- Angular diameter of the Sun, $\theta_{\odot} = 0.5547°$
- Rotational speed of the Sun, $v_{\odot} = 1.86 \frac{km}{s}$
- Rotational period of the Sun, $P_{\odot} = 26.1$ days
But what we actually want to calculate is the astronomical unit (AU). How can we find it though?
Well, let's draw a diagram that relates the astronomical unit to the angular diameter:
Based on this diagram, we can relate the astronomical unit to the angular diameter of the Sun, $\theta_{\odot}$ and the radius of the Sun, $R_{\odot}$:
\begin{align}
\tan \left(\frac{\theta_{\odot}}{2}\right) &= \frac{R_{\odot}}{AU}\\
AU &= \frac{R_{\odot}}{\tan \left(\frac{\theta_{\odot}}{2}\right)}
\end{align}
Okay, so we have an equation for calculating AU, but it requires us to have the radius of the Sun, $R_{\odot}$. We don't have that value, but we can calculate it using the Sun's rotational speed, $v_{\odot}$ and the rotational period, $P_{\odot}$ using the equation for rotational velocity:
\begin{align}
v_{\odot} &= \frac{2\pi R_{\odot}}{P_{\odot}}\\
R_{\odot} &= \frac{v_{\odot} \times P_{\odot}}{2\pi}\\
R_{\odot} &= \frac{1.86 \frac{km}{s} \cdot \frac{10^5 \text{ cm}}{1 \text{ km}}\times 26.1 \text{ days} \cdot \frac{24 \text{ hours}}{1 \text{ day}} \cdot \frac{60 \text{ minutes}}{1 \text{ hour}} \cdot \frac{60 \text{ seconds}}{1 \text{ minute}}}{2\pi}\\
R_{\odot} &= 6.68 \times 10^{10} \text{ cm}
\end{align}
Now that we have the radius of the Sun, we can plug this value into the first equation to solve for the AU:
\begin{align}
AU &= \frac{R_{\odot}}{\tan \left(\frac{\theta_{\odot}}{2}\right)}\\
AU &= \frac{6.68 \times 10^{10} \text{ cm}}{\tan \left(\frac{0.5547°}{2}\right)}\\
AU &= 1.379 \times 10^{13} \text{ cm}
\end{align}
We have calculated the AU to be $1.379 \times 10^{13} \text{ cm}$, which is very close to the actual answer of $1.5 \times 10^{13} \text{ cm}$, which is within 8% error. This can be due to the inaccuracies in measuring the sunspot latitudes, since the spherical grid was not very precise since it lacked small unit divisions. Also, measuring the angular diameter could be inaccurate due to human error of stopping the stopwatch at the accurate time.
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