How far to the earth?

Abstract

By taking visual data from the TRACE satellite, which tracked the position of Mercury as measured against the sun, we will determine the distance between earth and the sun (the AU) and other qualities about the sun, Earth, Mercury.

Introduction

For the purposes of this problem, we will consider the angles from the satellite (which we assume is at a distance equal to Earth’s radius from the center of the planet) to some point on the sun (the lines defining these angles are black on the diagram). We will also consider the shadow that Mercury casts on the sun (shown as the blue lines on the diagram). The angle these two lines make from the earth is ‘a’, while the angle from the centroid of earth to the centroid of the sun to the satellite is ‘b’.

Solution

We know that the angle ‘a’ refers to one-half the amplitude of Mercury’s oscillations as seen transposed against the sun. TRACE’s orbit is what causes this oscillation, so the angle it makes with the sun is based on the radius of its orbit (which we defined as equal to the radius of the earth).

We define theta, q, as seen on the diagram.

Consider this section of the diagram, bringing special attention to the triangle outlined in blue:

The angles in this triangle must sum to p. So, we can write the expression:

p=a+b+p-q/2

Which simplifies to:

a+b=q/2

This expression relates one-half the amplitude of Mercury’s oscillations (a) to angles b and q/2. We can measure the oscillations using the image given in class. On his image, we measure the amplitude of the oscillations as about .4 cm (on the paper). Meanwhile, the scale of the sun on the paper is about 11 cm across. We know the angular diameter of the sun from earth is about 0.5° (the same as the moon), so this means that the amplitude of the orbit, as an angle, is 3*10^-5 (.4/11*.05/360*2*p). So a = 3*10^-5

Now, we can look at yet another sub-triangle of the Mercury-Earth-Sun system.

Here, we consider the right triangle that has as its legs the radius of the earth and the distance from earth to mercury. We can set up this triangle as follows:

tan(q/2) = R_earth / R_{earth-mercury}

By a small angle approximation:

(q/2) = R_earth / R_{earth-mercury}

We earlier solved for q/2! So, we can substitute, as follows:

a+b = R_earth / R_{earth-mercury}

We found a, we know R_earth. Based again on our triangles, we can define b as follows:

tan(b) = R_earth / AU

Again, by the small angle approximation:

(b) = R_earth / AU

Substitute!

a +R_earth / AU = R_earth / R_{earth-mercury}

R_earth/( a +R_earth / AU )= R_{earth-mercury}

AU - R_earth/( a +R_earth / AU ) = R_mercury

(P_mercury)²/(P_earth)² = (A_mercury)³/(A_earth)³

We know P_earth=365 days and P_mercury=87 days. Also, we solved for R _mercury =A_mercury earlier. So:

(P_mercury)²/(P_earth)² = (AU - R_earth/( a +R_earth / AU ) = R_mercury)³/(A_earth)³

(87)²/(365)² = (AU - 6378/( 3*10^-5 +6378 / AU ))³/(AU)³

What an equation! Plugging this into Wolfram Alpha (or solving for an approximate solution) give the value:

AU = 1.33 * 10⁸ km

The real value for an AU is (a quick Google search can give us an approximate value):

AU = 1.496 *10^{8 km}

Not only are these values on the same order of magnitude, they are actually pretty close – so you can see that even rough measurements such as measuring the period of Mercury’s projections function using a ruler and a print-out image are sufficient to estimate the AU fairly accurately.

For a final answer, we can convert this value to cm, as the lab asked:

One AU = 1.33 * 10¹³ cm

We can find more values using these same few equations, as well!

(Pearth)²/( A_earth)³ = 4p²/(MG)

(365*24*60*60)²/( 1.33 * 10¹³)³ = 4p²/(M*6.7*10^-8)

So, M = 1.39*10³³ grams (this figure is a little off…)

Actual value: 5.97*10²⁷ grams

Can this be attributed to error? It seems like way too big of an error… could someone else who did the problem tell me what I did wrong here? Thanks!

Epilogue

We have determined one AU to be 1.33 * 10¹³ cm. Additionally, we have found the mass of the sun to be 1.39*10³³ grams. The orbital period of mercury is roughly 87 days.

Acknowledgements

Thanks to Iryna and Monica for working with me on this problem!