Discover the JNow RA and Dec of L4 and L5 in the Sun/Earth System

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SUMMARY

The JNow Right Ascension (RA) and Declination (Dec) of the L4 and L5 points in the Sun/Earth system are determined by their positions relative to the Sun, forming equilateral triangles with the Earth. Specifically, L4 is located 60 degrees East and L5 60 degrees West of the Sun along the plane of the ecliptic. To calculate their exact RA and Dec, one must start with the Sun's coordinates, which can be obtained from the JPL Horizons website, and apply trigonometric principles to find the positions of L4 and L5. This calculation assumes a circular orbit of the Earth, which is an approximation.

PREREQUISITES
  • Understanding of celestial mechanics and Lagrangian points
  • Familiarity with Right Ascension (RA) and Declination (Dec) coordinates
  • Basic trigonometry for angular calculations
  • Knowledge of the JPL Horizons system for astronomical data
NEXT STEPS
  • Research how to calculate celestial coordinates using trigonometric methods
  • Explore the JPL Horizons website for real-time astronomical data
  • Study the properties and applications of Lagrangian points in orbital mechanics
  • Learn about the elliptical nature of planetary orbits and its effects on calculations
USEFUL FOR

Astronomers, astrophysicists, and students of celestial mechanics who are interested in the dynamics of the Sun/Earth system and the calculation of Lagrangian points.

Valareos
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Does anyone know the JNow Ra and Dec of the L4 and L5 points in the Sun/Earth System?
 
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Hi,
interesting question, can't find any references on the web, so here's a couple of thoughts...

My understanding of the Earth/Sun L4 & L5 points is that they lie on the Earths orbit with the lines linking them to the Sun making 60° angles with the Earth-Sun line. So Earth/Sun/L4 and Earth/Sun/L5 are equilateral triangles (when viewed from above the plane of the ecliptic). So, L4 must be 60 degrees East of the Sun and L5 60 degrees West, along the plane of the ecliptic. So if you know the RA and Dec of the Sun, you've got a good start! (Have a look at JPL Horizons website,that should give the Sun RA and Dec for given time). Then the question becomes one of trigonometry...how to calculate the position of a point 60 degrees East(or West) of the Sun along the plane of the Ecliptic which is inclined at 23.45 degrees to the celestial sphere equator.
My trigs not that good, unfortunately! Anyone?...

NB This assumes the Earth's orbit to be circular, which it isn't, so this is only an aproximation, I guess...

Good luck,
BrianJ.
 
Lagrangian points are actually ellipses around the Sun. They are always over an imaginare line from the Sun to our planet. ¿Can have RA and DEC, those imaginary orbits? Difficult to calc...
 

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