Hall's solution for Mercury's precession

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SUMMARY

Hall's solution for Mercury's precession demonstrates that a slight increase in the radius exponent in Newton's equation influences orbital precession. Specifically, Hall presents the formula \theta = {\pi \over {\sqrt{n+3}}}, where \theta represents the angle between the minimum and maximum radius vector for orbits with small eccentricity, and n is the exponent in the equations of motion. When n equals -2, the result aligns with Newton's law of gravitation. This insight provides a mathematical framework for understanding deviations from classical mechanics in celestial mechanics.

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Shaw
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Hall showed that a slight increase in the radius exponent in Newton's equation would cause precession. What would the result be if the exponent was slightly less than 2.
 
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FWIW:
http://books.google.com/books?id=eQY-AQAAMAAJ&pg=RA1-PA49&cad=2#v=onepage&q&f=false

In which Hall gives a formula, attributed to Bertrand,

[itex]\theta = {\pi \over {\sqrt{n+3}}}[/itex]

where [itex]\theta[/itex] is the angle between minimum and maximum radius vector for an orbit of small eccentricity, and [itex]n[/itex] is the exponent to be used in the equations of motion. He says, "If [itex]n = -2[/itex] we have the Newtonian law." So I imagine you can figure it out from there.
 
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Thanks for this. I must be thick, but I'm still a bit puzzled.
 

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