Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years in a circular orbit around its star, which has a measured mass of 3.03*1030 kg. Find the radius of the exoplanet's orbit.

r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

F = ma

G m M / r² = m a

The centripetal acceleration is given by

a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

v = d / t

The distance traveled Esla orbits, in a circle the distance is

d = 2 π r

Time in time to complete the orbit, called period

v = 2π r / T

Let's replace

G m M / r² = m a

G M / r² = (2π r / T) ² / r

G M / r² = 4π² r / T²

G M T² = 4π² r3

r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

T = 1.03 10⁸ s

Let's calculate

r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

r = ∛ (21.44 10³⁵ / 39.478)

r = ∛ (0.0543087 10 36)

r = 0.3787 10¹² m

r = 3.787 10¹¹ m