# Labs from YSI 93 : Photometry of Stars and Clusters.

Dr. Jim Sweitzer
CARA Yerkes Summer Institute, August 1993

This is meant to be given to the students.

## Introduction

Tonight you will actually measure the distance to the stars! Of course the Sun is a star and you have measured the distance to it, but tonight you will really discover the distant to the stars of the night. First, we will determine the distance to a bright star in the summer sky (Arcturus). Then, we will measure the distance to a globular cluster (A globular cluster is a round [looks like a "globe", hence the name], dense cluster of stars. Globulars are ancient clusters of stars that orbit our Galaxy, like satellites). When you are done, you will then have discovered about how far the stars in the night sky are and derived a distance scale for our Galaxy of stars, the Milky Way.

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)^2
```
where 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:
D = D (W / W)

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.

## Measuring the Distance to Arcturus

Arcturus is one of the brightest stars in our summer evening sky. It's high in the western sky and easy to find in the early evening. Notice how it has a yellowish glow. The great brightness of this star, coupled with its color, make astronomers refer to it as a "orange giant" star. Our Sun is, by comparison a "yellow dwarf" star. We are interested in giant stars like Arcturus tonight, because they are the ones that typically make up most of the light in the globular cluster you will be observing with the telescope. As you will see, if we can measure the distance to one of these stars, we can get the distance to an entire cluster of them.

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....

```

```

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!)

_______________________light years

## Measuring the Distance to Globular Clusters

The last thing we'll do tonight is measure the distance to an entire cluster of stars like Arcturus. We will have to do this with a telescope, since the cluster is too faint to be seen with your unaided eye. We will also use a special camera to actually measure the difference in brightness between Arcturus and the cluster. Unfortunately, we can't move Arcturus until it is exactly the same brightness as the cluster. Instead, we will have to measure the ratio of apparent brightnesses of the star and the cluster. If you were to go back to the equation we had for the inverse square law on page one, then this factor would introduce a term called R, which is the ratio of the brightness of Arcturus to the globular cluster. The equation would then look like:
```     W(1)         W(2)
-------  = R  ------
D(1)^2       D(2)^2
```
Without 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.

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