Chapter 1, Problem 1.1P

To determine

The relationship between a planet’s synodic period and its sidereal period. Consider both superior and inferior planets.

Answer to Problem 1.1P

The relationship between a planet’s synodic period and its sidereal period is $\frac{1}{S}=\frac{1}{P_{\oplus }}-\frac{1}{P}$ for superior planet and $\frac{1}{S}=\frac{1}{P}-\frac{1}{P_{\oplus }}$ for inferior planet.

Explanation of Solution

Interior planets are the planets whose orbits are inside the orbits of the earth and superior planets are those planets whose orbits are outside the orbits of the earth.

Synodic period $\left(S\right)$ is the former time interval of planet and Sidereal period $\left(P\right)$ is the latter time interval of planet.

Conclusion:

The required diagram is shown in Figure 1.

Figure 1

The above figure shows the orbital motion of the earth and mars.

The Earth makes $\frac{S}{P_{\oplus }}$ orbits during the time required for another planet to make $\frac{S}{P}$ orbits. If the other planet is superior, then to overtake the planet earth must make one extra orbit around the sun and it is expressed as,

    $\frac{S}{P_{\oplus }}=\frac{S}{P}+1$

By rearranging the above equation.

    $\begin{array}{c}\frac{S}{P_{\oplus }}-\frac{S}{P}=1\\ S\left(\frac{1}{P_{\oplus }}-\frac{1}{P}\right)=1\\ \frac{1}{S}=\frac{1}{P_{\oplus }}-\frac{1}{P}\end{array}$

In the case of inferior planet, the planet has to make one extra trip to overtake the earth.

    $\frac{S}{P}=\frac{S}{P_{\oplus }}+1$

By rearranging the above equation

    $\begin{array}{c}\frac{S}{P}-\frac{S}{P_{\oplus }}=1\\ S\left(\frac{1}{P}-\frac{1}{P_{\oplus }}\right)=1\\ \frac{1}{S}=\frac{1}{P}-\frac{1}{P_{\oplus }}\end{array}$

Thus, the relationship between a planet’s synodic period and its sidereal period is $\frac{1}{S}=\frac{1}{P_{\oplus }}-\frac{1}{P}$ for superior planet and $\frac{1}{S}=\frac{1}{P}-\frac{1}{P_{\oplus }}$ for inferior planet.