Eccentricity & GR precession; Mercury vs. GPB

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SUMMARY

The discussion centers on the relationship between the eccentricity of Mercury's orbit and its gravitational precession, specifically the 43 ArcSec/century observed. Gravity Probe B (GP-B), which has an eccentricity of zero, confirmed General Relativity (GR) precession, while Mercury's eccentricity of 0.205 significantly contributes to its noticeable precession. The precession angle per orbit can be approximated by the formula 6π(Gm/c²)/(a(1-e²)), indicating that for fixed semi-major axes, precession is inversely proportional to the eccentricity. The conversation also explores hypothetical scenarios of varying eccentricities and their effects on precession.

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  • Understanding of General Relativity (GR) principles
  • Familiarity with orbital mechanics and eccentricity
  • Knowledge of gravitational precession concepts
  • Basic proficiency in mathematical expressions related to physics
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  • Research the mathematical derivation of the precession angle formula in General Relativity
  • Explore the implications of eccentricity on orbital dynamics in celestial mechanics
  • Study the results and methodologies of the Gravity Probe B experiment
  • Investigate the effects of varying eccentricities on gravitational phenomena
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Astronomers, physicists, and students of gravitational physics who are interested in the intricacies of orbital mechanics and the effects of eccentricity on gravitational precession.

RandallB
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Eccentricity & GR precession; Mercury vs. GP-B

How big a factor is the Eccentricity of Mercury orbit in contributing to its precession?

GR precession has been confirmed in the Gravity Probe B orbit even though it has an eccentricity of zero. The GP-B used a gyroscope to take the place of a detectable aphelion, while Earth has a nearly circular orbit with an eccentricity of 0.016 it gives measurable aphelion that allow modern observations the ability to see the small precession undetectably to 19 century astronomers.

Is it the large orbit eccentricity of 0.205 that gives Mercury such a large and easier to notice precession of 43 ArcSec/century?
I assume by moving up and down between different levels of GR curved space the effect on precession is magnified.

Does anyone have a reference to show or explain what how or what part of the GR formulas account for the eccentricity contribution?

If Mercury orbit was a more circular eccentricity of 0.01 what would the precession be?

If eccentricity is doubled to 0.41 is there some function to describe how much the precession would increase?
 
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RandallB said:
How big a factor is the Eccentricity of Mercury orbit in contributing to its precession?

GR precession has been confirmed in the Gravity Probe B orbit even though it has an eccentricity of zero. The GP-B used a gyroscope to take the place of a detectable aphelion, while Earth has a nearly circular orbit with an eccentricity of 0.016 it gives measurable aphelion that allow modern observations the ability to see the small precession undetectably to 19 century astronomers.

Is it the large orbit eccentricity of 0.205 that gives Mercury such a large and easier to notice precession of 43 ArcSec/century?
I assume by moving up and down between different levels of GR curved space the effect on precession is magnified.

Does anyone have a reference to show or explain what how or what part of the GR formulas account for the eccentricity contribution?

If Mercury orbit was a more circular eccentricity of 0.01 what would the precession be?

If eccentricity is doubled to 0.41 is there some function to describe how much the precession would increase?

As far as I know (backed up by a quick look at Rindler "Essential Relativity"):

For a given orbital angular momentum, GR precession is not affected by the eccentricity. However, if you consider alternative orbits with varying eccentricity which have fixed semi-major-axis a instead of a fixed angular momentum, the precession is proportional to 1/a(1-e2).

The general (approximate) expression for the precession angle per orbit is:

6 pi (Gm/c^2)/(a(1-e2)).

(I tried to show the general expression in TeX but that seems to be broken at the moment, in that preview processing showed something completely different).
 


Jonathan Scott;1968996The general (approximate) expression for the precession angle per orbit is: 6 pi (Gm/c^2)/([I said:
a[/I](1-e2)).

Thought there was a pi^3 in the precession angle formula; or is that the relativistic correction to this...
 

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