Answer :
Answer:
Part A: 1.47 * 10^ 20 kg
Part B: 5.22 * 10^30 kg.
Explanation:
The acceleration due to gravity on a planet of mass M and radius r is given by
[tex]a=G\frac{M_{}}{r^2}[/tex]where G = 6.67 *10^-11 m^3 / kg *s^2 = gravitational constant.
Now solving for M gives
[tex]M=\frac{ar^2}{G}[/tex]Putting in a = 13.1 m /s^2 and r = (1.5 * 10^9 / 2) gives
[tex]M=\frac{13.1\times(1.5\times10^9)/2}{6.67\times10^{-11}}[/tex]which evaluates to give
[tex]\boxed{M=1.47\times10^{20}\operatorname{kg}\text{.}}[/tex]which is the mass of the planet.
Part B.
Here we use Kepler's third law, which says
[tex]T^2=\frac{4\pi^2}{GM}a^3[/tex]
where
T = orbital period of the planet
M = mass of the star
a = distance between the star and the planet.
G = gravitational constant
Now solving for M gives
[tex]M=\frac{4\pi^2}{GT^2}a^3[/tex]Now putting a = 2.20 * 10^11 , G = 6.67 * 10^-11, and T = 402 * 24 * 60 *60 gives
[tex]M=\frac{4\pi^2}{(6.67\times10^{-11})\times(402\cdot24\cdot60\cdot60)^2}\cdot(2.20\times10^{11})^3[/tex][tex]\boxed{M=5.22\times10^{30}\operatorname{kg}\text{.}}[/tex]Hence, the mass of the star is 5.22 * 10^30 kg.