Summer Astronomy Assignment #2

Text Problems:

Unit Number Review Questions Quantitative Problems 6 5, 8, 10, 11 15See note below, 16 7 4, 5 12, 14

Regarding Problem 15 in Unit 6: To answer this question you must complete the 2-D Local Horizon Map of the Sun’s maximum altitude when it is at the winter solstice (solar declination =-23.5), equinoxes (solar declination =0), and the summer solstice (solar declination =+23.5) in order to determine the Sun’s maximum altitude as it crosses the meridian. Additionally, you should address where around the horizon the sun rises and sets and how long it will be above the horizon.

Unit 6 Problems

15. Describe the motion you would see on the solstices and the equinoxes if you were observing the Sun from the Arctic Circle at a latitude of 66.5°N.

This problem is really three problems; the apparent motion of the Sun on the Summer Solstice, Winter Solstice and on the Equinoxes. The declination of the Sun on those three times is +23.5°dec, -23.5°dec and 0°dec, respectively. Let’s examine one at a time.

Before looking at the apparent motion of the Sun, let’s first get to know the sky at 66.5°N latitude. Examine the 2-D Local Horizon Map below drawn for 66.5°N latitude.

 You can see that the NCP (dec = 90) boundary of the circumpolar region is at 23.5° declination. Thus Sun on the Summer Solstice. +23.5°dec 23.5 any star or other celestial  object that has a Celestial Equator declination of23.5° or (dec = 0) 66.5 greater will be above the 66.5 Horizon Line 23.5 Horizon Line horizon all the time at (dec = -23.5) S N (dec = 23.5) The Southernmost 66.5 The Boundary of the 66.5 this latitude. Visible Star Circumpolar Region  The southernmost visible star has a declination of -23.5°. Thus, any star or 23.5 other celestial object that has a declination of -23.5° or less (meaning farther south) will never SCP (dec = -90) appear above the horizon at this latitude.

Now for the Sun. On the Summer Solstice the Sun will have a declination of +23.5°. Thus the Sun will be circumpolar at the arctic circle on that day. The Sun will not set that day. We can figure out the maximum altitude of the Sun by plotting the Sun on the above diagram and adding some angles together to get 47° maximum altitude at transit.

On the Winter Solstice the Sun will have a declination of -23.5°. Thus the Sun will be never rise at the arctic circle on that day.

On the Equinoxes the Sun will have a declination of 0°. The Sun will be on the celestial equator those two days. It will rise due east and set due west and be above the horizon for 12 hours. We can figure out the maximum altitude of the Sun by determining the maximum altitude of the celestial equator which you can see from the diagram is 23.5°.

16. If you wished to observe a star with a right ascension of 12h, what would be the best time of the year to observe it? What would be the best time of the year to observe a star with a right ascension of 6h?

This question is best answered by first sketching the Whole Sky Map as shown below

The Whole Sky Map

Star at 6h RA Sun 12h RA from Star Ecliptic Star at 12h RA

Celestial Equator 0°dec Celestial Equator 

Sun 12h RA from Star  Ecliptic

12h 6h 0h 18h 12h

The principle is this: Stars are best seen when they are farthest from the Sun – in opposition to the Sun. That is a star is easiest to observe when the Sun is on the opposite side of the celestial sphere, at least opposite in RA. Thus, the best observing of a star occurs when the Sun is 12h of RA away from the Sun.

So for a star at 12h RA, the Sun must be at 0h RA or at the Spring Equinox on Mar 22.

So for a star at 6h RA, the Sun must be at 18h RA or at the Winter Solstice on Dec 22. Instructor Assigned Topic. You have been hired by the City of Syracuse to make a scale model of the Milky Way Galaxy such that the Sun would be placed on a stone marker at the center of Clinton Square in downtown Syracuse and the center of the Milky Way would be marked on a similar stone marker at the Regional Market on Park Street in the north side of Syracuse. The concept being that visitors would start at Clinton Square and walk down North Salina Street passing similar stone markers that denote the position of significant objects as they walk to the center of the Milky Way Galaxy. The distance of the walk would be 3.3 km (about 2 miles). See map below. You will need to look up the distance the Sun of from the center of the Milky Way Galaxy.

Using proportions answer the following questions:

1. If the first stone marker in Clinton Square were to contain a pictorial representation of the Sun, what would be its diameter in cm? 2. How far from that first marker would a marker indicating the edge of the Solar System (actual distance 40 AU from the Sun) be? 3. How far away from that first marker would you place a third marker that represents the distance from the Sun to the nearest star, Proxima Centauri (actual distance from the Sun 4.3 ly)? 4. How far away would you place a fourth marker that would represent the distance to Polaris (actual distance to Polaris is 434 ly)? 5. The bright star Deneb is one of the most distant bright stars in the summer sky in the constellation of Cygnus. Deneb is 2,600 ly away from the Sun. How far away would you place a fifth marker that would represent the distance to Deneb? (Aside: You can see the first magnitude star Deneb at the zenith at 4:00 a.m. all this week) 6. The most distance visible star to the naked- eye may be Rho Cassiopeiae at a distance of 9,000 ly. How far away would you place a sixth marker that would represent the distance to Rho Cassiopeiae? 7. What would be the distance on the walk from this sixth marker of the most distant visible star from the Earth to the center of the galaxy? (Note this is not a proportion problem.)

Extra Credit: Mark the position of the stoneThe markersscale of this by map number is about on 0.167 the actualabove kilometers map of the per walk.1 cm on the map.