It is over complicating it a bit to worry about the influences of the other bodies. They do influence, but that influence is in the noise.
So, let’s see if we can calculate the mass of the Sun based on observational data…
In the early 17th century, Kepler spent a lot of time analyzing observations of the planets by people like Brahe. From from that analysis he determined a mathematical relationship.
That is that the period of a planet (time it takes to travel once around the sun) squared is equal to the distance between the sun and that planet, cubed. That is, if the units are years and astronomical units — with an astronomical unit (AU) being the distance between the sun and the Earth. From that relationship he concluded the following distances:
Kepler had found the geometric relationships for the distances in the solar system, but he didn’t know the scale — he didn’t know how far an AU actually was.
About half a century later, Cassini figured out a way to figure it out by taking advantage of a concept called parallax.
Cassini decided to use parallax to determine the distance to Mars. He had a buddy named Richer agree to observe Mars at the same time as he did, but Richer would do it from French Guiana while Cassini did it from Paris. They would conduct the experiment when Mars was at its closest to Earth, basically putting the Sun, Earth, and Mars on a straight line. They observed something like this:
While the extremely far away stars appeared static, Mars, much closer, appeared to be in a slightly different place, because of parallax.
Because Cassini knew how far apart he and Richer were, and he knew the angles involved in the observation, he could use trigonometry to determine the distance between the Earth and Mars.
That distance is about 78 million kilometers. Kepler told us that the Earth-Mars distance was 0.52 AU.
That means that the Sun-Earth distance of 1.0 AU would be about 150 million kilometers.
Now that we know the distance we can solve for the mass.
The Earth’s orbit is slightly elliptical, but we can simplify it and pretend it’s a circle for our purposes. For a circular orbit, the equation to figure out what the appropriate velocity would be, is:
Where G is the gravitational constant. M is the mass of the Sun. R is the distance from the center of the Sun to the Earth. Let’s reorganize that equation to solve for M.
Cavendish figured out G for us:
G = 6.67384 × 10–11 m3 kg-1 s-2
We determined R to be:
R = 150,000,000,000 meters
So we need to get v (orbital velocity) to figure out M.
The circumference of a circle is:
Therefore the circumference of Earth’s orbit is: 9.4245E+11 meters
There are 31,557,600 seconds in a year (which is one orbit), so dividing those two gives an average velocity of 29,864.43 meters per second.
Plugging in the V, R, and G, gives us a mass for the sun of:
2.005738E+30 kg
The actual accepted mass is 1.989E+30 kg. We were off by 0.841% — not bad. If we’ed taken the time to use the elliptical equation and more precise values for some of the variables we’d have been closer.
