Lesson 3: Newtonian Gravitational Fields

Introduction to Lesson 3

With the tools of vector calculus now firmly in hand, we are ready to begin our study of actual physical fields. Up to this point, we have worked primarily with the formal structure of fields and the operations that can be performed on them. We now apply these tools to one of the most important and historically significant field theories in physics: Newtonian gravitation.

Gravity was the first force to be successfully described as a field. Newton’s law of universal gravitation, originally expressed as a force between two point masses, can be reformulated in terms of a gravitational field that exists at every point in space. This field is generated by matter and, in turn, acts on other matter. This shift in perspective—from action at a distance to a field that mediates the interaction—marks a profound conceptual advance and serves as the foundation for all later field theories.

In this lesson, we will develop the Newtonian theory of gravity using the mathematical framework we have built. We begin with Newton’s law of gravity and show how it leads naturally to the concept of a gravitational field. We will then introduce the gravitational potential, a scalar field from which the gravitational field can be derived. This potential satisfies one of the most important equations in physics—Poisson’s equation—which relates the field directly to the distribution of matter.

A significant portion of the lesson is devoted to understanding the gravitational field of a spherical shell and, by extension, of spherical bodies. This classic problem not only illustrates the power of the field concept but also reveals deep symmetries that simplify the mathematics considerably. We will also examine the geometry of the gravitational field through lines of force and equipotential surfaces, both of which provide intuitive pictures of how gravity behaves in space.

Finally, we will apply these ideas to a familiar but subtle phenomenon: the tides. By treating the Earth–Moon system within the framework of Newtonian gravity, we will see how the differential action of the gravitational field across an extended body produces observable effects.

Throughout this lesson, you will see the mathematical tools developed in the previous lessons — particularly the gradient, divergence, and the integral theorems—put to direct physical use. Newtonian gravity serves as an ideal first example of a classical field theory: it is conceptually clear, mathematically tractable, and rich with physical consequences.

By the end of this lesson, you will have a solid understanding of how matter generates a gravitational field, how that field can be described using a potential, and how the field equations encode the relationship between matter and geometry in Newtonian physics. This foundation will prove invaluable when we later move to more advanced field theories, including electromagnetism and, eventually, general relativity.

Newton’s Law of Gravity

Those of you who have studied classical mechanics will already be familiar with Newton’s Law of Universal Gravitation. You may wonder why we revisit it here. The reason is that we now wish to understand it from the perspective of field theory. Before proceeding, it is worth asking a deeper question, “Why do we regard Newton’s law as a fundamental law of nature?”

Prior to Newton, Johannes Kepler analyzed the meticulous observational records of Tycho Brahe and discovered three empirical laws that described the motion of the planets. These laws were descriptive—they summarized patterns in the data but offered no explanation for why the planets moved as they did. Newton took a decisive step forward. By combining his proposed law of universal gravitation with Kepler’s laws, he was able to predict the motion of the planets and other astronomical bodies. This predictive power suggested a deeper level of universality.

One of Newton’s most striking achievements was demonstrating that the same law governing the fall of an apple near the surface of the Earth also governs the motion of the Moon in its orbit. This was the first major unification in physics—showing that terrestrial and celestial phenomena obey the same fundamental principle. The law was further supported by Henry Cavendish’s famous experiment, that measured the gravitational attraction between laboratory masses and thereby verified the law at human scales.

To review, the gravitational force exerted on a mass m by another mass M separated by a distance r is given by

(3.1)

Traditionally, we put the origin of our frame at M. Where we have the gravitational constant in SI units.

Exercise 3.1: Explain, from (3.1), why the force is directed at the origin. Based on this, what kind of force is ?

What happens when we have a system of n masses? What is it that we are asking? We have not clearly defined a question. How about this one, “What is the force exerted on a mass, m, surrounded by a system of n masses all at fixed distances from m?” We can write the system of masses as . where i=1,2,...,n, and there will be a position vector for each, . This allows us to write,

(3.2)

Newton’s second law tells us that the net force on a mass equals its mass times its acceleration

(3.3)

Substituting the expression for the gravitational force, we obtain

(3.4)

We see that the test mass m appears on both sides of the equation and therefore cancels, leaving

(3.5)

Exercise 3.2: What does this suggest about the roll of m in accelerating a particle gravitationally?

