Question

A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average distance from the sun?

Answer 1

Answer: $72.66 AU=1.089(10)^{10} km$

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ of its orbit”:

$T^{2}\propto a^{3}$ (1)  

Now, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$), equation (1) becomes:  

$T^{2}=a^{3}$ (2)  

So, knowing $T=619.36 years$ and isolating $a$ from (2) we have:  

$a=\sqrt[3]{T^{2}}$ (3)  

$a=\sqrt[3]{(619.36 years)^{2}}$ (4)  

Finally:

$a=72.66 AU$ T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

$a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km$