Physics Help Forum Mercury's Orbit
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 Dec 17th 2017, 07:10 AM #1 Junior Member   Join Date: Dec 2017 Posts: 22 Mercury's Orbit I have been trying to understand Mercury's orbit and find statements to the effect that its orbit can't be defined by a single equation. If it isn't an ellipse, then what is it?
Dec 17th 2017, 07:27 AM   #2
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 Originally Posted by wanderer I have been trying to understand Mercury's orbit and find statements to the effect that its orbit can't be defined by a single equation. If it isn't an ellipse, then what is it?
Those statements while correct can be misleading. There's no simple word to name the orbit. You can think of it as an ellipse that precesses.

See: Precession of the perihelion of Mercury

at

Precession of the perihelion of Mercury

 Dec 17th 2017, 08:38 AM #3 Junior Member   Join Date: Dec 2017 Posts: 22 Mercury's orbit Thank you Pmb. I have looked at several sources of information regarding this subject and found 'space.com/36-mercury-the-suns-closest-planetary-neighbor.html' to be the best and most complete. I concluded that if the classical effects of all other solar system bodies, except the Sun and Mercury, were removed from Mercury's orbital equation, then Mercury's orbit would be a perfect ellipse with the equation x^2/A^2 + y^2/B^2 = 1 Where A and B are ellipse Parameters and 2*pi*B = ellipse perimeter length. And that equation could be used to determine the amount relativistic effects contribute to the precession of Mercury's orbit. Am I on the right path? Last edited by wanderer; Dec 17th 2017 at 08:41 AM.
Dec 17th 2017, 08:43 AM   #4
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 Originally Posted by wanderer Thank you Pmb. I have looked at several sources of information regarding this subject and found 'space.com/36-mercury-the-suns-closest-planetary-neighbor.html' to be the best and most complete. I concluded that if the classical effects of all other solar system bodies, except the Sun and Mercury, were removed from Mercury's orbital equation, then Mercury's orbit would be a perfect ellipse with the equation x^2/A^2 + y^2/B^2 = 1 Where A and B are ellipse Parameters and 2*pi*B = ellipse perimeter length. And that equation could be used to determine the amount relativistic effects contribute to the precession of Mercury's orbit. Am I on the right path?
No. You are on the wrong path. This is a general relativistic effect and you'd need general relativity to solve it. The equation of an ellipse that you posted cannot be used to find the precession of Mercury.

Please understand that without knowing what your math skills are I'm not able to help you to the best of my ability. Plus if you really want to understand this you should learn general relativity. Not an easy task to say the least but well worth the effort.

 Dec 17th 2017, 03:29 PM #5 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,071 As I understand the history of this; The basic Newtonian equations for the Sun and Mercury alone will give an elliptical orbit. With a bit of extra effort the small perturbations from the gravitational effects of the other planets can be added, still using the Newtonian equations. It was expected that this would give what should have been a perfect description of the orbit. However after completing the calculations(and checking, and checking again, and again), it was found that it did not give a perfect description. An extra planet (Vulcan) was hypothesised, which orbited between Mercury and the Sun, to explain these discrepancies, but was (of course) never found. It was not until Einstein's theory of relativity provided an alternative mathematical model (to Newton's) that the discrepancies could be successfully explained. __________________ ~\o/~ Last edited by Woody; Dec 17th 2017 at 03:32 PM.
 Dec 17th 2017, 04:31 PM #6 Junior Member   Join Date: Dec 2017 Posts: 22 Mercury's orbit Pmb and Woody, apparently i have been too vague about what i want to do. Obviously the ellipse equation i posted will not directly produce the 43 arc-seconds per century for the precession of Mercury's orbit attributed to GR. However, the posted equation can be used to describe Mercury's orbit in the absence of all Solar system bodies except the Sun and Mercury. Then from that base, the changes due only to GR can be determined. I need to know 43" from what? Regards; wanderer.
 Dec 18th 2017, 11:25 AM #7 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 I presume its the angle of "rotation" to the previous elliptical orbit (but I'm not an expert in this). __________________ Burn those raisin muffins. Burn 'em all I say.
Dec 18th 2017, 01:16 PM   #8
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 Originally Posted by wanderer Pmb and Woody, apparently i have been too vague about what i want to do. Obviously the ellipse equation i posted will not directly produce the 43 arc-seconds per century for the precession of Mercury's orbit attributed to GR. However, the posted equation can be used to describe Mercury's orbit in the absence of all Solar system bodies except the Sun and Mercury. Then from that base, the changes due only to GR can be determined. I need to know 43" from what? Regards; wanderer.
??? 43 arc-sec from where the axis of the ellipse was the previous year. That is what 'precession' means.

 Dec 18th 2017, 02:43 PM #9 Junior Member   Join Date: Dec 2017 Posts: 22 Thanks kiwiheritec. What i meant when I asked "43 arc-seconds from what" I was pointing out that the value of any GR impacted orbit parameter would be meaningless unless the same parameter not impacted by GR was known. The ellipse equation I posted was meant to be to establish mercury's orbital baseline before it is impacted by GR. Of course I will need to plug in some actual numbers for Mercuty's orbit first. Also, the 43 arc-seconds is the amount of precession attributed to GR, is the amount accumulated over 100 years. Regards; wanderer Last edited by wanderer; Dec 18th 2017 at 02:55 PM. Reason: add a sentence
 Dec 18th 2017, 03:18 PM #10 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,071 43 arc-seconds in 100 years is the best part of not a lot. Newtons equations are fine for most purposes, it is only when very strong gravitational fields are involved, or when extreme precision is necessary, that we have to resort to Einstein. Mercury is in a fairly strong gravitational field, being close to the Sun, but even so it took extremely precise observations to detect the anomaly. __________________ ~\o/~

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