Tuesday, October 18, 2011

Determining the Astronomical Unit from Mercury's Transit and Period

Acknowledgements to Eric and Daniel


We will determine the astronomical unit (the distance between the Earth and Sun) using Mercury's period with a time series of images of the transit of Mercury across the Sun taken by NASA's TRACE satellite.

In the image above, the blue circle is the Earth, the grey circle is Mercury, the yellow circle is the Sun, and the pink circle is TRACE.  The red lines are drawn from TRACE (at the two different points shown), through the center of Mercury, to the surface of the sun.  The green lines are drawn from TRACE to the apparent center at the surface of the sun.  The white line divides the diagram symmetrically.  Three angles are labelled (αβ, and ɵ), as well as the distance between Earth and Mercury, Δa.  

Using basic geometry, we can state the following:
ɵ = α + β


Using trigonometry, this becomes
(Earth Radius)/Δa = α + (Earth Radius)/a   (a being the astronomical unit)
assuming that the distance between TRACE and the surface of the Earth is negligible.


This can be simplified to the following:
*α=(Earth Radius)(1/Δa - 1/a)


Looking at the image of the transit above, we can see Mercury crossing with a sinusoidal path.  This is the result of parallax, given TRACE's polar orbit.  Geometrically, we can state that the ratio between this path's amplitude and the radius of the sun is equal to the ratio between our angle α and the angular diameter of the Sun (about half of a degree).


After gathering extremely precise measurements of the aforementioned amplitude relative to the radius, we can calculate α = 4e-5 radians.


In addition, using the relationship between the period of a planet and the distance from the Sun,
We can calculate the ratio between a and (a-Δa).  Doing the math, we then get that Δa=0.62a.

Substituting this and our value of α into the asterisked equation from above, we calculate the astronomical unit a to be 9.7e7 km.

This is quite close to the actual astronomical unit, which is 1.5e8 km.



1 comment:

  1. Good job getting the angle correct. That is very tricky.

    What periods are you using to calculate the ratio between a and (a-delta_a)? Make sure to write your blog posts so that someone who hasn't seen the assignment could understand them.

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