Gravitational Forces

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Gravitation· 14 min read

1. Newton’s Law of Universal Gravitation★★☆☆☆⏱ 3 min

📘 Definition

Newton’s Law of Universal Gravitation

$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$

Describes the magnitude of the gravitational force between two point masses or spherically symmetric extended masses. Force is always attractive and acts along the line connecting the centers of the two masses.

Example:

Doubling the center-to-center distance between two masses reduces the gravitational force to 1/4 of its original value.

Gravitational force is always attractive (there is no negative mass in classical mechanics, unlike electrostatic charge). By Newton’s third law, the force that mass 1 exerts on mass 2 is equal in magnitude and opposite in direction to the force mass 2 exerts on mass 1. This formula only applies directly to point masses and spherically symmetric extended masses; for non-spherical masses, we use superposition and integration.

📐 Worked Example

What is the magnitude of the gravitational force between a 10 kg solid lead sphere and a 50 kg solid lead sphere whose outer surfaces are separated by 10 cm? Each sphere has a radius of 7.5 cm.

1
First calculate the center-to-center separation $r$ by adding the two radii and the surface separation:
2
$r = 7.5 \text{cm} + 7.5 \text{cm} + 10 \text{cm} = 25 \text{cm} = 0.25 \text{m}$
3
Identify given values: $m_{1} = 10 kg$ , $m_{2} = 50 kg$ , $G = 6.67 \times 1 0^{- 11} N \cdotp m^{2} / kg^{2}$ .
4
Substitute into Newton’s law:
5
$F_g = (6.67 \times 10^{-11}) \frac{(10)(50)}{(0.25)^2} = (6.67 \times 10^{-11})(8000)$
6
Calculate the final magnitude:
7
$F_g \approx 5.3 \times 10^{-7} \text{N}$
8
By Newton’s third law, the force each sphere exerts on the other has this same magnitude.

Exam tip:

On multiple-choice questions, use proportional reasoning ( $F_{g} \propto m_{1} m_{2} / r^{2}$ ) to eliminate wrong options far faster than full numerical calculation.

2. Superposition of Gravitational Forces★★★☆☆⏱ 4 min

Gravitational force is a vector quantity, so when a test mass interacts with multiple source masses, the net gravitational force on the test mass is the vector sum of the individual forces exerted by each source. For AP problems, this process follows four steps:

Calculate the magnitude of each individual force using Newton’s law
Assign a coordinate system and resolve each force into $x$ and $y$ components
Add corresponding components to get net force components
Calculate the magnitude and direction of the net force if required

📐 Worked Example

Three point masses are arranged on the x-axis: $m_{A} = 2 kg$ at $x = 0$ , $m_{B} = 4 kg$ at $x = 3 m$ , and a 1 kg test mass $m$ at $x = 1 m$ . Find the net gravitational force on the test mass.

1
Set positive $x$ to the right. Force is always attractive, so $m_{A}$ pulls $m$ left (negative direction) and $m_{B}$ pulls $m$ right (positive direction).
2
Calculate separations: $r_{A} = 1 m$ (distance from $m_{A}$ to $m$ ), $r_{B} = 2 m$ (distance from $m_{B}$ to $m$ ).
3
Calculate individual force magnitudes:
4
$F_A = G \frac{m_A m}{r_A^2} = G \frac{(2)(1)}{1^2} = 2G, \quad F_B = G \frac{m_B m}{r_B^2} = G \frac{(4)(1)}{2^2} = G$
5
Add components to get net force:
6
$F_{net,x} = -F_A + F_B = -2G + G = -G = -6.67 \times 10^{-11} \text{N}$
7
The net force has magnitude $6.67 \times 1 0^{- 11} N$ directed left toward $m_{A}$ .

📐 Worked Example

Two identical point masses $M$ are fixed at $(- d, 0)$ and $(d, 0)$ on the x-axis. A third mass $m$ is placed at $(0, d)$ . What is the magnitude of the net gravitational force on $m$ ?

