Quote:
Originally Posted by Ram Iyer
for example, Saturn on 20 Mar will have RA 23 48 36 and DEC -02 01 00
how can one use this data to locate the planet for example at 9pm Melbourne
There is no problem locating the planet or the constellation (by star chart or azimuth/Latitude) but the query is about how to use declination data
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Forgive me but I believe your co-ordinates are in error. The co-ordinates for Saturn on the evening of Saturday, March 20, 2010 are RA=12h 09m and DEC=+1deg 47m (source SkyTools3Pro & Stellarium).
Look at declination this way: face due south and look at the south celestial pole (SCP). It is a point in the sky 37deg 50m above the south horizon (latitude of Melbourne). Point to this location with one arm. The celestial equator (declination zero degrees) is 90 deg away from the SCP due north. (90-38=52deg above the north horizon) The celestial equator arcs east to west through a point 52deg above the north horizon. Because Saturn has a DEC of +1deg 47min it is close to zero which means it will rise ever so slightly north of due east. However, due to local horizon variations, ie mountains and trees you may not see it for a good half hour after that slightly more north of east and a little higher.
THE RA is the longitude of a celestial object and to work out the position in the sky you need another parameter and that is Local Sidereal Time (LST). If you have a star chart with co-ordinates you can plot the position of Saturn on the chart and you see what is nearby and match it up with what is in the sky. The LST comes in handy to work out where a given RA is. Stated simply the LST = the RA of an object on the north south axis from the northern horizon to the SCP. The current LST = 19h 00m at 8h 13m Australian Eastern Daylight Time (AEDT). In 5 hours it will be 00h 00m LST and 1:13pm here. Another 8 hours approx (9:13pm) and the LST will be 8h 00m. Saturn is 12h 09m so it is 4h 09m behind. ie. at 1:22am it will be due north and 50 degrees high (50-1d47m approx).
Hope this helps.