b) On the same graph, sketch a qualitative diagram illustrating the times during which AY Sixteenus is observable from a northern latitude like Mt. Hopkins as a function of time of year (x-axis). This should appear as a band in your diagram.
c) Consider what happens to that band if you observe from di erent sites located at latitudes of -10 and +60 degrees.
d) At what declination (higher or lower) than AY Six, are stars more frequently observable from a northern site like Mt. Hopkins?
So here we are with a bunch more questions about our favorite definitely-not-made-up star, AY Sixteenus (located, as you'll recall, at RA = 18:00, Declination = +32). We'll start by figuring out when the sun will be down at our observatory, located at Mount Hopkins, AZ (conveniently located at latitude = +32). As we know, days get longer in winter and shorter in summer, but by how much? To figure out the exact times of sunrise and sunset at +32 degrees latitude, I turned to my trusty friend the online Nautical Almanac, which gives the sunrise and sunset times in UT for any given latitude and longitude (set longitude to zero so that UT would be the same as local time), then plotted it in Excel and fitted sinusoidal trendlines to them--the blue line is sunset, the red one sunrise, and the duration of night is the time between the two lines.
The times on the left look pretty confusing at first, which is understandable. Excel can be finicky about time (it "thinks" in units of days, with all smaller units being expressed as fractions of a day), so the best way to visually express it with midnight in the middle was to just measure time in minutes before and after midnight. So 1 AM would be 60, 6 AM would be 360, and noon would be 720. Looking backward, 11 PM would be -60, 6 PM would be -360, and the preceding noon would be -720. Months are measured numerically along the x-axis (January = 1, February = 2, etc.) and repeat in order to show how times change periodically (so 13 is the following January, 14 the following February, etc.). As you can see, the nighttime gets noticeably longer in midwinter and shorter in midsummer (though less so than they do here in Cambridge, ten degrees farther north).
Now we need to figure out when AY Sixteenus will be visible. There are two things that need to be the case for us to be able to see it, both of them fairly obvious. The first is that the star must be above the horizon--we're going to have a hard time seeing the star if our planet is in the way. The second is that the sun needs to be down--it's usually pretty difficult to see more than one star during the day (though if you count exploding stars, it it's not always impossible).
Since the declination of AY Sixteenus is equal to our latitude, we know that it's going to reach a maximum altitude of 90 degrees--in astronomy-speak, it will be at the zenith when it crosses our meridian. Every object moves 360 degrees each day (not accounting for variation due to precession and the Earth's movement around the sun, which has negligible effects for objects outside of the solar system). Some of that angular movement is altitudinal and some azimuthal, depending on the relationship between the object's declination and the observer's latitude. At the poles, for instance, objects experience no daily change in altitude--they simply circle at the same height in the sky. However, for an object to reach the zenith, its declination must equal the observers latitude, and it must rise straight up in the east and set straight down in the west, moving exactly 180 degrees. This means that it is completing exactly half of its daily 360 degree movement while above the horizon, and will be up for exactly half of the day. For the more mathematically inclined out there, the relationship looks like this (letting $d$ = the angular distance covered on the sky, and $t$ = the time an object spends above teh horizon):\[\frac{d}{360} = \frac{t}{24} \: (1)\] This is a very useful equation, which we'll come back to later, and it tells us that $t$ = 12 hours. Thus, it will rise six hours before it crosses the meridian, and set six hours later.
Now that we know how long before and after meridian passage AY Sixteenus will spend above the horizon, we need to figure out when meridian passage will occur each month. To do this, we just need to figure out when our LST will be 18:00 over the course of the year. LST = 18:00 is noon on the meridian at noon on the northern hemisphere's winter solstice (when the RA of the sun is also 18:00). Knowing that the RA of the sun changes by two hours every month, we know that the time we're looking for will also change by two hours every month--all we need to know is whether it's going up or down. As it turns out, since LST = 18:00 will occur at 6 AM on the vernal equinox, it's going down. We can extrapolate this for the entire year, giving us this neat improvement on the graph from earlier, where the shaded region represents the time AY Sixteenus will be above the horizon:
Thus, the star will be visible in the parts of the shaded region that fall between sunset and sunrise.
But what if we aren't in the neat scenario where our latitude is the same as AY Sixteenus' declination? Let's fly down to -10 degrees and see how things look there. The sunrise and sunset times will be different, for one--we're much closer to the equator, so the variation is much less significant (once you're on the equator itself, the length of days doesn't really change over the course of the year). It's also the opposite hemisphere, so days will be longer from September to March than they are from March to September. However, we can pull the times from the Nautical Almanac, so once we know that it'll be different, we can figure out exactly how different pretty easily. What's trickier is gauging how long the star will be up for. For that, I'm going to introduce a scary-looking, but incredibly helpful equation which ties together an object's declination $\delta$, its right ascension $\alpha$, its altitude $h$, your latitude $\theta$, and another value which I'll explain in a moment, $a$. Here it is:\[sin(h) = sin(\delta)sin(\theta)+cos(\delta)cos(\theta)cos(a) \: (2)\]The mystery variable, $a$, is something called the semi-diurnal arc--the angular distance an object travels from rising to meridian passage, and again from meridian passage down to setting (the diurnal arc itself is the full angular distance covered between rising and setting). You can probably see why this is a nice value to know--we knew that the semi-diurnal arc when observed from Mt. Hopkins was 90 degrees, and we used that to figure out how long it was above the horizon. Since that's all we're looking for, we can simplify the equation a bit by setting $h = 0$ to put the object on the horizon and solving for $cos(a)$, giving us a much more manageable version.\[cos(a) = tan(\delta)tan(\theta)\] Solving again for $a$ gives us 83.67 degrees. We can plug that all the way back into equation $(1)$ and solve again for $t$ to find that, at our new position, AY Sixteenus will spend 11:09 above the horizon every day. Since meridian passage will occur at the same time as at Mt. Hopkins (we've only changed our latitude, not our longitude), all we need to do is calculate the rise and set times (5:35 before and after meridian passage), plot when AY Sixteenus will be up over a plot of sunset and sunrise times, and we're set.
The difference isn't incredibly dramatic (though you can easily see how much flatter the sunset and sunrise lines are). However, since AY Sixteenus is rising half an hour earlier and setting half an hour later, the band is a bit more narrow here than it was at +32 degrees.
Moving back north, up to +60 degrees latitude, yields some pretty interesting changes. At that latitude it's possible to see objects from declinations of -30 up to +90 degrees--you can figure this out just by adding and subtracting 90 to your latitude to determine the declinations of objects on your northern and southern horizons. Obviously +90 isn't 60+90, but since declination doesn't go past +/- 90, you end up actually seeing "over" the North Pole and looking at objects from +90 down to +30 degrees on the far side of the Earth. Since AY Sixteenus' declination is +32 degrees, it never actually drops below the horizon. Instead, it'll circle the North Celestial Pole in the sky, still 2 degrees above the northern horizon at its lowest point.
AY Sixteenus on the far side of the Earth--it will cross the meridian twice, once at an altitude of 58 degrees (in the South) and once at an altitude of 2 degrees (in the North)
This means that AY Sixteenus is, for an observer at +60 degrees latitude, a circumpolar star. This gives us a pretty useful formula:\[|\delta_{min}| = 90-|l|\:(3)\] Give $\delta_{min}$ the same sign as your latitude and it tells you the minimum declination for a star to be circumpolar. Give it the opposite sign as your latitude and it tells you what stars will never rise above your horizon.
Wonderful post Tom, very clear, far beyond what was expected for this problem!
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