Einstein and Gravitation — Question 8

Question 8

Relativistic periapsis advance accumulates over many orbits

For a weak-field bound orbit, use the supplied leading relativistic advance per orbit $\Delta\varpi=6\pi GM/[a(1-e^2)c^2]$. Consider $GM/(ac^2)=10^{-8}$ and $e=.60$. Neglect all other sources of precession.

1. Find the advance per orbit and after 100 orbits, using $1\,\mathrm{rad}=206265$ arcseconds.

2. Compare with the small-eccentricity limit at the same $a$, and explain what differs from the ideal Newtonian two-body ellipse.

Question 8 - Solution

Relativistic periapsis advance accumulates over many orbits

(a) Derivation and calculation

With $1-e^2=1-.60^2=.64$, $$\boxed{\Delta\varpi=\frac{6\pi(10^{-8})}{.64} =2.94524\times10^{-7}\ \mathrm{rad/orbit}}.$$ Accumulating 100 identical leading-order advances, $$\Delta\varpi_{100}=100\Delta\varpi=2.94524\times10^{-5}\ \mathrm{rad},$$ $$\boxed{\Delta\varpi_{100}=2.94524\times10^{-5}(206265) =6.07501\ \mathrm{arcsec}}.$$

(b) Result and interpretation

At fixed $a$, relative to the small-eccentricity limit, $$\boxed{\frac{\Delta\varpi(e=.60)}{\lim_{e\to0}\Delta\varpi} =\frac1{1-e^2}=1.5625}.$$ The ideal Newtonian two-body ellipse has a fixed apsis direction; here $$\varpi_N\simeq\varpi_0+N\Delta\varpi.$$ A truly circular orbit has no unique periapsis to track, so the comparison is a small-eccentricity limit.

Physical check

The angle per orbit and total accumulated angle are both small, and other precession sources were explicitly excluded.