Answer:
8 years
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between square of the orbital period and the cube of the orbital radius of any object revolving around the Sun is constant:
[tex]\frac{T_e^2}{r_e^3}=\frac{T_a^2}{r_a^3}[/tex]
where:
[tex]T_e = 1 y[/tex] is the orbital period of Earth
[tex]r_e [/tex] is the orbital radius of Earth
[tex]r_a=2 r_e[/tex] is the orbital radius of the asteroid (which is twice that of Earth)
[tex]T_a = ?[/tex] is the orbital period of the asteroid
Substituting and re-arranging the equation, we find
[tex]T_a^2 = \frac{r_a^3}{r_e^3} T_e^2 = \frac{(2 r_e)^2}{r_e^3} T_e^2=8 \frac{r_e^3}{r_e^3} T_e^2=8 T_e^2 = 8 (1y)^2=8 y[/tex]
so, the orbital period of the asteroid will be 8 years.
