Gravitational forces — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Newton’s law of universal gravitation, gravitational force as a vector, superposition of gravitational forces for multiple masses, field-based force calculation, and rules for extended mass distributions.

You should already know: Vector component addition for multi-vector systems. Newton's second and third laws of motion. Inverse-square proportional relationships.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Gravitational forces?

Gravitational force is the always-attractive long-range fundamental force that acts between any two objects with non-zero mass. It is the foundation of all topics in Unit 7 Gravitation for AP Physics C: Mechanics, and accounts for roughly half of the unit’s 6–14% total exam weight, meaning it typically contributes 3–8% of your total exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a building block for orbital mechanics and gravitational potential energy questions later in the unit.

Standard notation conventions for AP C: $G=6.67\times 10^{-11}{\text{Ncdotpm}}^2/{\text{kg}}^2$ is the universal gravitational constant, $m_1,m_2$ refer to the masses of the two interacting objects, $r$ is the center-to-center distance between the objects, and ${\vec{F}}_g$ denotes the gravitational force vector. Gravitational force is sometimes referred to as universal gravitational force or Newtonian gravitational force to distinguish it from relativistic gravity, which is not tested on the AP exam. Unlike electrostatic force, which can attract or repel, gravitational force is always attractive because there is no experimental evidence for negative mass in classical mechanics.

2. Newton’s Law of Universal Gravitation

Newton’s law of universal gravitation describes the magnitude of the gravitational force between two point masses (or spherically symmetric extended masses). The magnitude is given by: $$F_g=G\frac{m_1{m}_2}{r^2}$$ The force always acts along the line connecting the centers of the two masses, pulling each mass toward the other. By Newton’s third law, the force that $m_1$ exerts on $m_2$ is equal in magnitude and opposite in direction to the force $m_2$ exerts on $m_1$.

Key intuition: The force scales linearly with each mass (more mass means more gravitational pull) and falls off with the square of the distance between centers, so doubling the distance reduces the force to 1/4 its original value. A critical point for AP problems: this formula only applies directly to point masses and spherically symmetric extended masses. For non-spherical masses, we need to integrate or use superposition, which we cover later. For spherically symmetric masses, we can treat all the mass as concentrated at the center of the sphere, a result Gauss’s law for gravitation confirms, that AP examiners test regularly.

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 $r$: add the two radii and the surface separation: $r=7.5\text{cm}+7.5\text{cm}+10\text{cm}=25\text{cm}=0.25\text{m}$.
2. Identify given values: $m_1=10\text{kg}$, $m_2=50\text{kg}$, $G=6.67\times 10^{-11}{\text{Ncdotpm}}^2/{\text{kg}}^2$.
3. Substitute into Newton’s law: $F_g=(6.67\times 10^{-11})\frac{(10)(50)}{(0.25{)}^2}=(6.67\times 10^{-11})(\frac{500}{0.0625})=(6.67\times 10^{-11})(8000)$.
4. Calculate the final magnitude: $F_g\approx 5.3\times 10^{-7}\text{N}$. By Newton’s third law, the force each sphere exerts on the other is equal in magnitude.

Exam tip: On MCQs, you almost never need to calculate the full numerical value of $F_g$; use proportional reasoning ($F_g\propto m_1{m}_2/r^2$) to eliminate wrong options 10x faster than full calculation.

3. Superposition of Gravitational Forces

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 mass. This is the principle of superposition, which holds for all classical forces, including gravity.

For AP problems, the most common setup is 2–3 point masses arranged on a line or 2D grid, with the question asking for the net force on one test mass. The step-by-step process is: (1) calculate the magnitude of each individual force using Newton’s law, (2) assign a coordinate system and resolve each force into x and y components, (3) add the corresponding components to get net force components, (4) calculate the magnitude and direction of the net force if required. For continuous extended mass distributions, we replace the vector sum with an integral over infinitesimal mass elements, which is a common FRQ skill.

Worked Example

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

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

Exam tip: Always draw a coordinate system and confirm the direction of each force before adding components—AP examiners intentionally place masses to test if you mix up signs for attractive forces.

4. Gravitational Force from Gravitational Field

Gravitational force can also be described using the gravitational field $\vec{g}$, defined as the gravitational force per unit mass at a given point in space: $\vec{g}={\vec{F}}_g/m$. Rearranging this definition gives the general relation for gravitational force: $${\vec{F}}_g=m\vec{g}$$ This formulation simplifies calculations when you already know the net gravitational field at a point: you can just multiply by the test mass to get net force, instead of adding individual force vectors. A familiar special case is the weight of an object near Earth’s surface: the gravitational field is approximately uniform near the surface, with magnitude $g\approx 9.8{\text{m/s}}^2$, so $F_g=mg$, which is just an approximation of Newton’s universal law. For any point a distance $r$ from the center of a spherical mass $M$, the gravitational field magnitude is $g=GM/r^2$, which varies with position.

Worked Example

The net gravitational field at a point in space is $\vec{g}=(-1.8\hat{i}+4.2\hat{j}){\text{m/s}}^2$. 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 ${\vec{F}}_g=m\vec{g}$ to get the vector force.
2. Multiply each component by the mass $m=15\text{kg}$: ${\vec{F}}_g=15(-1.8\hat{i}+4.2\hat{j})=(-27\hat{i}+63\hat{j})\text{N}$.
3. Calculate the magnitude using the Pythagorean theorem: $|F_g|=\sqrt{(-27{)}^2+(63{)}^2}=\sqrt{729+3969}=\sqrt{4698}\approx 68.5\text{N}$.

