Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years

T = 1.4696 10⁴ years

Explanation:

For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law

               F = ma

               G m M / r² = m a_c = m v² / r

                G M / r = v²

the speed of the circular orbit is

               v = 2π r / T

we substitute

               G M / r = 4π² r² / T²

               T² = (4π² / G M)  r³

Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse

               T² = (4π² /GM)  a³

the constant is worth

                (4π² / GM) = 2.97 10⁻¹⁹    s² / m³

let's reduce the distance to SI units

AU is the distance from the Earth to the Sun

               a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)

               a = 8.976 10¹³ m

               T² = 2.97 10⁻¹⁹ (8.976 10¹³)³

               T² = 21.4786 10²²

               T = 4.63 10¹¹ s

Let's reduce to years

               T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)

               T = 1.4696 10⁴ years