The full power series for the Schwarzschild portion of perihelion shift is given in Mathpages as:(adsbygoogle = window.adsbygoogle || []).push({});

where [itex]L = a(1-\epsilon^2)[/itex], [itex]a[/itex] the semi-minor axis and [itex]\epsilon[/itex] the eccentricity. This implies that as [itex]\epsilon[/itex] tends to zero, the perihelion shift tends to a non-vanishing [itex]6\pi m/a[/itex] + some much smaller higher order terms.

If we do an angular acceleration derivation for a Schwarzschild orbit ([itex]d^2\phi/dt^2[/itex]), I have arrived at

[tex]\frac{d^2\phi}{dt^2} = \frac{-2v_r v_\phi}{r} \frac{r-3m}{r-2m}[/tex]

where [itex]v_r=dr/dt[/itex] and [itex]v_\phi=d\phi/dt[/itex] (geometric units: c=G=1)

Pervect and CarlB have arrived at equivalent results in this old orbital acceleration thread.

My question is that since [itex]d^2\phi/dt^2[/itex] goes to zero when [itex]v_r[/itex] goes to zero for the circular orbit, how is this compatible with the non-vanishing perihelion shift above?

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# Perihelion shift in (very) near-circular orbits

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