## Saturday, October 22, 2011

### Finding the Astronomical Unit

Given this image of Mercury's orbit around the sun, we need to calculate the Astronomical Unit (The distance between the Earth an the Sun). We are given that Mercury's period is 87 days and that the Sun has an angular diameter of about .5 degrees.

The diagram of this problem is represented below. The Earth is the sphere on the left, Mercury is the point M, and the sun is the right edge.

We can start off by calculating the angular amplitude of Mercury's orbit, alpha.  Using a ruler and print out of the first image, we can see that the amplitude of Mercury's orbit is about 4mm and that the diameter of the sun is roughly 11cm.  Since we know that the sun's angular diameter is 0.5 degrees, we can set up a proportion to find the angular amplitude.
$\small \frac{\alpha}{0.5}=\frac{0.4}{11}\mathrm{, }\;\;\;\;\; \alpha = 1.8\times10^{-3}\mathrm{degrees}=3.2\times10^{-5}\mathrm{radians}$

Doing a little bit of triangle geometry (the pink triangle), we can see that
$\small \pi = \alpha + \beta + (\pi-\frac{\theta}{2})$ therefore, $\small \alpha +\beta =\frac{\theta}{2}$

Now, let's look at the blue triangle. We know that
$\small tan\Big(\frac{\theta}{2}\Big)=\frac{R_{\oplus}}{x}$   where R is the radius of the Earth and x is the distance from the Earth to Mercury.
Since theta is such a small angle, we can use the sin small angle approximation to find an expression for theta
$\small \theta = \frac{2R_{\oplus}}{x}$.

Now let's consider the superposition of the two triangles. From this we f ind the expression:
$\small tan(\beta)= \frac{R_{\odot}}{AU}$ Once again, using the small angle approximation, we can rewrite this as: $\small \beta= \frac{R_{\odot}}{AU}$

We can plug this result back in to our previous equation to show that
$\small \alpha + \frac{R_{\oplus}}{AU} = \frac{R_{\oplus}}{x}$   or $\small x=\frac{2R_{\oplus}}{\alpha+\frac{R_{\oplus}}{AU}}$

Because we know the period of Mercury (87 days) and the period of the Earth (365 days), we can use that information to find the distance from the sun to Mercury, a_m
$\small \Bigg(\frac{P_m}{P_{\oplus}}\Bigg)^2=\Bigg(\frac{a_m}{AU}\Bigg)^3$
we know that a_m is just AU - x, so we can rewrite this proportion as:
$\small \Bigg(\frac{P_m}{P_{\oplus}}\Bigg)^2=\Bigg(\frac{AU-\frac{R_{\oplus}}{\alpha+\frac{R_{\oplus}}{AU}{}}}{AU}\Bigg)^3 = \Bigg(1-\frac{R_{\oplus}}{AU\alpha+R_{\oplus}{}}\Bigg)^3$

we know the radius of the Earth to be 6.38*10^8 cm and alpha = 3.2*10^-5. Plugging in this messy expression into Wolfram Alpha, we find that the astronomical unit,

$\large {\color{Red}AU = 1.25 \times10^{13}cm }$

The real value, however, is known to be about 1.5*10^13 cm. The errors in my calculations come from my estimates in measuring alpha.

We can also use this information to find the mass of the sun.
Using Kepler's third law we have:
$\small \frac{P_{\oplus}^2}{AU^3}=\frac{4\pi^2}{M_{\odot}G}$ where G is the gravitational constant, which is 6.7*10^-8 in cgs units.

Using WolframAlpha once again, we find that the mass of the sun is:

$\large {\color{Red}M_{\odot} = 1.16 \times10^{33}g}$
This is a little off from the real value of 1.98*10^33g, but this is because our value for the AU was a little off.

Thank you Juliette and Monica for all the help!

#### 1 comment:

1. Excellent job! Your answers are quite accurate, especially considering that you had to guess at the size of the Sun in the image, which is no mean feat.