Answer:
3 AU
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
[tex]\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}[/tex]
where
[tex]r_a[/tex] is the distance of the asteroid from the sun (orbital radius)
[tex]T_a=5.2 y[/tex] is the orbital period of the asteroid
[tex]r_e = 1 AU[/tex] is the orbital radius of the Earth
[tex]T_e=1 y[/tex] is the orbital period the Earth
Solving the equation for [tex]r_a[/tex], we find
[tex]r_a = \sqrt[3]{\frac{r_e^3}{T_e^2}T_a^2} =\sqrt[3]{\frac{(1 AU)^3}{(1 y)^2}(5.2 y)^2}=3 AU[/tex]
So, the distance of the asteroid from the Sun is exactly 3 times the distance between the Earth and the Sun.
