## Thursday, January 13, 2022

### Solutions To Planetary Sidereal Period Computation Problems

1) Recall we defined the synodic period to mean the time interval for the revolution of a planet as determined from the same phases, or the same geometrical configurations, i.e. "western elongations". Then the sidereal period of an inferior planet that appeared at greatest western elongation exactly once a year would be from:

1/ S = 1/P1 - 1/P2

Bear in mind if the planet is inferior then P1 refers to the planet's sidereal period and P2 refers to Earth's.

So:

1/P1 = 1/S + 1/P2

where: S = 1 yr., P2 = 1 yr.

then:

1/P1 = 1/1 + 1/1 = 2

so: P1 = 1/2 yr.

2) Saturn is a superior planet 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.

3) From the diagrams in Fig. 1(a)-(b) of previous (Jan. 11)  blog post it is evident that an inferior planet's maximum elongation is still less than 90 degrees - i.e. when its geocentric radius vector is tangential to its orbit configuration. By contrast, the elongation of a superior planet can vary from 0 deg at conjunction to 180 degrees at opposition.

Hence the planet must be a superior one.

4) Again, we employ:

1/ S = 1/P1 - 1/P2

Bear in mind, that since Mercury is inferior then P1 refers to the planet's sidereal period and P2 refers to Earth's.

Then: P1 = 88 d and P2 = 365¼ d

So:

1/S = 1/88 - 1/ (365¼ days)

1/S = 0.01136 - 0.00273 = 0.00863

S = 1/ (0.00863) = 115.8 days