Chapter 7, Problem 34A

To determine

The length of the Jupiter’s orbital period in terms of Earth years.

Answer to Problem 34A

  $11.86\text{Earth years}$

Explanation of Solution

Given:

Jupiter is 5.2 times farther from the Sun than Earth.

Formula used:

Kepler’s Third Law,

  ${\left(\frac{T_A}{T_B}\right)}^2={\left(\frac{r_A}{r_B}\right)}^3$

Where, $T_A$ and $T_B$ are the periods of the planets A and B around a planet, and $r_A$ and $r_B$ are their average distances from the planet.

Calculation:

It is known that Earth takes 1 year to orbit the Sun, thus $T_E=\text{1 year}$ .

And, Uranus takes 84 years to orbit the Sun, thus $T_U=\text{84 years}$ .

Let, the average distance of Earth from the Sun be $r_E$ , and that of Jupiter from the Sun be $r_J$ . Then, it can be written as:

  $r_J=5.2r_E$

Now, using Kepler’s law,

  $\begin{array}{c}{\left(\frac{T_J}{T_E}\right)}^2={\left(\frac{r_J}{r_E}\right)}^3\\ {\left(\frac{T_J}{1\text{ Earth year}}\right)}^2={\left(\frac{5.2r_E}{r_E}\right)}^3\\ \frac{T_J}{1\text{ Earth year}}=11.86\\ T_J=11.86\text{Earth years}\end{array}$

Conclusion:

Thus, Jupiter’s orbital period around the Sun is $11.86\text{Earth years}$ .