Monday, July 25, 2011

Tackling Simple Astronomy Problems (5)


In this blog we return to basic astronomy in the form of simple time-keeping problems using very basic estimation and approximation procedures. For much of the layout we will use a timing diagram, such as that in Fig. 1 which shows the observer ('Obs') at longitude L with respect to Greenwich (G, or the Greenwich meridian at 0 degrees). The L.S.T. or local sidereal time is 6h 00m as dnoted by the fact that 6h Right Ascension (RA) is on the observer's meridian. Then the hour angle (HA) for the star will be as shown. (Note: the view shown is a POLAR one, i.e. the Earth as seen from above Earth's north pole).

The Right Ascension (RA) is measured in units (hours, minutes, seconds) associated with time, and the Right Ascension of the star is clearly equal to the local sideral time (L.S.T.) plus the hour angle (which in this case is negative). Thus, we can write:

HA = RA of observer meridian - RA of object

Again, the hour angle is the angular distance in hours separating your local meridian and the RA of the object. It increases westwards, unlike RA - which increases eastwards. As may be seen, the RA of the observer's meridian (6 h in Figure 1) is none other than the local sidereal time. If you know what Right Ascension is on your meridian you know the L.S.T.

Another useful expression used often is:

Local time of transit (L.T.T.) = Star's RA - Sun's RA

where L.T.T. is the local mean solar time of transit (or L.M.T.) of the star or other object and Sun's RA is the Sun's Right Ascension on the particular date. A simple interpolation method is adequate (for most purposes) to obtain the latter, given the base data for the specific dates of the solstices and equinoxes below:

Date: ----- March 21---------June 22-----------Sept. 23--------Dec. 22

Sun R.A. ----( 0 hr.)--------(6 hr.)-----------(12 hr.)----------(18 hr.)

March 21 is just the (approx.) date of the Vernal Equinox, June 22 is the summer solstice, Sept. 23 the autumnal equinox and Dec. 22 the winter solstice. Since the Earth complete one circuit of the Sun in 365 days, the Sun (as seen from the Earth) will be apparently displaced along the ecliptic (the plane of the Earth's orbit projected onto the celestial sphere) by about one degree per day or equivalent to a 4 minute time difference. (Since 24 hours of Right Ascension corresponds to a 360 deg angular difference).

Thus, using either of the four known RA dates for the Sun - and knowing the Sun's RA changes by 4 mins/day, the new solar RA on any day can be found.

Example (1):

Find the Right Ascension of the Sun on July 1st.

We note that 8 days separates July 1st from June 22 when the Sun's R.A. is known to be 6h 00m.

The time difference is: 8 days x (4 min/day) = 32 mins.

So the Sun's RA on July 1st = 6h 00m + 32 min = 6h 32 m


Example (2):

Aldebaran (RA = 4h 34m) is found to be 45 degrees east of your meridian on a particular date. What is your L.S.T.?

Solution


First, find the hour angle (HA) of Aldebaran. We have 45 deg = 45 deg/(15 deg/h) = 3 h. But since the HA is measured east of the meridian then one must use the negative value (-3 h) so:

24 h + (- 3h) = 21 h 00m

We know:

HA = RA of the meridian - RA (object)

and RA of the meridian = L.S.T.

Therefore:

L..S.T.= HA + RA of object

L.S.T. = 21h 00m + 4 h 34 m = 25h 34m = 1h 34m (25h 34m - 24h 00m)

Other problems for industrious readers:

(1) The Right Ascension of Dubhe in Ursa Major is 11h 02m. Its declination is 61 deg 56 minutes. From a ship at sea it's observed to transit 42 degrees 30 minutes north of the zenith on April 8, at 1 a.m. according to the ship's chronometer (which reads Greenwich Mean time or G.M.T.)

Find the position of the ship.

(2)(a) Using a clear diagram, show that the hour angle of a star is related to the Star's RA, the Sun's RA and L.M.T. (local mean time) by:

RA (star) = Sun's RA - (HA) + L.M.T.


(b)Shaula (RA = 17h 31m) transits at 10 p.m. on June 29th. Find the hour angle it exhibits from your location (local hour angle).

(3) The star Canopus (RA = 6h 20m) is observed to have a local hour angle = 45 deg on Feb. 10th for a given location.

(a) What is the local sidereal time.

(b) At what local mean time and standard time would Canopus transit?

(c) What is the approximate LST at noon on the same date?

(4) Study the diagram for Fig. 1. Using the diagram and inferences regarding time - including Sun's changes in Right Ascension, give the date for which the diagram is referred. If the star in Fig. 1 has an RA = 7h 10m then find its hour angle for the observer (Obs). Hence, or otherwise, obtain the longitude of the observer.

(5)(a) Regulus (10h 07m) is observed to transit on a given date. Given the same date, what would be the local hour angle for: (i) Denebola (RA = 11h 47m), and (ii) Arcturus (RA = 14h 14m)?

(b) Find the LST one half way between the transits of Denebola and Arcturus. How much later would these two stars set than Regulus?

(c) What RA circle must be rising when Arcturus is setting?

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