To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.
Kepler's third law tells us that the period is defined as
[tex]T^2 = \frac{4\pi^2 d^3}{2GM}[/tex]
The given data are given with respect to known constants, for example the mass of the sun is
[tex]m_s = 1.989*10^{30}[/tex]
The radius between the earth and the sun is given by
[tex]r = 149.6*10^9m[/tex]
From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun
Therefore:
[tex]m = 8.2*1.989*10^{30}[/tex]
[tex]d = 6.2*149.6*10^6[/tex]
Substituting in Kepler's third law:
[tex]T^2 = \frac{4\pi^2 d^3}{2}[/tex]
[tex]T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}[/tex]
[tex]T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}[/tex]
[tex]T = 120290789.7s[/tex]
[tex]T = 120290789.7s(\frac{1year}{31536000s})[/tex]
[tex]T \approx 3.8143 years[/tex]
Therefore the period of this star is 3.8years
