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

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 $\mathrm{s}$

  • $r$ = mean orbital radius, measured in $\mathrm{m}$

• 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

$T^2=\frac{4{\mathrm{\pi }}^2{r}^3}{GM}$

• Where:

  • $T=$ time period of the orbit, measured in $\mathrm{s}$

  • $r=$orbital radius, measured in $\mathrm{m}$

  • $G=$ Gravitational Constant

  • $M=$ mass of the object being orbited, measured in $\mathrm{kg}$

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

$T^2=\frac{4{\mathrm{\pi }}^2{r}^3}{GM}$

Derivation:

Step 1: Identify the fundamental principles

• Uniform circular motion equation:

$T=\frac{2\mathrm{\pi r}}{v}$

• Where:

  • $T$ = time period, measured in $\mathrm{s}$

  • $r$ = radius of orbit, measured in $\mathrm{s}$

  • $v$ = tangential velocity of object in orbit, measured in $\mathrm{m}/\mathrm{s}$

• Newton's law of gravitation:

  $\left|\overrightarrow{F_g}\right|=G\frac{m_1{m}_2}{r^2}$

• Where:

  • $\left|\overrightarrow{F_g}\right|=$ magnitude of the gravitational force between the two objects, measured in $\mathrm{N}$

  • $G=$universal gravitational constant = $6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2$

  • $m_1=$mass of object 1, measured in $\mathrm{kg}$

  • $m_2=$mass of object 2, measured in $\mathrm{kg}$

  • $r=$the distance between the center of mass of the two objects, measured in $\mathrm{m}$

• Centripetal force equation:

$\overrightarrow{F_c}=\frac{m{v}^2}{r}$

• Where:

  • $\overrightarrow{F_c}$ = centripetal force, measured in $\mathrm{N}$

  • $m$ = mass of orbiting object, measured in $\mathrm{m}$

  • $v$ = tangential speed of orbiting object, measured in $\mathrm{m}/\mathrm{s}$

  • $r$ = orbital radius, measured in $\mathrm{m}$

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

$\overrightarrow{F_c}=\left|\overrightarrow{F_g}\right|$

$\frac{m_2{v}^2}{r}=G\frac{m_1{m}_2}{r^2}$

$\frac{\cancel{m_2}{v}^2}{r}=G\frac{M\cancel{m_2}}{r^2}$

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

$v^2=G\frac{M·r}{r^2}$

$v^2=\frac{GM}{r}$

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

$T=\frac{2\mathrm{\pi r}}{v}\to v=\frac{2\pi r}{T}$

$v^2=\frac{GM}{r}={\left(\frac{2\pi r}{T}\right)}^2$

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

$\frac{GM}{r}=\frac{4{\pi }^2{r}^2}{T^2}$

$\left(\frac{GM}{r}\right)·T^2=\left(\frac{4{\pi }^2{r}^2}{T^2}\right)·T^2$

$\left(\frac{GM}{r}\right)·T^2=\left(\frac{4{\pi }^2{r}^2}{\cancel{T^2}}\right)·\cancel{T^2}$

$T^2=4{\pi }^2{r}^2÷\left(\frac{GM}{r}\right)$

$T^2=4{\pi }^2{r}^2·\left(\frac{r}{GM}\right)$

$T^2=\frac{4{\pi }^2{r}^3}{GM}$