Answer:
Mass of the planet = 1.48 × 10²⁵ Kg
Mass of the star = 5.09 × 10³⁰ kg
Explanation:
Given;
Diameter = 1.8 × 10⁷ m
Therefore,
Radius = [tex]\frac{\textup{Diameter}}{\textup{2}}[/tex] = [tex]\frac{\textup{1.8}\times10^7}{\textup{2}}[/tex]
or
Radius of the planet = 0.9 × 10⁷ m
Rotation period = 22.3 hours
Radius of star = 2.2 × 10¹¹ m
Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s
free-fall acceleration = 12.2 m/s²
Now,
we have the relation
g = [tex]\frac{\textup{GM}}{\textup{R}^2}[/tex]
g is the free fall acceleration
G is the gravitational force constant
M is the mass of the planet
on substituting the respective values, we get
12.2 = [tex]\frac{6.67\times10^{-11}\times M}{(0.9\times10^7)^2}[/tex]
or
M = 1.48 × 10²⁵ Kg
From the Kepler's law we have
T² = [tex]\frac{\textup{4}\pi^2}{\textup{G}M_{star}}(R_{star})^3[/tex]
on substituting the respective values, we get
35164800² = [tex]\frac{\textup{4}\pi^2}{6.67\times10^{-11}\timesM_{star}}(2.2\times10^{11})^3[/tex]
or
[tex]M_{star}[/tex] = 5.09 × 10³⁰ kg
