Third kepler's law error on sun mass calculation

[undergraduate]

Homework Statement

We consider that at the third Kepler's law the Earth mass is 0, that way we are making an error at the Sun mass calculation in comparison to the real Sun mass. In the same calculation we make another error by taking the common error at the gravitational constant, G. Which one of the two errors is the most significant? We are given the Earth mass= 5.9736 × 10^27gr, and the today accepted gravitational constant, G = 6.67384 ± 0.0008 ×10^-8 cgs(cm,g, s).

Homework Equations

third Kepler's law: P^2=4*π^2*α^3/G(sun mass+earth mass)

The Attempt at a Solution

It turns out that the constant in Kepler's Third Law depends on the total mass of the two bodies involved. Kepler himself, studying the motion of the planets around the Sun, always dealt with the 2-body system of Sun-plus-planet. The Sun is so much more massive than any of the planets in the Solar System that the mass of Sun-plus-planet is almost the same as the mass of the Sun by itself. Thus, the constant in Kepler's application of his Third Law was, for practical purposes, always the same.The constant G in the equations above is known as the Gaussian gravitational constant. If we set up a system of units with:
period P in days,
semimajor axis a in AU,
mass Mtot in solar masses,
we find: G=6.707 in cgs units.
This number indicates me that by taking the Earth mass=0 is an unconsidered error.But how am I going to prove that the G error is the significant one?

 

[undergraduate]

Should I try calculate the solar mass throught the formula: M(i)=4*π^2*α^3/P^2*G and then do it again like this:M(ii)= (4*π^2*α^3/P^2*G)-m(earth mass).
In the M(i) if I include the ± 0.0008 error for G I get an error σ= ±1/0,0008= ±1250 at my solar mass(cgs units). Is this correct? In the M(ii) I get the incorrect G(without the ±error and I am getting M(ii)=2.01*10^33gr which is similar with the real solar mass(1.989*10^33gr). Is this all I have to do to solve this problem?