Day Lab Report

The Measurement of the Astronomical Unit

The purpose of this lab is to determine the length of the Astronomical Unit (AU), which is the distance between the Earth and the Sun. We will use geometry principles and an array of tools from the lab. The AU is important because it's the fundamental unit used in astronomy - if we have error, it would affect other measurements such as parallax measurements to nearby stars. This lab consists of three steps to gather the three pieces of information we need: angular size of the sun, rotational speed, and rotational period.

Step 1: Determine the Angular Size of the Sun

Procedure

The angular size of the Sun is how much of the sky the Sun covers, in degrees. From our perspective on the Earth, it looks like the sun moves around the entire Earth (360 degrees) in a day (24 hours). By measuring the amount of time it takes for the Sun to move its diameter, we can determine the angle that the Sun occupies in the sky.

We focus the Sun's light through a lens on a window onto an easel, producing a bright projected spot on the paper. First, we mark the right edge of the bright spot on the paper, and at the same time start timing on a stopwatch. Then, we wait for the Sun to move across the paper until the left edge crosses the first mark made on the paper, and stop the stopwatch. We repeat this to get 3 measurements to take an average and estimate the uncertainty.

Analysis

Because the sky was cloudy on our lab day, we used the Friday lab's measurements:

Trial 1: 02:12.65 min = 2.2108 min

Trial 2: 02:12.41 min = 2.2068 min

Trial 3: 02:10.99 min = 2.1832 min

We get a mean time of 2.200 min. A day is 24 hours (1440 minutes) and it's 360 degrees all around a circle. We can use the ratio of this time over a day to find the angular diameter.

$$\frac{\text{movement time}}{\text{length of day}} = \frac{\theta}{360^{\circ}}$$

$$\theta = \frac{360^{\circ} \times \text{movement time}}{\text{length of day}}$$

$$\theta = \frac{360^{\circ} \times 2.200 \text{ min}}{1440 \text{ min}}$$

$$\theta = 0.55^{\circ}$$

Error

The error comes from the reaction time of marking the line and pressing the stopwatch. We're using this neat reference for propagation of error by multiplication of measured quantities. According to Human Benchmark, the average reaction time is 284 ms.

$$\delta \theta = |\theta| \cdot \sqrt{\left(\frac{\delta t}{t}\right)^2}$$

$$\delta \theta = 0.55^{\circ} \cdot \sqrt{\left(\frac{0.284 \text{ s}}{2.200 \text{ s} \times 60 \text{ s/min}}\right)^2}$$

$$\delta \theta = 0.0255^{\circ}$$

Step 2: Determine the Rotational Speed of the Sun

Procedure

We can use Doppler shift to measure radial velocity. We measure NaD emission lines from the East limb and then the West limb of the Sun. Using the CCD (charge-coupled device, a digital imaging instrument) we take measurements of the Doppler shifts of these two emission lines relative to that of a Telluric (meaning terrestrial, not the element) absorption line which comes from water in the Earth's atmosphere. Since we don't know how the Sun is oriented, we take 8 measurements around the edge of the Sun in pairs: left, right; top, bottom; top right, bottom left; top left, bottom right.

The setup is tedious because we need to align the CCD with the NaD lines. We use the Sodium lamp to adjust the slit width to get the sharpest lines possible without losing out on brightness. Once everything is aligned, we remove the lamp and allow sunlight into the slit. We take images at 8 locations of the Sun's limb because we don't know the orientation of the Sun - the Sun is tilted, the Earth is tilted, and the image is reversed after being reflected off all the mirrors. The two opposite locations with the biggest shift are the ones closest to the equator of the Sun.

Analysis

We import the image files using MaximDL. Then, we draw a box from the very left edge to the very right edge, to ensure that the data will be 765 pixels wide. We have 4 pairs of datasets to compare because we want to choose the pair that will be closest-aligned to the solar equator. We can find the index of the minimum value for each dataset, and determine the pair with the biggest difference between the indices of their respective minimum values. We ended up choosing the Bottom Left/Top Right pair.

