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

Study Guide

Written by: Ann Howell

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

$T^{2} \propto r^{3}$

Where:
- $T$ = orbital time period, measured in $s$
- $r$ = mean orbital radius, measured in $m$

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

$T^{2} = \frac{4 π^{2} r^{3}}{G M}$

Where:
- $T =$ time period of the orbit, measured in $s$
- $r =$ orbital radius, measured in $m$
- $G =$ Gravitational Constant
- $M =$ 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

$T^{2} = \frac{4 π^{2} r^{3}}{G M}$

Derivation:

Step 1: Identify the fundamental principles

Uniform circular motion equation:

$T = \frac{2 πr}{v}$

Where:
- $T$ = time period, measured in $s$
- $r$ = radius of orbit, measured in $s$
- $v$ = tangential velocity of object in orbit, measured in $m / s$
Newton's law of gravitation:
$|\vec{F_{g}}| = G \frac{m_{1} m_{2}}{r^{2}}$
Where:
- $|\vec{F_{g}}| =$ magnitude of the gravitational force between the two objects, measured in $N$
- $G =$ universal gravitational constant = $6.67 \times 10^{- 11} N \cdot m^{2} / {kg}^{2}$
- $m_{1} =$ mass of object 1, measured in $kg$
- $m_{2} =$ mass of object 2, measured in $kg$
- $r =$ the distance between the center of mass of the two objects, measured in $m$

Centripetal force equation:

$\vec{F_{c}} = \frac{m v^{2}}{r}$

Where:
- $\vec{F_{c}}$ = centripetal force, measured in $N$
- $m$ = mass of orbiting object, measured in $m$
- $v$ = tangential speed of orbiting object, measured in $m / s$
- $r$ = orbital radius, measured in $m$

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, $m_{1} = M$ and object 2 the smaller mass

$\vec{F_{c}} = |\vec{F_{g}}|$

$\frac{m_{2} v^{2}}{r} = G \frac{m_{1} m_{2}}{r^{2}}$

$\frac{m_{2} v^{2}}{r} = G \frac{M m_{2}}{r^{2}}$

Step 3: Rearrange the equation to make $v^{2}$ the subject

$v^{2} = G \frac{M \cdot r}{r^{2}}$

$v^{2} = \frac{G M}{r}$

Step 4: Substitute for $v$ from the uniform circular motion equation

$T = \frac{2 πr}{v} \to v = \frac{2 π r}{T}$

$v^{2} = \frac{G M}{r} = {(\frac{2 π r}{T})}^{2}$

Step 5: Expand the brackets, rearrange to make $T^{2}$ the subject and simplify

$\frac{G M}{r} = \frac{4 π^{2} r^{2}}{T^{2}}$

$(\frac{G M}{r}) \cdot T^{2} = (\frac{4 π^{2} r^{2}}{T^{2}}) \cdot T^{2}$

$T^{2} = 4 π^{2} r^{2} \div (\frac{G M}{r})$

$T^{2} = 4 π^{2} r^{2} \cdot (\frac{r}{G M})$

$T^{2} = \frac{4 π^{2} r^{3}}{G M}$

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 $\frac{orbital period of planet A}{orbital period of planet B}$ ?

A $\frac{1}{\sqrt[3]{6}}$

B $\frac{1}{\sqrt{6}}$

C $\frac{1}{\sqrt[3]{6^{2}}}$

D $\frac{1}{\sqrt{6^{3}}}$

The correct answer is D

Answer:

Step 1: Analyze the scenario and identify the known relationships

Kepler's third law states $T^{2} \propto r^{3}$
The orbital period of planet A: $T_{A} \propto \sqrt{r^{3}}$
The orbital period of planet B: $T_{B} \propto \sqrt{{(6 r)}^{3}}$

Step 2: Determine the ratio $\frac{orbital period of planet A}{orbital period of planet B}$

Therefore the ratio is equal to:

$\frac{T_{A}}{T_{B}} \propto \frac{\sqrt{r^{3}}}{\sqrt{{(6 r)}^{3}}} \propto \frac{1}{\sqrt{6^{3}}}$

Last updated: 7 October 2024

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

Study Guide

Kepler's third law

Derived equation

Derivation:

Worked Example

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Author: Ann Howell

Author: Caroline Carroll