The period of the planet is 58.1 years
Explanation:
We can solve this problem by using Kepler's third law, which states that:
"The square of the orbital period of a planet is proportional to the cube of its semimajor orbital axis"
Translated into equations, we can write:
[tex]\frac{T_x^2}{r_x^3}=\frac{T_e^2}{r_e^3}[/tex]
Where, taking the orbits of the planets as almost circular, we have:
[tex]T_x[/tex] is the orbital period of the unknown planet
[tex]r_x = 15.0 au[/tex] is the radius of the orbit of the planet
[tex]T_e = 1 y[/tex] (1 year) is the period of the orbit of the Earth
[tex]r_e = 1 AU[/tex] is the orbital radius of the Earth
Solving for [tex]T_x[/tex], we find the orbital period of the unknown planet:
[tex]T_x = T_e \sqrt{\frac{r_x^3}{r_e^3}}=(1y)\sqrt{\frac{(15.0)^3}{(1.0)^3}}=58.1 y[/tex]
Learn more about Kepler's third law:
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