This is meant to be given to the students.
You probably could have guessed that we would use our old friend for this: THE INVERSE SQUARE LAW. Recall that when comparing two sources of light that are of equal brightness to your eye, that it can be written:
W(1) W(2) ------- = ------ D(1)^2 D(2)^2where the W's are the power ratings (in watts) of the sources of light you are comparing and the D's are the distances the sources are away in centimeters. To use this law in the following investigations we must solve for D(2). The algebra is straightforward, so let's not take the time to go through it here. The result is:
W(2) D(2)^2 = D(1)^2 ------- W(1)We can solve this for D(2) by taking the square root of both sides of the equation. Let's also put the indexes down to the lower right. We get the final equation that we will need in our observations. It is:
The previous equation simply says that the distance you want to derive (when you have compared two sources of light of equal brightness) is the same as the distance to the first source of light times the square root of the ratio of the wattages. Now, let's use this very powerful equation.
As you might guess, a giant star has a much higher wattage than a dwarf star. It turns out that Arcturus puts out about 100 times more light. Recall that the wattage we derived for the Sun (after we take into account the inefficiency in our light bulb and that the bulb is not a point source of light) in our investigations in Chicago is 4 x 10 watts. Thus, the power rating of Arcturus is = 4 x 10 x 10 = 4 x 10 watts.
So, let's set W(2) = Wa = 4 x 10 watts. (The "a" is for Arcturus.) We have to compare it with something we also know the wattage of at a distance we can measure, if we are going to derive Da. A 100 watt light bulb is obviously too bright. In fact, we would have to move it about 20 miles away for it to be the same brightness-- obviously an impractical distance!
Instead, we will use the little lamp we measured in the lab. Even it,
however, must be dimmed down a lot, so that we can make the
comparison. Recall that the wattage we measured for it (if we round
to one place) was about 0.1 watt. This was when we compared it to
a 100 watt light bulb. Do you remember how we knew we had to take
into account that incandescent light bulbs, like this one and the 100
watt bulb, emit most of their energy as heat? I believe the
efficiency factor for this type of bulb is typically that only 0.4 (or
40%) of their energy is emitted as visible light! If we multiply this,
the real power rating for thie bulb is then
0.1 watt x 0.4 = 0.04 watts = 4 x 10
Even this power output is too large for the measurement we need to make. So, we have to dim this lamp down. In the Camera Obscura activity you will measure a filter similar to the N.D. 4.00 we will be using (An N.D. 4.00 fileter is ust like two N.D. 2.00 filters -- 10 x 10 = 10). It only passed 1/10,000 (= 10) of the light striking it.
If we were to put this in front of the little bulb, how many watts will this little light emit?
_________________________watts (=W(1) )
Next, go outside and measure how far the little light with the filter in front of it has to be to get it to be the same brightness as Arcturus in the sky.
The distance you get (be sure to put it into centimeters) is:
___________________________centimeters (=D(1) )
You finally have all you need to determine the distance to Arcturus. Plug the numbers that you have derived above into the Powerful Distance Equation. Use the space below to compute your answer:
D = D (W / W)
Plug in and solve below....
Write your final answer here:
Now convert this number to light years (a light year is the distance light travels in one year. It is equal to 9.5 x 10 cm! This is six trillion trillion miles!)
W(1) W(2) ------- = R ------ D(1)^2 D(2)^2Without going through all the previous steps, this makes our Powerful Distance Equation come out to be the following, when solved for D(2) and when we change the numbers to little indexes (a = Arcturus and 92 is for M92).
D = D (R W / W)
We need one more ingredient from a previous lab. If the cluster is made up of stars identical to Arcturus, then all you need to know is the number of stars in the cluster to determine W right? You did that in the Globular Cluster activity during the day.
For the cluster you are going to observe tonight, how many stars did you count?
_________________________ stars.(=N ).
Now, think of the cluster as if it is a chandelier way off in space. Its wattage is then simply equal to the wattage of each bulb multiplied by the number of the bulbs. Right? So, in our case, each of the "bulbs" in the chandelier (the globular cluster) is equal to W, the wattage of Arcturus. The wattage of the cluster is then = W = N x W. If you plug this into the last equation, you will have the W terms cancel and you get:
D = D (RN)
Let's call this the Powerful Cluster Distance Equation.
Now we're all set to get the distance to the cluster.
What was D from the previous part of this lab?
Da = _________________________ light years(easier to use than cm)
Using the telescope and the CCD camera on it, find out what the number R is. Remember it is how much brighter Arcturus appears than the cluster.
R = _____________________________
Now plug in the three numbers [Da, R, & N]you have either measured or derived into the Powerful Cluster Distance Equation and compute the answer. Here's some work space.
So, the answer is........
Congratulations, you made it! You have just determined the size scale of the Galaxy. Big, isn't it?
The important points to remember, after all these labs, is that in science and especially astronomy:
Important Disclaimers and Caveats: