Answer:
a) [tex]M_{planet} = 5.416\times 10^{25}\,kg[/tex], b) [tex]M_{star} = 1.325\times 10^{31}\,kg[/tex]
Explanation:
a) Gravity constant on the surface is equal to:
[tex]g = \frac{G\cdot M_{planet}}{r_{planet}^{2}}[/tex]
The mass of the planet can be determined by clearing its variable in the expression:
[tex]M_{planet} = \frac{g\cdot r_{planet}^{2}}{G}[/tex]
[tex]M_{planet} = \frac{(44.6\,\frac{m}{s^{2}} )\cdot(0.9\times 10^{7}\,m)^{2}}{6.67\cdot 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} }[/tex]
[tex]M_{planet} = 5.416\times 10^{25}\,kg[/tex]
b) By using Newton's Laws, the equation of equilibrium of the planet with respect to the star is:
[tex]\Sigma F_{r} = G\cdot\frac{M_{planet}\cdot M_{star}}{r^{2}} = M_{planet}\cdot \frac{v^{2}}{r}[/tex]
[tex]G\cdot \frac{M_{star}}{r} = v^{2}[/tex]
[tex]M_{star} = \frac{v^{2}\cdot r}{G}[/tex]
The translation speed of the planet is:
[tex]v = \frac{2\pi\cdot (3\times 10^{11}\,m)}{(402\,days)\,(\frac{24\,h}{1\,day} )\,(\frac{3600\,s}{1\,h} )}[/tex]
[tex]v = 54270.188\,\frac{m}{s}[/tex]
The mass of the star is:
[tex]M_{star} = \frac{(54270.188\,\frac{m}{s} )^{2}\cdot (3\times 10^{11}\,m)}{6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} }[/tex]
[tex]M_{star} = 1.325\times 10^{31}\,kg[/tex]
