The final exam consists mainly of partial versions of the previously assigned exercises, but it will also contain some questions based on the underlying theory, as explicitly discussed in each exercise. Below are some sample questions like what will appear on the exam. You will need a calculator and planisphere, and you may bring one 8 1/2 x 11 piece of paper, both sides filled with notes, to the exam.

Sidereal Day
1) Astronauts on Mars observe that it takes 51.26 minutes for Sirius (RA=6h 43m) to replace Betelgeuse (RA=5h 53m) at the meridian (assuming that RA and Dec are measured the same way as from Earth). What is the sidereal rotational period of Mars in hours and minutes?
 

5h 103 m  -5h 53 m  _______   50m = 0.833 h

51.26 min               SD  _______     =     ________  0.833 h(RA)       24 h(RA)   SD = 1476min = 24.61hr

 

2) What would be Mars' surface speed (km/hr) at latitude 60^o S?

Mars' radius=3398 km.

  Cos 60^o = 0.5 = 1-L / 3398 km

  1-L = 0.5 x 3398 km = 1699 km

  V = distance / time = 2 pi x 1699 km / SD = 6.28 x 1699 km / 24.61 hr

 3) Your latitude is 20 degrees north. A star crosses the meridian above Polaris at an altitude of 55^o. Is it circumpolar?

Altitude Polaris = 20^o

Altitude star = 55^o

Only stars within a 20^o radius (equal to it's altitude) of Polaris will always remain above the horizon. A star which crosses the meridian above Polaris at an altitude greater than 40^o will lie outside that circle and not be circumpolar.

55^o > 40^o, therefore, it is not circumpolar.

Starcharts
4) Assume that you are situated in the town of Quito, Ecuador, directly on the equator. What would be the angle and direction (from south, north, etc., of the zenith) of the sun's rays, relative to vertical, at noon on March 21?

Zenith

5) Given the SCE starchart, on what date would Sirius be on the meridian at 7 PM?

• Find RA of Sirius on chart: approx 6h 45'.
• Go back 7 hrs in time and RA to get RA of sun on meridian (noon): 6h 45' - 7h = 23h 45'.
• Look up date on ecliptic corresponding to sun at that position. Corresponding date on ecliptic is March 16.

• Sirius
• Arcturus
• Betelgeuse

• Time zone = - 8 hrs. Time is 05:00 - 08:00 = 21:00 previous day, Dec31, 1999.
• Look up RA of sun on ecliptic: 18h 40'.
• Add number of hours since noon to sun's RA, then look on chart: 18h 40' + 9h = 27h 40' or 3h 40'.
• Betelgeuse is closest, east of the meridian.

• Rises in east
• Sets in west
• Look up date corresponding to 8 PM

(8 - 10): Above is a representation of the sky, facing south on Sept 10, 2001 at 8 PM Daylight Time.

8) What is the approximate sidereal time?
Read RA at the arrow that marks South.

 9) When did Pluto cross the meridian?
Pluto is 1 hour of RA west of the meridian. Since Earth rotates approximately 1 hour of RA for each hour of time, Pluto was on the meridian one hour earlier.

10) When will Pluto set that night?
We will assume, as we did in Part 2 of the Starcharts exercise, that the visible sky extends 6 hours of RA east and west (+ and -) from the meridian. Since Pluto is 1 h of RA west of the meridian, it has 5 h of RA = 5 h time left before setting. Therefore, it will set at 1 AM the next morning.

• Read V - Mv on vertical axes with B-V's aligned.
• Log r = (V - Mv +5) / 5 = (6 + 5) / 5 = 2.2
• Use 10^x  function
• r = 158.5 pc

12) Age = ?

• Read B-V of turn-off point.
• Read age on vertical axis for B-V of turn-off point.

13) Which is farthest, closest, youngest, oldest ?

• Longest is youngest.
• Highest is nearest.

• e = d / a = 0.6
• 0.6 = d / 10,000 mi
• d = 6,000 mi
• Distance from Earth's center = 10,000 mi - 6,000 mi = 4,000 mi
• Earth's radius = 4,000 mi.
• Therefore, perigee altitude = 0 mi.

• mvr = 4.0 x 10¹¹ kg km²/hr = (2,000 kg) V [10,000 + (0.6)10,000] km
• V = 12,500 km/hr

Given a composite image of the moon and Aldebaran before and after occultation, the dimensions shown, and the times corresponding to the positions of the moon as shown, determine the following:

16) Plate scale in min/mm.

21:00 - 18:30 = 2hr 30 min = 150 min

150 min / 100 mm = 1.5 min/mm

17) Universal Time (UT) of occultation start.

