Unit 7 Gravitation in AP Physics C: Mechanics — Forces, Fields, and Potential Energy

Newton's Law of Universal Gravitation

What the law says (and what it really means)

Newton's Law of Universal Gravitation states that any two masses attract each other with a force whose magnitude is proportional to the product of the masses and inversely proportional to the square of the distance between their centers.

In equation form:

$F_g = G\frac{m_1 m_2}{r^2}$

Here:

• $F_g$ is the magnitude of the gravitational force between the masses (newtons, N)
• $G$ is the **universal gravitational constant** (approximately $6.67\times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$)
• $m_1$ and $m_2$ are the masses (kg)
• $r$ is the distance between the masses’ centers (m)

The phrase “universal” matters: this same law describes apples falling near Earth, the Moon orbiting Earth, and planets orbiting the Sun. In AP Physics C: Mechanics, you use this law not just to compute forces, but also to build the ideas of gravitational field and gravitational potential energy.

Why the inverse-square form matters

The $1/r^2$ dependence is not arbitrary. It captures how something that “spreads out” in three dimensions gets weaker with distance. Imagine the gravitational influence of a mass spreading through space. The surface area of a sphere grows like $4\pi r^2$, so the “influence per area” decreases like $1/r^2$. You are not required to re-derive this from geometry on the AP exam, but understanding the idea helps you remember the form and reason about what happens when distances change.

A quick scaling consequence you should be able to do mentally:

• If $r$ doubles, $F_g$ becomes one-fourth.
• If one mass triples, $F_g$ triples.

Direction: gravity is attractive and acts along the line joining centers

The equation above gives the magnitude. The force is always attractive, meaning each mass pulls toward the other. The force acts along the line connecting the two centers.

In vector form, a common way to write the force on mass 2 due to mass 1 is:

$\vec{F}_{2\leftarrow 1} = -G\frac{m_1 m_2}{r^2}\hat{r}_{2\leftarrow 1}$

• $\hat{r}_{2\leftarrow 1}$ is a unit vector pointing from mass 1 toward mass 2.
• The negative sign indicates the force on 2 points back toward 1 (opposite that unit vector).

You can also write this using a position vector difference. If $\vec{r} = \vec{r}_2 - \vec{r}_1$ and $r = |\vec{r}|$, then:

$\vec{F}_{2\leftarrow 1} = -G\frac{m_1 m_2}{r^3}\vec{r}$

This form is handy because it automatically gives the correct direction when you know the displacement vector between the objects.

Superposition: gravity adds like vectors

A major skill in gravitation problems is superposition: when multiple masses act on a target mass, the net gravitational force is the vector sum of the individual gravitational forces.

For several point masses $m_i$ pulling on a test mass $m$:

$\vec{F}_{\text{net}} = \sum_i \vec{F}_i$

This matters because many real situations involve more than two bodies (Earth and Moon acting on a satellite, or multiple masses in a line). The key is that gravity is a vector force, so directions matter.

When you are allowed to treat objects as point masses

Newton’s law is exact for point masses. Real objects have size, so you need a model:

• If an object is spherically symmetric (like a planet) and you are outside it, you may treat its entire mass as if it were concentrated at its center. Then $r$ is measured from the center of the sphere.
• For two spherically symmetric bodies with non-overlapping volumes, you can treat both as point masses at their centers.

This is not just a convenient approximation; for an ideal spherical mass distribution, it is an exact result outside the sphere.

A common pitfall is using the distance to the surface instead of the distance to the center. If a satellite is at altitude $h$ above Earth’s surface, then the correct center-to-center distance is:

$r = R_E + h$

where $R_E$ is Earth’s radius.

Example 1: Comparing gravitational forces at different distances

Suppose a satellite of mass $m$ orbits Earth at a distance $2R_E$ from Earth’s center. Compare the gravitational force there to the force on the same mass at Earth’s surface (distance $R_E$ from the center).

Step 1: Write the force at each location

At the surface:

$F_1 = G\frac{M_E m}{R_E^2}$

At $2R_E$:

$F_2 = G\frac{M_E m}{(2R_E)^2}$

Step 2: Take the ratio

$\frac{F_2}{F_1} = \frac{G\frac{M_E m}{4R_E^2}}{G\frac{M_E m}{R_E^2}} = \frac{1}{4}$

So the gravitational force is one-fourth as large at twice the distance.

