Chapter 7, Problem 6STP

To determine

The distance of the satellite $S_2$ from the center of the planet.

Answer to Problem 6STP

  $8\times {10}^5\text{}\text{km}$

Explanation of Solution

Given:

Satellite $S_1$ takes 20 days to orbit the planet at a distance of $2\times {10}^5\text{}\text{km}$ from the center of the planet while satellite $S_2$ takes 160 days to orbit the planet.

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:

Satellite $S_1$ takes 20 days to orbit the planet, thus $T_1=20\text{days}$ .

Satellite $S_2$ takes 160 days to orbit the planet., thus $T_2=\text{160 days}$ .

And, the distance of satellite $S_1$ from the center of the planet is $2\times {10}^5\text{}\text{km}$ , thus $r_1=2\times {10}^5\text{}\text{km}$ .

Let, the average distance of satellite $S_2$ from the planet be $r_2$ . Then, using Kepler’s law,

  $\begin{array}{c}{\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{r_2}{r_1}\right)}^3\\ {\left(\frac{160}{20}\right)}^2={\left(\frac{r_2}{2\times {10}^5\text{}\text{km}}\right)}^3\\ 4=\frac{r_2}{2\times {10}^5\text{}\text{km}}\\ r_2=8\times {10}^5\text{}\text{km}\end{array}$

Conclusion:

Thus, the distance of the satellite $S_2$ from the planet is $8\times {10}^5\text{}\text{km}$ .