Trigonometry related to Kepler's Equation

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Discussion Overview

The discussion revolves around the derivation of Kepler's Equation, specifically focusing on the trigonometric relationships involved in calculating the true anomaly (nu). Participants express curiosity about the transition between specific equations in the derivation process and seek clarification on the use of trigonometric identities.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes the use of trigonometry in deriving Kepler's Equation and questions how one equation transitions into another, specifically regarding the true anomaly (nu).
  • Another participant suggests that writing down the equations in LaTeX would facilitate checking the notation and providing answers.
  • A third participant shares a link to a resource that contains the relevant derivation they were looking for.
  • One participant challenges the use of a specific equation as the source, proposing an alternative approach involving the computation of trigonometric functions related to nu.
  • There is a request for clarification regarding which source is being referred to, indicating some confusion about the materials being discussed.
  • A participant clarifies that they were referring to an attachment in a previous post, expressing satisfaction with the longer attachment provided.
  • Areas of Agreement / Disagreement

    Participants express differing views on which equations or sources should be used in the derivation, indicating that multiple competing perspectives remain unresolved.

    Contextual Notes

    There are references to specific equations and trigonometric identities, but the discussion lacks clarity on the assumptions and definitions being used, which may affect the understanding of the derivation process.

solarblast
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In deriving Kepler's Equation, there's a little trig work that gets at nu (see figure), which is the true anomaly. I'm curious how the third equation makes it into the 4th (last) equation. It's been a very long time since I've played with trig. nu of course is the atan of the last equation.
 

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solarblast said:
In deriving Kepler's Equation, there's a little trig work that gets at nu (see figure), which is the true anomaly. I'm curious how the third equation makes it into the 4th (last) equation. It's been a very long time since I've played with trig. nu of course is the atan of the last equation.


You should write down, with LaTeX, directly the equations here so that it'll be possible to check the notation and give, eventually, and answer, otherwise

it is way too cumbersome.

DonAntonio
 
See <http://rip94550.wordpress.com/2011/05/02/elliptical-orbits-%E2%80%93-deriving-keplers-equation/>. Attached is the part I was looking for.
 

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Last edited by a moderator:
You're using the wrong equation as the source. Use the first two. From the second, compute [itex]1+\cos\nu[/itex]. Next compute [itex]sin\nu/(1+\cos\nu)[/itex]. This is just [itex]\tan(\nu/2)[/itex] per the half angle formula. Use the same formula to simplify the right hand side.
 
What source are you referring to? The link or one of the equations in the attachment? The link is 10 pages long.
 
Last edited:
I'm referring to the attachment in your first post.
 
Ah, that was produced by a CM "pdf book" on the web. I'm satisfied with the longer attachment. I may contact the author of the pdf.
 

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