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Wikipedia on Bertrand's theorem, when discussing the deviations from a circular orbit says:

...The next step is to consider the equation for $u$ under small perturbations ${\displaystyle \eta \equiv u-u_{0}}$ from perfectly circular orbits.

(Here $u$ is related to the radial distance as $u=1/r$ and $u_0$ corresponds to the radius of a circular orbit ) ...

The deviations are as

The solutions are ${\displaystyle \eta (\theta )=h_{1}\cos(\beta \theta ),}$

For the orbits to be closed, $β$ must be a rational number. What's more, it must be the same rational number for all radii, since β cannot change continuously; the rational numbers are totally disconnected from one another

Why does $\beta$ have to be the same rational number for all radii at which a circular orbit is possible?

I understand why it should be rational, but why the same number for all radii?

Link:https://en.wikipedia.org/wiki/Bertrand%27s_theorem

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The variable $\beta$ must vary continuously with the radius because it is defined in terms of another function $J$ that varies continuously with the radius. Now, suppose there are radii $r_1$ and $r_2$ such that $\beta(r_1)=3$ and $\beta(r_2)=3.2$. Because $\beta$ is continuous, there must be a radius $r_3$ between $r_1$ and $r_2$ such that $\beta(r_3)=\pi$. This cannot happen because $\beta$ must be rational, and there is no such thing as varying continuously over the rational numbers. So, $\beta$ cannot vary and must be constant.

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  • $\begingroup$ But β is defined only at specific radii where we've a circular orbit. So β isn't a continuous function to begin with as is defined in the article : ${\displaystyle \beta ^{2}\equiv 1-J'(u_{0})}$ $\endgroup$
    – Kashmiri
    Commented Oct 27, 2021 at 17:08
  • $\begingroup$ @Kashmiri Circular orbits can occur at any radius, so $u_0$ is a continuous variable. If you can calculate different $\beta$s at different radii, then $\beta$ must vary continuously. $\endgroup$
    – Mark H
    Commented Oct 27, 2021 at 17:19
  • $\begingroup$ So basically we are interested in such a force that produces closed orbits for small perturbations for all radii where circular orbits are possible and for that β has to be constant rational? $\endgroup$
    – Kashmiri
    Commented Oct 28, 2021 at 7:25
  • $\begingroup$ @Kashmiri Right. For central forces that only depend on distance, it is always possible to create a circular orbit at any radius by correctly setting the initial velocity. $\endgroup$
    – Mark H
    Commented Oct 28, 2021 at 9:41
  • $\begingroup$ Thank you so much. :) $\endgroup$
    – Kashmiri
    Commented Oct 28, 2021 at 10:35

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