Circular Motion and Gravitation — AP Physics 1 Phys 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Centripetal acceleration and force, universal gravitation, Kepler's three laws of orbital motion, and apparent weight in rotating reference frames.

You should already know: Algebra 2, basic trig, no calculus required.

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 1 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Circular Motion and Gravitation?

This unit unifies two interrelated physical phenomena: the motion of objects along curved circular paths, and the attractive gravitational force that governs the orbital motion of planets, satellites, stars, and other celestial bodies. Per the 2024-25 AP Physics 1 CED, this topic makes up 10-16% of your total exam score, with a mix of conceptual multiple-choice questions, quantitative calculations, and multi-part free-response questions (FRQs). All circular motion relies on a net force pointing toward the center of the curved path, and gravity is the most common natural source of this force for large-scale astronomical systems.

2. Centripetal acceleration $a_c=\frac{v^2}{r}$

When an object moves at constant speed along a circular path, its velocity is constantly changing direction, even if its speed stays the same. Since acceleration is defined as a change in velocity (not just speed), the object is always accelerating, and this acceleration points directly toward the center of the circle: this is called centripetal (center-seeking) acceleration. The magnitude of centripetal acceleration is derived from the vector difference of velocity over a small time interval, leading to the formula: $$a_c=\frac{v^2}{r}={\omega }^{2}r$$ Where:

• $a_c$ = centripetal acceleration (m/s²)
• $v$ = tangential speed of the object (speed along the edge of the circle, m/s)
• $r$ = radius of the circular path (m)
• $\omega$ = angular speed of the object (rad/s, related to tangential speed by $v=\omega r$)

Worked Example: A delivery truck drives around a circular roundabout with radius 35 m at a constant speed of 12 m/s. What is its centripetal acceleration? $$a_c=\frac{(12{)}^2}{35}=\frac{144}{35}\approx 4.11{\text{ m/s}}^2$$ This is roughly 42% of the acceleration due to gravity, so you would feel a noticeable push toward the side of the truck as it turns. Exam tip: Examiners frequently test that centripetal acceleration is always perpendicular to tangential velocity, so it never changes the speed of the object, only its direction of motion.

3. Centripetal force $F_c=\frac{m{v}^2}{r}$

From Newton's second law ($F_{net}=ma$), any acceleration requires a net force. The net force that causes centripetal acceleration is labeled centripetal force. Substituting centripetal acceleration into Newton's second law gives the formula: $$F_c=m{a}_c=\frac{m{v}^2}{r}=m{\omega }^{2}r$$ Where $m$ = mass of the moving object (kg), and all other terms are the same as defined for centripetal acceleration. A critical note: centripetal force is not a new fundamental force (like tension, friction, or gravity). It is simply the label for the vector sum of all real forces acting on the object that point toward the center of the circular path. You should never draw "centripetal force" on a free-body diagram (FBD), only the actual physical forces acting on the object.

Worked Example: A 0.15 kg yo-yo is tied to a 0.8 m long string and spun in a horizontal circle at a constant speed of 5 m/s. What is the tension in the string (ignore gravity for horizontal motion)? Tension is the only force providing centripetal force, so $F_c=T$: $$T=\frac{0.15\ast (5{)}^2}{0.8}=\frac{0.15\ast 25}{0.8}=\frac{3.75}{0.8}\approx 4.69\text{ N}$$

4. Universal gravitation $F_g=G\frac{Mm}{r^2}$

Isaac Newton first proposed that the same gravitational force that pulls objects toward Earth's surface also keeps the Moon in its orbit around Earth. The universal law of gravitation states that the attractive gravitational force between any two point masses is proportional to the product of their masses and inversely proportional to the square of the distance between their centers of mass: $$F_g=G\frac{Mm}{r^2}$$ Where:

• $G$ = universal gravitational constant = $6.67\times 10^{-11}{\text{ Ncdotpm}}^2/{\text{kg}}^2$
• $M,m$ = masses of the two objects (kg)
• $r$ = center-to-center distance between the two objects (m)

This formula also explains the value of $g$ (acceleration due to gravity) on a planet's surface: set $mg=G\frac{Mm}{r^2}$, cancel the mass of the smaller object $m$, and you get $g=\frac{GM}{r^2}$.

Worked Example: What is the gravitational force between a 65 kg student and Earth? Earth's mass $M=5.97\times 10^{24}\text{ kg}$, Earth's radius $r=6.37\times 10^6\text{ m}$. $$F_g=6.67\times 10^{-11}\ast \frac{5.97\times 10^{24}\ast 65}{(6.37\times 10^6{)}^2}\approx 637\text{ N}$$ This matches the expected weight of the student: $mg=65\ast 9.8=637\text{ N}$, confirming the formula works.

