An Astronomically Difficult Cache Mystery Cache

There are two parts to this Astronomically Difficult cache. First, you need to go to the posted coordinates at the correct time and record the star visible in the stated position. Once you know the identity of the star, you will need to do some research to figure out all the clues to the final location. This is meant to be a tough puzzle cache and will take quite a bit of effort, but you should learn some things about Astronomy if you put in the time to solve it.

Part 1 – The Observation

MUST be done on a clear night
If you can't see the stars, you will not be able to accomplish the first task.

The first stage of this cache is to go to the posted coordinates and observe a particular star. The posted coordinates take you to a public sidewalk just west of St. Hubert’s Church in Chanhassen. The building to your west is a senior living center, so please be respectful and quiet when you are there at night to make the observation.

You will find a lamppost very close to the coordinates with a sewer grate just south of the lamppost and a tree just north of it. Stand with your back to the lamppost and look at the cross at the top of the church steeple. There should be a bright star very close to the cross. Identify that star using a star map, planetarium software like Google Sky, etc. The star you need to identify is one of the 50 brightest stars in the sky, so if there are also some faint stars close to the cross ignore them and identify the bright one.

Of course, it isn’t that easy. The sky is constantly changing as the Earth rotates on its axis and revolves around the Sun. So for the same star to appear next to the cross as it was when I did the initial observation at Midnight on November 15th, you’ll have to be there at the correct time on the day you make the observation. Here’s a short Astronomy lesson to explain why.

The Earth spins once on its axis with respect to the Sun in 24 hours and 0 minutes. That is the basis of our day. However, because the Earth moves around the Sun during that day, a star will reach the same position in the sky about 4 minutes earlier. This gives the Sidereal day; 23 hours, 56 minutes and 4 seconds, which is the time it takes the Earth to rotate once with respect to the background stars. The diagram below shows this graphically.

As you can see in the graphic, because the Earth has moved during one sidereal day, it has to turn just a little more to get the Sun to come back to the same position in the sky than to get the distant star back to its same position. That is why the solar day is a little longer than the sidereal day. Thus for you to make the observation and find the correct star you must arrive 4 minutes per day (or 2 hours per month) earlier than I did. The following table will help:

See the note posted 7/28/14 in the logs below for the time to observe for dates before November 15th.

If you want to make the observation on a day between listed dates, simply pick the closest date and adjust 4 minutes per day (for example December 12th you’d want to observe at 10:12PM).

Note: After February 15, the Sun doesn’t set until after the star has passed the appropriate position, making this observation impossible. Even in Early February it might be twilight, but the star should still be bright enough to see even if the sky isn’t completely dark.

It would be possible (but much more difficult) to identify the correct star at other times by going to the correct location and noting the position of the cross. Then using planetarium software, set the sky to one of the correct Date/Time combinations and decide what star is at the cross location. Depending on how accurate you are, you may be able to come up with the correct star this way. I believe it will be much easier to do the real observation however.

Once you have identified the correct star, you are ready to use that information to decode the clues to the final cache location.

Part Two - The Cache Location

A few notes about the final cache location. It is in a public park within 2 miles of the starting coordinates. Please obey all park rules INCLUDING park hours of 6am to 10pm. Do not go there at nighttime directly after making the observation in part one. The cache is in some semi-rough terrain between the mowed grass of the park and some private homes. Please be sure to enter the trees from the park side and stay on park property. In the summer there will be plenty of weeds and brush making the terrain even tougher. There is a small but fairly steep hill to negotiate as well. I rated it a 2.5, but it’s probably a half point easier in winter, tougher in summer. Finally, I found a single strand of barbed wire near the southwest side of the beacon very near the ground, in fact it looked mostly buried with just a few feet sticking out. If you are circling be careful not to run into it.

The cache container is fairly unusual. The log is in a bison tube attached to the larger object. The larger object is what it looks like, and it does work (at least it did when first hidden). Feel free to give it a try if you want. Just don’t point it at the Sun!

The cache is located at North 44 AB.CDE and West 093 FG.HIJ. Solve the clues below to find the final location. “This star” or “the target star” in the clues below refer to the star you identified in part 1 above

A - Roughly half of all stars are members of multiple star systems, groups of 2 or more stars that are gravitationally bound to each other and orbit each other like the Earth orbits the Sun. This star is no exception. It has a companion faint enough that it is impossible to see without a fairly large telescope. Research the time it takes this faint companion to orbit the main star in years, divide by 10, and that result is A.

