Worksheet 2.1, Problem 4: Dimensional Analysis to Derive Kepler's Third Law Using

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 the period, total mass, and separation of a two-body gravitational orbit? Use only the units and a tad of physics, e.g. what constants are likely involved? (Note: there are many paths to this solution)

We are given that Kepler's Third Law relates period \((P)\), total mass \((M_{tot})\), and the distance between the two objects \((d)\). The units of each are seconds \((s)\), grams \((g)\), and centimeters \((cm)\) respectively. We can already see that in order to relate the period to mass and distance, another constant is needed. This constant is the gravitational constant, \(G\), whose units can be found using the equation for the gravitational force between to objects.

\(F=\dfrac{GM_1M_2}{d^2}\)

Solving for \(G\) we get:

\(G=\dfrac{Fd^2}{M_1M_2}\)

This means that the units of \(G\) are:

\([G]=\dfrac{(g\cdot cm/s^2)(cm^2)}{g^2}\)

\([G]=\dfrac{cm^3}{g\cdot s^2}\)

Now, in finding the form of Kepler's Third Law, we should consider what is actually happening between these two masses. The orbiting mass is gravitationally bound to the object it is revolving around. 

Let's imagine two scenarios of an object orbiting another object. In each scenario, the distance between the orbiting mass and the central mass is the same, but in one scenario, the total mass of the system is greater than that of the other scenario. We know by laws of gravity that objects with larger mass will have stronger gravitational force and therefore stronger gravitational attraction. This means that in the scenario with greater total mass, the orbiting object would have to revolve at a faster pace in order to keep from falling into the central mass, meaning that the period would be smaller than that of the other scenario. This implies that mass is inversely proportional to the period (so it should go in the denominator of Kepler's equation).

Similarly, imagine two scenarios of an object orbiting another object, but this time the total mass of the system is the same in both, and the distances differ. Again, by our understanding of gravity, we know that the force of gravity is weaker at larger distances, so in the scenario where the separation of the masses is larger, the orbiting mass will revolve at a slower rate, meaning that the period will be larger. This implies that the separation is directly proportional to the period (so it should go in the numerator of the equation).

So we have:

\([P]= [G]^a\dfrac{[d]^b}{[M_{tot}]^c}\)

\(s=\bigg(\dfrac{cm^3}{g\cdot s^2}\bigg)^a\dfrac{(cm)^b}{(g)^c}\)

In order to make units cancel, we need \(G\) to be in the denominator to take care of the \(cm\) in the numerator and \(g\) in the denominator. But \(G\) has \(cm^3\), so we must cube the distance \(d\) so that \(cm\) no longer remain. The grams in \(G\) and \(M_{tot}\) cancel each other out, and we are left with \(s^2\), so we need to take a square root and we will be left with seconds, which is what is desired.

So \(a=-1\), \(b=3\), and \(c=1\). This leaves us with the form:

\(P\propto\sqrt{\dfrac{d^3}{GM_{tot}}}\)

Which can be rewritten as:

\(P^2\propto\dfrac{d^3}{GM_{tot}}\)

And to check our answer:

\([P]^2=\dfrac{[d]^3}{[G][M_{tot}]}\)

\(s^2=\dfrac{cm^3}{\dfrac{cm^3}{g\cdot s^2}\cdot g}=s^2\)

This answer agrees with the full form of Kepler's Third Law, which is given as:

\(P^2=\dfrac{4\pi ^2d^3}{GM_{tot}}\)

I would like to thank Charles Law for his help in solving this problem and in proofreading this blog post.