dimario@ihlpg.UUCP (Michael J. DiMario) (08/23/85)
I am looking for a clear explanation of determining the R.A. of an object given its R.A. at 0h UT for a meridian at a given time. I seem to have difficulties w/ this in the field. For example, given that Jupiter is at R.A. 20 53.0 on Aug 16th at 0h UT, what is its R.A. at 88 degrees W. longitude and 2200 hours CDT locally? Is this the Hour Angle and if so, how do you use it? Is there anyone that can advise me on this? Thank you in advance. mike dimario ihnp4!ihlpg!dimario 312/979-5397
anita@utastro.UUCP (Anita Cochran) (08/27/85)
> I am looking for a clear explanation of determining the R.A. of an > object given its R.A. at 0h UT for a meridian at a given time. I seem to > have difficulties w/ this in the field. For example, given that Jupiter > is at R.A. 20 53.0 on Aug 16th at 0h UT, what is its R.A. at 88 degrees > W. longitude and 2200 hours CDT locally? Is this the Hour Angle > and if so, how do you use it? Is there anyone that can advise me on this? > > Thank you in advance. > > mike dimario ihnp4!ihlpg!dimario > 312/979-5397 I think a bit of confusion exists in the questioner's mind. The right ascension (RA) is based on a celestial globe, i.e. it is the east-west position of a celestial object, measured eastwards in hours (1 hour=15 degrees) from the place where the sun crosses the celestial equator heading north (the vernal equinox). Thus, except for a planetary object (which moves relative to the background stars), the RA of an object does not change with time (well, technically, with a large enough time it changes due to proper motion, etc). What I think that Mike is trying to find out is the hour angle, i.e. how far off the meridian an object is. What you really need to know with changing longitude is the local sidereal time (LST). The sidereal time is the time based on one complete rotation of the earth on its axis with respect to the fixed stars. The normal (solar) day is the time it takes the earth to make 1 complete rotation on its axis with respect to the sun but since we are in orbit around the sun, this is not the same as the sidereal day (1 sidereal day = 23h 56m). There is not just one sidereal time since we have broken the earth into time zones and the sun passes overhead at different times within 1 time zone. The local sidereal time tells us what RA an object on the meridian (a line going through the zenith, or overhead point, down to both the north and south horizons at the same longitude) has at that moment. Thus, in Mike's example, Jupiter would be on the meridian when the LST was 20 53.0. The hour angle (HA) is then just the LST-RA. So if the LST is 18 53.0, Jupiter would have an HA of -2 hours. What does this mean? Well HA is measured with 0 hours on the meridian and negative HA towards the east, positive towards the west. Recalling that 1 hour=15 degrees, an HA of -2 hours means that Jupiter will be 30 degrees east of the meridian. So knowing the LST and the RA, you know where in the sky to look. -- Anita Cochran uucp: {noao, ut-sally, ut-ngp}!utastro!anita or seismo!ut-sally!utastro!anita arpa: anita%utastro@UTEXAS.ARPA snail: Astronomy Department The University of Texas at Austin Austin, TX 78712 at&t: (512) 471-4461