Homework 2

Homework 2 

Shannon Dulz

1)        Midnight at winter solstice, LST =6

+2 months, LST=10

9 pm, LST =7

HA = LST-RA = 7-5.5 = 1.5 hr *15= 22.5 degrees

The star crossed the meridian 1.5 hours ago and is in the western part of the sky

When crossing the meridian: Alt= 90-(34-12)= 68 degrees above the horizon

The star is in the southern part of the sky since it’s Dec < Latitude.

The star is in the southwestern sky about 68 degrees above the horizon. It is a good target for observing.

2)        Midnight at autumnal equinox, LST =0

-1 month, LST=22

2 am, LST =0

HA = LST-RA = 0-22 = -22 hr +24 = 2 hr *15= 30 degrees

The star crossed the meridian 2 hours ago and is in the western part of the sky

When crossing the meridian: Alt= 90-(-34-(-83))= 41 degrees above the horizon

The star is in the southern part of the sky since it’s Dec < Latitude.

            The star is in the southwestern sky about 41 degrees above the horizon. It is a good target for observing.

3)        Midnight at vernal equinox, LST =12

+2 weeks, LST=11

11 pm, LST =10

HA = LST-RA = 10-15 = -5 hr *15 = -75 degrees

The star crossed the meridian 21 hours ago and will cross again in 5 hours and is in the eastern part of the sky.

When crossing the meridian: Alt= 90-(37-24)= 77 degrees above the horizon (but well below this now)

The star is in the southern part of the sky since it’s Dec < Latitude.

            The star is in the southeastern sky but far from the meridian and thus not a good target for observing although it will be in several hours.

4)        Springfield longitude: -93.2861 degrees

            CDT +5 hr = UT

            UT = 3 am Nov 16, 2011

            d= 4016.5+305+(16-1)+(3/24)=4336.625

            LST = 100.46 + 0.985647*4336.625-93.2861+15*3 = 4326.555

            LST – (multiples of 360) = 6.555 degrees = 0.437 hr

5)        RA= 23:13:42 = 23.228 degrees

            Dec= 52:27:12 = 52.4533 degrees

            HA = LST-RA = 0.437 – 23.2283 = -22.79 +24 = 1.208 hr *15 = 18.13 degrees

            Alt = sin^-1(sin(52.45)*sin(37.195)+cos(52.45)*cos(18.13)*cos(37.195))

            Alt=70.16 degrees

            A=cos^-1((sin(52.45)-sin(70.16)*sin(37.195))/(cos(70.16)*cos(37.195)))

            A = 0.9989 degrees

            Az=360-A = 359.00 degrees