Kepler's Laws of Planetary Motion/Third Law

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Physical Law

Kepler's third law of planetary motion is one of the three physical laws of celestial mechanics deduced by Johannes Kepler:

The square of the period of the orbit of a planet around the sun is proportional to the cube of its average distance from the sun.


Proof

Consider a planet $p$ of mass $m$ orbiting a star $S$ of mass $M$ under the influence of the gravitational field which the two bodies give rise to.

From Kepler's First Law of Planetary Motion, $p$ travels in an elliptical orbit around $S$:

$(1): \quad r = \dfrac {h^2 / k} {1 + e \cos \theta}$

where $k = G M$.

From Equation of Ellipse in Reduced Form: Cartesian Frame, the equation of the orbit can also be given as:

$(2): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$

where the foci are placed at $\tuple {\pm c, 0}$.

From Focus of Ellipse from Major and Minor Axis:

$a^2 - b^2 = c^2$

and also:

$e = \dfrac c a$



Thus:

\(\displaystyle e^2\) \(=\) \(\displaystyle \dfrac {a^2 - b^2} {a^2}\) $\quad$ $\quad$
\((3):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle b^2\) \(=\) \(\displaystyle a^2 \paren {1 - e^2}\) $\quad$ $\quad$


EllipsePolarCartesian.png


Then mean distance $a$ of $p$ from the focus $F$ is half the sum of the least and greatest values of $r$.

So $(1)$ and $(3)$ give:

\(\displaystyle a\) \(=\) \(\displaystyle \dfrac 1 2 \paren {\dfrac {h^2 / k} {1 + e} + \dfrac {h^2 / k} {1 - e} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {h^2} {k \paren {1 - e^2} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {h^2 a^2} {k b^2}\) $\quad$ $\quad$
\((4):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle b^2\) \(=\) \(\displaystyle \dfrac {h^2 a} k\) $\quad$ $\quad$


Let $T$ be the orbital period of $p$.

From Area of Ellipse, the area $\mathcal A$ of the orbit is given by:

$\mathcal A = \pi a b$

From Kepler's Second Law of Planetary Motion it follows that:

$\dfrac {h T} 2 = \pi a b$

and so from $(4)$:

\(\displaystyle T^2\) \(=\) \(\displaystyle \dfrac {4 \pi^2 a^2 b^2} {h^2}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\dfrac {4 \pi^2} k} a^3\) $\quad$ $\quad$

We have that:

$k = G M$

where $G$ is the gravitational constant and $M$ is the mass of $S$.

Hence the result.

$\blacksquare$


Astronomical Units

When distance is measured in astronomical units and time is measured in years, Kepler's Third Law of Planetary Motion can be expressed as:

$T^2 = a^3$

These units are derived specifically from the nature of the revolution of the Earth in its orbit, for which $T = 1 \, \mathrm{yr}$ and $a = 1 \, \mathrm {au}$.


Examples

Let $P$ be a planet orbiting the sun $S$.

Let $P$ be:

$\text{(a)}: \quad$ Twice as far away from $S$ as the Earth
$\text{(b)}: \quad$ $3$ times as far away from $S$ as the Earth
$\text{(c)}: \quad$ $25$ times as far away from $S$ as the Earth.


Then the orbital period of $P$ is:

$\text{(a)}: \quad$ approximately $2.8$ years
$\text{(b)}: \quad$ approximately $5.2$ years
$\text{(c)}: \quad$ $125$ years.


Also see


Source of Name

This entry was named for Johannes Kepler.


Historical Note

Kepler derived his three laws of planetary motion in the early $1600$s from a concentrated study over the course of $20$ years of the colossal wealth of observational data which had been made previously by Tycho Brahe of the behavior of the planets of the solar system, and in particular Mars.

The first two of these results he published in his gigantic work Astronomia Nova.

The third appears some ten years later in his Harmonices Mundi of $1619$.

It was Isaac Newton who managed to interpret these three laws and so work out what is now known as Newton's Law of Universal Gravitation, from which Kepler's laws can straightforwardly be derived.


Sources