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:


`(GMm)/(r^2) = (mv^2)/(r)`


`M=rv^2/G`

We compute
`v = (2 \pi r)/(T)`


`M=(4 \pi^2 r^3)/(G T^2)`

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


`M=(4\pi^2(3 \cdot 1.496 * 10^(11)) ^3)/(6.674* 10^(-11) ( 1200 \cdot 24 \cdot 60 \cdot 60 )^2 )`


`M=4.9744* 10^(30)` kg

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