Answer: 5.808 AU
Explanation:
According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
[tex]T^{2}\propto a^{3}[/tex] (1)
Talking in general, this law states a relation between the orbital period [tex]T[/tex] of a body (moon, planet, satellite, comet) orbiting a greater body in space with the size [tex]a[/tex] of its orbit.
However, if [tex]T[/tex] is measured in years, and [tex]a[/tex] is measured in astronomical units (equivalent to the distance between the Sun and the Earth: [tex]1AU=1.5(10)^{8}km[/tex]), equation (1) becomes:
[tex]T^{2}=a^{3}[/tex] (2)
Knowing [tex]T=14 years[/tex] and isolating [tex]a[/tex] from (2):
[tex]a=\sqrt[3]{T^{2}}=T^{2/3}[/tex] (3)
[tex]a=(14 years)^{2/3}[/tex] (4)
Finally:
[tex]a=5.808 AU[/tex]
