This is the first post in a series of four posts on a lab in which we are measuring the distance to the Sun, known as the Astronomical Unit (AU). We will do this using only basic geometry and a few tools in the laboratory. The accuracy of this measurement is important since the AU is the basis of the cosmic distance ladder and has been used to calibrate other methods of distance measurement. An error in the measurement of the AU would throw off many of our estimations for other objects in the Universe.
In our calculation of the AU, we will use rotational speed found by the Doppler shift due to the Sun's rotation, the rotational period found by the motion of sunspots across the Sun, and the angular size of the Sun found using a "sundial" lens. In this post, we will discuss the measurement of the angular size of the Sun, which is the angle that the diameter of the Sun subtends in the sky from Earth. The image below shows the angular size of the sun, represented as \(\theta\).
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| Angular Size of the Sun Image Credit: http://www.ifa.hawaii.edu/~barnes/ast110_06/homework/hw08_fig1.gif |
Equipment:
For this portion of the lab, we only need a lens attached to a window facing the Sun, which focuses the Sun's light to a sharper image on an easel. We call this lens a "sundial." The image below shows the setup, and you can see the image of the sun projected onto the easel. It goes without saying that this lab is best done on a sunny day.
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| Lab Setup with Image of Sun through Sundial Image Credit: http://www.fas.harvard.edu/~astrolab/angulardiameter.html |
To solve for the angular size of the Sun, we need to measure the time that it takes for the image of the Sun to traverse its own diameter. Knowing that the Earth completes one rotation in 24 hours (86,400 seconds), we can therefore say that the Sun makes a complete rotation (360 degrees) in the sky with respect to the Earth in 86,400 seconds. We can solve for the diameter of the Sun by creating a ratio of the angular distance traveled by the Sun to the time it takes the Sun to move this angular distance.
\(\dfrac{360^\circ}{86400\text{ secs}}=\dfrac{\theta}{t}\)
Where \(\theta\) is the angular size of the Sun, and \(t\) is the time it took the image of the Sun to traverse it's diameter. So solving for the angular size, we get:
\(\theta = \dfrac{360^\circ t}{86400\text{ secs}}\)
Procedure:
The sundial lens and easel were positioned so that there is a circular and focused image of the Sun on the easel. We could see that the image was moving slowly to the right as its position in the sky was changing with the Earth's rotation. We made a mark on the easel to the right of the Sun's image and began a timer when the right side of the image reached the mark. Then, we stopped the timer when the left side of the Sun's image crossed the mark (after the image traversed its diameter). This was done three times and their average was taken in order account for errors in measurement. The picture below shows the motion of the Sun's image.
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| The circle to the left of the line is where the timer is started, and the circle to the right is where the timer is stopped. The arrow indicates direction of motion. |
Results and Analysis:
The traversal times from our three trials are shown in the table below:
From this data, we can find that there is an average traversal time of 131.84 seconds. Now, putting this into our equation for the angular size, we get:
\(\theta = \dfrac{360^\circ t}{86400\text{ secs}}\)
\( = \dfrac{360^\circ \times 131.84\text{ secs}}{86400\text{ secs}}\)
\(\theta = 0.55^\circ\)
So we have found that the Sun has an angular diameter of 0.55 degrees, or \(9.59\times 10^{-3}\) radians.
Error Analysis and Discussion:
According to Wikipedia, the angular size of the Sun is about 32 arcminutes, which is equal to about 0.53 degrees or \(9.31\times 10^{-3}\) radians. So our answer is only off by about 0.02 degrees or \(2.8\times 10^{-4}\) radians, which is a percent error of approximately 3%.
There are several factors that may have caused this error, including human error in timing, or perhaps imperfect alignment of the sundial and the easel. We can calculate the standard error, which is an estimate of the standard deviation of some sample that describes the variability of the individual measurements. The standard error is given by:
\(SE_{\overline{x}}=\dfrac{s}{\sqrt{n}}\)
Where \(s\) is the sample standard deviation, \(n\) is the sample size, and \(\overline{x}\) is the sample mean. We already know that \(\overline{x}=131.84\) seconds, and \(n=3\).
The sample standard deviation is given by:
\(s=\sqrt{\dfrac{1}{n-1}\displaystyle\sum _{i=1}^n (x_i -\overline{x})}\)
\(s=\sqrt{\dfrac{(135.90-131.84)^2+(130.78-131.84)^2+(128.83-131.84)^2}{3-1}}\)
\(s=3.65\)
Plugging values in, we get:
\(SE_{\overline{x}}=\dfrac{3.65}{\sqrt{3}}=2.11\) seconds
So we have a standard error of \(1.53\times 10^{-4}\) radians.
I would like to thank Charles Law, and the TFs Allyson and Andrew for their help in this lab.
Citations:
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