Why does the orbital period not depend on the semi-minor axis?

In Kepler's Third Law, the orbital period is described by

$$T=2\pi\sqrt \frac{a^3}{\mu}$$

where $$a$$ is the lenght of the semi-major axis. I wonder, why does the lenght of the semi-minor axis not affect the orbital period? In this image, we have a semi-major axis ($$a$$) with lenght $$2$$, while the semi-minor axis $$b$$ has the lenght $$1$$. The circumference of an ellipse can be simplified to $$U=\pi \times \sqrt {2 \times (a^2 + b^2)}$$

For our ellipse, this is about $$9.935$$.

If we stretch the semi-major axis to $$1.8$$, we get an ellipse closer to a circle: This ellipse has a circumference of about $$11.955$$ which is remarkably longer than the circumference of the ellipse.

So why is the orbit for, let's say a planet orbiting a star, still the same regardless of the lenght of the semi-minor axis?

• The velocity is different at different points of the orbit. Jul 30 '20 at 8:57

Although the expression for the third Kepler law you showed is the most common one, you can use the fact that $$b = a\sqrt{1-e^2}$$, and with that, Kepler law reads $$T = 2\pi \sqrt{\frac{b^3}{\mu (1-e^2)^{3/2}}}$$ and you could change your question to "why it doesn't deppend on semi-major axis?"
I think the key point is using the second Kepler law as well. The velocity along the orbit is not constant, the body is slower farther from the central body, and this makes the energy to be conserved. It can be shown that the total orbital energy only depends on semi-major axis: $$\frac{E}{m}=\epsilon=-\frac{\mu}{2a}$$ And with this, the period is $$T = \sqrt{-\frac{\pi^2 \mu^2}{4\epsilon^3}}$$ So the fact that the period is the same for two orbits with the same semi-major axis but different eccentricity comes from the fact that their energies are the same.