Solution for timelike orbits and precession

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The solution to the Newtonian equation give the well known conics

where and e is the eccentricity of the orbit [ if e=0 it is circular ].

We can obtain an approximate solution to the exact orbit equation [ valid for ], if we substitute the Newtonian solution into the term quadratic in U, that is if we solve

where we have put [ for timelike orbits ]. This equation is solved by choosing a particular integral of the form [ Assignment 7 ]:

This gives

The most important term is the one that is linear in because it is the only one which in the course of time [ with many revolutions of the planet ] becomes larger and larger. We therefore ignore the other corrections to and obtain

Now

But we have so , and this simplifies our solution to

The orbit of the planet is thus only approximately an ellipse. The solution for U is still a periodic function, but no longer with a period . The point at which the orbit is closest to the sun is reached again only after an additional rotation through the angle

This is the famous perihelion precession  of planetary orbits. For Mercury we obtain 43.03'' per century.