Answer:
[tex]9.4\cdot 10^{10} m[/tex]
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_o[/tex] is the distance of the new object from the sun (orbital radius)
[tex]T_o=180 d[/tex] is the orbital period of the object
[tex]r_e = 1.50\cdot 10^{11} m[/tex] is the orbital radius of the Earth
[tex]T_e=365 d[/tex] is the orbital period the Earth
Solving the equation for [tex]r_o[/tex], we find
[tex]r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m[/tex]
