Find the angle $\Phi$ for an orbit close to a circle of radius $r$

Question

There is an odd identity I cannot figure out in the titular exercise from V. I. Arnold's book on mechanics. He claims in an answer to the exercise (page 37) that: \begin{align} \frac{M}{r^2\sqrt{V''(r)}} = \sqrt{\frac{U'}{3U'+rU''}} \end{align} when $V(r)=U(r)+\frac{M^2}{2r^2}$. This possibly involves a change of variables $x=M/r.$

I've tried to prove this but have no leads. I figure it might have something to do with the form he claims U has, although part of this chain of exercises is to find what $U$ can be. My numerical tests of the claim fail too so I must be missing something.

Answer 1

$$\frac{M}{r^2\sqrt{V''(r)}} = \sqrt{\frac{U'}{3U'+rU''}}$$ $$\frac{M^2}{r^4V''(r)} = \frac{U'}{3U'+rU''}$$ $$\frac{r^4}{M^2}V''(r)=3+r\frac{U''}{U'}$$ Squaring might introduce additional solutions. At end of this calculus one must check and keep only the valid solutions. $$V(r)=U(r)+\frac{M^2}{2r^2}$$ $$V''(r)=U''(r)+\frac{3M^2}{r^4}$$ $$\frac{r^4}{M^2}(U''(r)+\frac{3M^2}{r^4})=3+r\frac{U''}{U'}$$ $$(\frac{r^4}{M^2}-r\frac{1}{U'})U''(r)=0$$ First set of solutions , from $U''(r)=0$ : $$\boxed{\begin{cases} U(r)=c_1r+c_2\\ V(r)=c_1r+c_2+\frac{M^2}{2r^2} \end{cases}}$$ Second set of solutions, from $\frac{r^4}{M^2}-r\frac{1}{U'}=0\quad\implies\quad U'=\frac{M^2}{r^3}$ $$\begin{cases} U(r)=-\frac{M^2}{2r^2}+c_3\\ V(r)=-\frac{M^2}{2r^2}+c_3+\frac{M^2}{2r^2}=c_3 \end{cases}\quad\text{rejected because}\quad \begin{cases} V''(r)=0\\ 3U'+rU''=0 \end{cases}$$ The original ODE becomes $\quad\frac{M}{r^2\sqrt{0}} = \sqrt{\frac{U'}{0}}=\infty$

This is the conclusion in a mathematical sense. Nevertheless one should check whether the solution $\quad \begin{cases} U(r)=-\frac{M^2}{2r^2}+c_3\\ V(r)=c_3 \end{cases}\quad$ has a physical meaning.

Answer 2

First one has to remark that the question, while connected to the differential equation of the motion under a central force field, is not itself a differential equation but the relation between Taylor coefficients of the expansion of two potential functions.

Central force field

$\newcommand{\fx}{{\mathfrak{x}}}\newcommand{\pd}{\partial}\newcommand{\D}{{\mathit\Delta}}$

The section that this exercise is from is about the motion in the plane under a central force field $$ \ddot\fx=-\frac{\pd U(|\fx|)}{\pd \fx}.\tag1 $$ Due to the rotational symmetry of this situation, the angular momentum is a conserved quantity, first integral, constant of motion. After fixing the angular momentum (here $M$ instead of the more usual $L$) $$M=r^2\dot \phi,\tag2$$ where $(r,\phi)$ are the polar coordinates of the plane point $\fx=(x,y)$, the dynamic for the radius is independent of the angle, it is $$ \ddot r=-V'(r)\text{ where }V(r)=\frac{M^2}{2r^2}+U(r).\tag3 $$

With that as back-story, compute the derivatives of $V$ \begin{align} V'(r) &= -\frac{M^2}{r^3}+U'(r)\tag{4a} \\ V''(r) &= \frac{3M^2}{r^4}+U''(r)\tag{4b} \end{align}

Perturbation of a circular orbit

The exercise in question now asks to find the angle $\Phi$ between pericenter and apocenter for an orbit close to a circular orbit of radius $r^*$ (and the same angular momentum).

Previously it was explained that the radius of such a circular orbit is at a (regular) minimum of $V$, so that $V'(r^*)=0$, $V''(r^*)>0$. The perturbed orbit has an energy $$\frac12\dot r^2+V(r)=E=V(r^*)+\D E$$ so that at the extremal points with $\dot r=0$ one gets in the Taylor expansions around $r^*$ the equation $$V(r^*)+\D E=V(r^*+ \D r)=V(r^*)+\tfrac12 V''(r^*)(\D r)^2+O((\D r)^3).$$ Simplifying and solving for the increment gives $$r_{\max},r_{\min}\approx r^*\pm\D r$$ with $$\D r=\sqrt{\frac{2\D E}{V''(r^*)}}. $$

Half-period

By the previous exercise, the half-period is computed as \begin{align} \Phi&=\int_{r_{\min}}^{r_{\max}}\frac{M/r^2}{\sqrt{2(E-V(r))}}dr\tag{5a} \\ &\approx\int_{r_{\min}}^{r_{\max}}\frac{M/r^2}{\sqrt{2(\D E-\frac12V''(r^*)(r-r^*)^2)}}dr \\ &=\int_{-1}^{1}\frac{M/(r^*+s\D r)^2}{\sqrt{V''(r^*)}\sqrt{1-s^2}}ds =\frac{M/(r^*)^2}{\sqrt{V''(r^*)}} \int_{-1}^{1}\frac1{\sqrt{1-s^2}}ds+O((\D r)^2) \\ &\approx\Phi_{\rm cir}=\pi\frac{M/(r^*)^2}{\sqrt{V''(r^*)}}\tag{5b} \end{align} Now use $V'(r^*)=0\implies U'(r^*)=\frac{M^2}{(r^*)^3}$ to eliminate the auxiliary variables $M$ and $V$ from this formula (not completely, as $r^*$ depends on $M$) to get $$ V''(r^*)=\frac{3}{r^*}U'(r^*)+U''(r^*) \\ \frac{M}{(r^*)^2}=\sqrt{\frac{U'(r^*)}{r^*}} \\ \implies \Phi_{\rm cir}=\pi\frac{M/(r^*)^2}{\sqrt{V''(r^*)}}=\pi\sqrt{\frac{U'(r^*)}{3U'(r^*)+r^*U''(r^*)}}.\tag6 $$