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How do I map a celestial object's declination ?

  1. Mar 22, 2010 #1
    to my local meridian at various times of the year. An object's declination will be fixed - relatively speaking - but its altitude on my local horizon varies from season to season because of earth's movement, doesn't it?
    Also, if I'm at a lat. 47.3 N shouldn't celestial objects w/dec. of -{90 - (Lat. + 22.5*)} be, at some given time, on my local horizon?
  2. jcsd
  3. Mar 22, 2010 #2


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    Yes, only stars with a declination greater than 90-lat are visible all year = circumpolar stars
  4. Mar 22, 2010 #3

    Filip Larsen

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    You can imagine that circles of equal latitude is projected into the sky and become circles of equal declination. The great circle of equator, for instance, corresponds to the great circle in the sky with declination zero; the small circle of latitude 45 north corresponds to the small circle with declination +45. So if you are a latitude 45 north, the circle of declination +45 will pass through your zenith (directly overhead), the zero declination circle (equator) will be at maximum elevation of 90-latitude+declination = 90+45+0 = 45 deg directly south (at the local meridian), and the -45 declination circle with just touch your horizon from below at elevation 90-45+(-45) = 0 deg. As you can see, each circle for a fixed declination remain fixed in your sky at all times at a maximum elevation determined by your latitude.

    An object with a certain declination and right ascension will now be at a certain point on its declination circle in the sky and so from your point of view it will describe a full traversal of that declination circle in the course of a single sidereal day. Since we measure the length of our day as solar day which are a bit longer than the sidereal day, each solar day the object will describe a full circle plus a tiny bit more so that after a full year all the tiny extra bits add up to a full circle and we can start all over again. It is this difference in solar and sidereal day that makes an object with a fixed RA and declination to culminate at different times during a year.

    So, to map a celestial object to the instantaneous elevation at your latitude you would need to know both RA and declination (and time), but if you are only interested in the maximum elevation then that is given as mentioned above (i.e. 90-latitude+declination).

    At a given latitude, objects with a declination of latitude-90 or above will have elevation at zero or more during the year. Note this has nothing to do with the ecliptic.
  5. Mar 23, 2010 #4
    OK. I'm a little fuzzy about the plane of the "elliptic" and the celestial equator, but, if I calculated correctly, NGC 4594 RA12h39m Dec-11*37' should be @ ~38 alt. due south @ ~13.30pm EST for Lat. 47.3N. Right?
  6. Mar 23, 2010 #5

    Filip Larsen

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    No, because the maximum elevation for a celestial object at that declination is 90-47.3+(-11.6) = 31,1 deg which is less than the 38 deg you mention.

    Normally one would in practice use an astronomy program, spread sheet or similar to transform equatorial coordinates RA and Dec into horizontal coordinates elevation and altzimuth for a given geographic latitude and longitude and time. You would also normally need to know the longitude of the observer to make this transformation (I assume you do know that and just forgot to include it in your post).
  7. Mar 24, 2010 #6
    Right; the ~38* was an arithmatic error. But to calculate the time the object appears on my local meridian, I used the local sidereal time for my longitude. Is this correct? Also, is there an equation to calculate an estimate of your LST from your local standard time?
  8. Mar 24, 2010 #7

    Filip Larsen

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    Montenbruck [1] gives the mean sidereal time at Greenwich in hours as

    GMST = 6.656306 + 0.0657098242 * (JD0 - 2445700.5) + 1.0027379093 * UT

    where JD0 is the Julian date for the date in question at 0 UT and UT is universal time in hours (which is within 0.9 sec of UTC). Once you have GMST you can get the local mean sidereal time in hours as

    LMST = GMST - longitude * (1 h / 15 deg).

    If you want more precision, Montenbruck gives a higher order approximation in [2]. Unfortunately he does not seem to quote the accuracy of the two approximations (the one above and the one in [2]) so it can't tell you how accurate they are. If you do not require any particular accuracy, approximations like that on http://en.wikipedia.org/wiki/Sidereal_time will probably be sufficient.

    [1] Practical Ephemeris Calculations, Montenbruck, Springer 1989.
    [2] Astronomy on the Personal Computer 4th, Montenbruck, Pfleger, Springer 2000.
  9. Mar 27, 2010 #8

    Filip Larsen

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    Let me emphasise, that JD0 is the Julian date for the same date, but at 0 UT. Thus, if you have an arbitrary Julian date JD you can get JD0 by using a calculation like JD0 = floor(JD-0.5)+0.5. Alternatively you can insert the above rounding into the original formula and get

    GMST = 6.656306 + 0.0657098242 * floor(JD - 2445700.5) + 1.0027379093 * UT
    Last edited: Mar 27, 2010
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