The Distance to the Sun

The above image is a series of pictures of Mercury passing between the Earth and the Sun. It was taken from a satellite in polar orbit around the Earth. Is there any way we can use information from this picture, and some other basic astronomy knowledge, to find the distance from the Earth to the Sun?

The above diagram can help us find everything we need to know. We want to find a, a variable that stands for the astronomical unit, or the distance from the Earth to the Sun. Supposing we have only the orbital periods of the Earth and Mercury, and the two pictures above, can we find it? Kepler's Laws and trigonometry say yes.

First, we can use the picture of Mercury's motion to find the angle alpha in the upper left of the diagram. If you look at the picture, you may notice that Mercury's "height" relative to the Sun oscillates up and down. This is due to the movement of the satellite around the Earth: when it's near the North Pole, Mercury's position relative to the Sun appears different compared to when viewed from the South Pole. The angle between Mercury's high and low points is alpha.

We know that the height of the Sun (the diameter) is 0.5 degrees relative to the total sky. By comparing Mercury to the Sun in the above picture, one can estimate Mercury's oscillation in degrees. It's not very big: about 0.0005 degrees. This is alpha. In radians, alpha is 0.00000876. 

OK, what next? Take a look at the top triangle, whose angles are alpha, beta, and pi minus theta. Since the sum of these angles is equal to pi (using radians), we have

Using trigonometry and small-angle approximations, we have



Solving for delta-a, we get



Since the total distance a is the sum of delta-a and the distance from Mercury to the Sun, we can write



So, how does this help us? Fortunately, Kepler's Third Law of Planetary Motion says that the square of a planet's orbital period is proportional to the cube of the planet's semimajor axis, which we call a. So we have



We know the radius of the Earth (using 6300 km), and that the Earth's orbital period is 365 days, so if we also happen to know that Mercury's orbital period is 87 days, we can solve for a. Though I must warn you, it involves some pretty hideous algebra. Do not attempt at home.

But, if you do, you get around 200 million kilometers. How much is it really? About 150 million kilometers. Cool.

Given this, we can use a refined version of Kepler's Third law to find the mass of the Sun:



Assuming we're using Earth, we know everything in this equation now except for the mass of the Sun (We can neglect the Earth's mass, as it is ridiculously tiny compared to that of the Sun). Using the values we've calculated, I get



for the mass of the Sun. What is it really?



So we end up a bit under on our estimate, coming up with only about 30% of the mass of the Sun.

Worked on with Mee and David.