Radius of the Sun from Doppler Shift

Since the Sun is rotating on its axis, half of its body is rotating towards us while half of it is rotating away from us. This results in a measurable Doppler shift in the emission spectra from the two sides of the sun. Such a spectrum is shown below. The red line is the spectrum from a point on the Sun rotating away from Earth, the black line is from the center of the Sun, and the green line is from a point on the Sun rotating towards Earth. Note the trough at 6224.75 wavenumbers. This indicates the wavelength that is most readily absorbed by the Earth's atmosphere, and has been used to align the three curves.

Our reference wavelength will be the one from the center of the Sun (black line), because the center of the Sun is not rotating and thus this wavelength is not Doppler shifted. Our change in wavelength is approximately the same for the parts of the sun that are moving towards and away from us (as we would expect). We will calculate the values of the reference wavelength and the change in wavelength:

Using the Doppler shift equation, we can solve for the speed at the surface of the Sun:

If we know the angular frequency of the Sun's rotation, we can find the radius. Consider these two images of a particular sunspot (labeled "George") taken 7 days apart. 

We see that the labeled sunspot rotates about a quarter of a turn in this much time. Thus, we can find the angular frequency of the Sun's rotation as follows:

Now, we can solve for radius:

This is the same order of magnitude as the accepted value of 

Transit of Mercury & Distance to Sun

The Transit of Mercury across the Sun allows us to approximate the value of the Astronomical Unit (AU), which is defined as the distance between the Sun and the Earth. As Mercury progressed across the sun, it was photographed periodically by the Transition Region and Coronal Explorer (TRACE) telescope, which was revolving around the earth. An image of the trace is shown below.

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,

Measuring Earth's Radius from Angle to Horizon

Here, we will determine how to measure the radius of the Earth by measuring the angle to the horizon. Assume we are at the beach and can see a clear horizon line where the sky meets the water. We will call the radius of the Earth r. We stand a height h above sea level. We measure a distance θ to the horizon. Since the tangent line is perpendicular to the radius of the Earth, we can deduce the following geometry:

We can see that θ is equal to the angle between the two radius vectors. Taking the cosine of this angle, we can deduce the following:

Since the angle down to the horizon will be very small, this 2nd order Taylor approximation for θ is valid. 

Knowing that the true radius of the earth is between 10^6 and 10^7 meters, we can find h as a function of θ.  This will tell us how high we must elevate ourselves, given the minimum angle we are capable of measuring. Solving for h,

Using this equation, we can construct a table for the value of h needed, corresponding to the minimum θ capable of being measured, for r = 10^6 m and for r = 10^7 m:

Assuming that the radius of the earth is 10^7 m (worst case scenario), and that we are capable of measuring 0.5° of angular displacement to the horizon, we see that we must achieve an elevation of 380 m to accurately perform the measurement.