Question

The mean distance from Saturn to the sun is 9 times greater than the mean distance from Earth to the sun. Calculate the time period of Saturn in terms of Earth’s time period
(A) 18 Earth years
(B) 27 Earth years
(C) 81 Earth years
(D) 243 Earth years
(E) 729 Earth years

Answer 1

Hint
The mean distance of Saturn from the sun is 9 times greater than the mean distance of the Earth from the sun’, means that the radius of Saturn is given 9 times the radius of Earth for its orbit around the sun. So by using Kepler’s third law of planetary motion, which is given by ${T^2} \propto {R^3}$ we can find the ratio of the time period of Saturn to that of Earth.
To solve this problem we will use the following formula,
${T^2} \propto {R^3}$
Where $T$ is the time period of the revolution of a planet around the sun and $R$ is the radius of the orbit of that planet.

Complete step by step answer
Kepler's third law of planetary motion states that the square of the time period of revolution of a planet in an orbit around the sun is proportional to the third power of the length of that orbit.
So mathematically we can write the Kepler’s law as,
$\Rightarrow {T^2} \propto {R^3}$
Let us consider the time period of revolution for Earth is given by ${T_1}$ and the radius be $\Rightarrow {R_1}$. So from the Kepler’s law, we have,
$\Rightarrow {T_1}^2 \propto {R_1}^3$
And let us consider the radius of Saturn be ${R_2}$ and its time period of revolution around the sun be given by ${T_2}$, then from Kepler’s law er have,
$\Rightarrow {T_2}^2 \propto {R_2}^3$
Now, in the question, we are given that the mean distance of Saturn from the sun is 9 times the mean distance of Earth from the sun. The mean distance means radius, so this statement means that the radius of the Saturn is 9 times the radius of Earth.
Since ${R_1}$ and ${R_2}$ are the radius of Earth and Saturn respectively, so
$\Rightarrow {R_2} = 9{R_1}$
Therefore the ratio of the square of the time period of the Saturn to that of the Earth is,
$\Rightarrow \dfrac{{{T_2}^2}}{{{T_1}^2}} = \dfrac{{{R_2}^3}}{{{R_1}^3}}$
Here in the place of ${R_2}$ we can write ${R_2} = 9{R_1}$.
So we get,
$\Rightarrow \dfrac{{{T_2}^2}}{{{T_1}^2}} = \dfrac{{{{\left( {9{R_1}} \right)}^3}}}{{{R_1}^3}}$
$\Rightarrow \dfrac{{{T_2}^2}}{{{T_1}^2}} = \dfrac{{{9^3}{R_1}^3}}{{{R_1}^3}}$
We can cancel the ${R_1}^3$ from both the numerator and denominator, and we get,
$\Rightarrow \dfrac{{{T_2}^2}}{{{T_1}^2}} = {9^3}$
Now we can take ${T_1}^2$ from the denominator of L.H.S to the R.H.S and get,
$\Rightarrow {T_2}^2 = {9^3}{T_1}^2$
Now we take square root on both the sides,
$\Rightarrow {T_2} = \left( {{9^{{3 \mathord{\left/
 {\vphantom {3 2}} \right.} 2}}}} \right){T_1}$
The value of $\left( {{9^{{3 \mathord{\left/
 {\vphantom {3 2}} \right.} 2}}}} \right)$ is 27.
So we get, ${T_2} = 27{T_1}$
Hence the time period of Saturn is 27 times of the time period of Earth.
So the correct answer is option (B).

Note
The Kepler’s third law which we have used in this question is also known as the law of harmonics as it compares the orbital period and the radius of orbit of a planet. Unlike the first and the second law, it compares the motion characteristics of different planets.