Solved Problem on Circular Motion

The International Space Station (ISS) moves around the Earth in 1.5 hours. Find:
a) The period of rotation in minutes and seconds;
b) The frequency of rotation of the station around the Earth in Hertz.

Problem data:

Period of rotation of the station around the Earth: T = 1.5 h;

Problem diagram:

Figure 1

Solution

a) Converting the rotation period, given in hours to minutes

\[ \begin{gather} T=\left(1.5\;\mathrm{\cancel h}\right)\times\left(\frac{60\;\mathrm{min}}{1\;\mathrm{\cancel{h}}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T=90\;\mathrm{min}} \end{gather} \]

The period in seconds will be

\[ \begin{gather} T=\left(1.5\;\mathrm{\cancel h}\right)\times\left(\frac{3600\;\mathrm s}{1\;\mathrm{\cancel h}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T=5400\;\mathrm{s}} \end{gather} \]

Note: to get the rotation period in seconds. we could also take the result obtained in minutes and multiply by 60, \( T=\left(90\;\mathrm{\cancel{min}}\right)\times\left(\frac{60\;\mathrm s}{1\;\mathrm{\cancel{min}}}\right)=5400\;\mathrm s \) , obtaining the same result.

b) To obtain the frequency, in Hertz, we use the value of the period in seconds, obtained in the previous item, and calculate

\[ \begin{gather} \bbox[#99CCFF,10px] {f=\frac{1}{T}} \end{gather} \]

\[ \begin{gather} f=\frac{1}{5400\;\mathrm s} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {f=0.000185=1.85\times 10^{-4}\;\mathrm{Hz}} \end{gather} \]

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