Using this pair of datasets, we want to derive the most accurate minimum positions of the NaD lines. We achieve this by fitting the curves to a quadratic $ax^2 + bx + c$ and then use the $a$ and $b$ values to calculate the minimum pixel value, using the equation $min = \frac{-b}{2a}$.

Between the left and right NaD lines, there are 343.0319 pixels separating them. We know in fact that there are 5.97 angstroms separating the two lines, which means that we have a conversion factor of $5.97 \text{ ang}/343.0319 \text{ pixel} = 0.017404 \text{ ang/pixel}$.

For the left emission line, we have a 2.478508 pixel separation, which converts to 0.043135 angstroms. For the right emission line, we have a 2.083325 pixel separation, which converts to 0.036257 angstroms.

We can take these distances and convert them into a velocity, using the equation:

$$\Delta v = \frac{\Delta \lambda \cdot c}{\lambda}$$

For the left side, we get:

$$\Delta v_{left} = \frac{\Delta \lambda \cdot c}{\lambda} = \frac{0.043135 \text{ ang} \cdot c}{5895.94 \text{ ang}} = 2.19 \text{ km/s}$$

For the right side, we get:

$$\Delta v_{right} = \frac{\Delta \lambda \cdot c}{\lambda} = \frac{0.036257 \text{ ang} \cdot c}{5889.96 \text{ ang}} = 1.85 \text{ km/s}$$

We have to be careful not to double count the rotational velocity, since one side is going away from us, while the other is going towards us. To account for this, we first take the average of the two sides, and then halve the result.

$$v_{rot} = \frac{\Delta v_{left} + \Delta v_{right} }{2} \cdot \frac{1}{2} = \frac{2.19 \text{ km/s} + 1.85 \text{ km/s}}{4} = 1.01 \text{ km/s}$$

Error

To account for error, we use the Telluric absorption line, which, as its name suggests, is purely terrestrial. It arises from water in the Earth's atmosphere. Again, we fit the curves to a quadratic to find the pixels with the minimum values.

We determine the offset to be 0.2656 pixels, so we can convert to angstroms using our conversion factor:

$$0.2656 \text{ pixels} \times 0.017404 \text{ ang/pixel} = 0.004622 \text{ angstroms}$$

Meanwhile, the average separation between left and right is as follows:

$$\Delta \lambda_{avg} = \frac{0.043135 \text{ ang} + 0.036257 \text{ ang}}{2} = 0.039696 \text{ ang}$$

Let's use the propagation of error from multiplied quantities:

$$\delta v = |v| \sqrt{\left(\frac{\delta \lambda}{\Delta \lambda_{avg}}\right)^2}$$
$$\delta v = 1.01 \text{ km/s} \cdot \sqrt{\left(\frac{0.004622 \text{ ang}}{0.039696 \text{ ang}}\right)^2}$$
$$\delta v = 0.118 \text{ km/s}$$

Step 3: Determine the Rotational Period of the Sun

Procedure

We want to determine how long it takes for the Sun to rotate about its axis. We can use sunspots as markers of the sun's surface to measure how long it takes for the sun to rotate a certain number of degrees. This relies on the assumption that the spots are carried across the Sun by its rotation.

Ideally, we would track the movement of sunspots ourselves using a heliostat. However, since the weather was cloudy, we ended up using archival data. We opened the sunspot movie on the computer and resized the window to fit in the transparency grid of the Sun. We marked the transparency as the sunspot moved across the Sun, noting the angle the sunspot traveled and the amount of time it took. We measured 3 different sunspots to take an average.

Analysis

The following are the measurements for the 3 sunspots we tracked.

