Kepler's third law states that the ratio between the period of a planet in orbit round a star squared over the average distance to that star cubed, is constant. And in the case of an elliptical orbit, we take the length of the semi-major axis as the distance. But why don't we take the average between the semi-major and semi-minor axis as the distance, that seems more intuitive?
4 Answers
Your intuition is failing you because the star is not at the center of the ellipse. The star is located at one of the two foci.
When you realize this you will find that the semi-major axis is the average between the planet's maximum and minimum distances from the star. Intuition restored.
There is no “average distance” in statements of Kepler’s Third Law that I have seen. This law says that the square of the period is proportional to the cube of the semi-major axis. The semi-minor axis is simply irrelevant to the period.
Is this intuitive? Not to me! I don’t think it’s intuitive that an orbit of two bodies under an attractive inverse square force between them is a closed ellipse with one focus at the center of mass, much less that the period depends only on the semi-major axis of that ellipse.
This is is why math is so indispensible in physics... intuition gets you almost nowhere, and when you do get someplace with intuition, it’s often wrong.
Kepler's laws are empirical and thus not intuitive at all, as they preceded Newton's theory of gravitation by quit a few years.
The period $T$ of a planet motion (ellipse) is the area of the ellipse $A_{\text{el}}$ divided by area velocity $A_v$.
$$T=\frac{A_{\text{el}}}{A_v}=\frac{\pi\,a\,b}{h/2}$$
with:
$\frac{h^2}{\mu}=a\,(1-e^2)$
$e^2=\left(1-\frac{b^2}{a^2}\right)$
we get for the period $T$:
$$\boxed{T=2\,\pi\frac{a^{3/2}}{\sqrt{\mu}}}\tag 1$$
with:
$\mu=G\,(M_s+m_p)$
The period $T$ is a function of the semi-major axis $a$ and $\mu$
where:
$a\quad$ semi-major axis
$b\quad$ semi-minor axis
$M_s\quad$ sun mass
$m_p\quad $ planet mass