question_answer

The moon has a period of 28 days and an orbital radius\[\text{3}.\text{8}\times \text{1}0\text{5 km}\]. What is the orbital radius of a satellite that has a period of one day?

Answer:

According to Kepler?s third law, \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{r_{1}^{3}}{r_{2}^{3}}\] or \[{{r}_{2}}={{r}_{1}}{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2/3}}\] Here\[{{r}_{1}}=3.8\times 105\,km;\,\,{{T}_{1}}=28\,\,days;\,\,{{T}_{2}}=1\,\,day.\] \[\therefore {{r}_{2}}=3.8\times {{10}^{5}}\times {{\left( \frac{1}{28} \right)}^{2/3}}=4.1\times {{10}^{4}}km\]