Unexpected findings in need of an explanation

[Sylvain9595]

TL;DR

Why does Mars show greater retrograde arc amplitude when farther from Earth (2012) than when closer (2003)?

Why does Mars show greater retrograde arc amplitude when farther from Earth (2012) than when closer (2003)?

I've been comparing Mars retrograde motion data from JPL Horizons for the 2003 and 2012 oppositions and noticed something unexpected:

• 2003 opposition: Earth-Mars distance = 0.373 AU (55.8 Mkm), retrograde arc ≈ 40 arcminutes in RA (over 61 days)
• 2012 opposition: Earth-Mars distance = 0.674 AU (100.8 Mkm), retrograde arc ≈ 72 arcminutes in RA (over 81 days)

In the heliocentric model, retrograde motion results from Earth overtaking Mars, creating an apparent parallax effect. Geometrically, I would expect the angular amplitude of retrograde motion to be inversely proportional to the Earth-Mars distance—greater amplitude when closer, smaller when farther.


However, the data shows the opposite: nearly twice the retrograde arc amplitude when Mars is almost twice as far away.


I understand that Mars was near perihelion in 2003 (moving faster) and near aphelion in 2012 (moving slower), which affects the duration of retrograde motion. But duration and amplitude are geometrically distinct—like overtaking a car on a highway: regardless of relative speeds, the angular displacement during overtaking should depend primarily on distance, not velocity.


What is the standard astronomical explanation for this counter-intuitive observation? Are there geometric factors (orbital inclination, 3D projection effects, etc.) that I'm missing, or is there published literature addressing this specific phenomenon?


I'm seeking a rigorous geometric explanation, ideally with references to technical literature.

 

[Baluncore]

Welcome to PF.

—like overtaking a car on a highway: regardless of relative speeds, the angular displacement during overtaking should depend primarily on distance, not velocity.

The path of Mars, as seen from Earth, is mapped onto the celestial sphere behind Mars. Maybe it has something to do with the sign of the velocity. When Earth is travelling on the opposite side of the Sun to Mars, the differential velocity has a different sign, to when Earth and Mars are close together on the same side of the Sun.

 

[Sylvain9595]

Thank you for the welcome, and happy new year.

I believe there may be a misunderstanding about when retrograde motion occurs. Retrograde motion happens specifically during opposition, when Earth and Mars are on the same side of the Sun—not on opposite sides. This is precisely when Earth overtakes Mars, like one car passing another on the same highway.

During opposition:

• Both planets are moving in the same general direction (counterclockwise as seen from north ecliptic pole)
• Earth is moving faster than Mars
• The differential velocity creates the apparent retrograde motion
• The sign of the relative velocity doesn't change during the retrograde period

The geometrical puzzle I'm pointing to is this: regardless of how fast or slow the overtaking happens, the angular extent of the retrograde arc traced against the background stars should primarily depend on the geometric separation between Earth and Mars during the overtaking maneuver.

To use the highway analogy more precisely: whether I overtake a car quickly or slowly, the angular displacement of that car relative to distant mountains behind it depends mainly on how close I pass to it, not on how fast I'm going relative to it.

The 2003 and 2012 data seem to show the opposite of this geometric expectation, which is why I'm curious if there's a known explanation.

 

[DaveC426913]

Are you accounting for Earth's peri/aphelion at the times of opposition? (green arrows in diagram)

For example, if Earth happens to be at it perihelion during opposition, Mars' retrograde velocity will appear greater.

I think I need to see 2003 and 2012 visually to compare them.

 

[mfb]

Mars being slower at aphelion wins over the other effects.

Earth's orbital velocity is 29.3 km/s to 30.3 km/s with an average of 29.8 km/s. Mars' velocity varies between 26.5 km/s and 22 km/s. Relative to Earth and using Earth's average, that's 3.3 km/s when Mars is at perihelion and 8.2 km/s when it is at aphelion. That is more than a factor 2, while the difference in distance is less than a factor 2.

The 2012 opposition also happened close to Earth's perihelion - while Mars was unusually slow, Earth was unusually fast.

 

[Sylvain9595]

@mfb and Dave,
You're absolutely right that the relative velocity differs by more than a factor of 2 between the two oppositions, and this clearly explains why the duration of retrograde motion is longer in 2012 (81 days vs 61 days).

But my question is specifically about the angular amplitude of the retrograde arc, not its duration. These are geometrically distinct quantities.

Consider this analogy: If I'm overtaking a car on a highway, the time it takes to complete the overtaking maneuver depends on our relative velocity. But the angular displacement of that car as seen from my vehicle—the total angle through which it appears to move backward relative to distant background objects—depends primarily on the geometry: how close I pass to the car, not how fast I'm going relative to it.

If I pass the same car at the same distance twice—once at high relative speed, once at low relative speed—the angular displacement should be the same both times. Only the duration changes.