Example 2: Net gravitational force from two masses (superposition)

Two equal masses $M$ are fixed on the x-axis at $x=-a$ and $x=+a$. A test mass $m$ is located at $x=0$. What is the net gravitational force on $m$?

Each mass attracts $m$ with equal magnitude:

$F = G\frac{Mm}{a^2}$

But the left mass pulls left and the right mass pulls right with equal magnitude. The forces cancel, so:

$\vec{F}_{\text{net}} = \vec{0}$

A classic misconception is thinking “two masses means double the force.” Only the magnitudes add if the directions match; here they do not.

Exam Focus

• Typical question patterns:
  • Compare gravitational forces at two distances using ratios (often avoiding heavy arithmetic).
  • Find a net gravitational force from multiple masses using vector addition and symmetry.
  • Use $r = R + h$ correctly for satellites and altitude problems.
• Common mistakes:
  • Using altitude $h$ as the distance in $1/r^2$ instead of center-to-center distance $R+h$.
  • Forgetting gravity is a vector and adding magnitudes when directions differ.
  • Confusing $G$ (universal constant) with $g$ (local gravitational acceleration).

Gravitational Field and Acceleration

From force to field: separating “source” from “test object”

A gravitational field is a way to describe how a mass creates a “pull” throughout space without having to specify the mass of the object being pulled each time. It is defined as the gravitational force per unit mass on a small test mass.

Mathematically, the gravitational field (also called gravitational field strength) is:

$\vec{g} = \frac{\vec{F}_g}{m}$

• $\vec{g}$ is measured in $\text{N/kg}$, which is equivalent to $\text{m/s}^2$.
• The direction of $\vec{g}$ is the direction a freely falling object would accelerate.

This definition matters because it makes clear why all objects fall with the same acceleration in a gravitational field (ignoring air resistance): since $\vec{F}_g$ is proportional to $m$, dividing by $m$ cancels the test mass.

Field due to a point mass (or spherical planet)

Start from Newton’s law for a mass $M$ attracting a test mass $m$:

$F_g = G\frac{Mm}{r^2}$

Divide by $m$ to get the field magnitude:

$g = G\frac{M}{r^2}$

Vector form (pointing toward the mass):

$\vec{g}(r) = -G\frac{M}{r^2}\hat{r}$

where $\hat{r}$ points radially outward from the mass, so the negative sign indicates the field points inward.

A key conceptual connection: the familiar near-Earth value $g \approx 9.8\ \text{m/s}^2$ is just the gravitational field magnitude at Earth’s surface:

$g_0 = G\frac{M_E}{R_E^2}$

So $g$ is not a new constant; it changes with location.

How gravitational acceleration relates to Newton’s 2nd law

In mechanics you often start with Newton’s 2nd law:

$\sum \vec{F} = m\vec{a}$

If gravity is the only force acting (free fall far from air resistance and without other forces), then:

$\vec{F}_g = m\vec{a}$

But $\vec{F}_g = m\vec{g}$ by definition of $\vec{g}$, so:

$\vec{a} = \vec{g}$

That is why gravitational field and gravitational acceleration are often used interchangeably in this unit.

Superposition for fields (often easier than forces)

Because $\vec{g}$ is force per unit mass, it also obeys superposition:

$\vec{g}_{\text{net}} = \sum_i \vec{g}_i$

This is especially convenient because you do not have to carry the test mass through the calculation.

Real-world interpretation: weight as a gravitational force

Near Earth’s surface, you often use the approximation $F_g = mg$ with constant $g$. In gravitation problems, you should recognize that this is an approximation to the universal gravitation law when $r$ does not change much compared with $R_E$.

At altitude, “weight” (gravitational force magnitude) becomes:

$F_g(r) = mG\frac{M_E}{r^2}$

So astronauts in orbit are not “beyond gravity”; they are in continuous free fall where gravity provides centripetal acceleration.

Example 1: Gravitational acceleration at altitude

Find the gravitational acceleration at an altitude $h = R_E$ above Earth’s surface (so the distance from Earth’s center is $r = 2R_E$) in terms of $g_0$.

At Earth’s surface:

$g_0 = G\frac{M_E}{R_E^2}$

At $2R_E$:

$g = G\frac{M_E}{(2R_E)^2} = G\frac{M_E}{4R_E^2} = \frac{g_0}{4}$

So gravitational acceleration is one-fourth its surface value at that altitude.

