Question

The mean distance of Jupiter from the sun is nearly 5.2 times the corresponding distance between earth and sun. Using Kepler's law, find the period of revolution of Jupiter in its orbit.

Answer 1

Hint: We are given the relation between Distance between earth and sun, and the distance between Jupiter and sun. Using Kepler’s law we can form the relation between time period and the distance to the sun. By equating this we can find the time period of revolution of Jupiter.
Formula used:
${{T}^{2}}\propto {{r}^{3}}$

Complete step-by-step solution:
In the question, it is said that the mean distance of the planet Jupiter from the sun is 5.2 times the corresponding distance between the earth and the sun.
We are asked to find the time period of Jupiter in its orbit, with the help of Kepler’s law.
Let us consider Kepler’s third law; according to the law, we know that the square of a planet’s time period is proportional to the cube of the distance between the sun and the planet.
${{T}^{2}}\propto {{r}^{3}}$, where ‘T’ is the time period of the planet and ‘r’ is the distance from the sun to that planet.
Now, let $'{{T}_{1}}'$ be the time and $'{{r}_{1}}'$ be the radius of Earth.
Then, according to Kepler’s third law we have,
${{T}_{1}}^{2}\propto {{r}_{1}}^{3}$
Let $'{{T}_{2}}'$ be the time period and $'{{r}_{2}}'$ be the radius of Jupiter.
Then we have,
${{T}_{2}}^{2}\propto {{r}_{2}}^{3}$
Dividing these two equations together, we get
${{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}$
It is given that the distance from sun to Jupiter is 5.2 times the distance from sun to Earth.
${{r}_{2}}=5.2\times {{r}_{1}}$
Therefore, we can write that
${{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \dfrac{{{r}_{1}}}{5.2\times {{r}_{1}}} \right)}^{3}} \\
 \Rightarrow {{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \dfrac{1}{5.2} \right)}^{3}} \\
 \Rightarrow \left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)={{\left( \dfrac{1}{5.2} \right)}^{{3}/{2}\;}} \\
 \Rightarrow {{T}_{2}}={{\left( 5.2 \right)}^{{3}/{2}\;}}\times {{T}_{1}} $
We know that time period of earth is 365 days = 1 year
${{T}_{1}}=1year$
Therefore,
$
 \Rightarrow {{T}_{2}}={{\left( 5.2 \right)}^{{3}/{2}\;}}\times 1 \\
 \Rightarrow {{T}_{2}}=11.858years $
Hence the time period of Jupiter in its orbit is 11.858years.

Note: Kepler’s laws were proposed as the laws for the motion of a planet in a sun-centered solar system.
There are three laws in Kepler’s laws of planetary motion.
The \[{{1}^{st}}\] law or the law of ellipses stated that all the planets move around the planets in an elliptical orbit rather than in a circular orbit with the position of the sun at the focus of the ellipse.
The ${{2}^{nd}}$ law or the law of areas stated that an imaginary line drawn from the center of the sun to the planet will cover out equal areas in equal time intervals.
Kepler’s ${{3}^{rd}}$ law or the law of harmonies stated that the ratio of the square of the time period of any planets will be equal to the ratio of the cubes of the average distance of that planet from the sun.
Or simple, the square of the time period of a planet is proportional to the cube of the distance of that planet from the sun.