Question

A planet requires 1200(Earth) days to complete it's circular orbit around it's hosting star. The planet's orbital radius is about 3 AU. What is the mass of the hosting star?

Answer 1

Let's call $M$ the mass of the star and $m$ the mass of the planet, which should cancel.  The gravitational force should equal the centripetal force:

$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$

$M=rv^2/G$

We compute $v = \dfrac{2 \pi r}{T}$

$M=\dfrac{4 \pi^2 r^3}{G T^2}$  

This is Keppler's Law.  We convert to MKS as we plug in and get

$M=\dfrac{4\pi^2(3 \cdot 1.496 \times 10^{11}) ^3}{6.674\times 10^{-11} ( 1200 \cdot 24 \cdot 60 \cdot 60 )^2 }$

$M=4.9744\times 10^{30}$ kg

which is about two and a half solar masses if I haven't made any errors.