The Astronomical Unit

Ay 20 – Lab 2:  The Distance from the Earth to the Sun

By:  Joanna Robaszewski

Purpose of Experiment:  Using images of the transit of Mercury across the sun from the Transition Region and Coronal Explorer (TRACE) and some prior knowledge of the solar system, determine the value of the Astronomical Unit, the distance from the Earth to the Sun, in centimeters.  Additionally, find values for the semi-major axis of Mercury’s orbit, the mass of the sun, and the mass of Earth.

Procedure:  We were given the following image of the transit of Mercury across the sun:

The image is from:  http://trace.lmsal.com/POD/images/Mercury2003_combo.gif

We were also given the following information:

•     The angular width of the sun as see from Earth is 0.5 degrees.
•         The period of Mercury is P_M = 87 days
•     The period of Earth is P_⊕ = 365 days.
•        TRACE is in a polar orbit around Earth
•        Kepler’s third law is P² = (4*pi² *a²) / G (M₁ + M₂), where a is the semi-major axis of a planet’s orbit.

We know the radius of the Earth from our previous lab, though we can use a more accurate value than the one we calculated.

Looking at the solar system from a side view we get the following set-up:

Where delta_a is the distance between Earth and Mercury.

We know that the oscillations seen in the image of Mercury’s transit are caused by the change of position of TRACE.

Using Kepler’s third law we can get a ratio of Mercury’s period and semi-major axis to Earth’s period and semi-major axis:

We can substitute a_⊕ - delta_a for a_M which gives us:

We know the periods of Mercury and Earth and we want a_⊕ so all we need to find is Δa.  To find Δa we needed to find theta.

We did this by estimating the amplitude of the wave seen in the transit image and then finding how many of these amplitudes could fit into the diameter of the sun.  We had to finish drawing the circular cross-section of the sun and then divided the diameter into segments that were as long as the amplitude.  We found approximately 50 amplitude-long segments fit into the diameter.  Since the sun has an angular width of 0.5 degrees we could find the angular width of the amplitude of the transit and that would be equal to theta.

From the diagram depicting the partial solar system from the side we know:

Theta is small enough that we can apply the small angle approximation:

We can now put this into the ratio from Kepler’s third law:

We can now find a_M:

Using Kepler’s third law we can find the mass of the sun:

We can neglect the mass of Mercury because it is so small in comparison to the sun.

Now we can find the mass of the Earth by using Kepler’s third law and the mass of the sun:

Inserting values and evaluating gives the result:

Error Analysis and Results

We want to compare our calculated values for the semi-major axes of Earth and Mercury and the mass of the Earth and the sun with accepted values.

We found the semi-major axis of Earth, and therefore the Astronomical Unit, to be

Using NASA’s planetary facts sheet, available at http://nssdc.gsfc.nasa.gov/planetary/factsheet/, the accepted value for the AU is 1.5 * 10¹³ cm.  So our percentage error for the AU is:

We found Mercury’s semi-major axis to be:

The accepted value is 5.8 * 10¹² cm.  The percentage error for Mercury’s semi-major axis is:

We found the mass of the sun to be 5*10³¹ g.  The accepted value is 2*10³³ g.

Our percent error for the mass of the sun is:

We found the mass of the Earth to be 9.5 * 10³¹ g.  The accepted value is 6 *10²⁷ g. Our percent error for the mass of the Earth is:

The values we found for the semi-major axes are reasonable, considering the approximations we used.  The values we found for the masses are not accurate.  This is most likely due to error propagation, as we used the experimental values of the semi-major axes to calculate the masses.