Solutions to Intermediate Astronomy Problems (2)

Recall the problems from the previous instalment (2):


(1) What would be the sidereal period of an inferior planet that appeared at greatest western elongation exactly once a year?

(2) If the synodic period of Saturn is 1.03513 years, find its sidereal period.

(3) A planet's elongation is measured as 125 degrees. Is it an inferior or superior planet?

(4)If the sidereal period of Mercury is 88 days what is its synodic period?

(5) Calculate the ratio of the Earth's mean motion, n, to that of the planet Neptune, given the distance of Neptune is 30.06 AU. (Hint: You will need to apply Kepler's 3rd or harmonic law, see 'Tackling simple Astronomy problems (6)' to first get Neptune's period!)

(6) A newly discovered planet in the Zeta Reticuli system is found to have a semi-major axis of 10 AU. Compare its mean motion n(z) to that of Saturn (hint: reference solution to Problem #2)


Solutions:

1) Recall we define 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, ie. "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

And 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 blog it is evident that an inferior planet's maximum elongation is still less than 90 degrees - i.e. when its geocentric radius vecotr is tangential to its ornit 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

And 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


5) We are comparing the mean motions, viz.

n1 = 360 deg/ P1

n2 = 360 deg/P2

and let P1 = 365¼ days = 1 yr. for Earth (hence n1 = 0.985647 deg/day) and P2 for Neptune is to be found. But we know: a = 30.06 AU

By use of Kepler's 3rd or harmonic law:


(P2/ P1)^2 = k(a2/ a1)^3

where we let P1, a1 have Earth values (a1 = 1 AU, P1 = 1 yr.) then if a2 = 30.06 AU:

(P1/ 1 yr)^2 = k(30.06 AU/ 1 AU)^3


P = [(30.06)^3]^½ = 164.8 yrs.

Then Neptune's mean motion is:

n2 = 360 deg/P2 = 360 deg/ (164.8 yrs.) = 2.18 deg/ year

or, in terms of degrees per day:

n2 = 360 deg/ (60193 d) = 0.006 deg/day (approx.)

Comparing the two by ratio:

n1/ n2 = (0.985647 deg/day)/ 0.006 deg/day = 164.8 x greater

Which could also have been arrived at by taking the ratio (P2/ P1).

6) We know already that Saturn's sidereal period = 29.4 yrs. (solution to #2)

For the planet in Zeta Reticuli we must use the same Kepler 3rd law approach used in #5, and we know a2 = 10 AU - let a1, P1 be Earth values) so:

(P(z)/ 1 yr)^2 = k(30.06 AU/ 1 AU)^3

P(z) = [10)^3]^½ = 31.62 yrs.

so the planet's mean motion is:

n(z) = 360 deg/ P(z) = 360 deg/ 31.6 yr = 11.4 deg/yr.

and for Saturn:

n(S) = 360 deg/P(S) = 360 deg/ 29.4 yrs. = 12.2 deg/yr.