Answer:
4 AU
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the square of the period of revolution of a planet around the Sun and the cube of its average distance from the Sun is constant for every planet orbiting the Sun:
[tex]\frac{T^2}{r^3}=k[/tex]
where
T is the orbital period
r is the average distance of the planet from the Sun
If we take two planets 1 and 2, the equation can be rewritten as
[tex]\frac{T_1^2}{r_1^3}=\frac{T_2^2}{r_2^3}[/tex]
In this problem, we have:
[tex]T_1 = 1 y[/tex] is the orbital period of the Earth
[tex]r_1 = 1 AU[/tex] is the distance of the Earth from the Sun
[tex]T_2 = 8 y[/tex] is the orbital period of the second planet
Therefore, we can re-arrange the equation to calculate r2, the averag distance of the other planet from the Sun:
[tex]\frac{r_2^3}{T_2^2}=\frac{r_1^3}{T_1^2}\\r_2 = \sqrt[3]{\frac{r_1 ^2 T_2^2}{T_1^2}}=\sqrt[3]{\frac{(1 AU)^3(8 y)^2}{(1 y)^2}} =4 AU[/tex]
