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 $$T^{-2}L M$$ and Newton's gravitation law that $$F=Gm_1m_2/r^2$$. The following notation can be used: 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: 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.