Example 2: Net gravitational field on a line (direction matters)

Two masses, $M$ and $4M$, are fixed on the x-axis at $x=0$ and $x=3a$. Find the point on the x-axis between them where the net gravitational field is zero.

Let the point be at position $x$ with $0<x<3a$.

Field magnitude from $M$ at distance $x$:

$g_1 = G\frac{M}{x^2}$

Direction: toward $M$, i.e. toward $x=0$, which is negative x-direction.

Field magnitude from $4M$ at distance $3a-x$:

$g_2 = G\frac{4M}{(3a-x)^2}$

Direction: toward $4M$, i.e. toward $x=3a$, which is positive x-direction.

For net field zero, magnitudes must be equal:

$G\frac{M}{x^2} = G\frac{4M}{(3a-x)^2}$

Cancel $G$ and $M$:

$\frac{1}{x^2} = \frac{4}{(3a-x)^2}$

Take square roots (distances are positive):

$\frac{1}{x} = \frac{2}{3a-x}$

Solve:

$3a - x = 2x$

$3a = 3x$

$x = a$

So the zero-field point is closer to the smaller mass, which makes sense because you must be closer to $M$ to match the stronger pull from $4M$.

Exam Focus

• Typical question patterns:
  • Derive or use $g = GM/r^2$ and compare $g$ values via ratios.
  • Find locations where net gravitational field is zero using superposition and symmetry.
  • Interpret free-fall motion in terms of $\vec{a} = \vec{g}$.
• Common mistakes:
  • Treating $g$ as constant for large altitude changes (it is only approximately constant near Earth’s surface).
  • Setting forces equal instead of fields equal when asked for a point where net field is zero (fields are simpler because the test mass cancels).
  • Getting directions wrong: fields point toward masses, so signs matter on a line.

Gravitational Potential Energy

Why introduce potential energy for gravity?

When gravity is the only force doing work (or when it’s one of several conservative forces), energy methods often make problems far easier than force-and-acceleration methods. Instead of tracking forces at every point, you track how energy changes between two positions.

Gravity is a conservative force. That means:

• The work done by gravity depends only on the initial and final positions, not on the path taken.
• You can define a gravitational potential energy function $U(r)$ so that changes in $U$ account for gravitational work.

This is especially important in orbital and space problems because the gravitational force changes significantly with distance. The familiar near-Earth formula $U = mgh$ is only an approximation for small height changes compared with Earth’s radius.

Work done by gravitational force (radial motion)

Consider a mass $m$ moving in the gravitational field of a mass $M$. The gravitational force magnitude is:

$F(r) = G\frac{Mm}{r^2}$

The force points inward (toward smaller $r$). If the object moves radially from $r_1$ to $r_2$, the work done by gravity is:

$W_g = \int_{r_1}^{r_2} \vec{F}\cdot d\vec{r}$

For radial motion, $\vec{F}\cdot d\vec{r}$ becomes negative when you move outward because force inward and displacement outward are opposite. Carrying out the standard integral gives:

$W_g = GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$

Check the sign with a physical test:

• If you move outward, $r_2 > r_1$, then $1/r_2 < 1/r_1$, so $W_g$ is negative. Gravity removes mechanical energy as you move away (you must supply energy).

Defining gravitational potential energy

For conservative forces, the change in potential energy is the negative of the work done by the force:

$\Delta U = -W_g$

So for gravity:

$\Delta U = -GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right) = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

To define an absolute potential energy function, you choose a reference point where $U=0$. For gravity, the standard (and most useful) choice is:

$U(\infty) = 0$

With that reference, the gravitational potential energy at distance $r$ from mass $M$ is:

$U(r) = -G\frac{Mm}{r}$

This negative sign is not a mathematical annoyance; it encodes the physics that two masses attract and form a bound system. Since you set $U=0$ at infinity, any finite separation corresponds to a state where energy would be required to separate them completely, so $U$ must be negative.

Connecting potential energy, potential, and field

A related quantity is gravitational potential $V$, defined as potential energy per unit mass:

$V(r) = \frac{U(r)}{m}$

For a point mass (or spherically symmetric mass):

$V(r) = -G\frac{M}{r}$

This can simplify problems because the test mass cancels, much like using $\vec{g}$ instead of $\vec{F}$.