Applying this to Mars: regardless of relative velocities, the total angular arc traced out during retrograde motion sshould be determined only by the Earth-Mars distance during opposition. Why would a slower relative velocity produce a larger angular displacement when Mars is farther away?

What geometric mechanism allows velocity to affect the amplitude rather than just the duration?

I'm genuinely trying to understand the geometric explanation here. Thank you for your patience!

 

[DaveC426913]

But my question is specifically about the angular amplitude of the retrograde arc, not its duration. These are geometrically distinct quantities.

Yes, you did mention that. Altough surely the magnitude is affected by the relative velocities because they affect when, in their cycles, the retrograde motion reverts to prograde.

A shorter duration of retrograde phase equals a shorter magnitude - which is what you seek.


And you also appear to recognize that this is a straight up geometry issue. So why don't we resolve it with geometry? Is there any way we can visualize both set of opposition data in an illustration? I'm happy to do the illustration if we can get the relevant data.

 

[Sylvain9595]

@Dave,
Thank you for your thoughtful response and for offering to create a visualization—that would be extremely helpful and could indeed resolve this issue!

I appreciate your point about relative velocities affecting when the retrograde motion reverts to prograde. However, I'm still trying to understand how this affects the total angular displacement rather than just the timing. In the highway analogy: whether I overtake a car quickly or slowly, the angular displacement of that car relative to my reference frame during the complete overtaking maneuver should be geometrically determined by our relative positions, not our relative speeds. The speeds determine how long the overtaking takes, but not how much angular displacement occurs.

That said, I'm very open to being wrong about this! A geometric visualization would be perfect for clarifying where my reasoning breaks down.

I would be very happy to see your simulation—it could indeed resolve our problem. If you could model both scenarios (2003 and 2012 oppositions), that would be invaluable.

The key data points from JPL Horizons would be:

• 2003: Opposition around August 28, Earth-Mars distance ~0.373 AU, retrograde arc ~40 arcmin in RA over 61 days
• 2012: Opposition around March 3, Earth-Mars distance ~0.674 AU, retrograde arc ~72 arcmin in RA over 81 days

Thank you again for taking this seriously and offering to help visualize it!

 

[DaveC426913]

I appreciate your point about relative velocities affecting when the retrograde motion reverts to prograde. However, I'm still trying to understand how this affects the total angular displacement rather than just the timing.

OK, I am hoping that you will intuit that, for a given speed, a shorter time spent in retrograde would result in a smaller observed displacement, no?

In the highway analogy: whether I overtake a car quickly or slowly, the angular displacement of that car relative to my reference frame during the complete overtaking maneuver should be geometrically determined by our relative positions, not our relative speeds. The speeds determine how long the overtaking takes, but not how much angular displacement occurs.

I get the analogy, but I think it is not sufficient to capture the nuances of the planetary orbits, precisely because there are more factors at-play.

That said, I'm very open to being wrong about this! A geometric visualization would be perfect for clarifying where my reasoning breaks down.

I would be very happy to see your simulation—it could indeed resolve our problem. If you could model both scenarios (2003 and 2012 oppositions), that would be invaluable.

Not simulation - since there's no need for a time axis. Merely a 2D diagram (similar to the one in my post 4). Two diagrams - one for 2003 and one for 2012.

The key data points from JPL Horizons would be:

• 2003: Opposition around August 28, Earth-Mars distance ~0.373 AU, retrograde arc ~40 arcmin in RA over 61 days
• 2012: Opposition around March 3, Earth-Mars distance ~0.674 AU, retrograde arc ~72 arcmin in RA over 81 days

We would also need to know when perihelion and aphelion are for both bodies.

 

[Sylvain9595]

"OK, I am hoping that you will intuit that, for a given speed, a shorter time spent in retrograde would result in a smaller observed displacement, no?"

Well, not in the highway analogy of course. And my quest is precisely to understand what are the "other factors" at play in the case of planetary orbits.

I guess two diagrams in 2D could do the trick, but not like the one you posted as it doesn't show the actual retrograde loop as seen from Earth.

