# A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is...

## Question:

A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is {eq}1.50 \times 10^8 {/eq} km away from the sun?

## Kepler's Law:

Kepler's law is the laws given by the Johannes Kepler, which gives information about the motion of planets around the sun. There are three laws given by the Johannes Kepler. One of them says that the square of time period taken by the planet to revolve around the sun is comparatively equal to the cube of the distance between the planet and the sun.

Given data

• The distance between Saturn and earth is: {eq}{r_e} = 1.50 \times {10^8}\;{\rm{Km}} {/eq}.
• The Saturn year is 29.5 times the earth year.
• The time period of earth is: {eq}{T_e} = T {/eq}
• The time period of Saturn is: {eq}{T_s} = 29.5\;T {/eq}.

The relation between time period and radius is given by keplers law,

{eq}{T^2}\alpha \;{r^3} {/eq}

The distance between Saturn and sun can be calculated by above relation.

{eq}{\left( {\dfrac{{{T_s}}}{{{T_e}}}} \right)^2} = {\left( {\dfrac{{{r_s}}}{{{r_e}}}} \right)^3} {/eq}

Here, {eq}{r_s} {/eq} is distance between Saturn and sun.

Substitute the known value.

{eq}\begin{align*} {\left( {\dfrac{{29.5\;T}}{T}} \right)^2} &= {\left( {\dfrac{{{r_s}}}{{1.50 \times {{10}^8}\;{\rm{km}}}}} \right)^3}\\ {r_s}^3 &= {\left( {1.50 \times {{10}^8}\;{\rm{km}}} \right)^3}{\left( {\dfrac{{29.5\;T}}{T}} \right)^2}\\ {r_s} &= \left( {1.50 \times {{10}^8}\;{\rm{km}}} \right){\left( {\dfrac{{29.5\;T}}{T}} \right)^{\dfrac{2}{3}}}\\ &= 1.43 \times {10^9}\;{\rm{km}} \end{align*} {/eq}

Thus, the distance between the Saturn and sun is {eq}1.43 \times {10^9}\;{\rm{km}} {/eq}. 