5. Orbital motion — Kepler's laws

Johannes Kepler derived three empirical laws of planetary motion from observational data decades before Newton developed the law of universal gravitation, which later explained why Kepler's laws hold. All three laws are tested on AP Physics 1:

1. Law of Ellipses: All planets orbit the Sun in elliptical paths, with the Sun at one focus of the ellipse. Most orbits in our solar system are nearly circular, so you can approximate them as circles for most AP Physics 1 problems.
2. Law of Equal Areas: A line connecting an orbiting object to its central mass sweeps out equal areas in equal time intervals. This means objects move faster when closer to the central mass (perihelion for solar orbits) and slower when farther away (aphelion), because angular momentum is conserved.
3. Law of Periods: The square of the orbital period $T$ (time to complete one full orbit) is proportional to the cube of the semi-major axis $a$ of the orbit: $T^2\propto a^3$. For circular orbits, $a=r$, so you can derive the exact formula by setting gravitational force equal to centripetal force: $$G\frac{Mm}{r^2}=m\frac{v^2}{r}\text{and}v=\frac{2\pi r}{T}$$ Rearranging gives: $$T^2=\frac{4{\pi }^2}{GM}{r}^3$$ For two objects orbiting the same central mass, you can use the proportionality form to avoid calculating with $G$ and $M$: $\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}$.

Worked Example: The Moon orbits Earth at a radius of $3.84\times 10^8\text{ m}$ with a period of 27.3 days. What is the orbital period of a satellite orbiting Earth at a radius of $4.2\times 10^7\text{ m}$? Use the proportionality form of Kepler's third law: $$T_{sat}=T_{moon}\ast {\left(\frac{r_{sat}}{r_{moon}}\right)}^{3/2}=27.3\ast {\left(\frac{4.2e7}{3.84e8}\right)}^{1.5}\approx 1.14\text{ days}$$

6. Apparent weight in circular motion

Your apparent weight is equal to the normal force exerted on you by a supporting surface, not your actual gravitational weight $mg$. When you are moving in a circular path, the net force required for centripetal acceleration changes the magnitude of the normal force, so your apparent weight shifts:

• Top of a vertical circular path (e.g. top of a roller coaster loop, upside down): Forces acting on you are gravity $mg$ pointing down (toward the center of the circle) and normal force $N$ from the seat pointing down (toward the center). Net force: $mg+N=\frac{m{v}^2}{r}$, so $N=\frac{m{v}^2}{r}-mg$. If $v=\sqrt{gr}$, $N=0$, and you feel weightless (this is the minimum speed to stay on the track without falling).
• Bottom of a vertical circular path: Forces are gravity $mg$ pointing down, and normal force $N$ pointing up (toward the center). Net force: $N-mg=\frac{m{v}^2}{r}$, so $N=mg+\frac{m{v}^2}{r}$, making you feel heavier than your actual weight.
• Orbiting astronauts: Apparent weight is zero, not because there is no gravity in space, but because the astronaut and their spacecraft are in constant free fall: gravity provides all the centripetal force needed for orbit, so there is no normal force acting on the astronaut.

Worked Example: A 70 kg rider is on a roller coaster loop of radius 12 m, moving at 14 m/s at the top of the loop. What is their apparent weight? $$N=\frac{70\ast (14{)}^2}{12}-70\ast 9.8=\frac{70\ast 196}{12}-686=1143.3-686=457.3\text{ N}$$ This is ~67% of their actual weight of 686 N, so they feel lighter.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Drawing "centripetal force" as a separate force on free-body diagrams. Why students do it: They mistakenly think centripetal force is a new fundamental force. Correct move: Never label $F_c$ on FBDs; only draw real physical forces (tension, friction, gravity, normal force), then write that the sum of forces pointing toward the center of the circle equals $\frac{m{v}^2}{r}$.
• Pitfall 2: Using surface distance instead of center-to-center distance for gravity/orbital problems. Why students do it: They forget $r$ is defined as the distance between centers of mass, not surfaces. Correct move: Always add the radius of the central planet to the altitude of the orbiting object to get $r$.
• Pitfall 3: Thinking centripetal acceleration changes the speed of an object in circular motion. Why students do it: They associate all acceleration with changes in speed. Correct move: Centripetal acceleration is perpendicular to tangential velocity, so it only changes direction, not speed. Tangential acceleration (if present) changes speed.
• Pitfall 4: Believing orbiting astronauts are weightless because there is no gravity in space. Why students do it: They see astronauts floating and assume gravity is zero. Correct move: Gravity at the ISS altitude is ~90% of surface gravity; apparent weight is zero because the station and astronauts are in free fall, with no normal force acting on them.
• Pitfall 5: Misremembering Kepler's third law as $T\propto r^3$ instead of $T^2\propto r^3$. Why students do it: They mix up the exponents. Correct move: For objects orbiting the same central mass, use the ratio form $\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}$ to avoid mistakes.

