Chapter 9, Problem 100A

To determine

Theradius of the circular orbit of the satellite

Answer to Problem 100A

The radius of the circular orbit of the satellite = 42984 km

Explanation of Solution

Given:

The period of revolution of the satellite $T_s =1\text{day}$

The period of revolution of moon $T_m =27.33\text{days}$

Radius of moon $r_m =3.9\times {10}^8\text{m}$

Formula Used:

According to Kepler’s law, the time period of revolution ( T ) of asatellite around a planet and the distance between planet to satellite( r ) are related as,

  $T^2\alpha r^3$

Calculation:

From Kepler’s law,

  $\begin{array}{l}\text{For Earth- Moon system , }{T}_m{}^2\alpha r_m{}^3\\ \text{For Earth- Satellite system , }{T}_s{}^2\alpha r_s{}^3\end{array}$

  $\begin{array}{c}{\left(\frac{T_m}{T_s}\right)}^2={\left(\frac{r_m}{r_s}\right)}^3\\ {\left(\frac{\text{27}\text{.33 day}}{\text{1 day}}\right)}^{\text{2}}\text{= }{\left(\frac{\text{3}{\text{.9×10}}^{\text{8}}\text{m}}{r_s}\right)}^{\text{3}}\\ r_s{}^3=\frac{{\left(\text{3}{\text{.9×10}}^{\text{8}}\text{m}\right)}^{\text{3}}}{\text{746}\text{.93}}\\ r_s\text{ = 42983}\text{.8 km}\\ \approx 42984\text{km}\end{array}$

Conclusion:

The radius of the circular orbit of the satellite obtained is 42984 km