Ay1: The Evolving Universe
Solutions to Observing Lab 1: Basic Sky Observations

1. The Static Sky

For this section, you will be learning how to use the star chart posted on the web or contained in the back of the textbook to answer useful observing questions. Many computer programs and programs on the web will give you very accurate rise/set times, etc. for stars and planets; however, for this assignment, we ask that you use the star charts from class or from the textbook.

Be alert regarding Daylight Savings Time. The star charts in the textbook take Daylight Savings Time into account. Please indicate whenever listing a time in this assignment whether you are using LDT (Local Daylight Time) or LST (Local Standard Time).

1.1. Objects on the meridian (1 Point for Each Table Cell = 8pts Total)

Which constellations are at zenith (straight overhead) and on the northern horizon at the following dates and times? You will have to be a little bit clever, since we ask about some times of day that aren't explicitly listed on the charts. Recall that the sky rotates around the North Star, with objects rising in the east and setting in the west, at a rate of (almost) 1 hour of RA per 1 hour of time.

ZenithNorthern Horizon
15 April, 8:00 PM (LDT)
(10pm LDT 15 March = 9pm LST 15 March = Pg. 814)		 Gemini, Cancer, LynxCepheus, Draco, Ursa Minor
10 March, 6:00 PM (LST)
(6pm LST 10 March = 8pm LST 10 Feb, 10pm LST 10 Jan = In Between Pgs. 812 and 813)		 Gemini, Taurus, Auriga, PerseusDraco, Ursa Minor
01 October, 3:00 PM (LDT)
(3pm LDT 01 Oct = 11pm LDT 01 June = Pg. 816)		 Virgo, Leo, Bootes, Ursa Major Cepheus, Cassiopeia
30 June, 7:00 AM (LDT)
(7am LDT 30 June = 9pm LDT 30 Nov = 8pm LST 30 Nov = Pg. 822)		 AndromedaUrsa Major, Ursa Minor, Draco

1.2. Rise and shine (1 Point for Each Table Cell = 8pts Total)

Now, using the star charts determine approximately when the following constellations will rise and set on April 15th.

To determine rise times, use the star charts to find a particular date and time when the constellation first appears over the eastern horizon. Then convert this date and time into the rise time on April 15th remembering that each month objects rise two hours earlier. Use the same technique at the western horizon for set times.

rise timeset time
HerculesRise around 9pm LST at the beginning of April
= 8pm LST on April 15th 		Sets around 9pm LST mid November
= 11am LST on April 15th
PiscesRise around 10pm LDT at middle of August
= 5am LST on April 15th 		Sets around 9pm LST on March 1st
= 6pm LST on April 15th
CygnusRise around 10pm LDT at middle of May
=11pm LST on April 15th 		Sets around 9pm LST in mid January
= 3pm LST on April 15th
CassiopeiaAlways visibleAlways visible

1.3 Finding planets (1 Point for Each Table Cell = 12pts Total)

Sky and Telescope's excellent website states that 4 naked-eye planets--Mercury, Mars, Jupiter, and Saturn--are currently visible during the evening and morning hours.

On April 15th, Mars is located at RA = 21h29m and declination = -16o18', Jupiter is located at RA = 12h48m and declination = -3o26', and Saturn is at RA = 07h30m and declination = +21o58'. Your challenge is to determine the times that these planets will rise and set on April 15th as well as the time they cross the meridian and their elevation at meridian crossing. While the rise and set times will be fairly approximate, you should be able to get the meridian time to within 15 minutes or so. For reference, Polaris, the North Star, is at the same altitude (elevation about the horizon) as your latitude (i.e) 34o in Pasadena and thus has maximum elevation 34o.

In order to determine the rise and set times for the planets we need to look at the star chart with RA and Declination from the web and find reference constellations for each planet and find when these reference constellations rise and set using the book's star charts. On April 15th Mars is located in Capella, Jupiter is located in Virgo, and Saturn is in Gemini. So we can use the same proccedure as above to find their rise and set times.

To calculate the time at meridian crossing, remember that RA=12hr is directly overhead at midnight at the spring equinox, which is roughly March 22nd. The next day, RA=12hr04m will be directly overhead at midnight. This means that, taking 24 days between the spring equinox and April 15th and an hour for Daylight Savings, about RA=12hr36m will be directly overhead at midnight. Therefore, to find when a planet will be overhead, calculate the difference between the planet's RA and 12hr36m - if planet's RA is greater than 12h36m then the difference will be the number of hours after midnight that the planet will cross the meridian. If planet's RA is less than 12h36m then the difference will be the number of hours before midnight that the planet will cross the meridian. All times listed are PDT.

