Due to the effect of parallax, mercury appears to be following a sinusoidal trajectory. We will use this image to determine the ratio of the amplitude of Mercury's motion to the diameter of the sun. By drawing two chords across the arc portion of the Sun, and finding the intersection of their perpendicular bisectors, we can find the center point of the Sun. Measuring the distance from this center to any point on the arc gives us the radius of the Sun, which we double to get the diameter. Additionally, the amplitude Mercury's motion can be found by drawing two lines through the highest points and the lowest points, and halving the distance between them. These measurements were done using CAD Software, as shown below.
Thus, we can now define the ratio
The geometry of this situation is given by the following diagram.
Using Kepler's third law, we have that
Where Pm and Pe, the periods of the orbits of Mercury and Earth, are 87 days and 365 days respectively. To find Δa we will approximate that α ≈ θ. This is a valid approximation because the radius of the Earth is much smaller than the distance from the Earth to the Sun, and so the lines from the surface of the Sun to the center of the Earth and to the surface of the Earth are ractically parallel. We can find the angle α using the quotient we found from Mercury's transit across the Sun. Since the angle across the diameter of the Sun when viewed from Earth is approximately 0.5°, we have that
Using the small angle approximation, we solve for Δa:
Now, we can solve for the semi-major axis of the Earth:
This is the same order of magnitude of the accepted value,
Which way was TRACE orbiting the Earth?
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ReplyDeleteThis is a very nice writeup of the solution to this problem. Your figurs are easy to read, and your writing is very clear.
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