Day Lab Final Analysis

After collecting data and making calculations on the angular size of the sun, the rotational speed of the sun, and the rotational period of the sun, we are now ready to determine the distance to the sun!

The first step in completing this is two calculate the solar radius:
\[V_{radial} = \frac{2 \pi R}{P}\] \[R = \frac{V_{radial} \times P}{2 \pi} \] Let's consider our knowns: \[ P = 27.57 \, days = 2.38 \times 10^6 \, seconds\] \[V_{radial} = 1.36 \, km/s = 1.36 \times 10^5 \, cm/s\] \[ \theta = 0.56 \, degrees  = .0098 \, radians\] Now, solving for R, we get: \[ R = \frac{1.36 \times 10^5 \, cm/s \times 2.38 \times 10^6 \, seconds}{2 \pi} = 5.15 \times 10^{10} \, cm \]

Next, can use the small angle approximation to determine the distance to the sun. Below is a diagram of what we are measuring:

Using basic trigonometry, we know that the above relationship holds: \[sin( \theta ) = \frac{2R}{AU}\] Additionally, since theta is such a small angle, we can use the following approximation: \[sin( \theta ) \approx \theta \] We now have the following relationship" \[ \theta = \frac{2R}{AU} \] Now, solving for AU, we get: \[ AU = \frac{2R}{ \theta} = \frac{2 \times 5.15 \times 10^{10} \, cm}{.0098 \, radians} = 1.05 \times 10^{13}  \, cm \pm 1 \times 10^{12} \, cm\]
Compared to the actual measurement of the AU, which is \(1.495 \times 10^{13} \, cm \) we had a percentage error of 30%. In the lab, there were several potential sources of error that could help account for this. In the section where we determined the rotational speed of the sun, we had to round several values, and make some eyeball approximations on the plots. Additionally, the Telluric offset may not have been completely accurate, which would completely effect the doppler effect measurement. Another source of error could likely come from our tracers on the sunspots. To mark the spots, we used a market with a slightly broad tip, which would potentially give inaccurate locations.