Earth and Saturn both revolve in definite orbits around sun due to gravitational force of attraction due to sun now when a planet revolves in orbit around sun due to attractive gravitational force between planet and sun, from Kepler’s third law of planetary motion we know that square of time period one complete rotation of planet T around sun is proportional to cube of mean distance (average distance) between planet and sun R

T²∝ R³

Le the Time taken by the earth for one complete revolution be T_e

T_e = 1 Year

Let mean distance of the earth from the sun or orbital radius be Re

Time is taken by the Saturn for one complete revolution or its time period of revolution around the sun be T_s

We are given that a Saturn’s year is 29.5 times the earth year i.e. time period of Saturn is 29.5 times that of earth

i.e. T_s = 29.5 T_e

Let, the orbital radius of this Saturn be R_s

Now, according to the Kepler’s third law of planetary motion, we have

And

Using both equations we get the relation

;

Or we can say

Simplifying we get the relation for the orbital radius of Saturn as

now we have

T_s = 29.5 Te

i.e. T_s/T_e = 29.5

i.e. radius of planet is

Putting the value of the orbital radius of the earth

R_e = 1.50 × 10⁸ km

we get the distance of Saturn from the sun as

R_p = 9.54 × 1.50 × 10⁸ km = 1.43 × 10⁹ Km = 1.43 × 10¹² m

So orbital radius of Saturn is 1.43 × 10⁹ Km or we can say it is at a distance of 1.43 × 10¹² m from sun