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