(a) [tex]2.7\cdot 10^{25} kg[/tex]
The acceleration due to gravity on the surface of the planet is given by
[tex]g=\frac{GM}{R^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
Here we know:
[tex]g=22.4 m/s^2[/tex]
[tex]d=1.8\cdot 10^7 m[/tex] is the diameter, so the radius is
[tex]R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m[/tex]
So we can re-arrange eq.(1) to find M, the mass of the planet:
[tex]M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg[/tex]
(b) [tex]4.8\cdot 10^{31}kg[/tex]
The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:
[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex] (1)
where
m is the mass of the planet
M is the mass of the star
v is the orbital speed of the planet
r is the radius of the orbit
The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:
[tex]v=\frac{2\pi r}{T}[/tex]
where
[tex]T=402 days = 3.47\cdot 10^7 s[/tex]
Substituting into (1) and re-arranging the equation
[tex]m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}[/tex]
And substituting the numbers, we find the mass of the star:
[tex]M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg[/tex]
