In Class Problems, September 11, 2009

Rise and Set

Assume that you are observing from KPNO, located just outside of Tucson, AZ. KPNO is located at 111o36.0' W longitude and 31o57.8' N latitude. Arizona is in the Mountain Standard Time zone all year. MST is (by definition) 7 hours W of Greenwich. The GMST at 0h UT tonight is 23h24m31.4s. Remember that 1 sidereal hour = 0.99727 solar hours.

1. Calculate the LST at midnight at KPNO.
KPNO is at 111o36.0' W = 111.6o= 7.44 hr W.
GMST at 0h UT = 23h 24m 31.4s = 23.4087222
GMST at 7h UT = (GMST at 0h UT) + 7 (solar) x 1 (sidereal) / 0.99727 (solar)
= 23.408722 + 7 (solar) x 1 (sidereal) / 0.99727 (solar)
= 23.408722 + 7.01916
= 30.4278822 - (24.) = (6.42788222)
LST = GMST + longitude
= 6.4278822 + (24.) - 7.44
= 22.9878822
= 22h 59m 16.4s

2. Calculate the hour angle (HA) at midnight for one of the following astronomical targets:
Object RA(2009.5) Dec(2009.5)
h m s deg m s
NGC 6333 (M9) 17 19 45.2 -18 31 32
NGC 7078 (M15) 21 30 25.8 +12 12 32
NGC 1068 (M77) 02 43 09.9 +00 01 37

HA = LST -RA

NGC 6333 : 17h 19m 45.2s = 17.32916667
HA = 22.9878822 - 17.32530556 = 5.658715534
HA = 5h 39m 31.4s and object is in the west.

NGC 7078: 21h 30m 25.8s = 21.5071666
HA = 22.9878822 - 21.5071666 = 1.480715534
HA = 1h 28m 50.6s and object is in the west.

NGC 1068: 02h 43m 09.9s = 2.71941667
HA = 22.9878822 - (2.71941667 + 24) = -3.73153447
HA = -03h 43m 53.5s and object is in the east.

3. Calculate the transit time (MST) for the above object.
Remember that 1 sidereal hour=0.99727 solar hours.

delta_t (solar) = 0.99727 x delta_t (sidereal)

NGC 6333 transits before midnight.
delta_t = 5.658715534 x 0.99727 = 5.643267241 hr before midnight.
= 24 - 5.643267241 = 18.35673276
transit = 18:21:24.2 MST

NGC 7078 transits before midnight.
delta_t = 1.480715534 x 0.99727 = 1.476673181 hr before midnight.
= 24. - 1.476673181 = 22.52332682
transit = 22:31:24.0 MST

NGC 1068 transits after midnight.
delta_t = 3.73153447 x 0.99727 = 3.72134738 hr after midnight.
transit = 03:43:16.8 MST

4. Calculate the hour angles at which this object will rise and set.
Remember that cos h = - tan (declination) * tan(latitude)

h = +/- cos-1 [- tan(dec.) tan(lat.)]

NGC 6333: h = +/- cos-1[- tan(-18.525555)tan(31.96333)]
= +/- 77.9309559o x 1 hr/15o= +/- 5.195397 hr
rise = -5h 11m 43.4s
set = +5h 11m 43.4s

NGC 7078: h = +/- cos-1[- tan(12.208889)tan(31.96333)]
= +/- 97.75922944o x 1 hr/15o= +/- 6.517282 hr
rise = -6h 31m 02.2s
set = +6h 31m 02.2s

NGC 1068: h = +/- cos-1[- tan(0.0268889)tan(31.96333)]
= +/- 90.0167781o x 1 hr/15o= +/- 6.0011185 hr
rise = -6h 00m 04.0s
set = +6h 00m 04.0s