Kepler's Laws

(Kepler's laws of planetary motion)

Kepler's 1st Law

K1L

The orbit of every planet is an ellipse with the Sun at one of the two foci.

Kepler's 2nd Law

K2L

A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

The same (blue) area is swept out in a fixed time period. The green arrow is velocity. The purple arrow directed towards the Sun is the acceleration. The other two purple arrows are acceleration components parallel and perpendicular to the velocity.

Kepler's 3rd Law

K3L

The ratio of the square of an object's orbital period with the cube of the semi-major axis of its orbit is the same for all objects orbiting the same primary.

now effectively what this means is:

\[ T^2 \propto a^3 \]

where \(T\) is the period and \(a\) is the semi-major axis of the orbit.

You can use the proportionality as such:

\[ \frac{T_1^2}{T_2^2} \propto \frac{a_1^3}{a_2^3} \]

you can use \(T_2 = 1yr\), \(a_2 = 1AU\) and get \(T^2 = a^3\) with \(T\) in years and \(a\) in AU (AU is the distance from earth to sun)

Question

(I07 -T06 - A)Solution

A Sun-orbiting periodic comet is the farthest at 31.5 AU and the closest at 0.5 AU. What is the orbital period of this comet?

\[\begin{align*} a &= \frac{r_{pe}+r_{ap}}{2} \\ &= \frac{31.5+0.5}{2} \\ &= 16AU\end{align*}\]

We can use Kepler's 3rd law to solve this.

Since the units are in AU and years, \(T^2 = a^3\). Therefore:

\[\begin{align*}T &= \sqrt{a^3} \\ &= \sqrt{16^3} \\ &= 64.0 years\end{align*}\]