__Question__: Can you please explain the difference between the sidereal and synodic periods for a planet and how they can be computed? Thanks!

__Answer__:

The

*sideral period*is the time interval for a planet's revolution about the Sun reckoned against the fixed stars. We photograph a planet at point 'x' in its orbit, with the stars determining a specific background format, then again when it reaches the same point x, relative to that background.

Of particular interest also in these computations are what we call the 'mean motions'. For reference I use the diagram below:

Let P1 and P2 be the

*sidereal periods*of revolution of two planets (p1 and p2) around the Sun. Then the mean motions are defined:

n1 = 360 deg/ P1

n2 = 360 deg/P2

where we assume for example, that n1 > n2

in which case the radius vector for the (interior) planet p1 gains on the radius vector for planet p2.

As an example, let us consider the earth (P2 = 365¼ days) vs. Venus (P1 = 224.7 days).

Then:

n2 = 360 deg/ (365¼ days) = 0.985647 deg/day

n1 = 360 deg/(224.7 days) = 1.602130 deg/day

which confirms that a planet nearer the Sun has a smaller period of revolution than one further away so that P1 < P2, and also: n1 > n2.(By virtue of the

*SMALLER PERIOD*it has a more rapid mean motion.)

And, in effect, in the case above, Venus' radius vector gains on the Earth's by (n2 - n1) degrees per day, or [(1.602130 deg/day) - (0.985647 deg/day)] = 0.61646 deg/day.

By inspection of the diagram let the alignment Sp1p2 denote the positions of the Sun, Venus and Earth at a particular time or "epoch" in the parlance of celestial mechanics. Then, it is clear on the same inspection that

*one synodic period*S will have elapsed by the time the two are in the new alignment Sp1'p2'. So, during the time interval, Venus' radius vector (Sp1) will have gained 360 degrees on the Earth's.

Now, since Sp1 gains on Sp2 by 0.61646 deg/day then it gains 360 degrees in a time S =

S x (0.61646 deg/day) = 360 deg

Or: S = 360 deg/ (0.61646 deg/day) = 583.9 days

Which is exactly the

*synodic*period for Venus!

Hence, in general:

S x (n1 - n2) = 360 deg or

S = 360 deg/ (n1 - n2)

Or, by reference to the earlier formulations for the n1, n2 mean motions:

S (360/P1 - 360/ P2) = 360 deg

And:

S = 360 deg/ [360/P1 - 360/ P2]

Or:

1/ S = 1/P1 - 1/P2

In most problem -solving scenarios a planet's period is sought when Earth's is known. The key to the solution lies in determining whether the unknown planet is an inferior one (e.g. interior to Earth's orbit) or a superior one, e.g. exterior to the Earth's orbit. Thus there emerge two cases:

(a) The planet is inferior (like Venus), then P1 refers to the planet's sidereal period and P2 refers to Earth's.

(b) The planet is superior (e.g. Jupiter) then P1 refers to the Earth's sidereal period and P2 to the planet's.

__Example Problem__:

The synodic period of Venus is found to be 583.9 days. If the length of the Earth's year is 365¼ days, find the

*sidereal period of Venus*:

1/S = 1/P1 - 1/P2

Venus is an inferior planet, so that Earth's period is inserted for P2.

Then:

1/(583.8 d) = 1/P1 - 1/(365¼ d)

Or:

1/P1 = [365¼ + 583.9 d]/ [583.9 x 365¼ ]

yielding P1 = 224.7 days

__Another problem with the roles of P1, P2 reversed__:

If the synodic period of Saturn is 1.03513 years, find its

*sidereal period*.

*:*

__Solution__Saturn is a superior planet (orbit exterior to Earth's) so we apply:

1/ S = 1/P1 - 1/P2

but

__now__with P1 =

*Earth's*sidereal period, and P2 =

*Saturn's*. Thus we have:

1/P2 = 1/P1 - 1/S

where S = 1.03513 years

1/P2 = 1/1 - 1/1.03513 = 1 - 0.966 = 0.0339

P2 = 1/ 0.0339 = 29. 4 yrs.

yielding P1 = 224.7 days

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