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

User Ashlyn
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1 Answer

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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

User MoMoney
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