Kepler's Third Law (College Board AP® Physics 1: Algebra-Based)

Study Guide

Ann Howell

Written by: Ann Howell

Reviewed by: Caroline Carroll

Kepler's third law

  • For a satellite in circular orbit around a central body, the satellite’s centripetal acceleration is caused only by gravitational attraction

  • Kepler's Third Law states:

For planets or satellites in a circular orbit about the same central body, the square of the time period is proportional to the cube of the radius of the orbit

  • This law describes the relationship between the time of an orbit and its radius

T squared space proportional to space r cubed

  • Where: 

    • T = orbital time period, measured in straight s

    • r = mean orbital radius, measured in straight m

  • The period and radius of the circular orbit are related to the mass of the central body using the equation for Kepler's third law

T squared space equals space fraction numerator 4 straight pi squared r cubed over denominator G M end fraction

  • Where:

    • T space equals time period of the orbit, measured in straight s

    • r space equals spaceorbital radius, measured in straight m

    • G space equals space Gravitational Constant

    • M space equals space mass of the object being orbited, measured in kg

Derived equation

  • The period and radius of the circular orbit are related to the mass of the central body using the equation for Kepler's third law

T squared space equals space fraction numerator 4 straight pi squared r cubed over denominator G M end fraction

Derivation:

Step 1: Identify the fundamental principles

  • Uniform circular motion equation:

T space equals space fraction numerator 2 straight pi straight r over denominator v end fraction

  • Where:

    • T = time period, measured in straight s

    • r = radius of orbit, measured in straight s

    • v = tangential velocity of object in orbit, measured in straight m divided by straight s

  • Newton's law of gravitation:

    open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space equals space G fraction numerator m subscript 1 m subscript 2 over denominator r squared end fraction

  • Where:

    • open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space equals magnitude of the gravitational force between the two objects, measured in straight N

    • G space equals spaceuniversal gravitational constant = 6.67 space cross times space 10 to the power of negative 11 end exponent space straight N times straight m squared space divided by space kg squared

    • m subscript 1 space equals spacemass of object 1, measured in kg

    • m subscript 2 space equals spacemass of object 2, measured in kg

    • r space equals spacethe distance between the center of mass of the two objects, measured in straight m

  • Centripetal force equation:

stack F subscript c with rightwards arrow on top space equals space fraction numerator m v squared over denominator r end fraction

  • Where:

    • stack F subscript c with rightwards arrow on top space = centripetal force, measured in straight N

    • m = mass of orbiting object, measured in straight m

    • v = tangential speed of orbiting object, measured in straight m divided by straight s

    • r = orbital radius, measured in straight m

Step 2: Combine the equations for centripetal and gravitational force

  • For an object in orbit in a uniform gravitational field around a larger mass, the centripetal force is created by the gravitational force

    • Where object 1 is the larger mass, m subscript 1 space equals space M and object 2 the smaller mass

stack F subscript c with rightwards arrow on top space equals space open vertical bar stack F subscript g with rightwards arrow on top close vertical bar

space fraction numerator m subscript 2 v squared over denominator r end fraction equals space G fraction numerator m subscript 1 m subscript 2 over denominator r squared end fraction

space fraction numerator up diagonal strike m subscript 2 end strike v squared over denominator r end fraction equals space G fraction numerator M up diagonal strike m subscript 2 end strike over denominator r squared end fraction

Step 3: Rearrange the equation to make v squared the subject

space v squared equals space G fraction numerator M times r over denominator r squared end fraction

space v squared equals space fraction numerator G M over denominator r end fraction

Step 4: Substitute for vfrom the uniform circular motion equation

T space equals space fraction numerator 2 straight pi straight r over denominator v end fraction space rightwards arrow space v space equals space fraction numerator 2 pi r over denominator T end fraction

space v squared equals space fraction numerator G M over denominator r end fraction space equals space open parentheses fraction numerator 2 pi r over denominator T end fraction close parentheses squared

Step 5: Expand the brackets, rearrange to make T squared the subject and simplify

fraction numerator G M over denominator r end fraction space equals space fraction numerator 4 pi squared r squared over denominator T squared end fraction

open parentheses fraction numerator G M over denominator r end fraction close parentheses space times space T squared space equals space open parentheses fraction numerator 4 pi squared r squared over denominator T squared end fraction close parentheses space times space T squared

open parentheses fraction numerator G M over denominator r end fraction close parentheses space times space T squared space equals space open parentheses fraction numerator 4 pi squared r squared over denominator up diagonal strike T squared end strike end fraction close parentheses space times space up diagonal strike T squared end strike

space T squared space equals space 4 pi squared r squared space space divided by space open parentheses fraction numerator G M over denominator r end fraction close parentheses

T squared space equals space 4 pi squared r squared space space times space open parentheses fraction numerator r over denominator G M end fraction close parentheses

T squared space equals space fraction numerator 4 pi squared r cubed over denominator G M end fraction

Worked Example

Planets A and B orbit the same star.

Planet A is located an average distance r from the star. Planet B is located an average distance 6 r from the star.

Which of the following correctly expresses the ratio fraction numerator orbital space period space of space planet space straight A over denominator orbital space period space of space planet space straight B end fraction?

A      fraction numerator 1 over denominator cube root of 6 end fraction

B      fraction numerator 1 over denominator square root of 6 end fraction

C      fraction numerator 1 over denominator cube root of 6 squared end root end fraction

D      fraction numerator 1 over denominator square root of 6 cubed end root end fraction

The correct answer is D

Answer: 

Step 1: Analyze the scenario and identify the known relationships

  • Kepler's third law states T squared space proportional to space r cubed

  • The orbital period of planet A:  T subscript A space proportional to space square root of r cubed end root

  • The orbital period of planet B:  T subscript B space proportional to space square root of open parentheses 6 r close parentheses cubed end root

Step 2: Determine the ratio fraction numerator orbital space period space of space planet space straight A over denominator orbital space period space of space planet space straight B end fraction

  • Therefore the ratio is equal to:

T subscript A over T subscript B proportional to space fraction numerator square root of r cubed end root over denominator square root of open parentheses 6 r close parentheses cubed end root end fraction space proportional to space fraction numerator 1 over denominator square root of 6 cubed end root end fraction

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Ann Howell

Author: Ann Howell

Expertise: Physics Content Creator

Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.