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3.1.3 Centripetal Force in the Motion of Satellites and Planets

SPM Physics Revision 3.1 Newton’s Law of Universal Gravity 3.1.3 Centripetal Force in the Motion of Satellites and Planets

Image of centripetal force experiment

Diagram above shows the circular motion of a ball attached to a string when it is rotated above the head.
The tension in the string provides the centripetal force for the circular motion of the ball
The ball moves in a circle at a constant speed. The velocity is constantly changing because the direction is changing.
The ball is accelerating towards the centre of circle because the velocity is constantly changing. This acceleration is called centripetal acceleration, \(a=\frac{v^2}{r}\).
For this acceleration to happen, there must be a resultant force. This force is called the centripetal force, \(F\).
From Newton’s Second Law of Motion, \(F=m a\)
\[
F=\frac{m v^2}{r}
\]
where \(m=\) mass
\(v=\) linear speed
\(r=\) radius of circle

Example

A cyclist goes round a circular track of radius \(10 m\) at a constant speed of \(5.0 m s ^{-1}\). Calculate
(a) the acceleration of the cyclist,
(b) the resultant force on the bicycle and cyclist if together they have a mass of \(80 kg\)
Solution
(a) Given \(r=10 m , v=5.0 m s ^{-1}\), \(a=\) ?
Using formula \(a=\frac{v^2}{r}=\frac{5^2}{10}\)
\[
=2.5 m s ^{-2}
\]
(b)
\[
\text { Using formula } \begin{aligned}
F & =m a \\
& =80 \times 2.5 \\
& =200 N
\end{aligned}
\]

Image of Satellite

Diagram above shows a satellite orbiting the Earth in a circular motion. The centripetal force is directed towards the Earth.
The centripetal force is equivalent to gravitational force for the satellite to maintain the motion.

Example

A satellite was launched into a circular orbit near the Earth. It travels around the Earth once every 97 minutes and orbits about \(547 km\) above the Earth. [Radius of the Earth \(=6.37 \times 10^6 m\) ]

(a) Calculate the linear speed of the satellite.
(b) The mass of the satellite is \(1.2 \times 10^4 kg\). Calculate the magnitude of the centripetal force that acts on it.
Solution
(a) Radius of orbit, \(r=R+h\)
\[
\begin{aligned}
r & =\left(6.37 \times 10^6\right)+547000 \\
& =6917000 m
\end{aligned}
\]
Period of orbit, \(T=97 \times 60\)
\[
=5820 s
\]
Linear speed \(v=\frac{\text { distance }}{\text { time }}\)
\[
\begin{array}{l}
=\frac{2 \pi r}{T} \\
=\frac{2 \pi(6917000)}{5820} \\
=7467 m s ^{-1}
\end{array}
\]
(b) Given \(m =1.2 \times 10^4 kg\) and \(v=7467 m s ^{-1}\) from (a).
Using \(F=\frac{m v^2}{r}\)
\[
\begin{array}{l}
=\frac{\left(1.2 \times 10^4\right)(7467)^2}{6917000} \\
=96729 N
\end{array}
\]

Mass of the Earth and the Sun

Image of Earth orbiting the Sun

Diagram 3.5 shows the attraction force between the Sun and a planet.
The centripetal force required to keep a planet in a circular orbit is the gravitational force between the Sun and the planet.
Therefore,
\[
\begin{aligned}
\frac{G M m}{r^2} & =\frac{m v^2}{r} \\
v & =\sqrt{\frac{G M}{r}}
\end{aligned}
\]
Speed of the planet around the Sun, \(v=\frac{2 \pi r}{T}\) where \(T\) is the period of orbit.
Equating equations in 2 and 3 ,
\[
\begin{aligned}
\sqrt{\frac{G M}{r}}= & \frac{2 \pi r}{T} \\
& M=\frac{4 \pi^2 r^3}{G T^2}
\end{aligned}
\]

Example

Given that

the orbital radius of the Moon around the Earth is about \(384000 km\),
the period of the orbit of the Moon around the Earth is about \(2 7 . 3\) days,
the orbital radius of the Earth around the Sun, \(r\), is \(1.5 \times 10^8 km\),
the period of the orbit of the Earth around the Sun is about 365 days and
\(G=6.67 \times 10^{-11} N m ^2 kg ^{-2}\)
1. Determine the mass of thi Earth.
2. Determine the mass of the Sun.

Solution
(a) Mass of the Earth, \(M\)
\[
\begin{align}
M & =\frac{4 \pi^2 r^3}{G T^2} \\
& =\frac{4 \pi^2\left(3.84 \times 10^8\right)^3}{\left(6.67 \times 10^{-11}\right)(27.3 \times 24 \times 3600)^2 }
\\
& =6.02 \times 10^{24} kg
\end{align}
\]

(b) Mass of the Sun,
\[
\begin{aligned}
M & =\frac{4 \pi^2 r^3}{G T^2} \\
& =\frac{4 \pi^2\left(1.5 \times 10^{11}\right)^3}{\left(6.67 \times 10^{-11}\right)(365 \times 24 \times 3600)^2 } & \\
& =2.01 \times 10^{30} kg
\end{aligned}
\]

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