Question

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

Answer 1

T = 1.4696 10⁴ years

Step-by-step 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