1) Mars reaches a stationary point 36 ½ days after opposition. Its elongation is then measured to be 136.2 deg. Given that the sidereal period is 687 days, find the distance of Mars from Earth in AU. Also, find the time to its next stationary point.
Solution:
Use: 1/ S = 1/P1 - 1/P2
And first obtain Mars synodic period, S:
with P1 = Earth's sidereal period (1 yr), and P2 = Mars (687/365.25) = 1.88 yr
1/S = 1 – 1/1.88 = 1 – 0.5319 = 0.468 yr
Then: S = 1/0.468 = 2.13 yrs. (or 779.9 days)
Mars’ distance from Earth is PE from the diagram, so 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 = [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)
= [(1.52)^ ^½ ( (1)^½ + (1.52)^½)]/ ((1)^3/2 + (1.52)^3/2)
cos Θ = 0.958
so: Θ = 16.7 deg
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 deg
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):
Then:
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 first obtain Jupiter’s synodic period, S:
Where P1 = Earth's sidereal period (1 yr), and P2 = Jupiter’s (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 motion in each synodic period is:
T_r =2 Θ/ 360 x S
Where:
cos Θ= [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)
and a =1, b = 5.2, so:
cos Θ= [(5.2)^½ ( (1)^½ + (5.2)^½)]/ ((1)^3/2 + (5.2)^3/2)
And: cos Θ = 0.582
So: Θ = arc cos(0.582) = 54.4 deg
and (time to move retrograde over each synodic per.))
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 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= [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)
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 = 1, so:
cos Θ= [(29.9)^½ ( (1)^½ + (29.9)^½)]/ ((1)^3/2 + (29.9)^3/2)
cos Θ= 0.215 and Θ= arc cos(0.215) = 77.6 deg
Meanwhile, by using a scaled diagram (e.g. see the example shown – but best done using large graph paper) we find: 90 + φ = 105 deg so:
φ = 105 – 90 = 15 deg
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!)
Solution:
Use: 1/ S = 1/P1 - 1/P2
And first obtain Mars synodic period, S:
with P1 = Earth's sidereal period (1 yr), and P2 = Mars (687/365.25) = 1.88 yr
1/S = 1 – 1/1.88 = 1 – 0.5319 = 0.468 yr
Then: S = 1/0.468 = 2.13 yrs. (or 779.9 days)
Mars’ distance from Earth is PE from the diagram, so 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 = [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)
= [(1.52)^ ^½ ( (1)^½ + (1.52)^½)]/ ((1)^3/2 + (1.52)^3/2)
cos Θ = 0.958
so: Θ = 16.7 deg
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 deg
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):
Then:
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 first obtain Jupiter’s synodic period, S:
Where P1 = Earth's sidereal period (1 yr), and P2 = Jupiter’s (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 motion in each synodic period is:
T_r =2 Θ/ 360 x S
Where:
cos Θ= [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)
and a =1, b = 5.2, so:
cos Θ= [(5.2)^½ ( (1)^½ + (5.2)^½)]/ ((1)^3/2 + (5.2)^3/2)
And: cos Θ = 0.582
So: Θ = arc cos(0.582) = 54.4 deg
and (time to move retrograde over each synodic per.))
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 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= [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)
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 = 1, so:
cos Θ= [(29.9)^½ ( (1)^½ + (29.9)^½)]/ ((1)^3/2 + (29.9)^3/2)
cos Θ= 0.215 and Θ= arc cos(0.215) = 77.6 deg
Meanwhile, by using a scaled diagram (e.g. see the example shown – but best done using large graph paper) we find: 90 + φ = 105 deg so:
φ = 105 – 90 = 15 deg
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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