PHYS 1061 Intro Astronomy (Stars and Galaxies)

Astronomy Today, 9th (and 8th) Ed.
	
Week 4.  
 Ch. 1 Prob 3,5-7
 Ch. 1 M-C 1-4,8-10       (2nd assignment from Ch. 1)

Ch. 1 Problems 

3. Scorpius is the correct answer (found using Stellarium and
  setting coord system to an equinox of the current time).
   You get Ophiuchus from stellarium if you keep the coord system
   on equinox 2000.   (acceptable)
   Libra, and Sagittarius are also acceptable, given the low precision 
   of using Fig 1.15.
   8000yrs/26000 yrs = 0.31 of way around zodiac.  
                       0.3x12-> shift of 3.6 constellations.
                       0.3x13-> shift of 4.0 constellations.

[ NOT ASSIGNED:
4. Relative to the stars, through how many degrees, arc minutes,
   or arcseconds does the Moon move in a) 1 hour, b) 1 minute,
   c) 1 second?  How long does it take for the Moon to move d) its 
   own diameter?
    The Moon moves at 360/27.3 = 13.2 deg/day relative to the stars.  
 This converts to a) 0.508 deg/hour, b) 0.0085 deg/min, c) 0.00014 deg/sec.
    But it's more convenient to use:
 a) 0.508 deg/hour, b) 0.508 arcmin/min, c) 0.508 arcsec/sec
 d) The Moon's angular diameter is 0.518 deg, so .518/.508 = 1.02 hrs.
]

5 At what distance is an object if its parallax, as measured from
  either end of a 1000-km baseline, is 
  a) 1 deg  
    ANS: D = 1000 km * 360 / 2*pi*1 deg 
           = 57,296 km (using book equation, assuming curved baseline)
         D = 57,294 km (using tan f'n, assuming straight baseline)
    Or let baseline B=LD in my small-angle formula A(deg)=57.3 LD/D.
    Then 1 deg = 57.3 B/D so D = 57.3 B/1 = 57.3 (1000km) = 57,300 km
  b) 1'   ANS: 3.44x10^6 km  (60x last answer)
  c) 1"   ANS: 2.06x10^8 km  (60x last answer)

   Using the correct (astronomical) def of parallax (as 1/2
   of the total angle) you get:
  a) 28,700 km         b) 1,719,000  km    c) 1.03x10^8 km

6. Given the angular size of Venus is 55" when the planet is
   45,000,000 km from Earth, calculate Venus's diameter (in km).
     A(deg) = 57.3 LD/D 
     A('') = 206265 LD/D, so (55*D)/206265 = LD = 11999~12000 km.

7. The Moon lies 384,000 km from Earth, and the Sun lies
   150,000,000 km away.  If both have the same angular size as
   seen from Earth, how many times larger than the Moon is the
   Sun?
   Using similar triangles: 
      150,000,000/384,000 = LD_Sun/LD_Moon= 391 times ~ 400.
   Or, use the class notes which mention the magic number 400.

Chap 1 Multiple Choice
1. (b)
2. (b)
3. (d) (actually, none of the above.  Sun is in Sag and Capric in Jan.
        But d (Aquarius) is closest to the right answer.)
4. (a) 
8. (c but b acceptable)
9. (a) smaller parallax angle
10.(d) geometry

[NOT ASSIGNED:
5. (c) waning phase
6. (a)
7. (a) (The book's answer, c, is wrong.  Best answer is a:  if the
       Moon's speed doubled, it's period would be halved (assuming a
       fixed orbit size), and the frequency of new moons would double.  
       So during the 34day eclipse season the number of solar eclipses 
       would approximately double.)
  ]