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.
Doing a little bit of triangle geometry (the pink triangle), we can see that
Now, let's look at the blue triangle. We know that
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
Now let's consider the superposition of the two triangles. From this we f ind the expression:
Once again, using the small angle approximation, we can rewrite this as:
We can plug this result back in to our previous equation to show that
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
we know that a_m is just AU - x, so we can rewrite this proportion as:
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,
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:
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:
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!