Consider the motion of a closed orbit in a central force field ${\bf F}({\bf r})$ other than the inverse square force i.e., $${\bf F}({\bf r})\neq -\frac{k}{r^2}\hat{\bf r}.$$

However, since ${\bf F}({\bf r})$ is still central, the total energy $E$ and the angular momentum vector ${\bf L}$ are conserved. The magnitude of the total energy $$E=\frac{\bf p^2}{2m}+V(r)$$ and both the magnitude and direction of ${\bf L}$ is fixated by the initial conditions ${\bf r}_0, {\bf p}_0$. Now let us look at the following picture:

enter image description here

From the figure above, it appears that the particle reaches the same point again and again (say, the common point where the orbits meet in the figure), and each time the particle returns to this location (say, ${\bf r}_0$), it must return with a different momentum ${\bf p}\neq {\bf p}_0$, for it to follow a different trajectory. This is because for orbits to differ (i.e. precession to occur) the initial condition must also differ in some way. But because of the conservation of ${\bf L}$, if ${\bf r}_0$ is reached, shouldn't the momentum ${\bf p}$ also reach ${\bf p}_0$? If so, that would prevent the orbits to differ i.e. precession to occur. What is the resolution to this apparent paradox?

  • $\begingroup$ $\mathbf L$ is definitely conserved during whole quasi-orbit, and its closedness and ellipticity are not a necessary condition for that. ($\mathbf r_0$ may move since its just a mathematical point with no mass.) Notoriously, this perihelion shift was observed with Mercury orbit about a century ago, although its actual reason is different from your example. $\endgroup$
    – dominecf
    Dec 30 '20 at 8:36
  • $\begingroup$ Do you agree that if ${\bf r}$ returns to ${\bf r}_0$, ${\bf p}$ must also return to ${\bf p}_0$ in order to conserve ${\bf L}$? @dominecf $\endgroup$
    – SRS
    Dec 30 '20 at 8:49
  • $\begingroup$ Without any meticulous check, I guess your assumption only holds for $F\propto r^{-2}$ and $F\propto r^{1}$ in Eucleidian space, i.e. the cases where closed elliptical orbits are always observed. $\endgroup$
    – dominecf
    Dec 30 '20 at 8:51
  • $\begingroup$ Your figure is wrong. You have drawn the orbits rotating about the perihelion point i.e. that point remains fixed, and this is not the case. The point of perihelion rotates so it is (approximately) opposite to the point of aphelion. $\endgroup$ Dec 30 '20 at 8:57
  • $\begingroup$ Is this a better representation? en.wikipedia.org/wiki/Rosetta_orbit @JohnRennie $\endgroup$
    – SRS
    Dec 30 '20 at 9:01

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