Example
The International Space Station orbits at an altitude of \(400 km\) above the surface of the Earth. What is the space station’s linear speed?
IG \(=6.67 \times 10^{-11} N m ^2 kg ^{-2}\), mass of the Earth \(=6.0 \times 10^{2+} kg\), radius o[ the Earth \(=6.37 \times 10^6 m\) ]
Solution
\[
\begin{array}{l}
M=6.0 \times 10^{24} kg , G=6.67 \times 10^{-11} \\
N m ^2 kg ^{-2}, v=\text { ? } \\
\left.r=R+h=\left(6.37 \times 10^6\right)+(400) \times 10^3\right) \\
=6370000+400000 \\
=6770000 m \\
\end{array}
\]
Using formula \(v=\sqrt{\frac{G M}{r}}\)
\[
\begin{array}{l}
=\sqrt{\frac{\left(6.67 \times 10^{-11}\right) \times}{\left.677010^{24}\right)}} \\
=7669 m s ^{-1}
\end{array}
\]
The linear speed of the International Space Station is \(7669 m s ^{-1}\).
Example
A satellite is orbiting the Earth with a linear speed of \(3200 m s ^{-1}\).
What is the orbital radius?
\(G=6.67 \times 10^{-11} N m ^2 kg ^{-2}\),
mass of the Earth \(=6.0 \times 10^{24} kg\),
radius of the Earth \(\left.=6.37 \times 10^6 m \right]\)
Solution
\[
\begin{array}{l}
v=3200 m s ^{-1}, \\
M=6.0 \times 10^{24} kg , \\
\begin{array}{l}
G=6.67 \times 10^{-11} N m ^2 kg ^{-2}, \\
r=? \\
\text { Using } v=\sqrt{\frac{G M}{r}} \\
r=\frac{G M}{v^2} \\
=\frac{\left(6.67 \times 10^{-11}\right)\left(5.97 \times 10^{24}\right)}{(3200)^2} \\
=3.897 \times 10^7 m
\end{array}
\end{array}
\]
The orbital radius for this satellite is \(3.897 \times 10^7 m\).
\[
\begin{array}{|l|l|l|}
\hline \text { Characteristic } & \text { Geostationary satellite } & {\begin{array}{c}
\text { Non-geostationary } \\
\text { satellite }
\end{array}} \\
\hline \text { Location } & \begin{array}{l}
\text { – The orbit must be } \\
\text { above the equator in } \\
\text { one plane, known as } \\
\text { Geostationary Earth } \\
\text { Orbit } \\
\text { – Above the same } \\
\text { geographical location }
\end{array} & \begin{array}{l}
\text { – Below or above } \\
\text { Geostationary Earth } \\
\text { Orbit } \\
– \text { Above different } \\
\text { geographical locations }
\end{array} \\
\hline \begin{array}{l}
\text { Direction of } \\
\text { motion }
\end{array} & \begin{array}{l}
\text { Same direction as } \\
\text { Earth’s rotation }
\end{array} & \begin{array}{l}
\text { Not the same direction as } \\
\text { Earth’s rotation }
\end{array} \\
\hline \text { Orbital period } & \text { The period is 24 hours } & \begin{array}{l}
\text { The period of orbit is less } \\
\text { or more than 24 hours }
\end{array} \\
\hline \text { Uses } & \text { Telecommunication } & \text { GPS, weather forecast } \\
\hline
\end{array}
\]
Example
Find the orbital radius of a geostationary satellite. What height is this satellite above the Earth’s surface? \(\left[G=6.67 \times 10^{-11} N m ^2 kg ^{-2}\right.\), mass of the Earth \(=6.0 \times 10^{24} kg\), radius of the Earth \(=6.37 \times 10^6 m\) ]
Solution
Using equation \(T^2=\frac{4 \pi^2 r^3}{G M}\)
\[
\begin{aligned}
r^3 & =\frac{G M T^2}{4 \pi^2} \\
& =\frac{\left(6.67 \times 10^{-11}\right)\left(6.0 \times 10^{24}\right)}{\times(24 \times 60 \times 60)^2} \\
& =7.567 \times 10^2 \\
r & =4.23 \times 10^7 m
\end{aligned}
\]
Height above Earth’s surface,
\[
\begin{aligned}
h & =r-R \\
& =\left(4.23 \times 10^7\right)-\left(6.37 \times 10^6\right) \\
& =3.59 \times 10^7 m
\end{aligned}
\]
Example
A satellite is placed in an orbit around Jupiter. The orbital period is 10 hours. What is the orbital radius?
\[
\left[G=6.67 \times 10^{-11} N m kg ^{-2}\right. \text {, mass of Jupiter} =1.9 \times 10^{27} kg ] \]
Solution
\[
\begin{array}{l}
G=6.67 \times 10^{-11} N m kg ^{-2} \text {, } \\
M=1.9 \times 10^{27} kg \\
\text { Using formula, } r^3=\frac{G M T^2}{4 \pi^2} \\
\left(6.67 \times 10^{-11}\right)\left(1.9 \times 10^{27}\right) \\
=\frac{\times(10 \times 60 \times 60)^2}{4 \pi^2} \\
=4.160 \times 10^{24} m ^3 \\
\end{array}
\]
Therefore, \(r=1.61 \times 10^8 m\)
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