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Kepler's Third Law: The Harmony of the Worlds 📂Classical Mechanics

Kepler's Third Law: The Harmony of the Worlds

Kepler’s Third Law: The Law of Harmonies

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.


Kepler’s third law among Kepler’s laws of planetary motion. When approximating the orbit of a planet as a circle, it becomes ‘The square of the orbital period is proportional to the cube of the distance to the sun’.

Proof1

Let the area of the planetary orbit be $A$ and the areal velocity be $\dot {A}$. From Kepler’s second law, we know that the areal velocity is a constant as follows.

$$ \begin{equation} \dot{A}=\frac{L}{2m}=\frac{l}{2}=\text{constant} \label{eq1} \end{equation} $$

$l=\frac{L}{m}$ is the angular momentum per unit mass. The period is the time it takes for a planet to make one complete orbit. Let us denote this as $\tau$. Then, by the definition of areal velocity and period, and by $\eqref{eq1}$, the following formula holds.

$$ A=\int_{0}^{\tau}\dot{A}dt=\frac{l\tau}{2} $$

Therefore, it is as follows.

$$ \tau=\frac{2A}{l} $$

However, the area of an ellipse is well known to be $ab\pi$. Hence, the period is as follows.

$$ \tau = \frac{2ab\pi}{l} $$

Now, let’s look at the figure below.

5F22517D0.png

The semi-major axis, semi-minor axis, and eccentricity of an ellipse satisfy formula $b=a\sqrt{1-\epsilon^{2}}$. Substituting into the period, we obtain the following formula.

$$ \tau = \frac{2\pi a^{2}\sqrt{1-\epsilon^{2}}}{l} $$

Squaring both sides of the above formula yields the following.

$$ \tau ^{2} = \frac{4\pi^{2}a^{4}}{l^{2}}(1-\epsilon^{2}) $$

Since the transverse axis of the ellipse is $\alpha=a(1-\epsilon^{2})$, the above formula is rewritten as follows.

$$ \tau^{2} = \frac{4 \pi^{2}\alpha}{l^{2}}a^{3} $$

Furthermore, since the orbit of a planet moving under the sun’s gravity has a transverse axis of $\alpha=\frac{l^{2}}{GM}$, it follows that.

$$ \tau ^{2} = \frac{4\pi^{2}}{GM}a^{3} $$ Here, $G$ is the gravitational constant, and $M$ is the mass of the sun. As you can see, the preceding proportionality constant, $\frac{4\pi^{2}}{GM}$, applies equally to all objects orbiting the sun under the influence of its gravity.


  1. Grant R. Fowles and George L. Cassiday, Analytical Mechanics (7th Edition, 2005), p238-239 ↩︎