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The full power series for the Schwarzschild portion of perihelion shift is given in Mathpages as:
where L = a(1-\epsilon^2), a the semi-minor axis and \epsilon the eccentricity. This implies that as \epsilon tends to zero, the perihelion shift tends to a non-vanishing 6\pi m/a + some much smaller higher order terms.
If we do an angular acceleration derivation for a Schwarzschild orbit (d^2\phi/dt^2), I have arrived at
\frac{d^2\phi}{dt^2} = \frac{-2v_r v_\phi}{r} \frac{r-3m}{r-2m}
where v_r=dr/dt and v_\phi=d\phi/dt (geometric units: c=G=1)
Pervect and CarlB have arrived at equivalent results in this old orbital acceleration thread.
My question is that since d^2\phi/dt^2 goes to zero when v_r goes to zero for the circular orbit, how is this compatible with the non-vanishing perihelion shift above?
where L = a(1-\epsilon^2), a the semi-minor axis and \epsilon the eccentricity. This implies that as \epsilon tends to zero, the perihelion shift tends to a non-vanishing 6\pi m/a + some much smaller higher order terms.
If we do an angular acceleration derivation for a Schwarzschild orbit (d^2\phi/dt^2), I have arrived at
\frac{d^2\phi}{dt^2} = \frac{-2v_r v_\phi}{r} \frac{r-3m}{r-2m}
where v_r=dr/dt and v_\phi=d\phi/dt (geometric units: c=G=1)
Pervect and CarlB have arrived at equivalent results in this old orbital acceleration thread.
My question is that since d^2\phi/dt^2 goes to zero when v_r goes to zero for the circular orbit, how is this compatible with the non-vanishing perihelion shift above?
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