I am struggling with the logic for completing the following problem.

The problem is part b of 3.19 in Goldstein's Classical Mechanics book.

A particle moves in a force field described by the Yukowa potential $$ V(r) = \frac{k}{r} exp (-\frac{r}{a}), $$ where k and a are positive.

Show that if the orbit is nearly circular, the apsides will advance approximately by $\pi r_0 / a$ per revolution, where $r_0$ is the radius of the circular orbit.

The following beautiful solution that I found online due to Professor Laura Reina at Florida State Uni has helped me get 75% of the way there.

The logic for solving this problem goes as follows:

Thinking about the following graph for $U_{eff}$ one can see that circular orbit (orbit at fixed radius) occurs when $U_{eff}$ is minimized. Solving $\frac{d U_{eff}}{dr}=0$ gives us the value of $r_0$ from the statement of the problem.

Next, since the problem states "nearly" circular, we let r deviate slightly from $r_0$. I.e. we write r as a function of $\theta$ (using r, $\theta$ polar coordinates) in the following way

$$ r(\theta) = r_0[1+\delta(\theta)] $$

where $\delta$ is a function of $\theta$ like r since $r_0$ is not allowed to vary.

The next step is to plug this equation for $r(\theta)$ into the so-called "orbit equation"

$$ \frac{d^2}{d\theta^2} \frac{1}{r(\theta)} + \frac{1}{r(\theta)} = -\frac{mr^2}{\ell^2}F(r) $$

where $F(r)$ can be found from the potential/force relation with the potential of the problem.

The usual substitution is then used $u=\frac{1}{r}$, and in our case, by a binomial approximation, we have $u=\frac{1}{r}\frac{1}{r_0}(1-\delta)$.

Via some algebra, one non-trivial part being expanding (one of!) the exponentials into its series expansion, we arrive at

$$ \frac{d^2}{d\theta^2} \delta(\theta) + \bigg(1-\frac{mkr_o{}^2}{\ell^2 a}e^{-\frac{r_0}{a}}\bigg)\delta(\theta) = 1-\frac{mkr_o}{\ell^2}e^{-\frac{r_0}{a}} $$

which, via Gert's great answer here, is clearing simple harmonic motion.

Using the definition of $r_0$ from our first equation, we can identify the frequency squared coefficient of SHO as

$$ \omega^2 = \frac{1}{1+\frac{r_0}{a}} $$

Here is where I lose the logical progression of the solution.

  1. Can someone offer some intuition for what is going on when the author of the solution says,

Now choose $\delta$ to be at maximum when $\theta=0$, then the next maximum will occur when...

  1. The author's next step is to find the change in $\theta$ via $\omega\theta=2\pi$. I am completely lost as to why we can use this for a non-circular orbit.

I'm convinced my confusion lies in something simple I could garner from here or something similar, but I've been stuck here for hours. Any tips appreciated.

Some drawings to expound on where my confusion lies are below.

Our particle/object does not follow a perfectly circular orbit at radius $r_0$ but rather one at radius $r(\theta)$ where again, $r(\theta) = r_0[1+\delta(\theta)]$,

enter image description here

Now, there is nothing stopping us from picking a maximum value of $r(\theta)$ and thus a max value of $\delta(\theta)$ since remember $r_0$ is fixed. But surely this is a maximum only for that particular revolution, no?! The question prompts us with

the apsides will advance approximately by $\pi r_0 / a$ per revolution

and so there is only natural that, after 20 revolutions let's say, the value of $r(\theta)$ and $\delta(\theta)$ are greater than they were previously.

enter image description here

  • 1
    $\begingroup$ I understand that your drawings are likely just exaggerations, but keep in the back of your mind that orbits can never look like this in "real life." The region bounded by the motion of the moving particle must always be a convex figure, as the force is attractive. Incorrect, concave drawings have been mistakenly put in all editions (and all cover illustrations) of Goldstein, and the problem got so bad, that it is even mentioned on the wikipedia page for the book $\endgroup$
    – najkim
    Jul 29, 2020 at 1:13
  • $\begingroup$ wow great pointer! I never thought to look for a wiki article ON the book itself! cheers $\endgroup$
    – Lopey Tall
    Jul 29, 2020 at 13:45

1 Answer 1


Recall that $\delta(\theta)$ is the deviation from the circular orbit. Choosing it to be maximum when $\theta = 0$ is a matter of mathematical convenience so you don't need to deal with an initial phase $\theta = \theta_0$ or a "maximum" phase $\theta = \theta_{\mathrm{max}}$.

  • $\begingroup$ Thank you for the answer. My confusion lies in how this relates back to the system at hand. Recall that we want to show that "the apsides will advance approximately by 𝜋𝑟0/𝑎 per revolution." By definition, how does one choose a maximum value of delta then? Does delta not continue to grow per revolution? I will edit my question with an illustration to more aptly convey my confusion. $\endgroup$
    – Lopey Tall
    Jul 24, 2020 at 13:27
  • $\begingroup$ No, $\delta(\theta)$ keeps the same maximum. I think what is meant by the "maximum" here is the local maximum for the range 0≤θ<2π. Notice that the solution comments "This is the equation for a simple harmonic oscillator (with a constant shift) ...". This constant shift is at the r.h.s. of the SHM equation. $\endgroup$
    – wyphan
    Jul 24, 2020 at 16:58

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