We can deduce Kepler’s third law by applying the orbit equation to two planets revolving around the Sun. Rather than directly deal with the orbit equation, we first work out a relation between the orbital period \(T\) of a planet and the planet’s distance \(r\) from the Sun. We work in polar coordinates with the planet in \(xy\) plane and the angular momentum pointed along \(z\) axis. Let \(l\) be the \(z\) component of the angular momentum. We have seen above that
\begin{equation*}
l = \mu r^2 \frac{d\theta}{dt}
\end{equation*}
Here \(\mu\) is the reduced mass of planet’s mass \(m_1\) and Sun’s mass \(m_2\text{.}\) Rearrange the equation, and divide both sides by 2 we get
\begin{equation*}
\dfrac{l}{2\mu} dt = \dfrac{1}{2} r (rd\theta)
\end{equation*}
The right side is a differential element of area enclosed in the ellipse centered about the focus. Therefore, upon integration over a whole period we will obtain the area of the ellipse, \(\pi a b\text{,}\) on the right side. On the left side, the integration will give a quantity proportional to the period \(T\text{.}\) Writing as
\begin{equation*}
\frac{l}{2\mu} T = \pi a b.
\end{equation*}
From Eq.
(12.51), we have
\(b = \sqrt{r_0 a}\text{.}\)
\begin{equation}
T = \alpha\, a^{3/2},\tag{12.62}
\end{equation}
where \(A\) is
\begin{equation*}
\alpha = \dfrac{2\pi \mu}{l} \sqrt{r_0}.
\end{equation*}
We can separate planet-independent part from the rest
\begin{equation*}
\alpha = \beta\, \sqrt{\dfrac{\mu}{m_1}},
\end{equation*}
where \(m_1\) is the mass of the planet, and
\begin{equation*}
\beta = \dfrac{2\pi}{\sqrt{G_N m_2}},
\end{equation*}
where \(m_2\) is the mass of Sun in the planet-Sun system. We now write this for two planets A and B.
\begin{align*}
\amp T_A/a_A^{3/2} = \beta\, \sqrt{\dfrac{\mu_A }{m_A}} \\
\amp T_B/a_B^{3/2} = \beta\, \sqrt{\dfrac{\mu_B }{m_B}}
\end{align*}
Taking the ratio of the two we get
\begin{equation}
\dfrac{T_A/a_A^{3/2}}{T_B/a_B^{3/2}} = \sqrt{\dfrac{\mu_A m_B}{\mu_B m_A}} .\tag{12.63}
\end{equation}
This shows that Kepler’s third law is not exact - it would have been exact if the right side of this equation were to equal 1.
Let us expand the right side of this equation for small \(m_{A,B}/M_\odot\) where \(M_\odot\) is mass of the Sun. Dropping even smaller terms than \(m_{A}/M_\odot\) or \(m_{B}/M_\odot\text{,}\) we find the deviation to depend on difference in the masses of the planets being compared and the mass of the Sun.
\begin{align*}
\sqrt{\dfrac{\mu_A m_B}{\mu_B m_A}} \amp = \left( \frac{m_B + M_\odot}{m_A + M_\odot} \right)^{1/2}\\
\amp = \left( \frac{ 1 + (m_B/ M_\odot)}{ 1 + (m_A / M_\odot)} \right)^{1/2}\\
\amp \approx \left( 1 + \frac{1}{2}\frac{m_B}{ M_\odot} \right) \left( 1 - \frac{1}{2}\frac{m_A}{ M_\odot} \right)\\
\amp \approx 1 + \frac{1}{2}\frac{m_B}{ M_\odot} - \frac{1}{2}\frac{m_A}{ M_\odot} \\
\amp = 1 + \frac{1}{2} \frac{m_B - m_A}{M_\odot}
\end{align*}
