Question

7. 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 , 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 6.76 AU from the sun? 15.23
17.58
154.46
3.58

Answer 1

In your question, this function is supposed to be in the little space after the word "function":

(square of the orbital period) = k · (cube of the distance from the sun)

' k ' is the same number for every solar-system object

To answer the question, we first have to find out what ' k ' is.  Since it's the same number for every planet in our solar system, we can find ' k ' for the Earth, and then use it for the mystery planet.

Period = 1 Earth year

Distance = 1 Astronomical unit

P² = k · D³

1² = k · 1³

k = 1  

Now for the mystery planet:  

P² = k · D³

P² = 1 · (6.76)³

P² = 308.9 Earth years²

P = √308.9

P = 17.58 Earth years

Answer 2

Answer: The planet will take 17.58 earth years

Explanation:

Mass of the sun = 1 Solar

Time period  of planet = T

Distance between the earth and planet = 6.67 AU

G = gravitation constant = $39.478 AU^3 y^{-2} MS^{-1}$

Using Kepler's third law:

$T^2=\frac{4\pi^2}{G M_{sun}}\times r^3$

$T^2=\frac{4\times 3.14\times 3.14}{39.478 AU^3 y^{-2} MS^{-1}\times 1 MS}\times (6.76 AU)^3$

$T^2=308.605 y^2$

T = 17.567156 years

The closest answer from the option is 17.58 years