It has been established experimentally that the inertial mass (which appears in Newton’s second law) and the gravitational mass (which appears in the law of gravitation) are equivalent. This equivalence is known as the Equivalence Principle and is one of the foundational ideas in our modern understanding of gravity.

Exercise 3.3: How is the gravitational force between two masses established experimentally?

The Gravitational Field

Newton’s original formulation of gravity leaves us with a deeply unsatisfying picture where a force that seems to reach across empty space and pull on a distant mass as if by magic. How can we move beyond this action-at-a-distance view?

The key insight is to shift our perspective. Instead of thinking only about the force on a particular test mass. Here a test mass is a hypothetical or real mass m that is so small compared to the source masses that it does not appreciably alter the gravitational force it is measuring. We use it to “feel” the strength and direction of the field at a particular location. Now consider what exists at every point in space around a source mass. At every location, there is a force that would act on a test mass if we placed one there. This creates a vector field—a quantity that experiences a force at every point in space.

We can make this idea precise. The gravitational force on a test mass m is independent of m itself. This means we can define a new quantity—the force per unit mass—that describes the gravitational influence at each point

(3.6)

This vector is the gravitational field. It tells us the strength and direction of the gravitational effect at any location, independent of any test mass we might place there. For a single point mass M, the gravitational field at a distance r is

(3.7)

where is the unit vector pointing from M to the point of interest.

If you have a system containing more than one mass, the total gravitational field at any point is simply the vector sum of the individual fields produced by each mass at that point. The contributions add together just like ordinary vectors. We call this idea superposition.

This field picture removes the mystery of action-at-a-distance. The field exists in space itself, and any mass placed in that field simply responds to the local value of . The mathematics now matches the physical reality we experience where the influence is mediated by a condition in the space around the source.

Gravity can be described by a vector field that exists at every point in space, rather than as a direct “magical” force between distant objects.

Exercise 3.4: A point mass kg (roughly Earth’s mass) produces a gravitational field at a distance m (roughly Earth’s radius).
    1) Calculate the magnitude of the gravitational field at this distance.
    2) If a test mass m=1 kg is placed there, what force does it experience?
    3) How would the field change if you moved twice as far away? Explain the physical meaning of this dependence.

Exercise 3.4: Consider two point masses, and , separated by a distance d.
    1) Write an expression for the total gravitational field at a point midway between them (assume the point is along the line joining them).
    2) Explain in your own words why we add the individual fields as vectors rather than scalars.
    3) What would happen to the total field at that midway point if one of the masses were removed? How does this illustrate the field concept replacing action-at-a-distance?

Matter Fields and the Gravitational Field

We have been working with simple collections of discrete particles so far. To describe reality more accurately, especially when mass is spread out continuously rather than concentrated in point particles, we need a richer picture.

The clue lies in the relationship between sums and integrals. A sum is discrete, matching the particle systems we considered in earlier lessons. If we let the number of particles increase without limit—so we can always imagine adding another one—the sum naturally becomes an integral. This lets us rewrite the gravitational force on a test mass m in a continuous distribution as

(3.8)

where is the mass density at the position vector within a continuous mass distribution. Such a continuous mass distribution is what we call a matter field.

We can then establish the field strength, or the field vector,

(3.9)

We can recall a Nabla-identity,

(3.10)

Exercise 3.4: Prove (3.10).

This allows us to rewrite the field as

(3.11)

Since the Nabla operator does not operate on , we can factor it out of our volume integral,

(3.12)

Recall from classical mechanics that we can write any conservative force as the gradient of a potential,

(3.13)

This potential is the gravitational potential,

(3.14)

What do we do with the constants of integration. By convention we have

(3.15)

This choice fixes the constant to zero. With this boundary condition the potential for a point mass (or a distribution whose total mass is M when viewed from far away) becomes simply

(3.16)

Exercise 3.5: Why the negative sign in (3.14)?