To find maximum elevation, remember that a star with a declination equivalent to our latitude will be directly overhead (elevation = 90o) when it crosses the meridian. Therefore, the difference in the planet's declination and our latitude will be the how much the elevation is less than 90o. For instance, Polaris the North Star is at declination=90o, so it differs from our latitude by 56o, which implies that the max. elevation is 90o - 56o = 34o.


rise time set time time at meridian crossing maximum elevation
Mars 3:30 14:00 8:50 39o42'
Jupiter 18:00 6:00 0:10 52o34'
Saturn 11:40 1:50 18:50 77o58'

Try to go out on April 15th or sometime near then and locate these planets! Could you find them? (1 Bonus Pt. for any attempt at locating the planets.)


1.4 Star Charts (2 Pts. Each Part = 4pts Total)

If you have traveled significantly to the north or to the south, you may have noticed that the positions of stars in the night sky change with your latitude. There are whole regions of the sky in the southern celestial sphere that someone in the Northern Hemisphere would never be able to see. This is one of the main reasons that astronomers are very keen on maintaining observatories in both the Northern and Southern Hemispheres.

Calculate the most southern declination that would be visible from Keck (20o north latitude) and from Pasadena (34o north latitude). For this problem we'll consider a star that just barely rises above the horizon as being visible, although this would never be observable by an observatory.

The southernmost declination visible will be one whose elevation is 0o, implying that the difference between the declination and the latitude of the observer is 90o. For Keck at 20o north latitude, this implies a southern declination limit of -70o declination. For Pasadena, this implies a southern declination limit of -56o declination.



Likewise, star charts, such as the one contained in the textbook are designed for a certain latitude. Calculate approximately the latitude that the star charts in our textbook are designed for. Be sure to explain how you arrived at this answer --- no credit will be given for answers that consist of only a latitude, even if it is the right latitude.

There are a number of different ways to approach this problem, all of which should yield approximately the same result. One approach is to note that Canopus appears only in 2 star charts (January and February) and when it does appear it is very close to the southern horizon. Now, from the star chart handed out in class, Canopus has a declination of approximately -52o. Accouting for the fact that Canopus comes a few degrees above the horizon, we can place the horizon at a declination of between -55o and -60o. The horizon is 90o away from zenith, and the declination of a star at zenith equals the latitude of the observer. So, zenith (and thus latitude) corresponds to 30o to 35o.

Other approaches include finding a star or constellation that appears at the zenith point on the star chart or calculating what fraction of the distance along the line from the zenith to the northern horizon Polaris, the North Star, lies at. You have to be a bit careful with this last approach, though, because of the projection effects of compressing a sphere to a 2-d structure (i.e. the star chart).

2. Measuring the Rotation Period of the Earth

The daily rising and setting of the Sun and stars is due to the Earth's rotation. The apparent motion of the Sun from east to west across the sky during the day is familiar to us all. But have you ever noticed that the stars complete a similar motion at night? The Sun is, after all, just a very nearby star. To estimate how fast the Earth spins on its axis, all we have to do is go outside and measure how fast stars appear to move across the sky. There's a complication here, though. The speed that stars appear to move across the sky depends upon where you look in the sky. For example, if you look near Polaris, the North Star, you'll notice no movement. The North Star is so named because it lies near where the Earth's North Pole projects upon the sky. When you spin a globe, the North Pole doesn't move.

Continuing the extension of the Earth onto the sky, there is a Celestial Equator which is just the projection of the Earth's equator on the sky. When you spin a globe, places near the equator spin fastest. The same can be said of the stars near the Celestial Equator -- they appear to move across the sky the fastest; these are the stars whose motions we want to measure.

What is the relation between the apparent angular speed of a star, and the speed it would have if it were at the celestial equator? Hint: it involves the star's declination i.e. stellar latitude, or delta.

 

vmeas = vequat x cos(dec) (1 point)

 

Pick a star which will be rising in the early evening (and preferably with declination close to 0o, and note its name and declination (to the nearest degree, as read from the star chart) here:

One point for listing a star and declination.