Exam tip: The near-Earth $g=9.8{\text{m/s}}^2$ is always an approximation. If a problem asks for force at a significant altitude above a planet’s surface, always calculate $g=GM/r^2$ instead of using 9.8.

5. Common Pitfalls (and how to avoid them)

• 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. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

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 $m$? A) $\frac{GMm}{2d^2}$ B) $\frac{\sqrt{2}GMm}{2d^2}$ C) $\frac{GMm}{d^2}$ D) $\frac{2GMm}{d^2}$

Worked Solution: First, calculate the distance from each $M$ to $m$: $r=\sqrt{d^2+d^2}=d\sqrt{2}$, so $r^2=2d^2$. The magnitude of the force from each $M$ is $F=\frac{GMm}{2d^2}$. Resolve forces into components: the x-components of the two forces cancel (equal magnitude, opposite direction), and the y-components add. Each force is at a 45° angle to the y-axis, so the y-component of each force is $F\cos (45^{\circ })=\frac{GMm}{2d^2}\cdot \frac{\sqrt{2}}{2}=\frac{GMm\sqrt{2}}{4d^2}$. Adding the two y-components gives $F_{net}=2\cdot \frac{GMm\sqrt{2}}{4d^2}=\frac{\sqrt{2}GMm}{2d^2}$. The correct answer is B.

Question 2 (Free Response)

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 $a>0$. (a) What is the linear mass density of the rod? (b) Set up, but do not evaluate, an integral for the magnitude of the net gravitational force on $m$ from the rod. (c) Evaluate your integral to find a closed-form expression for the net force, and show that it reduces to the point-mass result when $a\gg L$.

Worked Solution: (a) Linear mass density is total mass divided by total length for a uniform rod: $\lambda =\frac{M}{L}$.

(b) Take an infinitesimal segment of the rod at position $x$ with mass $dm=\lambda dx=\frac{M}{L}dx$. The distance between the segment and $m$ is $(L+a-x)$, so the infinitesimal force is $dF=\frac{Gmdm}{(L+a-x{)}^2}=\frac{GMmdx}{L(L+a-x{)}^2}$. Integrate over the full length of the rod: $$F={\int }_0^L\frac{GMm}{L(L+a-x{)}^2}dx$$

(c) Use substitution $u=L+a-x$, $du=-dx$, with limits $u=L+a$ at $x=0$ and $u=a$ at $x=L$: $$F=\frac{GMm}{L}{\int }_a^{L+a}\frac{du}{u^2}=\frac{GMm}{L}{\left[-\frac{1}{u}\right]}_a^{L+a}=\frac{GMm}{L}\left(\frac{1}{a}-\frac{1}{L+a}\right)=\frac{GMm}{a(L+a)}$$ When $a\gg L$, $L+a\approx a$, so $F\approx \frac{GMm}{a^2}=\frac{GMm}{(L+a{)}^2}$, which matches the point-mass result.

Question 3 (Application / Real-World Style)

The James Webb Space Telescope orbits at the L2 Lagrange point, 1.5 million km from Earth, on the line connecting Earth and the Sun. The Sun’s mass is $M_S=1.99\times 10^{30}\text{kg}$, Earth’s mass is $M_E=5.97\times 10^{24}\text{kg}$, and Webb has a mass of 6200 kg. The distance between the Sun and Earth is $1.50\times 10^{1}1\text{m}$. Calculate the net gravitational force on Webb from the Sun and Earth in this position.

Worked Solution: Set up coordinates with the Sun at $x=0$, Earth at $x=1.50\times 10^{11}\text{m}$, and Webb at $x=1.50\times 10^{1}1+1.5\times 10^9=1.515\times 10^{11}\text{m}$. The force from the Sun pulls Webb left toward the Sun (negative direction), and the force from Earth pulls Webb left toward Earth (also negative direction, since Earth is to the left of Webb). Calculate magnitudes: $F_S=G\frac{M_{S}m}{r_S^2}=(6.67\times 10^{-11})\frac{(1.99\times 10^{30})(6200)}{(1.515\times 10^{11}{)}^2}\approx 35.7\text{N}$ $F_E=G\frac{M_{E}m}{r_E^2}=(6.67\times 10^{-11})\frac{(5.97\times 10^{24})(6200)}{(1.5\times 10^9{)}^2}\approx 1.10\text{N}$ Net force: $F_{net}=-(35.7+1.10)=-36.8\text{N}$, so the net force has magnitude 36.8 N directed toward the Sun. This small net force is balanced by the centrifugal force in the rotating Earth-Sun frame to keep Webb stationary at L2.

7. Quick Reference Cheatsheet

8. What's Next

This chapter on gravitational forces is the foundational prerequisite for all remaining topics in Unit 7 Gravitation. Next, you will apply gravitational force to analyze orbital motion, where you equate 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 will not be able to set up the correct equations of motion for orbits, leading to errors on nearly all Unit 7 free-response questions. Gravitational forces also connect to earlier topics in the course, including circular motion, Newton’s laws of motion, 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.

Orbital motion Gravitational potential energy Kepler's laws