Determining the Astronomical Unit from Mercury's Transit

by Mee Wong-u-railertkun, John Pharo, David Vartanyan

We were given a picture of Mercury's transit taken by NASA's Transition Region and Coronal Explorer (TRACE) which is a polar low-Earth orbit satellite. The picture looks like this.

The path of Mercury is sinusoidal not because Mercury is really move that way, but because the parallax effect from different position of TRACE in the orbit of the Earth. The diagram below shows two greatest parallax.

The images taken by trace is a light travelled in the purple path in the diagram. Thus, in the image, we see mercury at the spot where purple lines intersect the sun. From the first picture, we can find the total angle Mercury appears to swing by completing the sun and measure the diameter of it. We know that the diameter of the sun is approximately 0.5 degree. Now, if we draw two lines, one connecting high point in Mercury's sinusoidal path and another one connecting crest of the path, we can measure the distance and compare to the diameter of the sun. From our measurement, that angle turn out to be 0.005 degree. But, what is that angle in our diagram?

Note that in the first picture, the sun as a background doesn't move, so we find a fix spot on the sun to compare and see the parallax effect. Let's say that point is the center of the sun. Now, if we connect the satellite with that spot, the angle between that line and the line that goes through mercury is alpha. Thus, alpha is the amplitude of the sinusoidal path. Since our distance from high to crest of the path is 0.005 degree, twice alpha is 0.005.

From trigonometry and small angle approximation, we get

Consider the blue triangle shown in diagram below.

The total angle in a triangle is 180 degree or pi in radian. Thus,

We can now substitute everything in and find delta a in term of a earth.

After that, we find a of Mercury in term of a earth.

From Kepler's Third law which relate the period with the distance between a star and a planet.

Thus, now we can get the astronomical unit.

We know that the Earth's period is 365 days and Mercury's period is 87 days. The earth's radius is approximately 6300 km or 6.3 * 10^8 cm. And don't forget to convert alpha from 0.005/2 degree in to radian. After substitute all variables in, we get the answer to be

The answer is surprisingly good because the "official" astronomical unit is approximately 1.5 * 10^13 cm, according to WolframAlpha.