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LESSON 22 / 22 · TOPIC 2.10

Circular orbits and Kepler’s third law

You will be able to: Derive the radius–period relationship from gravitational centripetal acceleration.

Calculus-based dynamicsFree study resourceReview editionTeacher review pending

Why do more distant circular orbits take longer?

A satellite in a larger circular orbit moves more slowly and has a longer path to complete. Together these effects make its period grow faster than its radius.

A useful starting point: A conical pendulum in space →

Words and symbols before equations

Central mass M
Mass dominating the gravitational field; much larger than the satellite mass here.
Orbital radius r
Distance from the central body’s center, not altitude.
Orbital period T
Time for one revolution.
Circular-orbit assumption
Constant-radius motion supplied by gravity alone, outside a spherical source.
Larger radius means a longer periodT/T₀r/r₀10233649512
Read this model snapshot. Radius ratio 4: speed ratio 0.5, period ratio 8. Squared period ratio 64 equals cubed radius ratio.
What this picture assumes

Small satellite on a circular orbit outside a spherical dominant central mass. r₀, v₀ and T₀ refer to one reference circular orbit. All plotted quantities are ratios, not SI values.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Radius ratio 4: speed ratio 0.5, period ratio 8. Squared period ratio 64 equals cubed radius ratio.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

Set gravitational force equal to the required inward net force: GMm/r² = mv²/r. The satellite mass cancels and v = √(GM/r). There is no extra centripetal force in addition to gravity.

Use T = 2πr/v to obtain T² = 4π²r³/(GM). For satellites around the same dominant central mass, T²/r³ is constant. Doubling r makes T larger by 2^(3/2), not by two.

A graph of T² versus r³ has slope 4π²/(GM); measured orbital data can infer the central mass. This lesson addresses circular orbits. Kepler’s first and second laws are not required by this unit’s stated boundary, and a circular formula should not be applied blindly to an arbitrary elliptical path.

A worked example, step by step

Satellite A orbits at r₀ with period 2 h. Satellite B has circular orbital radius 4r₀ around the same central mass. Find B’s period and speed ratio.

  1. From T² ∝ r³, T_B/T_A = (4)^(3/2) = 8.
  2. T_B = 8(2 h) = 16 h.
  3. From v ∝ r^(−1/2), v_B/v_A = 1/√4 = 1/2.
  4. The larger orbit takes longer both because the path is larger and because the speed is lower.
Common mix-up

Use orbital radius from the central body’s center. Altitude alone is not r.

CHECK THE IDEA

Does doubling the satellite mass change the circular orbital period in this model?

Compare with an explanation

No. Its inertial and gravitational mass cancel when the central mass dominates.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Increase r/r₀ from 1 to 4. Predict period and speed ratios. Then inspect the straight-line T²-versus-r³ relation.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Larger radius means a longer periodT/T₀r/r₀10233649512

Radius ratio 4: speed ratio 0.5, period ratio 8. Squared period ratio 64 equals cubed radius ratio.

Kepler linearization in normalized units(T/T₀)²(r/r₀)³0031.2531.2562.562.593.7593.75125125

Small satellite on a circular orbit outside a spherical dominant central mass. r₀, v₀ and T₀ refer to one reference circular orbit. All plotted quantities are ratios, not SI values.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus 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.

1. If orbital radius grows by a factor of 9, period grows by…

Show answer and reasoning

27. 9^(3/2) = 27.

2. The slope of T² against r³ equals…

Show answer and reasoning

4π²/(GM). Rearrange the circular-orbit period relation.

Original written challenge

4 points · self-check · not an official AP question

Two small satellites orbit the same central mass at radii r and 2r. Derive their speed ratio and period ratio, and explain why satellite mass cancels.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: Equate GMm/r² = mv²/r to obtain v = √(GM/r).
  2. 1 point: Outer/inner speed ratio = 1/√2.
  3. 1 point: Outer/inner period ratio = 2√2 from T ∝ r^(3/2).
  4. 1 point: The same satellite mass multiplies gravitational force and inertia, so it cancels.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1Which interaction supplies the inward force?

Gravity alone in the circular-orbit model.

RECALL 2What stays constant for a fixed central mass?

T²/r³.

RECALL 3What does orbital radius measure?

Distance from the central body’s center.

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) for circular gravitational orbits.
  • T² = 4π²r³/(GM).
  • For fixed M: T ∝ r^(3/2); v ∝ r^(−1/2).

Remember: Use orbital radius from the central body’s center. Altitude alone is not r.

Conditions: Small satellite on a circular orbit outside a spherical dominant central mass. r₀, v₀ and T₀ refer to one reference circular orbit. All plotted quantities are ratios, not SI values.

Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.10.B · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 2.10, objectives 2.10.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. 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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