Question

Calculate the period of revolution of Jupiter around the sun. The ratio of the radius of Jupiter’s orbit to that of the Earth’s orbit is 5.
(Period of revolution of the Earth is 1year.)

Answer 1

Hint: Kepler’s third law establishes a relationship between time period of revolution and distance between the planet and the sun(orbit radius of the planet). Use this to determine the period of revolution of Jupiter.

Complete step by step answer:
Kepler’s third law of planetary motion states that the ratio time squared to the cube of orbital radius of planets is a constant,
So $c=\dfrac{{{T}^{2}}}{{{R}^{3}}}$
Where $R=$ Orbital distance
            $T=$ Orbital period
For earth
$c=\dfrac{T_{e}^{2}}{R_{e}^{3}}$ ……..(i)
For Jupiter
$c=\dfrac{T_{j}^{2}}{R_{j}^{3}}$ ………..(ii)
Compare equation (i) and (ii)
$\dfrac{T_{e}^{2}}{R_{e}^{3}}=\dfrac{T_{j}^{2}}{R_{j}^{3}}$
$T_{j}^{2}=T_{e}^{2}\dfrac{R_{j}^{3}}{R_{e}^{3}}$
$T_{j}^{2}=T_{e}^{2}{{\left( \dfrac{{{R}_{j}}}{{{R}_{e}}} \right)}^{3}}$
Given in question $\left( \dfrac{{{R}_{j}}}{{{R}_{e}}} \right)=5$
 $T_{j}^{2}=T_{e}^{2}\times {{\left( 5 \right)}^{3}}$
Given that period of revolution of the earth is 1 year
Then $T_{j}^{2}=1\times 125$
$T_{j}^{2}=125$
Square root of both of the side
$\begin{align}
  & \sqrt{T_{j}^{2}}=\sqrt{125} \\
 & {{T}_{j}}=11.18years \\
\end{align}$
$\therefore$ The final answer is 11.8years.

Note: It is interesting to note that as distance from the sun increases the time taken to complete revolutions increases exponentially.