B – There are 88 officially recognized constellations. The official definition of a constellation is an area on the sky in a map maintained by the International Astronomical Union. Of course, the more common definition of a constellation is a recognizable grouping of stars. An asterism is a grouping of stars which isn’t officially recognized but is well-known to the public. For example, the Big Dipper is an asterism, but it is part of the constellation Ursa Major (the Great Bear). What constellation is this star a member of? The number preceding the correct answer below is B.
0) Canis Major
1) Canis Minor
2) Orion
3) Gemini
4) Taurus
5) Scorpio
6) Leo
7) Virgo
8) Ursa Major
9) Ursa Minor

C – Right Ascension and Declination are the celestial equivalent of Longitude and Latitude on Earth, something Geocachers should be very familiar with. They are used to specify exact positions on the sky. Declination measures north/south position and ranges from +90 degrees at the North Celestial Pole to 0 at the Celestial Equator to -90 degrees at the South Celestial Pole. Right Ascension measures east/west position and is measured in hours and minutes from the Vernal Equinox, the point on the sky where the Sun crosses the Celestial Equator moving North in March (the start of Spring for the Northern Hemisphere). It is measured in hours and minutes instead of degrees since the Earth rotates once in 24 hours and Earth's rotation is what causes object in the sky to move from east to west. Of course, since there are 24 hours in a circle and 360 degrees in a circle, one hour of Right Ascension corresponds to an angle of 15 degrees on the sky. To find the digit for C, look up the RA and Dec of the target star and use the ones digit of the RA hours.

D - Now that you’ve found the RA and Dec of the target star, let’s use that information and do a little geometry. The altitude of an object in the sky is its angular distance above the horizon. The maximum altitude an object can reach depends on its position on the sky as well as your location on Earth. From the starting coordinates of this cache, at N 44 degrees 51 minutes, what is the highest altitude this star will ever reach above the southern horizon to the nearest minute of arc? Choose the number before the correct answer below as your digit for D. Hint: This one will require some simple math, but consider how high the Celestial Equator is above the southern horizon as well as where this star is relative to the Celestial Equator.
0) 90 degrees
1) 60 degrees
2) 45 degrees, 09 minutes
3) 44 degrees, 51 minutes
4) 33 degrees, 33 minutes
5) 33 degrees, 27 minutes
6) 30 degrees
7) 28 degrees, 26 minutes
8) 28 degrees, 08 minutes
9) 0 degrees

E - And now for a little outside-the-box thinking. Imagine you are an Astronomer on a planet orbiting this star. And assume the constellations as seen from that planet are identical to the ones from Earth (maybe their Interstellar Astronomical Union stole a copy of our IAU’s map). What constellation would the Sun be in as seen from the planet orbiting this star? Hint: If I asked you what point on the Earth is exactly opposite from Minneapolis, how would you figure that out? This question is exactly analogous, so if you know how to do it on Earth use the same idea to solve this question.
0) Bootes
1) Ophiuchus
2) Corona Borealis
3) Crux
4) Hercules
5) Orion
6) Ursa Major
7) Delphinius
8) Virgo
9) Canis Major

F - Ok, time for an easy one. Take the number of letters in this star’s name and subtract 3. Use the result for F.

G - Stars are classified by their spectral type. The spectral types are OBAFGKM. O stars are the most massive and hottest stars, while M stars are the smallest and coolest. Within each base spectral type are subdivisions from 0 to 9 where 0 is hottest and 9 is coolest. The Sun is spectral type G2. The most massive and luminous stars known are spectral type O2 and are tens of thousands times brighter than the Sun. If Earth orbited one of those instead of the Sun it would be nothing but a ball of molten metals and rock with a surface temperature of thousands of degrees. On the other hand, if the Sun were a faint M star Earth would be a lifeless iceball only a few degrees above absolute zero. Look up the spectral type of the target star then research the characteristics of that type of star (brightness, temperature, size, etc.). If this star replaced the Sun, would Earth be habitable by humans? If you think it would, use 2 for G. If you think life as we know it would be impossible use 3 for G.

H - A light year is the distance light travels in a year. Stars are very far away. Other than the Sun, the nearest star is over 4 light years away. Look up how far away the target star is in light years, then calculate how many miles that would be. Round off to the nearest ten trillion miles. What is the tens digit in the number? (i.e. for 20 trillion miles use 2, for 780 trillion miles use 8, etc.) Use that number for H.

I - The Messier Catalog is a famous listing of deep-sky objects compiled by French comet hunter Charles Messier. It was first published in 1771. It contains 110 of the brightest galaxies, nebulae, and star clusters including famous ones like the Andromeda Galaxy, Orion Nebula, and Pleiades. How many Messier objects are in the same constellation as the target star? Use that number of Messier objects for I.

J - Stars come in a wide variety of masses. This star is more massive than the Sun. Find out how many times as massive as the Sun it is (rounding to the nearest integer), and multiply that by 4. The ones digit from the result is J.

You can check your answers for this puzzle on GeoChecker.com.

If you have significant difficulty solving this, feel free to email for hints. Tell us what star you think it is and which of the clues you're stuck on along with any work you have done so far on solving it. We'll try to provide a nudge in the right direction without just giving it away.