8. Practice Questions (AP Physics 1 Style)

Question 1 (MCQ, 1 point)

A 1500 kg car drives around a flat circular track of radius 75 m. The coefficient of static friction between tires and track is 0.65. What is the maximum constant speed the car can reach without sliding? A) 15.7 m/s B) 21.8 m/s C) 28.3 m/s D) 37.1 m/s

Solution: Static friction provides the centripetal force. Set ${\mu }_{s}mg=\frac{m{v}^2}{r}$, cancel $m$: $$v=\sqrt{{\mu }_{s}gr}=\sqrt{0.65\ast 9.8\ast 75}=\sqrt{477.75}\approx 21.8\text{ m/s}$$ Correct answer: B

Question 2 (FRQ, 5 points)

Two satellites orbit Jupiter. Satellite X has an orbital radius of $1.2\times 10^7$ m and a period of 10 hours. Satellite Y has an orbital radius of $3.0\times 10^7$ m. (a) State Kepler's third law of planetary motion. (2 points) (b) Calculate the orbital period of Satellite Y in hours. (3 points)

Solution: (a) (2 points total, 1 point for core statement, 1 point for same central mass condition) Kepler's third law states that the square of the orbital period of an object orbiting a central mass is proportional to the cube of the semi-major axis of its orbit. For all objects orbiting the same central mass, the ratio $\frac{T^2}{r^3}$ is constant. (b) (3 points total: 1 for correct formula, 1 for correct substitution, 1 for correct answer with units) Use the ratio form of Kepler's third law: $$T_Y=T_X\ast {\left(\frac{r_Y}{r_X}\right)}^{3/2}=10\ast {\left(\frac{3.0e7}{1.2e7}\right)}^{1.5}=10\ast (2.5{)}^{1.5}\approx 39.5\text{ hours}$$

Question 3 (FRQ, 7 points)

A 55 kg student rides a Ferris wheel with radius 14 m that rotates at a constant speed of 3.5 m/s. (a) Calculate the student's apparent weight at the top of the Ferris wheel. (3 points) (b) Calculate the student's apparent weight at the bottom of the Ferris wheel. (3 points) (c) Explain why the student feels lighter at the top than at the bottom. (1 point)

Solution: (a) (3 points: 1 for correct force equation, 1 for substitution, 1 for answer) At the top, net force points down toward the center: $mg-N=\frac{m{v}^2}{r}$ $$N=mg-\frac{m{v}^2}{r}=55\ast 9.8-\frac{55\ast (3.5{)}^2}{14}=539-48.125=490.9\text{ N}$$ (b) (3 points: 1 for correct force equation, 1 for substitution, 1 for answer) At the bottom, net force points up toward the center: $N-mg=\frac{m{v}^2}{r}$ $$N=mg+\frac{m{v}^2}{r}=539+48.125=587.1\text{ N}$$ (c) (1 point) At the top, part of the gravitational force is used to provide centripetal acceleration toward the center of the wheel, so the normal force (apparent weight) is less than the student's actual weight. At the bottom, the normal force must both counteract gravity and provide centripetal acceleration, so it is larger than actual weight.

9. Quick Reference Cheatsheet

10. What's Next

Mastery of circular motion and gravitation is a foundation for multiple later units in the AP Physics 1 syllabus, including rotational motion (where you will extend centripetal force concepts to rigid rotating bodies, torque, and angular momentum conservation) and work and energy (where you will calculate gravitational potential energy for orbital systems and energy changes for objects moving along curved paths). This topic is also frequently combined with kinematics and forces in 7-point multi-part FRQs, so strong performance here directly correlates with higher exam scores.

If you struggle with any concept, calculation, or practice problem in this guide, you can get personalized, 24/7 help from Ollie, our AI tutor, by visiting Ollie. You can also find more AP Physics 1 study guides, full-length practice tests, and exam strategy tips on the homepage to build your confidence ahead of test day.