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Tensors and Relativity: Chapter 8

Solution for timelike orbits and precession

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The solution to the Newtonian equation give the well known conics

equation1076

where tex2html_wrap_inline1320 and e is the eccentricity of the orbit [ if e=0 it is circular ].

We can obtain an approximate solution tex2html_wrap_inline1326 to the exact orbit equation [ valid for tex2html_wrap_inline1328 ], if we substitute the Newtonian solution into the term quadratic in U, that is if we solve

equation1078

where we have put tex2html_wrap_inline1307 [ for timelike orbits ]. This equation is solved by choosing a particular integral of the form [ Assignment 7 ]:

equation1080

This gives

equation1082

The most important term is the one that is linear in tex2html_wrap_inline1334 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 tex2html_wrap_inline1336 and obtain

equation1084

Now

equation1086

But we have tex2html_wrap_inline1338 so tex2html_wrap_inline1340 , and this simplifies our solution to

equation1088

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 tex2html_wrap_inline1344 . The point at which the orbit is closest to the sun is reached again only after an additional rotation through the angle

eqnarray1090

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