User:WalkingRadiance/Keplers Third Law and Dimensional Analysis

Dimensional analysis

Kepler's third law can be derived with dimensional analysis using the definition of force as $T^{-2}LM$ and Newton's gravitation law that $F=Gm_{1}m_{2}/r^{2}$. The following notation can be used:

Symbol	Representation
$F$	the gravitational force of the central body, in this case the Sun
$l$	semi-major axis of the orbit's ellipse
$m$	mass of the orbiting body, in this case the planet
$m_{s}$	mass of the central body, in this case the mass of the Sun
$t$	the time the body takes to complete its orbit, in this case the planetary orbit period

Then there is a function $t=f(F,m,l)$ that can be found with dimensional analysis. The following table shows how to find the nondimensional form using dimensional analysis with the Buckingham Pi theorem:

t	F	m	l
${\mathsf {T}}$	${\mathsf {T^{-2}LM}}$	${\mathsf {M}}$	${\mathsf {L}}$
	$F*m^{-1}$
	${\mathsf {T^{-2}L}}$
	$F*m^{-1}*t^{2}$
	${\mathsf {L}}$
	$F*m^{-1}*t^{2}*l^{-1}$
	1

This means the following equation is true:

$F*m^{-1}*t^{2}*l^{-1}=constant$

Newton's law of gravitation tells us that

$F\propto m*m_{s}*l^{-2}$

Then this can be written as an equation:

$F*m^{-1}*{m_{s}}^{-1}*l^{2}=constant$

The two equations can then be divided by each other. On the left hand side there is

$F*m^{-1}*t^{2}*l^{-1}*F^{-1}*m^{1}*{m_{s}}^{1}*l^{-2}$

Rearranging we have:

$F*F^{-1}*m^{-1}*m^{1}*{m_{s}}^{1}*t^{2}*l^{-1}*l^{-2}$

The variables $m$ and $m_{s}$ both have the same dimension of ${\mathsf {M}}$, which causes $m^{-1}*m^{1}*{m_{s}}^{1}$ to be dimensionally equivalent to ${\mathsf {M}}^{-1}{\mathsf {M}}^{1}={\mathsf {M}}^{-1+1}={\mathsf {M}}^{0}=1$. The force dimension also cancels out through $F*F^{-1}$. The only remaining terms now on the left hand side are $t^{2}*l^{-1}*l^{-2}=t^{2}*l^{-1+-2}=t^{2}*l^{-3}$. The equation is now $t^{2}*l^{-3}=constant$. This proves that $t^{2}\propto l^{3}$. This is how to use Newton's law of gravity and dimensional analysis to derive Kepler's third law.