Homework II

Joshua W. Kern

1) If its 9pm, 2 months after the Winter Solstice, then the time since noon of the meridian line is,

• 9 pm = 9 hrs past noon.

If we are 2 months after the Winter Solstice, then RA of the meridian line at noon would be,

• 18 hrs +4 hrs = 22hrs

Therefore, the RA of the meridian line at 9pm on this day is,

• RA= 7 hrs

Which means the hour angle is,

• RA(Star)- RA(Meridian Line)= 5:30 – 7:00 = -1:30 hrs

or

• Starting at the zenith, the star is 22.5 degrees due west

Now, if we are at 34 degrees latitude and the star has a declination of 12 degrees, then the altitude is,

• 90 degrees – (34 degrees – 12 degrees) = 68 degrees

So the star is 68 degrees above the southern horizon and 22.5 degrees west of the meridian line, meaning the star is observable.

2) If its 2am, 1 month before the Autumnal Equinox, then the time since noon of the meridian line is,

• 2 am  = 14 hrs past noon.

If we are 1 month before the Autumnal Equinonx, then RA of the meridian line at noon would be,

• 12 hrs -2 hrs = 10hrs

Therefore, the RA of the meridian line at 2am on this day is,

• RA= 0 hrs or 24 hrs

Which means the hour angle is,

• RA(Star)- RA(Meridian Line)= 22:00 – 24:00 = -2  hrs

or

• Starting at the zenith, the star is 30 degrees due west

Now, if we are at -34 degrees latitude and the star has a declination of -83 degrees, then the altitude is,

• 90 degrees – (-34 degrees + 83 degrees) = 41 degrees

So the star is 41 degrees above the southern horizon and 30 degrees west of the meridian line, meaning the star is visible but too low on the southern horizon to obtain good observations.

3) If its 11pm, 2 weeks  before the Vernal Equinox, then the time since noon of the meridian line is,

• 11 pm  = 11 hrs past noon.

If we are 2 weeks before the Vernal Equinonx, then RA of the meridian line at noon would be,

• 0 hrs -1 hrs = 23 hrs

Therefore, the RA of the meridian line at 11pm on this day is,

• RA= 10 hrs

Which means the azimuth angle is,

• RA(Star)- RA(Meridian Line)= 15:00 – 10:00 = 5 hrs

or

• Starting at the zenith , the star is 75 degrees due east

Now, if we are at 37 degrees latitude and the star has a declination of 24 degrees, then the altitude is,

• 90 degrees – (37 degrees - 24 degrees) = 77 degrees

So the star has an hour angle of 75 degrees towards the east and is 77 degrees above the southern horizon, meaning the star is too low on the horizon at this time but good observations will be available later in the evening.

4) If LST= 100.46 + 0.985647 * days + longitude + 15 * UT,

Where days = 4335.5 days, longitude= -93.2861 degrees, and UT = 3 hrs then,

• LST = 100.46 + 4273.2726 + (-93.2861) + 45 = 5.45 degrees

5) If,

RA= 348.425 degrees

DEC = 52.4533 degrees

LST =5.45 degrees

LAT = 37 degrees

HA= LST - RA

Then,

• Alt = arcsin (sin(DEC))*sin(LAT)) + (cos(DEC)*cos(HA)*cos(LAT) = 70.48 degrees

And,

• Az = arccos (sin(DEC)-(sin(Alt)*sin(LAT)))/ (cos(Alt)cos(LAT)) = 32.27 degrees