We can show how to calculate approximate synodic periods for the
planets easily. See the diagram below first.
Planet b is an inner planet (an inferior planet relative to planet a
in ancient astro jargon)
and planet a is an outer planet (a superior planet relative to planet b
in ancient astro jargon).
Now planet b because it is inferior has a greater angular velocity about
the Sun than planet a. Thus planet b will lap planet a eventually:
i.e., planet b will travel from one point of alignment with planet a
and the Sun (the vertical alignment in the diagram) and come into
a new alignment (the oblique alignment in the diagram) after traveling
more than 360 degrees.
The time it takes to lap is the synodic period itself.
If we measure angle from the vertical, then we have the following relations
where \theta is the angle of the second alignment, r_a is the angular
speed of planet a, r_b is the angular speed of planet b, and t is the
synodic period, our unknown.
Now we solve for t eliminating \theta (which is an unknown that can stay
360+\theta=r_a*t and \theta=r_b*t
where p_a is the sidereal period of planet a and p_b is the sidereal period
of planet b.
Now in fact the synodic period of planet b relative to planet a is the
same as that of planet a relative to planet b.
Thus if we use absolute value signs, we need make no distinction in
the formula between superior and inferior planets:
t=360/(r_a-r_b), and thus
If we assume p_a is constant, then we can see how t varies as p_b grows
from 0 to p_a to infinity.
- Intially, for p_b< When p_b goes to p_a, then t goes to infinity: the planets have the
same period and are on the same orbit: one can never lap the other.
- As p_b grows greater than p_a, t decreases from infinity and asymptotically
approaches p_a: in the limit, planet b is practically standing still and
the synodic period is just planet a's sidereal period.