Question

Kepler's Third Law of Orbital Motion states that you can approximate the period P (in Earth years) it takes a planet to complete one orbit of the sun using the function P = d^(2/3), where d is the distance (in astronomical units, AU) from the planet to the sun. How many Earth years would it take for a planet that is 6.66 AU from the sun?

a.) 14.79
b.) 17.19
c.) 3.54
d.) 147.7

Answer 1

Using Kepler's Third Law of Orbital Motion and the given distance of 6.66 AU, the approximate orbital period for a planet at this distance from the sun is calculated to be 17.19 Earth years. Therefore, option b) 17.19 is the correct answer.

Step-by-step explanation:

The question asks about Kepler's Third Law of Orbital Motion and its application to find the orbital period of a planet which is 6.66 astronomical units (AU) from the sun. According to Kepler's Third Law, the orbital period P (in Earth years) is approximate to the distance d (in AU) cubed and then taking the square root of that value (P = d^(3/2)). In order to find the orbital period for a distance of 6.66 AU, we raise 6.66 to the power of 3/2.

Calculation:

P = 6.66^(3/2)

P = (6.66^3)^(1/2)

P = (295.408)^(1/2)

P = √295.408

P ≈ 17.19 years

Therefore, the correct answer is b.) 17.19 Earth years.