Question

activity c (continued from previous page)

1. gather data: use the additional data display to find the orbital radius and period of each planet. record this data in the first two columns of the table below. include units.

planet	mean orbital radius (au)	period (earth years)	(r^{3})	(t^{2})
mercury
venus
earth
mars
jupiter
saturn
uranus
neptune

1. analyze: what happens to the period as the orbital radius increases?
2. calculate: kepler discovered a very interesting relationship between the cube of each planets orbital radius and the square of its period. use a calculator to find the cube of each planets orbital radius, and record these values in the “(r^{3})” column of the table. record the squares of the periods in the “(t^{2})” column.

how do the numbers in the “(r^{3})” and “(t^{2})” columns compare?
keplers third law states that the cube of the orbital radius is proportional to the square of the period for any orbiting body. if the orbital radius is measured in astronomical units and the period is measured in earth years, the numbers are nearly identical.

1. predict: pluto has an orbital radius of 39.529 au. based on keplers third law, what is the approximate period of plutos orbit?

(hint: find the cube of the orbital radius first, and then take the square root.)

1. confirm: look up plutos actual period in the gizmo. what is it, and how does it compare to the calculated value?

Explanation:

Step1: Recall Kepler's third - law formula

$R^{3}=kT^{2}$, where $R$ is the orbital radius, $T$ is the period, and $k$ is a constant. When $R$ is in astronomical units (AU) and $T$ is in Earth - years, $k = 1$.

Step2: Solve for the period of Pluto

Given $R_{Pluto}=39.529$ AU. First, find $R_{Pluto}^{3}=(39.529)^{3}$.
$(39.529)^{3}=39.529\times39.529\times39.529\approx61700.7$.
Since $R^{3}=T^{2}$ (from Kepler's third law with the given units), then $T=\sqrt{R^{3}}$.
$T=\sqrt{61700.7}\approx248$ Earth - years.

Answer:

The approximate period of Pluto's orbit is 248 Earth - years.