Exercise 3.6: What does (3.13) tell us about the gravitational field?

Exercise 3.7: Can you justify why is a scalar quantity?

Exercise 3.8: What are the dimensions of (3.14)?

Exercise 3.9: What implicit assumptions make (3.14) correct?

Exercise 3.10: Explain (3.15) and (3.16) correct?

Note that potential energy exists whenever something is placed in the field. We might speak of, “The potential energy of something in the field.” This is a mistake, the potential energy exists in the field itself. If you will, a more proper way of saying it would be, “The potential energy experienced by something placed in the field.”

Problem 3.1: How do we extend these results to an electrostatic potential?

The Spherical Shell

To find the gravitational potential of a spherical shell, we can use a powerful modeling technique: break the shell down into simpler pieces whose potentials we already know how to calculate, then add them all up.

Imagine a thin spherical shell of radius and total mass M. We can think of this shell as being made up of many thin rings stacked around its surface. If we can find the gravitational potential due to one such ring, we can integrate over all the rings to obtain the potential of the entire shell.

Here is a simple visualization of the idea in Wolfram Language

Figure 3.1: A spherical shell exhibiting a ring.

We begin by considering a point outside the shell, at a distance R from the center. The shell has total mass M and radius , then the mass density of the shell, , can be written,

(3.17)

Each ring will have an element of mass, dM. There is a corresponding differential area, dA,

(3.18)

Exercise 3.11: Derive (3.16).

This gives us the mass element of the ring

(3.19)

Exercise 3.12: Derive (3.17).

The distance from the ring to the external point is r, by the law of cosines this satisfies

(3.20)

Exercise 3.13: Derive (3.18).

By differentiation we get,

(3.21)

Exercise 3.14: Derive (3.19).

We can now express the mass element in terms of dr

(3.22)

Exercise 3.15: Derive (3.20).

The contribution to the potential from one such ring is

(3.23)

Integrating over the entire shell gives

(3.24)

To evaluate the limits, note that the maximum distance from the ring to the external point is the sum of the shell radius and the distance from the center to the point

(3.25)

while the minimum distance is

(3.26)

There are then two cases to consider. The first is when the point is outside the shell , in this case

(3.27)

The second case occurs inside the shell ,

(3.28)

The potential is then

(3.29)

Exercise 3.16: How can you extend this result to model a spherical mass distribution.

Gravitational Field Equations: Poisson’s and Laplace’s Equations

When you do Exercise 3.15 you will arrive at an expression similar to this, for the field strength and the surface element, dS,

(3.30)

Here we can write the solid angle subtended by the surface element,

(3.31)

This allows us to rewrite our surface integral,

(3.32)

Exercise 3.17: Derive this result.

Gauss’s divergence theorem allow us to rewrite the surface integral as a volume integral,

(3.33)

Substituting the earlier result gives

(3.34)

Since dVis arbitrary, we can equate the integrands to obtain the local relation

(3.35)

Recall that the gravitational field is the negative gradient of the potential, given in (3.13), . Substituting this in yields

(3.36)

This is called Poisson’s equation. When there is no matter, then ρ=0, so,

(3.37)

This is called Laplace’s equation. These second-order partial differential equations constitute the field equations of classical gravitation.

Problem 3.2 Explain a method to solve Poisson’s equation.

Lines of Force

Suppose we have a gravitational field described by the vector . We can imagine drawing a curve that starts at the source mass and extends outward to infinity such that, at every point along the curve, the direction of the curve is exactly parallel to the local field vector . Such a curve is called a line of force.

If you draw a line of force starting from every small surface element of the source mass, you can visualize the direction of the gravitational field at any point in space simply by looking at the pattern these lines create.

Exercise 3.18: What would the set of the lines of force from a single source look like?

These lines of force tell us the direction of the field everywhere. There is another important feature: if you imagine a small unit area placed perpendicular to the field at some point, the number of lines of force passing through that area is proportional to the magnitude of the field at that location. In this way, the density of lines gives a visual representation of both the direction and the strength of the vector field.