 

Now find a place from which to make your observations. This place should be located so that your star appears to be balancing on some structure, like a building, telephone pole, or electric power lines. Make sure this structure is rigid, not something flimsy like a tree branch which can be blown about by the wind. Also verify that this structure is far enough away that your star still appears to balance on it if you take a step to the right or left. Take careful note of this spot. You will need to come back to it periodically. You should choose a target somewhere towards the east, since you will need to repeat your observations a few hours later. You don't want your star to have set or gone behind something by then. Describe your place and reference point.

 

One point for a description of a reasonable reference point and place.

 

When you have the star balancing on your reference point, record the date and time to the nearest minute in the table below. Come back about an hour later and relocate your star. Carefully measure how far your star has moved from its original position using any means possible. For instance, your outstretched fist spans about 10 degrees at arm's length, and your fingertips each span 1 degree viewed from the side, so you can use them to estimate angles. Record the angle differential, along with the time to the nearest minute. Repeat an hour later.


time & date change in position
initial measurement   0
next measurement    
next measurement    

Compute the average angular speed of the star, in degrees/minute:



How long would it take for that star to travel all the way around the sky, i.e. 360 degrees?


One day. Given the uncertainties in the fist sextant, anything within a factor of two is fair game. 5 pt for a good answer (and good procedure), or 3 pt for good procedure that yielded a whacky answer.


Don't forget to include a correction for the declination, if need be. This period of time is how long it takes for the Earth to spin once. What is the common name for this time interval? How close did your estimate come to the "right answer" that we all learned in kindergarden? To what do you attribute any error? How might you restructure your observational techniques to reduce this error? Where possible, quantify error sourcees.


Up to 3 pts for good error thoughts - estimates of error bars, next time use instrumentation instead of inaccurate fist estimator, more data points, longer time baseline, for instance


Portions of this section were adapted from Introductory Astronomy Manual by Michael R. Collins, Craig Kulesa, and Connie Walker.

3. A bit on stars and constellations

This part of the lab is devoted to giving you a small tour of the constellations. It is entirely for your viewing pleasure, but it provides some good mnemonics for finding your way around the sky. For more detail, read any standard sky guide. The Audobon Society publishes The Field Guide to the Stars and Planets, and there is another retail book called SkyGuide that is quite good.

Face due North towards the mountains. You should see Polaris about 34 deg above the horizon. It will be relatively faint but visible without straining. It never moves noticeably to the naked eye and always is due North. The Celestial Equator is the circle located 90o away from Polaris, the North Star. Here are two easy ways to find Polaris. Find the Big Dipper (aka Ursa Major or the Big Bear). If you follow the two stars at the edge of the ladle, furthest from the handle, from bottom to top, they will point almost exactly at Polaris. These two stars in the dipper part of the Big Dipper are called the pointer stars.

If the Big Dipper is not visible to you, try finding Casseopeia. It is a giant fist-sized W on the sky. The W, from bottom to top, points towards Polaris as well. These two constellations are approximately on opposite sides of Polaris, so one of them should be up at ALL TIMES. Constellations that never rise or set because they are less than 34 deg away from the North Star are called "circumpolar".

The North Star is in the Little Dipper (Ursa Minor). Unless you are far from the city lights, the only other two stars you will see in the Little Dipper are the equivalent pointer stars. The other four stars are usually too faint. Go back to the Big Dipper, and follow the curve of the handle away from the ladle, while shouting, "Arc to Arcturus!" Arcturus is the third brightest star in the sky and is located in the constellation Bootes (pronounced Bow-ooh-tease). If you wait until later in the night or term you can continue this curve from Arcturus, slightly increasing the radius of curvature, and "Spike to Spica", in the constellation Virgo.

The Summer Triangle also should be visible later in the term or later at night. It is made of the three bright stars: Altair in Aquila, Deneb in Cygnus, and Vega in Lyra. Vega is clearly the brightest of the three. The Summer Triangle spans about 40 degrees on its longest side. Note that Altair is only 10 degrees north of the Celestial Equator. (It would make for a good star for the star motion portion of the lab, if you're a real night owl -- you will have to wait until after midnight to see these stars.) Also, all of these constellations are in or very near the Milky Way -- the fuzzy band of light that makes up the disk of our galaxy.

Last but not least on this mini-tour: finding Hercules. Between Arcturus and Vega lies the constellation Hercules. In the center of this constellation is an almost-trapezoid, known as the keystone, which forms the chest of this mythological hero.