1.5 min/mm x 15 mm = 22.5 min

18:30:00 + 22.5 min = 18:52:30 UT

Earth's Revolution
 

18) Determine the plate scale.   4282 - 4227 = 55  55 / 100 mm = 0.55 /mm

19) Using the plate scale found above and the 10x enlarged spectrum, what is the red displacement, delta lambda,, of the upper spectral line?  = 0.55 /mm x 5 mm/10 = 0.275   20) What is the radial velocity corresponding to the red-shifted upper spectral line?   V = C /= 3 x 105 km/sec x 0.275 / 4272  = 19.3 km/sec

21) The total elapsed time = 3 seconds; what is the frequency of the waves shown?

6 cycles / 3 sec = 2 cycles/sec = 2 hz

Hubble's Law

22) The distance between the lines of 3888 and 5015 is 130 mm. What is the plate scale?

PS = 1127 / 130 mm = 8.67/mm

23) The length of the arrow representing the displacement of the spectral line (enlarged 10 times) is 22 mm, and= 3951. Use the plate scale from the previous question. What is the recessional velocity of the galaxy cluster?

= 2.2 mm x 8.67/mm = 19.1

V = 3 x 10⁵ km/sec 19.1 / 3951 = 1444.6 km/sec

24) The pulsation period of the Cepheid variable is 20 days. What is the absolute magnitude M_v?

log P = 1.30

M_v = -5.25

25) H₀ =80 km/sec/Mpc. What is the age of the universe in years?

H0 = 80 km/sec   or
               Mpc

H0 = 80    km    
           sec Mpc

Substitute for Mpc and sec:

H0 =                   80 km            
          (3.17 x 10-8 yr) (3.057 X 1019  km)

H0 =                   80             
          (3.17 x 3.057) (1011) yr 

Now invert both sides to get 1/H0 in years:

 1  = (3.17 x 3.057) (1010) yr = 1.21 x 1010 yr = 12.1 billion yr.
 H0                8             

Quasar 3C 273

26) Given:

• plate scale = 15/mm
• distance between the reference and quasar H beta lines = 60 mm
• rest wavelength of the hydrogen beta line = 4861

= 15 /mm x 60 mm = 900

Red Shift = /= 900/ 4861= 0.185

 

27)   The distance to the quasar obtained from Hubble's law is 900 Mpc  The quasar's image size matches that of star C  Apparent Magnitude m
A= 10.8	B= 10.1
C= 13.2	D= 12.1
What is the quasar's absolute magnitude M?   R in pc  M = m + 5 - 5 log r = 13.2 + 5 - 5 log (9 X 108) = 13.2 + 5 - 5 (log 9 + log 108 )  M = - 26.57

28) If the quasar's absolute magnitude were -28, how many times as powerful as an elliptical galaxy with M= -21 would it be?

I _quasar / I _galaxy = 2.51 ^{|28 - 21|} = 628

Spectroscopy

29) What is the deflection angle shown on the vernier scale immediately above?

The index of the vernier is at 171^o 30' + what is read on the vernier. This would be the Angle _Left in the experiment. The 13' line on the vernier lines up with a line on the main scale, therefore, 13' is added to the 171^o 30', and the angle read is 171^o 43'. If the straight-through angle were 180^o, then the deflection angle would be 180^o - 171^o 43' = 8^o 17'.

(In the experiment, instead of assuming exactly 180^o for the straight-through angle, the angle to the right of straight-through, Angle_Right is also read. Then the average of the right and left deflection angles is 1/2 the total difference between Angle_Left and Angle_Right: (Angle_Right - Angle_Left) / 2. This is the deflection angle averaged over both sides).

30) Using the deflection angle from 29), find the corresponding wavelength using the calibration curve above.

Read the wavelength of approximately 4800 on the horizontal scale corresponding to the deflection angle 8^o 17' on the vertical scale.

Valles Marineris
 

29)   The Viking Orbiter PDS image of Valles Marineris measures 200 mm high.  The Viking Orbiter PDS image of Valles Marineris is 400 pixels high. The cliff face measures 10 mm high and runs horizontally across the image.  The scale is 0.9 km/pixel.  The cliff's face is 16 degrees from horizontal.  What is the vertical height H of the cliff face?   Cliff horizontal length = (10 mm / 200 mm) 400 pixels = 20 pixels  Scale = 0.9 km/pixel  Cliff horizontal length = (0.9 km/pixel) 20 pixels = 18 km   tan 16 = 0.287 = H / 18 km  H = (18 km) 0.287 = 5.16 km

30) If you weigh 50 pounds on Mars and want to land on Mars at 10 ft/sec, what should the area of your parachute be, assuming Cd=1.3, and =0.000034 slugs/ft³ ? 

 W = D = 1/2 V² A Cd

50 = (0.5) (0.000034) (100) A (1.3)

A = 22,624 ft ²