Lab Activity 2: The Astronomical Unit

Take a look at this image of Mercury passing in front of the Sun, as taken from the polar Earth-orbiting TRACE satellite.

From this image and a few other facts, it's possible to discover the distance from the Earth to the Sun, which is defined as one AU (astronomical unit). The other information necessary is that the Sun's angular width is 0.5°, Mercury's orbital period is 87 days, and the radius of the Earth is 6.738 * 10⁸ cm.


The slight sinusoidal "wiggle" comes from the fact that the TRACE satellite is orbiting the Earth as its taking images of Mercury, so the image of Mercury projected against the non-moving background of the sun moves up as the satellite moves down to the south pole, and down again as the satellite moves back up to the north pole. By extending the disk of the Sun into a full circle (using a printout of the image and a compass), it's possible to estimate how many "wiggles" would fit into the full height of the Sun, which we know to be 0.5°. From this, we can find the angular width of the "wiggle," which, through some tricky geometry, will give us the distance from the Earth to the Sun. We determined that about 50 "wiggles" would fit into the width of the sun, so the angular width of the sinusoid is .5/50 = 1/100th of a degree.

In the above, a_e is the Earth's orbital semi-major axis (1 AU), a_m is Mercury's orbital semi-major axis, and R_e is the radius of the Earth. The angles x, y, and z are also shown on the image above, and the angle y is the angular width of the sinusoid, which we found to be .001°. From considering the triangle formed from the center of the earth to the satellite to mercury and by using the small angle approximation, we find:

We know the radius of the radius of the Earth and the angle y, so we can find a_e – a_m. The small angle approximation requires the angle to be in radians, so y/2 = 8.725 * 10^-6 radians. We find that a_e – a_m = 7.723 * 10¹³ cm. Now we need Kepler's laws to obtain the astronomical unit. Kepler's Third Law states:

This means that the ratio of the periods of Mercury's and Earth's orbits squared is equal to the ratio of the distances from the Sun of Mercury's and Earth's orbits cubed. We don't have a_e or a_m, but we do have a_e - a_m. Plugging this in:

and rearranging for a_e:

Solving this, we finally find a value for the astronomical unit! 1 AU = 1.25 * 10¹⁴ cm. The actual value for the astronomical unit is 1.49 * 10¹³ cm, so this is a percent error of:

This is obviously a very high percent error, which is due to the not-very-exact method of estimating the number of "wiggle" widths in the full width of the sun, since that was the only estimated value. Since the calculated value for the astronomical unit is too big, this means that the value for a_e - a_m is too big, which means the estimated width y is too small.


Now we can obtain the distance Mercury is from the Sun from our value for a_e - a_m and our value for a_e. We find that a_m = 4.78* 10¹³ cm, and the actual distance between Mercury and the Sun is 5.79 * 10¹² cm. This percent error is 725%, which is to be expected since our percent error on the astronomical unit is so high.


Finally, we can obtain the mass of the Earth from Kepler's Third Law and the mass of the Sun.

Note that it is also possible to obtain the mass of the sun by neglecting the mass of Mercury in its orbital calculation in Kepler's Third Law,

but since it requires the distance from Mercury to the Sun and that distance has such a large percent error, I will just use the actual value for the mass of the Sun. We find, by rearranging Kepler's Third Law and plugging in the values for P_e, a_e, and M_s:

where the mass is measured in grams. The actual value for the mass of the Earth is 5.974 * 10²⁷ g, so my found value has a percent error of 2.5 * 10¹⁷ %. This is to be expected, since the value of the astronomical unit is used in the calculation, and if that has a percent error of 742%, then the value for the mass of the Earth will have a percent error of at least that cubed, since the astronomical unit is cubed in the equation for mass of the Earth. The actual percent error is higher because there are other factors in the equation that help in determining the value of the Earth's mass.


From just a few important facts, we've managed to find the astronomical unit, the distance from Mercury to the Sun, and the mass of the Earth. Our values are not very good, but they all stemmed from an estimate using a compass on a piece of paper, so this is to be expected.