AU Daytime Lab, Post 4: Measuring the Astronomical Unit

This is the final post regarding this lab. Now, we have found the angular size of the Sun, its rotational speed, and its rotational period. We have everything we need to calculate the distance to the Sun. 

Procedure:

First, we need to find the solar radius, which can be done using basic trigonometry.

Here, \(\alpha\) is half of the angular size of the Sun. We can solve for the radius, \(R_\odot\), by using the equation, distance equals rate times time, where the period is time, rotational speed is the rate, and the circumference is the distance.

\(2\pi R_\odot=v_\odot P_\odot\)

Solving for the radius:

\(R_\odot = \dfrac{v_\odot P_\odot}{2\pi}\)

We also know that:

\(\tan{\alpha}=\dfrac{R_\odot}{1\text{ AU}}\)

Plugging in our equation for \(R_\odot\) and setting \(\alpha\) equal to half of the angular size, \(\theta\), we get:

\(\tan{\frac{\theta}{2}}=\dfrac{\frac{v_\odot P_\odot}{2\pi}}{1\text{ AU}}\)

Now solving for the AU:

\(1\text{ AU}=\dfrac{v_\odot P_\odot}{2\pi\tan{\frac{\theta}{2}}}\)

Using the small angle approximation, \(\tan{\theta}\approx \theta\), we get:

\(1\text{ AU}=\dfrac{v_\odot P_\odot}{\pi\theta}\)

We can now plug in the values we solved for in the three previous posts:

\(\theta = 9.59\times 10^{-3} \pm 1.53\times 10^{-4}\) radians

\(v_\odot= 1.31\pm 0.04\) km/s

\(P_\odot = 27.6\pm 0.98\text{ days}= 2.38\times 10^6 \pm 8.47\times 10^4\) seconds

So we get a final answer of:

\(1\text{ AU}=\dfrac{(1.31\text{ km/s})( 2.38\times 10^6\text{ s})}{\pi \text{ }9.59\times 10^{-3}}\)

\(1\text{ AU}= 1.03\times 10^8\text{ km}\)

Error Analysis and Discussion:

Now we have to account for the error propagation due to the error in our variables that we plugged in.

When multiplying quantities X, Y, Z, with errors \(\delta X\), \(\delta Y\), \(\delta Z\):

\(R=\dfrac{X\cdot Y}{Z}\)

The error propagates in the following form:

\(\delta R= |R| \sqrt{\left(\dfrac{\delta X}{X}\right)^2+\left(\dfrac{\delta Y}{Y}\right)^2+\left(\dfrac{\delta Z}{Z}\right)^2}\)

Plugging in the variables, we get:

\(\delta R= (1.03\times 10^8\text{ km}) \sqrt{\left(\dfrac{0.04\text{ km/s}}{1.31\text{ km/s}}\right)^2+\left(\dfrac{0.98\text{ days}}{27.6\text{ days}}\right)^2+\left(\dfrac{1.53\times 10^{-4}\text{ rad}}{9.59\times 10^{-3}\text{ rad}}\right)^2}\)

\(\delta R= 5.10\times 10^6\) km

So we have a final answer of:

\(1\text{ AU}=1.03\times 10^8\pm 5.10\times 10^6\text{ km}\)

             \(=1.03\times 10^{13}\pm 5.10\times 10^{11}\text{ cm}\)

The Astronomical Unit is a value that has been very accurately measured. According to Google, \(1\text{ AU}=1.496\times 10^8\) km. So our answer has a percent error of about 30%. This is relatively high, but it is to be expected due to our large percent error in finding the rotational speed. Recall that there are several possible factors that might have contributed to this error. There was potential for human error in both the calculation of the angular size of the sun and the rotational period of the Sun. Additionally, the sensitivity of the spectrograph might have created error if it was bumped or shifted in any way. Moreover, the large shift in the Telluric line, where there should have been little or none, might be indication of a large uncertainty in our data.  It is likely that most of the error in our measurement is a result of the uncertainties in our calculations of rotational speed. Future experiments should work to improve this second step in the lab.

Conclusion: 

Now we have successfully measure the Astronomical Unit! The Astronomical Unit is generally used to describe stellar system scale distances. It is also very important in regards to measuring other, larger distances, since the AU is a basis in the cosmic distance ladder, as shown in the image below.

Cosmic Distance Ladder

Also, the parsec, which is often used to describe bigger distance scales, is defined by the AU. Furthermore, the AU was used to calibrate other methods for finding distances to objects in the universe. This means that if there were an error in our measurement of the AU, our estimations for many other distance scales would be off too, and we would have a false understanding of some of our universe. Fortunately, we know the distance to the Sun to a very small uncertainty of 0.000,000,003. Since the mid 1600s, Huygens already had a very good estimate for the Astronomical Unit, and by 1895, we had a fairly accurate measurement by Simon Newcomb. From this lab, you can see that solving for the AU is something that can be done using only simple geometry and a few tools from an astronomy laboratory.


I would like to thank Charles Law and the TFs Allyson and Andrew for their help in this lab.

Citations:
http://en.wikipedia.org/wiki/Astronomical_unit
http://en.wikipedia.org/wiki/Propagation_of_uncertainty