Answer:
8 years
Explanation:
We can solve the problem by using Kepler's third law, which states that the cube of the radius of the orbit of a planet is proportional to the square of its orbital period. We can rewrite this law as follows:
[tex]\frac{r_1^3}{T_1^2}=\frac{r_2^3}{T_2^2}[/tex]
where
[tex]r_1 = 1 AU[/tex] is the distance of the Earth from the Sun
[tex]T_1=1 y[/tex] is the Earth's orbital period
[tex]r_2 =4 AU[/tex] is the distance of the planet from the Sun
[tex]T_2[/tex] is its orbital period
By re-arranging the equation and substituting the numbers, we can find T2:
[tex]T_2 = \sqrt{\frac{r_2^3 T_1^2}{r_1^3}}=\sqrt{\frac{(4AU)^3(1y)^2}{(1 AU)^3}}=\sqrt{4^3}=\sqrt{64}=8 y[/tex]
