A221 HW #2 - Help
Problem 2
This problem requires numerous conversions from solar to
sidereal hours, and back again. Remember, 1 sidereal hour
equals 0.99727 solar hours (derived from
365.25 solar days/ 366.25 sidereal days in a year).
Thus, 4 solar hours = 4 / 0.99727 sidereal hours
LST = GMST + longitude
where GMST is the Greenwich Mean Sidereal Time at the
desired UT and longitude is the longitude measured in
hours (longitude in degrees divided by 15.). By convention,
western longitudes are negative.
Transit occurs when the LST = RA of the object.
Transit times (EDT) can be calculated by determining the
time difference between LST at midnight (0h EDT) and
the RA of the object. This time difference (converted
from sidereal to solar hours!) is the time difference
between midnight and the transit time of the source.
If the RA of the source is smaller than the LST at
midnight, the source transited prior to midnight. If
the RA of the source is larger than the LST at midnight,
the source transits after midnight. HOWEVER, RA and
LST are kept on 24 hour clocks. 0 h 0 m 0 s is 1 min
away from 23 h 59 m 0 s, not 23h and 59s.
I found it much easier to do the problems in "decimal
degrees" and "decimal hours". To convert from the
given values (X hour Y min Z sec) to decimal hours:
X + (Y/60.) + (Z/3600.)
However, the answers should be given in standard
time units (hours min secs), so remember to convert
back to these units after you have done the arithmetic.
For part c, your calculator will probably give
you the hour angle in degrees. To convert from degrees
to hours (the proper unit for the hour angle), remember
that there are 15 degrees in 1 hour.
For part d, you will want to convert the (sidereal)
hour angle into a (solar) time difference in order
to determine when (in EDT) the objects will rise and
set.
For part f, during the month of September,
CTIO is located within the Chilean
Standard Time Zone = 4 hr West.