Sunspot 1	Time	Degree
Start	2013-11-28_133000	30°
End	2013-12-04_210000	110°
Delta	151.5 hours (6 days, 7 hours, 30 mins)	80°
Sunspot 2	Time	Degree
Start	2013-11-27_090000	40°
End	2013-12-01_163000	90°
Delta	103.5 hours (4 days, 7 hours, 30 mins)	50°
Sunspot 3	Time	Degree
Start	2013-11-29_193000	40°
End	2013-12-03_000000	80°
Delta	76.5 hours (3 days, 4 hours, 30 mins)	40°

Assuming that time and angle scale linearly with each other, we can take the average of the times and angles to get the average rate of rotation.

$$\bar{t} = \frac{151.5 \text{ h} + 103.5 \text{ h} + 76.5 \text{ h}}{3} = 110.5 \text{ h}$$
$$\bar{\theta} = \frac{80^{\circ} + 50^{\circ} + 40^{\circ}}{3} = 56.67^{\circ}$$

We want to use the rate of rotation to determine the rotational period $P$, knowing that there are 360° in a circle.

$$\frac{P}{360^{\circ}} = \frac{t }{\theta}$$
$$P = \frac{110.5 \text{ h} \cdot 360^{\circ}}{56.67^{\circ}}$$
$$P = 702 \text{ h}$$

Therefore, it takes 702 hours for the Sun to rotate about its axis.

Error

A source of error comes from the measuring tool we used to measure angle, the transparency grid we placed on the computer monitor. The resolution we get comes from the width of the line marking 10° intervals, which are about 1° wide. Let's use the propagation of error from multiplied quantities:

$$\delta P = |P| \sqrt{\left(\frac{\delta \theta}{\theta}\right)^2}$$
$$\delta P = 702 \text{ h} \cdot \sqrt{\left(\frac{1^{\circ}}{55.67^{\circ}}\right)^2}$$
$$\delta P = 12.61 \text{ h}$$

Putting it all together

Calculation

Now we have everything we need to solve for the AU.

First, solve for the solar radius:

$$v = \frac{2 \pi R}{P}$$

$$R = \frac{vP}{2\pi} =  \frac{1.01 \text{ km/s} \times 702 \text{ h} \times 3600 \text{ s/h}}{2\pi}$$
$$R = 406239 \text{ km}$$

Then, using the skinny triangle approximation, we can determine the AU:

$$\sin(\theta/2) = \frac{R}{\sqrt{AU^2 + R^2}}$$

By approximation:

$$\theta/2 = \frac{R}{AU}$$
$$AU = \frac{R}{\theta/2} = \frac{406239 \text{ km}}{\frac{0.55^{\circ}}{2} \cdot \frac{\pi}{180^{\circ}}}$$
$$AU = 8.46392 \times 10^7 \text{ km} = 8.46392 \times 10^{12} \text{ cm}$$

Error

Let's use the propagation of error from multiplied quantities:

$$\delta AU = |AU| \cdot \sqrt{\left(\frac{\delta \theta}{\theta}\right)^2 + \left(\frac{\delta v}{v}\right)^2 + \left(\frac{\delta P}{P}\right)^@}$$
$$\delta AU = 8.46392 \times 10^{12} \text{ cm} \cdot \sqrt{\left(\frac{0.0255^{\circ}}{0.55^{\circ}}\right)^2 + \left(\frac{0.118 \text{ km/s}}{1.01 \text{ km/s}}\right)^2 + \left(\frac{12.61 \text{ h}}{702 \text{ h}}\right)^2}$$
$$\delta AU = 1.07468 \times 10^{12} \text{ cm}$$

Our final value of AU, including error, is $8.46392 \times 10^{12} \pm 1.07468 \times 10^{12} \text{ cm}$

Our calculated value of AU is within a factor of 2 of the actual measurement, $AU = 1.49597870 \times 10^{13} \text{ cm}$, which is pretty good. If we want to feel bad about ourselves, we notice that we're about 43% off.

There are many sources of error in the various procedures. In particular, the measurement of the rotational period of the Sun seemed particularly crude, because the transparency was very imprecise, and there is ambiguity as to the start or end of a sunspot. Overall, the lab managed to get a reasonably close estimate of the AU.