Answer: [tex]3.66(10)^{33}kg[/tex]
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity [tex]\omega[/tex] of the planet P1 with a period [tex]T=750years=2.36(10)^{10}s[/tex]:
[tex]\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}[/tex] (1)
Where:
[tex]V_{1}=40.2km/s=40200m/s[/tex] is the velocity of planet P1
[tex]R[/tex] is the radius of the orbit of planet P1
Finding [tex]R[/tex]:
[tex]R=\frac{V_{1}}{2\pi}T[/tex] (2)
[tex]R=\frac{40200m/s}{2\pi}2.36(10)^{10}s[/tex] (3)
[tex]R=1.5132(10)^{14}m[/tex] (4)
On the other hand, we know the gravitational force [tex]F[/tex] between the star S with mass [tex]M[/tex] and the planet P1 with mass [tex]m[/tex] is:
[tex]F=G\frac{Mm}{R^{2}}[/tex] (5)
Where [tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
In addition, the centripetal force [tex]F_{c}[/tex] exerted on the planet is:
[tex]F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (6)
Assuming this system is in equilibrium:
[tex]F=F_{c}[/tex] (7)
Substituting (5) and (6) in (7):
[tex]G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (8)
Finding [tex]M[/tex]:
[tex]M=\frac{V^{2}R}{G}[/tex] (9)
[tex]M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (10)
Finally:
[tex]M=3.66(10)^{33}kg[/tex] (11) This is the mass of the star S
