---
title: "Kepler's Third Law — AP Physics 1 Definition & Equation"
description: "Kepler's third law (T² = 4π²R³/GM) links a satellite's orbital period to its radius and the central body's mass. Learn how to derive and use it on the AP exam."
canonical: "https://fiveable.me/ap-physics-1-revised/key-terms/keplers-third-law"
type: "key-term"
subject: "AP Physics 1"
unit: "Unit 2"
---

# Kepler's Third Law — AP Physics 1 Definition & Equation

## Definition

Kepler's third law states that for a satellite in circular orbit, the square of the orbital period is proportional to the cube of the orbital radius: T² = (4π²/GM)R³, where M is the mass of the central body. In AP Physics 1, it falls under Topic 2.9 (Circular Motion), learning objective 2.9.B.

## What It Is

Kepler's third law connects two things you can measure about an [orbit](/ap-physics-1-revised/key-terms/orbit "fv-autolink"). How long one trip takes (the [period](/ap-physics-1-revised/unit-2/9-circular-motion/study-guide/phypMTqBWYSyW4Xd "fv-autolink"), T) and how far the satellite is from the center (the orbital radius, R). For a circular orbit, the relationship is T² = (4π²/GM)R³. Bigger orbit, much longer period. Squaring the period and cubing the radius means the relationship is not a simple straight line, which is exactly why the exam loves to test it with ratios.

The AP version isn't a memorize-and-move-on fact. The CED calls it a *derived equation*, and essential knowledge 2.9.B.1 tells you where it comes from. For a satellite in circular orbit, gravity is the only [force](/ap-physics-1-revised/unit-2/2-forces-and-free-body-diagrams/study-guide/jQ2Obd0dAU4QiTPN "fv-autolink"), so gravitational attraction supplies all of the centripetal acceleration. Set Newton's law of gravitation equal to the centripetal force (GMm/R² = mv²/R), substitute v = 2πR/T, and Kepler's third law pops out. Notice what survives the algebra. The satellite's mass m cancels completely, so the period depends only on the orbital radius and the mass of the central body.

## Why It Matters

Kepler's third law lives in Topic 2.9 (Circular Motion) in [Unit 2](/ap-physics-1-revised/unit-2 "fv-autolink"): Force and Translational Dynamics. It directly supports learning objective 2.9.B, "Describe circular orbits using Kepler's third law," and it's the payoff of 2.9.A, which builds centripetal acceleration (a꜀ = v²/r). It's also where two big Unit 2 ideas finally shake hands. Newton's [law of gravitation](/ap-physics-1-revised/key-terms/law-of-gravitation "fv-autolink") and circular motion combine into one clean, testable equation. If you can derive T² = (4π²/GM)R³ yourself instead of just quoting it, you've basically proven you understand how forces cause circular motion, which is the whole point of the unit.

## Connections

### Orbital period and orbital radius (Unit 2)

These are the two variables Kepler's third law ties together. The key insight is the mismatch in powers. Period is squared but [radius](/ap-physics-1-revised/key-terms/radius "fv-autolink") is cubed, so quadrupling the radius makes the period 8 times longer, not 4. Every ratio problem on this topic hinges on that T² ∝ R³ scaling.

### Centripetal acceleration (Unit 2)

Kepler's third law is what you get when gravity is the only thing providing centripetal acceleration. That's the whole derivation in one sentence. Gravity points toward the center, a꜀ = v²/r points toward the center, set them equal and solve.

### [Tangential speed (Unit 2)](/ap-physics-1-revised/key-terms/tangential-speed)

The bridge in the derivation. A satellite's [tangential speed](/ap-physics-1-revised/key-terms/tangential-speed "fv-autolink") relates to its period through v = 2πR/T (one circumference per orbit). Substituting that into the force equation is the step that turns Newton's gravitation into Kepler's third law.

### [Banked surface (Unit 2)](/ap-physics-1-revised/key-terms/banked-surface)

A useful contrast within Topic 2.9. On a banked curve, the [normal force](/ap-physics-1-revised/key-terms/normal-force "fv-autolink") (and maybe friction) supplies the centripetal force. In orbit, gravity does it alone. Same circular-motion framework, different force doing the centering job.

## On the AP Exam

This shows up almost exclusively as proportional reasoning, not plug-and-chug. A typical multiple-choice stem changes one variable and asks how another responds. For example, cutting a satellite's 24-hour period down to 6 hours means the radius must shrink by a factor of 4^(2/3), since T² ∝ R³. Another classic stem quadruples the star's mass at fixed radius, which halves the period because T ∝ 1/√M. Multi-planet ranking problems do the same thing. If planet Y orbits at 4 times planet X's radius, its period is 4^(3/2) = 8 times longer. No released FRQ has used the name "Kepler's third law" verbatim, but the derivation (gravity equals centripetal force, then substitute v = 2πR/T) is exactly the kind of symbolic, justify-your-equation work AP Physics 1 free-response questions reward. Practice doing the derivation cleanly and practice the fractional exponents until 4^(3/2) = 8 feels automatic.

## Kepler's third law vs Newton's law of universal gravitation

They're related but not the same statement. Newton's law (F = GMm/r²) describes the gravitational force between any two masses. Kepler's third law is a consequence of it, the specific result you get when that force is the only thing holding an object in a circular orbit. If a question gives you a force or asks about attraction between two objects, use Newton's law. If it gives you a period or asks how long an orbit takes, that's Kepler territory.

## Key Takeaways

- Kepler's third law for circular orbits is T² = (4π²/GM)R³, where M is the mass of the central body, not the satellite.
- The satellite's own mass cancels out of the derivation, so a tiny probe and a massive station at the same orbital radius have identical periods.
- You derive it by setting gravitational force equal to centripetal force and substituting v = 2πR/T for the tangential speed.
- Because T² ∝ R³, multiplying the radius by 4 multiplies the period by 8, and most exam questions test exactly this kind of fractional-power scaling.
- Period also depends on the central mass through T ∝ 1/√M, so quadrupling the star's mass cuts the orbital period in half.
- The CED (2.9.B.1) frames this as gravity being the sole cause of a satellite's centripetal acceleration, which is the conceptual sentence FRQ justifications are built on.

## FAQs

### What is Kepler's third law in AP Physics 1?

It's the relationship T² = (4π²/GM)R³ for a satellite in circular orbit, meaning the square of the orbital period is proportional to the cube of the orbital radius. It's covered in Topic 2.9 under learning objective 2.9.B.

### Does a heavier satellite orbit slower than a lighter one?

No. The satellite's mass cancels when you set GMm/R² equal to mv²/R, so the period depends only on the orbital radius and the central body's mass. Any object at the same radius around Earth has the same period.

### How is Kepler's third law different from Newton's law of gravitation?

Newton's law (F = GMm/r²) gives the force between two masses in any situation. Kepler's third law is what that force implies specifically for circular orbits, relating period to radius. Kepler's law is derived from Newton's, not a separate rule.

### Do I need to memorize the equation T² = 4π²R³/GM?

The CED lists it as a derived equation, so you're expected to be able to produce it by combining gravitational force with centripetal acceleration. Practicing the derivation is more valuable than memorizing it, because FRQs reward showing where equations come from.

### If the orbital radius quadruples, what happens to the period?

The period becomes 4^(3/2) = 8 times longer, because T² ∝ R³. This exact scaling shows up constantly in multiple-choice questions, like a planet at 9 times another's radius having a period 27 times longer.

## Related Study Guides

- [2.9 Circular Motion](/ap-physics-1-revised/unit-2/9-circular-motion/study-guide/phypMTqBWYSyW4Xd)

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