galaxy’s mass is concentrated in the central region, verify that this prediction does not match the observations, and show that the mass $M(R)$ distributed within distance $R$ from the galactic center must be proportional to $R$ to explain the observations.

First, if the galaxy’s mass $M$ is concentrated in the central region, the orbital velocity at distance $R$ is:

\[\frac{GMm}{R^2} = \frac{mv^2}{R}\] \[v = \sqrt{\frac{GM}{R}} \propto \frac{1}{\sqrt{R}}.\]

In this case, an orbital velocity decreasing as $1/\sqrt{R}$ is predicted, as shown by the dotted line in the two graphs above. However, according to observations, the orbital velocity $v$ is almost constant regardless of distance $R$, so the prediction does not match the observations. This observation can only be explained if $M(R)\propto R$.

If we set $M(R) = kR$ with a proportionality constant $k$:

\[v = \sqrt{\frac{GM(R)}{R}} = \sqrt{Gk}\ \text{(constant)}.\]

From this, astrophysicists conclude that many galaxies must contain undiscovered “dark matter,” and this dark matter must account for more than 90% of the universe’s mass. However, the nature of dark matter has not yet been clearly identified, and while not mainstream, there are attempts like Modified Newtonian Dynamics (MOND) that try to explain the observations without assuming the existence of dark matter. Today, this field of research is at the forefront of astrophysics.