Using : 1/ S = 1/P1 - 1/P2
With P1 = Earth's sidereal period (1 yr), and P2 = Mars sidereal (687/365.25) = 1.88 yr
Then:
1/S = 1 – 1/1.88 = 1 – 0.5319 = 0.468 yr
Whence: S = 1/0.468 = 2.13 yrs. (or 779.9 days)
Mars’ distance from Earth is PE from the diagram in the original blog post, so b may be obtained from:
b - a cos Θ = PE cos φ
and: PE = (b - a cos Θ)/ cos φ
Now (given a = 1 and b = 1.52):
cos Θ = [Ö a Ö b (Ö a + Öb)] / (a 3/2 + b3/2 )
= [Ö (1.52) ( Ö (1) + Ö (1.52))]/ ((1) 3/2 + (1.52) 3/2)
cos Θ = 0.958
so: Θ = 16.7 o
And the angle of elongation (136.2) is (in terms of the geometry):
136.2 = 180 – (Θ + φ)
So we can solve for φ:
φ = 180 – 136.2 – 16.7 = 27.1 o
Now:
PE = (b - a cos Θ)/ cos φ
= (1.520 – 0.958)/ cos (27.1) = (0.562)/0.890 = 0.631
So the distance of Mars from Earth denoted by PE (at stationary point) = 0.631 AU
The time to next stationary point:
T r/2 = Θ/ 360 x S
where: (given a =1, b = 1.52):
So:
T r/2 = Θ/ 360 x S = (16.7)/360 x (779.9 d) = 36.1 days
2) Find the length of time Jupiter has retrograde motion in each synodic period given its heliocentric distance is 5.2 AU and its sidereal period is 11.86 years.
Solution:
First, use: 1/ S = 1/P1 - 1/P2
And thereby obtain Jupiter’s synodic period, S:
Where P1 = Earth's sidereal period (1 yr), and P2 = Jupiter’s sidereal period (11.86 yrs)
1/S = 1 – 1/(11.86) = 1 – 0.0843 = 0.9156
Then: S = 1/ 0.9156 = 1.09 yrs. = 398.1 d
The time for retrograde (r) motion in each synodic period is:
T r =2 Θ/ 360 x S
Where:
cos Θ = [Ö a Ö b (Ö a + Ö b)] / (a 3/2 + b 3/2 )
and a =1, b = 5.2, so:
cos Θ= [Ö (5.2) (Ö (1) + Ö (5.2))]/ ((1) 3/2 + (5.2) 3/2)
And: cos Θ = 0.582
So: Θ = arc cos(0.582) = 54.4 o
and the time to move retrograde over each synodic period;
T r =2 Θ/ 360 x S = (2 x 54.4)/360 x (1.09 yr) = 0.329 yr. = 120.3 days
3) In the Epsilon Eridani star system a planet designated Epsilon Eridani III is determined to have the exact same orbital parameters as Earth (e.g. a, e, i etc.). In the same system, another exoplanet designated Epsilon Eridani IV is found to have V p = 4.5 km/s. a) Using your knowledge of the known parameters, plus the diagram shown, construct an appropriate parallelogram of velocities and hence obtain the angles: Θ and φ.
Solution:
We use V p = 4.5 km/s and V E = 47 km/s for the velocities.
Since Eridani III has the same orbital parameters as Earth we can employ Earth semi-major axis, etc. in the computations. Also, a check of tables (or previous problems from earlier sets) shows Eridani IV has the same orbital velocity as Neptune so will have approximately the same sidereal period of 163.73 yrs. Now to form the parallelogram we need to obtain the angles Θ and φ.
We must first obtain the synodic period (S) of Eridani IV:
1/ S = 1/P1 - 1/P2
Where P1 = 1 yr (for Eridani III), and P2 = 163.7 yrs. for IV
1/S = 1 – 1/163.7 = 1 – 0.0061 = 0.9939
Then: S = 1/0.9939 = 1.006 yr. = 367.7 d
To get Θ:
cos Θ = [Ö a Ö b (Ö a + Ö b)] / (a 3/2 + b3/2 )
where b is Eridani IV’s semi-major axis or:
b = {[P2] 2} 1/3 = { [163.7] 2} 1/3 = 29.9 AU
And we know a (semi-major axis for Earth) = 1 (AU) , so:
cos Θ= [Ö (29.9) (Ö (1) + Ö (29.9))]/ ((1) 3/2 + (29.9)3/2)
cos Θ = 0.215 and Θ = arc cos(0.215) = 77.6 o
Meanwhile, by using a scaled diagram (e.g. see the example shown ) we find:
90 + φ = 105 deg so:
φ = 105 – 90 = 15 o
b) Hence, or otherwise, estimate the time planet Epsilon Eridani IV will be moving retrograde relative to Epsilon Eridani III, and also the time between its opposition and the next stationary point.
Solution:
The time for retrograde is:
T r = 2 Θ/ 360 x S = (2 x 77.6)/ 360 x (367.7 d) = 158. 5 d
Between opposition and next stationary pt.:
T r/2 = (158.5 d)/2 = 79¼ d
c) Obtain the time during Epsilon Eridani IV's synodic period that it is moving direct.
Solution:
t(D) = (1 – Θ/ 180) x S = (1 - 77.6/ 180) (367.7 d) = 209 days
(N.B. Remember these are times referred to the synodic not sidereal periods!)
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