How can you calculate how large the perihelion shift should be under Schwarzschild conditions in general relativity for minimally disturbed circular orbits in the strong field limit?

I think there is a functioning expression that works well in the weak fields of our solar system.

What would the perihelion shift be for a minimally disturbed circular orbit just above the minimum stable circular orbit at $r=6GM/c^2$ and, for instance, at $r=9GM/c^2$, $r=12GM/c^2$ and $r=15GM/c^2$ ?


1 Answer 1


The periapsis precession for a circular geodesic with radius $r$ in the Schwarzschild metric is

$$ 2 \pi \left( \sqrt{\frac{r}{r-6M}}-1\right)$$

The answer for a generic bound geodesic with (dimensionless) semilatus rectum $p$ and eccentricity $e$ is:

$$ 4\sqrt{\frac{p}{p-6+2e}}K\left(\sqrt{\frac{4e}{p-6+2e}}\right)-2\pi,$$ where $K$ is the complete elliptic integral of the first kind.

These formula's are, of course, assuming that the orbit is a geodesic and therefore that the orbiting body is a test particle whose own gravitational influence can be neglected.

Calculation of corrections due to the mass of the orbiting body can be found here and here.

  • $\begingroup$ Thank you. This is basically true for all r > 6M and not just a weak field approximation? $\endgroup$
    – Agerhell
    Nov 16, 2023 at 22:09
  • 1
    $\begingroup$ @Agerhell - Since a truly circular orbit doesn't have a periapsis it must be an approximation for small ellipticities. You can calculate it numerically as I did here, but I think there was an analytical solution as well, if I find or remember it I'll come back at it. I think it might work if you replace r with r_max or r_min or something like that, but don't quote me on that (yet). $\endgroup$
    – Yukterez
    Nov 16, 2023 at 22:46
  • $\begingroup$ @Yukterez This is the exact circular orbit limit of the periapsis shift for Schwarzschild geodesics $\endgroup$
    – TimRias
    Nov 17, 2023 at 7:27

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