Chapter 7.1, Problem 2PP

To determine

The period of the asteroid in Earth’s years.

Answer to Problem 2PP

  $T_A=2.828\text{Earth years}$

Explanation of Solution

Given:

An asteroid revolves around the Sun with a mean orbital radius twice that of Earth’s.

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 the Sun, and $r_A$ and $r_B$ are their average distances from the Sun.

Calculation:

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

Let, the average distance of Earth from the Sun be $r_E$ , and that of asteroid from the Sun be $r_A$ . Then using the given information,

  $r_A=2r_E$

Then, using Kepler’s law,

  $\begin{array}{l}{\left(\frac{T_A}{T_E}\right)}^2={\left(\frac{r_A}{r_E}\right)}^3\\ {\left(\frac{T_A}{1\text{year}}\right)}^2={\left(\frac{2r_E}{r_E}\right)}^3\\ T_A=2.828\text{years}\end{array}$

Conclusion:

Thus, the period of the asteroid is 2.828 Earth years.