The Newton-Kepler law of planetary motion

In the previous posts, we discussed Kepler’s laws of planetary motion and the application of Kepler’s third law in the solar system.

\begin{equation}
T^2 = \frac{4\pi^2}{GM}a^3
\end{equation}

The reason why it works well within the solar system is because of the mass of the Sun. It dwarfs the planetary mass so much that the planet can be considered a point-like object.

\begin{equation}
\begin{aligned}
M_{\text{solar system bodies}} &= 2.66757168\times10^{27}\,\text{kg} \\
M_{\text{Sun}} &= 1.98847542\times10^{30}\,\text{kg} \\
\end{aligned}
\end{equation}

\begin{equation}
\text{mass ratio} = \frac{M_{\text{solar system bodies}}}{M_{\text{Sun}}} = 1.34151604\times10^{-3}
\end{equation}

However, this ratio does not hold everywhere. There are multi-body systems with comparable or even equal masses. To account for these problems, we must use the Newtonian counterpart of Kepler’s third law.

\begin{equation}
T^2 = \frac{4\pi^2}{G(m_1 + m_2)}a^3
\end{equation}

As can be seen from equation $(1)$ and $(4)$, Kepler’s third law and its Newtonian counterpart differ only in the inclusion of the secondary mass. To fully understand the differences, we need to understand the concept of the center of mass.

Center of mass

Consider a uniform rod such that the mass is distributed equally throughout its length. We can tell from our everyday experience that such a rod can be balanced at its midpoint. Let’s imagine the beam being balanced on its center point as shown below,

We now attach a mass $m_1$ to one end of the rod. The attached mass will produce a torque in the rod and start tilting it towards the direction of acting force (downwards).

As we know that the torque produced is the product of force and the moment arm so,

\begin{equation}
\begin{aligned}
\tau &= F\times r_1\\
\tau &= m_1g\times r_1
\end{aligned}
\end{equation}

If we add another mass $m_2$ at the opposite side, it will try to balance out the torque produced by the mass $m_1$.

Now both of these masses are producing torques on their respective ends. As evident from equation $(1)$, the amount of torque generated depends on the mass of the object and the arm length. In the above figure, for $r_1 = r_2$, the mass $m_2$ will generate more torque and hence the beam will tilt on its side. We can write the torque produced by the two masses as,

\begin{equation}
\tau_1 = m_1r_1g
\end{equation}

\begin{equation}
\tau_2 = m_2r_2g
\end{equation}

For the beam to be balanced, both the torques produced must be equal.

\begin{equation*}
\begin{aligned}
\tau_1 &= \tau_2\\
m_1r_1g &= m_2r_2g\\
m_1r_1 &= m_2r_2
\end{aligned}
\end{equation*}

Rearranging the equation for the arm length, we get

\begin{equation}
r_1 = \frac{m_2}{m_1}r_2
\end{equation}

Equation $(8)$ shows that if both the masses are equal, the center of the mass is midway between the two masses. However, two solve for two unknowns, we need two equations. We also know that the sum of arm lengths of both the masses is equal to the length of the rod itself.

\begin{equation} r = r_1 + r_2 \end{equation}

Substituting equation$(8)$ in equation $(9)$ we get

\begin{equation*}
\begin{aligned}
r &= \frac{m_2}{m_1}r_2 + r_2\\
r &= \left(\frac{m_2}{m_1} + 1\right)r_2
\end{aligned}
\end{equation*}

\begin{equation}
r_2 = \frac{m_1}{m_1 + m_2}r
\end{equation}

Similarly, for $r_1$ we can write the equation

\begin{equation}
r_1 = \frac{m_2}{m_1 + m_2}r
\end{equation}

Equation $(10)$ and $(11)$ give the arm length for mass $m_2$ and $m_1$ respectively.