3.2.1 Kepler’s Law

1. Kepler’s First Law (law of orbits): All planets orbit the Sun in a path described as an ellipse.
   

   

2. Kepler’s Second Law (law of equal areas): An imaginary line joining a planet and the Sun sweeps out an equal area of space in equal amounts of time.
   

   

3. Kepler’s Third Law: (law of periods) The square of period of a planet round the Sun is directly proportional to the cube of the radius of orbit.
   \[
   T^2 \propto r^3
   \]
   where \(T=\) orbital period of \(a\) planet
   \(r=\) radius of orbit
4. Mathematical derivation of Kepler’s Third Law Gravitational Centripetal force
   \[
   \begin{aligned}
   \frac{G M m}{r^2} & =\frac{m v^2}{r} \\
   \frac{G M}{r} & =v^2 . . .
   \end{aligned}
   \]
   Linear speed of planet,
   \[
   v=\frac{2 \pi r}{T}
   \]
   Substitute (2) into (1),
   \[
   \begin{array}{c}
   \frac{G M}{r}=\frac{(2 \pi r)^2}{T^2} \\
   T^2=\left(\frac{4 \pi^2}{G M}\right) r^3
   \end{array}
   \]
   \[
   \begin{aligned}
   T^2 & \propto r^3 \\
   \frac{T^2}{r^3} & =\text { constant }
   \end{aligned}
   \]

\[
\frac{T_1^2}{r_1^3}=\frac{T_2^2}{r_2^3}=\frac{T_3^2}{r_3{ }^3}
\]

Example

The average orbital distance of Mars is 1.52 times the average orbital distance of the Earth. Knowing that the Earth orbits the Sun in approximately 365 days, use Kepler’s Third Law to predict the time for Mars to orbit the Sun. [Given \(R_{\text {Mars }}=1.52 R_{\text {Earth }}\) and \(T_{\text {Earth }}\) \(=365\) days \(]\)
Solution
Use Kepler’s Third Law to relate the ratio of the period squared to the ratio of radius cubed.
\[
\begin{aligned}
\frac{\left(T_{\text {Mars }}\right)^2}{\left(T_{\text {Earth }}\right)^2} & =\frac{\left(R_{\text {Mars }}\right)^3}{\left(R_{\text {Earth }}\right)^3} \\
\left(T_{\text {Mars }}\right)^2 & =\left(T_{\text {Earth }}\right)^2 \times \frac{\left(R_{\text {Mars }}\right)^3}{\left(R_{\text {Earth }}\right)^3} \\
& =(365 \text { days })^2 \times(1.52)^3 \\
T_{\text {Mars }} & =684 \text { days }
\end{aligned}
\]
Note: \(\frac{R_{\text {Mars }}}{R_{\text {Earth }}}\) ratio is 1.52

Example

Given 2 satellites A and B are orbitting a planet. Satellite A takes 9 days to complete one full orbit around a planet, at a distance of 5.0 units. Satellite B orbits it in 7 days. How far is it from the planet?
Solution
\(T_A=9\) days, \(r_A=5\) units, \(T_B=7\) days, \(r_{ B }=\) ?

Using formula
\[
\begin{aligned}
\frac{\left(T_A^2\right)}{\left(r_A^3\right)} & =\frac{\left(T_B^2\right)}{\left(r_B^3\right)} \\
\frac{9^2}{5^3} & =\frac{7^2}{r_B^3} \\
r_B & =4.22 \text { units }
\end{aligned}
\]