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