[vsnet-chat 4017] Re: Heliocentric Correction

[Fraser Farrell <fraser@trilobytes.com.au>]

On Sat, 20 Jan 2001, M.Martignoni wrote:
>Could be someone so kind to let me have the exact formula of the so called
>"eliocentric correction" to insert in the above-mentioned worksheet, knowing
>that I have available as starting informations G.J.D. and R.A./DECL. of the
>star?.

That's "Heliocentric correction". Here's the formula I use:

(...assorted brackets used for clarity...)

HJD = GJD + K*R*
     { cos(th)*cos(RA)*cos(Dec)
     + sin(th)*[ sin(e)*sin(Dec)
     + cos(e)*cos(Dec)*sin(RA) ] }

where:
   HJD = heliocentric JD
   GJD = geocentric JD
   K = light-time in days from Earth to Sun (~0.000578 days)
   R = distance to the Sun in astronomical units
   th = true longitude of the Sun in degrees
   e = obliquity of the ecliptic (23d26'21.448" for J2000)
   RA, Dec = coordinates of star expressed as degrees


Most of the above quantities need to be calculated for the
actual time of observation. The following methods are
described in Jean Meeus' book "Astronomical Algorithms"
(published by Willmann-Bell). An excellent book; I recommend
it highly...anyway, on to the arithmetic:


(1) The time T (Julian centuries since J2000) is used in
many of these formulae:

   T = (JDE-2451545)/36525

where JDE is the Julian Ephemeris Day. JDE = JD+deltaT ,the
value of deltaT changes due to irregularities in the Earth's
rotation. I think it was about 67 seconds in 2000.

Note - you should calculate T to at least +/- 0.000001
precision.


(2) The mean obliquity of the ecliptic is given by:

  e = 23d26'21.448"
      - T * (46.815" - T*(0.00059" + T*0.001813"))


(3) The Sun's longitude needs a bit more work. First of all,
calculate the mean longitude (Lo) and mean anomaly (M) in
degrees:

Lo = 280.46645 + T*(36000.76983 + T*0.0003032))

M = 357.5291 + T*(35999.0503 - T*(0.0001559 + T*0.00000048))

  then calculate the eccentricity of the Earth's orbit:

Ecc = 0.016708617 - T*(0.000042037 + T*0.0000001236)

  then find the Sun's equation of centre in degrees:

C = sin(M)*((1.9146- T*(0.004817+ T*0.000014)))
    + sin(2*M)*((0.019993 - T*0.000101))
    + sin(3*M)*0.00029

The Sun's true longitude in degrees = Lo + C

The Sun's true anomaly = M + C


(4) Then we can calculate R (the distance from Earth to
Sun):

   R = 1.000001018 * [ (1 - Ecc*Ecc)/(1 + Ecc*cos(true
anomaly)) ]


After you get all those numbers for the time of observation,
feed them back into the heliocentric correction equation.
Repeat for every single observation. Aren't you glad we have
computers? ;-)

If you want to be really precise, then you also need to
consider the Earth's motion about the Earth-Moon barycentre
(I think we discussed this in vsnet-chat last year??),
perturbations by the planets....but for most of us these
formulae will be adequate. Unless you're timing pulsars.

I leave the translation into your favourite spreadsheet's
formulae as an exercise for the reader ;-)


cheers,
Fraser Farrell

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