Circular orbits and Kepler’s third law
You will be able to: Combine gravity and circular motion to compare orbital speeds and periods.
Why do larger circular orbits take longer?
An orbiting satellite is continually falling toward Earth while moving sideways. In a larger circular orbit, gravity is weaker, speed is lower, and the path around is longer.
A useful starting point: Inverse-square gravity →
Words and symbols before equations
- Orbit radius r
- Distance from the central body’s center, not altitude.
- Central mass M
- Mass producing the gravitational field; assumed much larger than the satellite mass.
- Orbital period T
- Time for one complete revolution.
What this picture assumes
Circular orbits about one dominant central mass. Satellite mass negligible. Speed, period and acceleration are ratios to the orbit at r₀.
Connect the picture to the physics
For a circular orbit dominated by one central body, gravity supplies the inward net force: GMm/r²=mv²/r. Cancel the satellite mass and rearrange to get v=√(GM/r).
Substitute this speed into T=2πr/v. Squaring gives T²=4π²r³/(GM). For satellites orbiting the same central mass, T²/r³ is constant. This is the circular-orbit form of Kepler’s third law.
A larger orbital radius means lower speed but longer period. These comparisons apply to circular orbits; do not apply the circular speed formula at every point of an elliptical orbit. Orbital energy is developed later in the course.
A worked example, step by step
Two small satellites orbit the same planet in circular paths of radii r and 4r. If the inner satellite has speed v and period T, find the outer speed and period.
- Speed ratio = √(r/(4r))=1/2, so outer speed is v/2.
- Circumference becomes four times larger.
- Period ratio = path ratio / speed ratio = 4/(1/2)=8.
- Equivalently, T_outer/T_inner=(4)^(3/2)=8. Both approaches agree.
Use the same central mass when comparing T²/r³. Different planets do not generally share that constant.
Does doubling satellite mass change this circular orbital speed?
Compare with an explanation
No, under the negligible-satellite-mass assumption. Its gravitational force and required inward force both scale with its mass.
Predict. Change one thing. Explain.
Change radius ratio from 1 to 4 at fixed central mass. Compare speed, period and gravitational acceleration ratios. Explain how slower speed and longer path together produce eight times the period.
v/v₀ = 0.5; T/T₀ = 8; a/a₀ = 0.06.
Circular orbits about one dominant central mass. Satellite mass negligible. Speed, period and acceleration are ratios to the orbit at r₀.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA satellite in a circular orbit has period 2 hours. Another orbits the same central body at four times the radius. (a) State the period-radius law, (b) find its period, (c) compare speeds, and (d) compare gravitational accelerations.
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Compare with the answer and four-point rubric
- 1 point: T²/r³ is constant for the same central mass.
- 1 point: Period is 2×8=16 hours.
- 1 point: Outer speed is one half.
- 1 point: Outer gravitational acceleration is 1/16, from the inverse-square law.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What keeps a circular satellite turning?
Gravity supplies the inward acceleration.
RECALL 2How does period depend on radius?
T is proportional to r^(3/2) for the same central body.
RECALL 3Does circular speed increase farther out?
No, it decreases as 1/√r.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Circular orbits and Kepler’s third law
- v=√(GM/r).
- T²=4π²r³/(GM); same central body → T²∝r³.
- Circular orbit, gravity alone, satellite mass negligible relative to central mass.
Remember: Use the same central mass when comparing T²/r³. Different planets do not generally share that constant.
Conditions: Circular orbits about one dominant central mass. Satellite mass negligible. Speed, period and acceleration are ratios to the orbit at r₀.
Refresh Kid · Unit 2 · Objectives 2.9.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.9, objectives 2.9.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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