Answer:
[tex]M= 4.8*10^{24}kg[/tex]
Explanation:
To solve this problem we need to apply the Kepler's third law, which say,
[tex]T^2 = \frac{4\pi^2R^3}{GM}[/tex]
Where ,
T= Period
R = Radius
G =Gravitational constant
M = Mass
We have all that values, then replacing,
[tex](3*10^5)^2 = 4\pi^2\frac{(0.9*10^8)^3}{6.67*10^{-11}M}[/tex]
Solving for M,
[tex]M = 4\pi^2\frac{(0.9*10^8)^3}{6.67*10^{-11}((3*10^5)^2 )}[/tex]
[tex]M= 4.8*10^{24}kg[/tex]
Note that [tex]0.9*10^8[/tex] is used because was provided the value of the diameter, not the radius which is equal to the half.