There is also a deep relationship between potential and field. In the radial direction:

$g(r) = -\frac{dV}{dr}$

If you start with $V(r) = -GM/r$ and take the derivative, you recover $g(r) = GM/r^2$ in magnitude (directed inward). You do not always need calculus on the exam to use this, but it helps you see that potential energy and field descriptions are two sides of the same physics.

The near-Earth approximation: recovering $mgh$

Near Earth’s surface, if you move a small height $h$ such that $h \ll R_E$, then the gravitational field is approximately constant at $g_0$. In that case, the change in potential energy is well approximated by:

$\Delta U \approx m g_0 h$

How does this connect to the exact expression? If you compute:

$\Delta U = GM_E m\left(\frac{1}{R_E} - \frac{1}{R_E+h}\right)$

and use the approximation $1/(R_E+h) \approx 1/R_E - h/R_E^2$ for small $h$, you get:

$\Delta U \approx GM_E m\left(\frac{h}{R_E^2}\right) = m\left(G\frac{M_E}{R_E^2}\right)h = mg_0 h$

So $mgh$ is not a different kind of gravitational energy; it is the small-height limit of the universal gravitation model.

Conservation of mechanical energy in gravitational systems

When only gravity does work (no thrust, no drag), mechanical energy is conserved:

$K + U = \text{constant}$

where kinetic energy is:

$K = \frac{1}{2}mv^2$

This lets you relate speeds at different radii without explicitly solving a differential equation.

Example 1: Energy required to move between two radii

A spacecraft of mass $m$ moves from radius $r_1$ to radius $r_2$ from Earth’s center, starting and ending at rest (so you only care about potential energy change). How much work must an engine do (ignore drag and assume slow movement so kinetic energy stays negligible)?

The change in gravitational potential energy is:

$\Delta U = GM_E m\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

If the craft is moved outward (so $r_2>r_1$), then $\Delta U > 0$. That means the engine must do positive work equal to the increase in potential energy:

$W_{\text{engine}} = \Delta U = GM_E m\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

This is a common exam theme: separating what gravity does (often negative work when moving outward) from what an external agent must do.

Example 2: Speed gained by falling from rest (energy method)

A mass $m$ is released from rest at radius $r_1$ and falls inward to radius $r_2$ (with $r_2 < r_1$). Find its speed $v$ at $r_2$ (ignore atmosphere).

Step 1: Set up conservation of energy

Initial energy:

$E_i = K_i + U_i = 0 - G\frac{Mm}{r_1}$

Final energy:

$E_f = K_f + U_f = \frac{1}{2}mv^2 - G\frac{Mm}{r_2}$

Conservation: $E_i = E_f$.

Step 2: Solve for $v$

$-G\frac{Mm}{r_1} = \frac{1}{2}mv^2 - G\frac{Mm}{r_2}$

Move the potential terms to one side:

$\frac{1}{2}mv^2 = GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$

Cancel $m$ and solve:

$v = \sqrt{2GM\left(\frac{1}{r_2} - \frac{1}{r_1}\right)}$

A typical misconception is to try $v^2 = 2g\Delta y$ using constant $g$ even when the drop is a significant fraction of the planet’s radius. The energy method above automatically accounts for changing gravitational force with $r$.

Sign conventions and what students often misread

• $U(r) = -GMm/r$ becomes **more negative** as $r$ decreases. Falling inward decreases potential energy.
• The change $\Delta U$ can be positive or negative depending on direction of motion. Moving outward increases $U$ (less negative), moving inward decreases $U$ (more negative).
• If you choose a different reference point than infinity, the absolute value of $U$ changes, but **differences** $\Delta U$ (and physics) remain consistent.

Notation reference (common symbols you’ll see)

Exam Focus

• Typical question patterns:
  • Use $U = -GMm/r$ to compute energy changes between two radii (often paired with conservation of energy).
  • Compare $mgh$ to the exact expression and decide when the approximation is valid.
  • Determine whether an external agent must add or remove energy based on the sign of $\Delta U$.
• Common mistakes:
  • Forgetting that $U$ is negative when the reference is at infinity and incorrectly concluding that “negative energy is impossible.”
  • Using $mgh$ for large altitude changes where $g$ is not approximately constant.
  • Dropping minus signs in $\Delta U = GMm(1/r_1 - 1/r_2)$ and getting the direction of energy transfer backward.