1
Calculate the distance from each $M$ to $m$ :
2
$r = \sqrt{d^2 + d^2} = d\sqrt{2}, \quad r^2 = 2d^2$
3
Magnitude of force from each $M$ is $F = \frac{GM m}{2 d ^{2}}$ .
4
The x-components of the two forces cancel (equal magnitude, opposite direction), and the y-components add. Each force has a y-component of $F cos (4 5^{\circ})$ :
5
$F_y = \frac{GMm}{2d^2} \cdot \frac{\sqrt{2}}{2} = \frac{GMm\sqrt{2}}{4d^2}$
6
Add the two y-components to get net force:
7
$F_{net} = 2 \cdot \frac{GMm\sqrt{2}}{4d^2} = \frac{\sqrt{2} GMm}{2d^2}$

Exam tip:

Always draw a coordinate system and confirm the direction of each force before adding components; AP examiners intentionally place masses to test sign errors for attractive forces.

3. Gravitational Force from Gravitational Field★★☆☆☆⏱ 3 min

📘 Definition

Gravitational Field

$g = \frac{F _{g}}{m}$

A vector field defined as the gravitational force per unit mass at a given point in space. This formulation simplifies force calculations when the net field is already known.

Example:

Near Earth’s surface, the gravitational field is approximately uniform with magnitude $g \approx 9.8 m/s^{2}$ .

Rearranging the definition gives the general relation between gravitational force and gravitational field: $F_{g} = m g$ , where $m$ is the mass of the test object. The familiar near-Earth weight formula $F_{g} = m g$ is just an approximation of Newton’s universal law, valid only for points close to the surface. For any point outside a spherical mass $M$ , the gravitational field magnitude is $g = \frac{GM}{r ^{2}}$ .

📐 Worked Example

The net gravitational field at a point in space is $g = (- 1.8 \hat{i} + 4.2 \hat{j}) m/s^{2}$ . What is the gravitational force on a 15 kg mass placed at this point, and what is the magnitude of the force?

1
Use the relation $F_{g} = m g$ to find the vector force.
2
Multiply each component by the mass $m = 15 kg$ :
3
$\vec{F}_g = 15(-1.8 \hat{i} + 4.2 \hat{j}) = (-27 \hat{i} + 63 \hat{j}) \text{N}$
4
Calculate the magnitude using the Pythagorean theorem:
5
$|F_g| = \sqrt{(-27)^2 + (63)^2} = \sqrt{4698} \approx 68.5 \text{N}$

Exam tip:

If asked for force at significant altitude above a planet’s surface, always calculate $g = GM / r^{2}$ instead of using the near-surface value of 9.8 m/s².

4. Extended Continuous Mass Distributions★★★★☆⏱ 4 min

For non-spherical continuous mass distributions, we use the principle of superposition with integration: we split the extended mass into infinitesimal mass elements, find the force from each element, then integrate over the entire distribution to get the net force. This is a common skill on AP Physics C Mechanics FRQs.

📐 Worked Example

A thin uniform rod of total mass $M$ and length $L$ lies along the x-axis from $x = 0$ to $x = L$ . A point mass $m$ is placed on the x-axis at $x = L + a$ , where $a > 0$ . Evaluate the net gravitational force on $m$ , and show it reduces to the point-mass result when $a ≫ L$ .

1
Linear mass density for the uniform rod is $λ = \frac{M}{L}$ . Take an infinitesimal segment at position $x$ with mass $d m = λ d x = \frac{M}{L} d x$ .
2
The distance between the segment and $m$ is $(L + a - x)$ , so the infinitesimal force is:
3
$dF = \frac{G m dm}{(L+a - x)^2} = \frac{GMm dx}{L(L+a - x)^2}$
4
Integrate over the full length of the rod, then use substitution $u = L + a - x$ , $d u = - d x$ :
5
$F = \frac{GMm}{L} \int_{a}^{L+a} \frac{du}{u^2} = \frac{GMm}{L} \left( \frac{1}{a} - \frac{1}{L+a} \right) = \frac{GMm}{a(L+a)}$
6
When $a ≫ L$ , $L + a \approx a$ , so $F \approx \frac{GM m}{a ^{2}}$ , which matches the point-mass result.

