Question

What is the semimajor axis (in AU) of a planet with an orbital period of 14 years?

Answer 1

Answer: 5.808 AU

Step-by-step 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”.

`T^(2)\propto a^(3)` (1)

Talking in general, this law states a relation between the orbital period
`T` of a body (moon, planet, satellite, comet) orbiting a greater body in space with the size
`a` of its orbit.

However, if
`T` is measured in years, and
`a` is measured in astronomical units (equivalent to the distance between the Sun and the Earth:
`1AU=1.5(10)^(8)km`), equation (1) becomes:

`T^(2)=a^(3)` (2)

Knowing
`T=14 years` and isolating
`a` from (2):

`a=\sqrt[3]{T^(2)}=T^(2/3)` (3)

`a=(14 years)^(2/3)` (4)

Finally:

`a=5.808 AU`