Answer:
The satellite has a period of 204.90 days
Explanation:
The period can be determine by means of Kepler's third law:
[tex]T^{2} = r^{3}[/tex] (1)
Where T is the period of revolution and r is the radius.
[tex]\sqrt{T^{2}} = \sqrt{r^{3}}[/tex]
[tex]T = \sqrt{r^{3}}[/tex] (2)
The distance between the moon and the Earth has a value of 384400 km, therefore:
[tex]r = (384400km)*(0.50)[/tex]
[tex]r = 192200 km[/tex]
Finally, equation 2 can be used:
[tex]T = \sqrt{(192200)^{3}}[/tex]
[tex]T = 84261672 km[/tex]
However, the period can be expressed in days, to do that it is necessary to make the conversion from kilometers to astronomical units:
An astronomical unit (AU) is the distance between the Earth and the Sun ([tex]1.50x10^{8} km[/tex])
[tex]T = 84261672 km \cdot \frac{1AU}{1.50x10^{8} km}[/tex] ⇒ [tex]0.561 AU[/tex]
But 1 year is equivalent to 1 AU according to Kepler's third law, since 1 year is the orbital period of the Earth.
[tex]T = 0.561 AU \cdot \frac{1year}{1AU}[/tex] ⇒ [tex]0.561 year[/tex]
[tex]T = 0.561 year \cdot \frac{365.25 days}{1year}[/tex] ⇒ [tex]204.90 days[/tex]
[tex]T = 204.90 days[/tex]
Hence, the satellite has a period of 204.90 days.