5. Common Pitfalls

Wrong move:

Using the surface-to-surface distance between two spheres instead of center-to-center distance in Newton’s law.

Why:

Students confuse the visible gap between objects with the separation required for the point-mass approximation.

Correct move:

Always add the radii of the two spheres to the surface gap to get the total $r$ for the formula.

Wrong move:

Claiming the larger mass exerts a larger gravitational force than the smaller mass.

Why:

Students confuse force magnitude with resulting acceleration, assuming more mass creates more force.

Correct move:

Always apply Newton’s third law: the force between two masses is equal in magnitude for both, regardless of their mass difference.

Wrong move:

Forgetting that gravitational force is attractive when assigning signs to vector components.

Why:

Students mix up gravitational force with electrostatic force, which can be repulsive, leading to reversed signs.

Correct move:

For any gravitational force from a source mass, the force on the test mass is always directed toward the source mass; set your coordinate system and assign signs accordingly.

Wrong move:

Applying Newton’s point-mass formula directly to non-spherical extended objects without integration.

Why:

Students memorize that spherical masses can be treated as point masses, so they incorrectly extend this to all shapes.

Correct move:

Only use the point-mass formula for point masses or spherically symmetric masses; for other shapes, use superposition or integration to find net force.

Wrong move:

Adding magnitudes of gravitational forces directly for 2D superposition problems, instead of adding vector components.

Why:

When all forces are along one line, adding magnitudes with sign works, so students incorrectly extend this to 2D problems.

Correct move:

Always break each force into x and y components, add components separately, then calculate the net force magnitude.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Newton's Law of Universal Gravitation	$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$	For point/spherical masses; $r$ = center-to-center separation
Gravitational Force from Field	$F_{g} = m g$	Applies to any mass distribution; $m$ = test mass
Near-Earth Surface Weight	$F_{g} = m g$	Approximation for points close to Earth's surface; $g \approx 9.8 m/s^{2}$
Superposition of Gravitational Force	$F_{n e t} = \sum_{i} F_{i}$	Net force is vector sum of individual forces
Gravitational Field from Spherical Mass	$g = \frac{GM}{r ^{2}}$	For any point outside a spherical mass $M$
Linear Mass Density (Uniform Rod)	$λ = \frac{M}{L}$	Used for integrals over continuous mass distributions
Universal Gravitational Constant	$G = 6.67 \times 1 0^{- 11} N \cdotp m^{2} / kg^{2}$	Provided on AP formula sheet; useful to memorize
Newton's Third Law for Gravity	$∣ F_{12} ∣ = ∣ F_{21} ∣$	Force magnitude is equal for both interacting masses

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Superposition of forces from 3 point masses
2022 · FRQ
Gravitational force from uniform rod

Going deeper

Official SyllabusAP Physics C: Mechanics Course and Exam DescriptionOfficial formula sheet and learning objectives

What's Next

Gravitational forces are the foundational prerequisite for all remaining topics in Unit 7 Gravitation. Next, you will apply gravitational force to analyze orbital motion, equating gravitational force to centripetal force to derive Kepler’s laws of planetary motion and calculate orbital speeds, periods, and energies. Without correctly calculating the magnitude and direction of gravitational force, you cannot set up correct equations of motion for orbits, leading to errors on nearly all Unit 7 free-response questions. Gravitational forces also connect to earlier topics including circular motion, Newton’s laws, and work and energy, and the inverse-square law of gravitation provides a template for understanding electric forces if you continue to AP Physics C: Electricity and Magnetism.

→Orbits of planets and satellites