Abstract
We present the solution to the Celestial Sphere and Observation Planning worksheet problem 4, from week 2, picking range of declination that both the Very Large Telescope (VLT) and The Keck Telescope can observe galaxies at visible wavelengths for at least 6 months a year. In order to choose a small area of the sky to be observed very deeply, it should be seen by as many telescopes as possible. We use the fact that stars, from Earth’s point of view, circle around the North Celestial Pole (NCP) and the South Celestial Pole (SCP) to determine behavior of stars at different declinations seen from VLT and Keck.
Introduction
Many factors affect visibility of a part of sky at some location – its declination and right ascension, weather, path of Milky Way galaxy or other dust objects, the Sun. In this problem, we care only the declination of a star. The star’s declination determines how long the star is in a sky during one sidereal day, if it even rises at all. The latitude of an observer also affect the path of stars; while some stars can be seen all year long at one location, they might never been seen at all at another location.
Solution
If an observer is in northern hemisphere, stars in the sky will circle around the NCP, where the Polaris locates. Figure 1 shows the path of stars over a night at some specific location. At the center of circles, NCP stays stationary over a night. The position of the NCP is depended on the latitude of an observer; at latitude 20 degree North, the NCP locates 20 degree above the horizon; at north pole (latitude 90 degree North,) the NCP locates 90 degree above the horizon (at the zenith.) For an observer in southern hemisphere, stars do the same pattern but around the SCP instead. Actually, those stars also circle around NCP but the NCP is below the horizon so we could not see the circular path.
Figure 1 - Circumpolar stars. Credit: www.grandeye.com.hk
The declination of a star is measured in degree from the celestial equator (figure 2.) So, the higher the declination, the closer is the star to the NCP. The radius of the circular path seen in figure 1 is the distance between the NCP and the star. In figure 3, an observer is at 30 degree North so the NCP is 30 degree above the horizon. The blue path shows a star with declination of +75 degree, so it circles the NCP at the radius of 90 – 75 = 15 degree. The purple path shows a star with declination of +60 degree, so it circles the NCP at the radius of 90 – 60 = 30 degree. The orange path shows a star with declination of +30 degree, so it circles the NCP at the radius 90 – 30 = 60 degree.
Figure 2 - Declination of a star. Credit: www.srrb.noaa.gov
Figure 3 - Path of a star with different declinations around NCP.
The stars on blue and purple paths can be seen at all time while the star on orange path can be seen only at some time because a portion of its path goes below the horizon. The stars that can be seen at all time are called the “circumpolar stars.”
Figure 4 - An observer is in northern hemisphere with latitude of L degree.
Figure 4 shows a more general picture of an observer locating at latitude L degree North, so the NCP is L degree above the horizon. The distance of circumpolar stars and NCP cannot be more than L degree. Let the declination of those stars be delta symbol.
From figure 4, the farthest star that can be seen (even only at some time), with an assumption that we can still see a star at the horizon, has the declination of 90 – L degree below the celestial equator. For example, at Palomar observatory, L is 30 degree so the farthest star that can be seen is 90 – 30 = 60 degree below equator or declination of -60 degree. Anything below that declination cannot be seen from the Palomar observatory. For an observer in southern hemisphere, it is similar to what we have done. From figure 5, the farthest star that an observer at latitude L degree South can see has the declination of 90 – L degree above the equator. Hence, an observer at the equator can see a star with all declination from -90 to +90.
Figure 5 - An observer is in southern hemisphere with latitude of L degree.
Back to the problem, VLT is located at latitude 25 degree South, so it can see a star that has declination from -90 to +75. For Keck Telescope, it is located at latitude of 20 degree North, so it can see a star that has declination from -70 to +90.
From intersecting those two ranges, the stars that can be seen by those telescopes must have declination between -70 to +75. At the extreme end of the range, either telescope will have difficulties looking at the star because the star is at the horizon. If the star has declination toward the +75 end, Keck telescope can see it for a very long time, but from VLT sky, the star barely rises over the horizon. On the other hand, if the star has declination toward the -70 end, VLT can see it for a very long time, but from Keck telescope’s sky, the star rises over the horizon for a short period only. Thus, it is the tug-of-war. Increase the declination of the star means more than for Keck but less time for VLT. To get the most time from both telescopes, we pick the part of sky with declination in between of the visibility range.
Summary and Discussion
The visibility of the star differs from one place to another because of the difference position of NCP or SCP due to the latitude on earth. In this problem, we only care about the visibility of the star due to declination. In real experiment, we have to care about the Sun since we want to observe at visible wavelengths, and the Milky Way galaxy, since visible light gets absorbed by dusts. Thus, the declination range can be much shorter than predicted in this problem.
Acknowledgement
We thank Google for finding us a picture for figure 1 and 2. Figure 3, 4, and 5 are drawn by me and scanned by SFL copy machine number 3. We thank Wikipedia for providing location for VLT and Keck Telescope. The equations were generated using CodeCogs online LaTex editor.
Really good explanation, and I love the illustrations!
ReplyDeleteAt a certain location on Earth, you will see a star trace the same path across the sky every day, but the time of day that occurs will change throughout the year as the Earth moves around the Sun. On average, you should get about six months a year where it is dark out when the star is at the peak of its path across the sky.
You are correct that we don't want to pick declinations where the star would always be low on the horizon for one of the observatories, because then the star would not be visible for very long. Can you think of another reason why it is bad to observe stars when they are low on the horizon?
Looking at stars around the horizon means looking through more layer of Earth's atmosphere. Since our atmosphere is not uniform, because of dust, or other particles, it could affect the light from the star.
ReplyDeleteActually, I could think of another reason. Stars near horizon have more probability to get covered (by building, mountain, tree etc.) than those at the zenith. Moreover, if they are in the direction of some cities, those stars will disappear because of light pollution. (I'm actually really amazed when I saw the light pollution from San Diego on Palomar.)