Kepler's laws for circular orbits

Question

Kepler's first law states that planets revolve around the sun in an ellipse with the sun at one focus of this ellipse. (a special case would be a circular orbit with the sun at the center). The second law states that the areal velocity is a constant. Thus we can write ($dA = c dt$). If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period. The third law states that the square of the time period is proportional to the cube of the semi-major axis of the elliptical orbit. My question is, should we swap out "semi-major axis" and replace it with "radius", or is there something missing? if we can, that leads to a contradiction with the second law. however, how can the result obtained from the second law be wrong for circular orbits? What am I missing?

Answer 1

My question is, should we swap out "semi-major axis" and replace it with "radius"...

We can always do that. As you noted, circle is a special case of ellipse.

The second law states that the areal velocity is a constant. Thus we can write (dA=cdt).

Correct. Here c is areal speed.

If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period.

Integrating we will get $\pi r^2 = cT$

From this we cannot conclude that $T$ is proportional to $r^2$. The reason is areal velocity $c$ is also dependent on $r$. I hope the source of confusion is clear.

Answer 2

For circular orbits, $$\frac{dA}{dt}=\frac{L}{2\mu}\Rightarrow \pi R^2=\frac{L}{2\mu}T$$ Further $$E=\frac{\mu C^2}{2L^2}\ \ \ \text{For circular orbits}$$ From now on I'm just going to keep track of proportionality. $$R^2\propto LT\propto\frac{1}{\sqrt{E}}T$$ But further we known that $R\propto 1/E$ $$R^{3/2}\propto T\Rightarrow \boxed{R^3\propto T^2}$$