Equipotential Surfaces

The gravitational potential is defined at every point in space (except exactly at the location of a point mass, where a Dirac delta function would be needed). This naturally leads to the idea of surfaces on which the potential takes the same value everywhere:

(3.38)

This surface is called an equipotential surface.

Since the gravitational field is the negative gradient of the potential, , it follows that the field cannot have any component tangent to an equipotential surface—it must be perpendicular to it everywhere.

Exercise 3.19: What does this tell us about the relationship between lines of force and equipotential surfaces?

If you think about this carefully, you will see that a body moving along an equipotential surface does no work against the gravitational field.

Exercise 3.20: Can equipotential surfaces intersect? Why?

Exercise 3.21: What would a set of equipotential surfaces of a single source look like?

For two gravitating particles of equal mass located at and , then the total potential is

(3.39)

This equation defines the equipotential surfaces for the two-body system.

Exercise 3.22: Derive (3.38).

The Tides

When a gravitating body is close enough to another object that the size of that object matters, the gravitational pull is no longer uniform across it. Different parts of the object feel slightly different forces. This difference produces tidal forces—a stretching or deformation of the body along the line connecting it to the source of gravity.

A familiar and striking example is the way the Moon raises tides in Earth’s oceans. In this section we explore how such tidal effects arise.

To decide whether tidal forces are significant, compare the gravitational field at the center of the body with the field at points on its surface—both on the side facing the source and on the opposite side. If there is a noticeable difference between these fields, a tidal force will appear. This force tends to stretch the body toward the source (and away from it on the far side).

Consider the gravitational field produced by the Moon and the center of the Earth. Let be the mass of the Moon, the distance between the centers of the Earth and Moon, and is the unit vector pointing from Earth’s center to the Moon’s center. The gravitational field at Earth’s center due to the Moon is

(3.40)

This central field sets the baseline. On the near side of Earth (closer to the Moon), the field is slightly stronger; on the far side, it is slightly weaker. The resulting difference in force across Earth’s diameter is what drives the tidal bulges.

The idea is simple but powerful: gravity is never perfectly uniform over any extended object. Whenever the size of the body is not negligible compared to the distance to the source, tidal forces appear and produce measurable deformations.

Here we see the situation.

Figure 3.2. A schematic of the Earth-Moon system with the gravitational field at the center of the Earth.

If we have a position vector to some point, P, on the Earth, , then we can write the field at that point, ,

(3.41)

The apparent field is

(3.42)

Here we see the field at different points.

Figure 3.3. The gravitational field at the center, two equatorial points (a and b), and two polar points (c and d).

Here is the apparent field at those points

Figure 3.4. The apparent gravitational field at the two equatorial points (a and b), and the two polar points (c and d).

The gravitational potential at any point described by the position vector is then,

(3.43)

We can write, where is the radius vector from the center to the surface of the earth at the point described by

(3.44)

Exercise 3.23: Derive this result.

We can then write the square of r,

(3.45)

We can then write

(3.46)

If we assume that , we can apply the binomial expansion to get

(3.47)

Exercise 3.24: Derive this result.

Now we need to approximate the term in (3.43)

(3.48)

Exercise 3.25: Derive this result.

Then we have,

(3.49)

Exercise 3.26: Derive this result.

We are free to choose our coordinate system, so we can make the choice such that .

(3.50)

From this we can approximate our tidal acceleration,

(3.51)

Exercise 3.27: Derive this result.

Since gravitational force is conservative, and is thus derived from a potential, we can write

(3.52)

So,

(3.53)

Exercise 3.28: Derive this result.

Since we can write x=r sin θ cos φ,

(3.54)

Exercise 3.29: Derive this result.

For Further Reading

Tai L. Chow, (2013), Classical Mechanics, CRC Press (Second Edition). This has a very nice chapter on gravitation.

Vernon Barger, Martin Olsson, (1973), Classical Mechanics A Modern Perspective, McGraw-Hill, Inc. (Second Edition was published in 1995). This is a nice little book. I know both authors personally, and I like the book.