When You're Bored on a Desert Island (2.1.04)

Question: Imagine you are on a desert island (without wireless) and for some reason you need Kepler's Third Law of motion. Using dimensional analysis, what is the form of this equation, which relates period, total mass, and separation of a two-body gravitational orbit? 

In order to derive Kepler's Third Law, you need to have a two-body gravitational orbit. Thankfully, since you are stuck in the middle of the desert without wireless (seriously?) you have all the time in the world to think about which heavenly body you can use. brightest object we can use, of course, is the Sun. Since the Earth revolves around the Sun, we have our two-body gravitational orbit. 

Now, here's what we know about the Sun-Earth system: 

• The Earth revolves around the Sun
• The Sun's gravitational force keeps the Earth in orbit
• The Earth's orbit is mostly circular 
• The period, $T$ of one Earth orbit around the Sun is 365 days
• The distance, $a$ between the Earth and the Sun is 1 AU
• The total mass of the Earth can be represented by $m_{Earth}$
• The total mass of the Sun can be represented by $M_{Sun}$
• The total mass of the system, $M_{Total}$

Okay, so we listed some hard numbers, such as the distance between the Earth and the Sun, as well as the period, $T$ of 365 days. However, since we are trying to come up with Kepler's laws with just algebraic physics and dimensional analysis, we'll ignore the numbers. 

Instead, let's look at some of the physics concepts we can use. We see that the Earth revolves around the Sun due to the gravitational force, and revolves around the Sun in mostly a circular fashion. The law of gravity is defined as follows:
\begin{align*}
F_G = \frac{GM_1 M_2}{r^2}
\end{align*}
Therefore we can use Newton's Second Law of motion to derive Kepler's Third Law: 

\begin{align*} F_{net} = ma_c\\
 \frac{GM_1 M_2}{r^2} = ma_c
\end{align*}

Since the Earth revolves around the Sun in an almost-circular orbit, we have to use centripetal acceleration, where $a_c = \frac{v^2}{r}$. We can also use the definition of velocity, where $v = \frac{distance}{time}$ to show the velocity of the Earth when it makes a full orbit around the Sun in a period, $T$. Therefore, velocity can be represented as $v = \frac{distance}{time} = \frac{2 \pi a}{T}$.
 \begin{align*}
 \frac{GM_1 M_2}{r^2} &= ma_c\\
\frac{G \cdot m_{Earth} \cdot M_{Sun}}{a^2} &= m_{Earth}\frac{v^2}{a}\\
\frac{GM_{Sun}}{a} &= v^2\\
\frac{GM_{Sun}}{a} &= \left(\frac{2 \pi a}{T}\right)^2\\
\frac{GM_{Sun}}{a} &= \frac{4 \pi^2 a^2}{T^2}
\end{align*}
Since we are looking for the relationship between the period, $T$ and the distance between two bodies, $a$, and the total mass, $M_total$, we can rearrange the previous equation to show this relationship.
\begin{align*}
T^2 &= \frac{4 \pi^2 a^3}{GM_{Sun}}
\end{align*}
Although the total mass of the system, $M_{total} = M_{Sun} + M_{Earth}$, since the $M_{Sun} \gg M_{Earth}$, we can assume that $M_{total} = M_{Sun}$, which leads to the equation:
\begin{align*}
T^2 &= \frac{4 \pi^2 a^3}{GM_{Total}}
\end{align*}
The equation above is actually Kepler's Third Law!!!

However,we are looking for the relation between period, distance, and total mass. Since $\frac{4 \pi^2}{G}$ is a constant, the relation between period, $T$, distance between an two bodies, $a$, and the total mass, $M_{Total}$ as follows:
\begin{align*}
T^2 &\propto \frac{a^3}{M_{Total}}
\end{align*}
To go a little further, in any given two-body gravitational system, the total mass, $M_{total}$ will also be a constant, so Kepler's Third Law actually shows the proportionality that:
\begin{align*}
T^2 &\propto a^3
\end{align*}
This is the profound relation in Kepler's third law, because it shows the relation between the distance between two bodies, and the orbital period of a smaller body revolving around a much larger body.