Here are the perihelion and aphelion data for both bodies during both oppositions:
2003 Opposition (August 28, 2003):
Mars:

Mars was at perihelion on August 28, 2003 (Ls ≈ 250°)
The opposition occurred almost exactly at Mars' perihelion
Mars perihelion distance from Sun: 1.381 AU (206.6 million km)

Earth:

Earth perihelion: January 4, 2003 at 05:02 GMT
Earth aphelion: July 4, 2003 at 05:40 GMT
During August opposition, Earth was ~1.5 months past its aphelion (farthest from Sun)

2012 Opposition (March 3, 2012):
Mars:

Mars orbital position at opposition: Ls = 163.29°
This is between Mars' aphelion (Ls = 70°) and autumn equinox (Ls = 180°)
Mars was closer to aphelion than to perihelion
Mars aphelion distance from Sun: 1.666 AU (249.2 million km)

Earth:

Earth perihelion: January 2, 2012 at 14:09 GMT
Earth aphelion: July 5, 2012 at 03:47 GMT
During March opposition, Earth was ~2 months past its perihelion (closest to Sun)

Summary:

2003: Mars at perihelion, Earth near aphelion → minimum Earth-Mars distance (0.373 AU)
2012: Mars near aphelion, Earth past perihelion → greater Earth-Mars distance (0.674 AU)

Source for Mars Ls values: SEDS Mars Oppositions table
Source for Earth apsides: Astropixels Earth Perihelion/Aphelion tables
Let me know if you need any other orbital parameters!

 

[DaveC426913]

Oof. I am becoming skeptical that a hand-drawn illustration will capture the details - I would have to intuit what dimensions can be minimized and what ones must be to-scale. The wrong ones might not produce the effect, or if they do it might be too large/tiny to be visible.

I wonder if this is better suited to a simulator after all, which takes these kinds of ephemera in-stride and has no troubles with scaling up and down. That's not my wheelhouse.

But I will pick away at the data and see what I can come up with.

 

[Ibix]

You can get a list of x,y,z coordinates day by day out of the Horizon page. Then you should be able to do almost anything with those. I might have time this weekend.

I would say that I expect that the issue is different relative velocities, so the points on the Earth's orbit where the retrograde motion starts, stops, and changes from increasing to decreasing will all be different.

 

[PeterDonis]

my question is specifically about the angular amplitude of the retrograde arc, not its duration. These are geometrically distinct quantities.

They may be "geometrically distinct", but they are not independent. You seem to be assuming that they are, which is a wrong assumption. Reasoning from wrong premises will lead to wrong conclusions.

Geometrically, I would expect the angular amplitude of retrograde motion to be inversely proportional to the Earth-Mars distance—greater amplitude when closer, smaller when farther.

The angular diameter of an object of fixed size decreases inversely as the distance. But that's not what you are trying to compare here. So this logic seems irrelevant to the actual thing you're trying to understand.

 

[DaveC426913]

Bah. Way too complex to do by-hand. I barely got three months done, and month-by-month is still not granular enough. Gave up half way through...

 

[PeterDonis]

@Sylvain9595 where are you getting your figures for the retrograde arc covered? Your numbers seem too small by roughly an order of magnitude.

 

[PeterDonis]

@Sylvain9595 where are you getting your figures for the retrograde arc covered? Your numbers seem too small by roughly an order of magnitude.

@Sylvain9595 you mentioned JPL Horizons, and I was able to find that app here:

https://ssd.jpl.nasa.gov/horizons/app.html#/

I pulled the data for the 2003 and 2012 Mars oppositions, and I see where you got your numbers now, but they're wrong, because you only looked at right ascension. You have to look at declination as well to get the total movement of Mars across the sky during the retrograde period.

Here's the data I found. (I'm taking the start and end of the retrograde period to be the dates on which the right ascension's time derivative changes sign, but that's actually not necessarily the correct measure, since the declination's time derivative also changes sign and not on the exact same day. But that's probably a whole separate discussion. The numbers below are sufficient to illustrate my point here.)

2003-Jul-31: RA 22 55 48.07 D -13 25 15.8
2003-Sep-30: RA 22 15 34.81 D -15 46 27.3

So yes, the RA change is about 40 arc minutes, but the declination change is about 140 arc minutes. The total angular movement using the Pythagorean theorem is about 146 arc minutes.

2012-Jan-25: RA 11 39 43.91 D +06 09 03.2
2012-Apr-16: RA 10 25 43.56 D +12 42 16.8

So here again the RA change is indeed about 80 arc minutes, but the declination change is a whopping 393 arc minutes. The total angular movement is 401 arc minutes.

 

[Sylvain9595]

@PeterDonis,

Thank you for taking the time to pull the ephemeris data and to spell out your reasoning so clearly. The numerical values you quote for the changes in right ascension and declination are not in dispute. Where we differ is in what quantity is being measured and what is meant by retrograde amplitude.

In astronomy, retrograde motion is defined specifically as a reversal of the apparent east–west motion of a planet against the background stars. Accordingly, the amplitude of the retrograde motion is conventionally defined as the total excursion in longitude, i.e. along the ecliptic, or equivalently in right ascension in the equatorial system. This is the coordinate in which the motion actually reverses sign. Declination does not reverse in any systematic way during a retrograde loop, and its extrema generally do not coincide with the start or end of retrograde motion.

What you are computing by combining changes in right ascension and declination via the Pythagorean theorem is the total angular path length of Mars on the celestial sphere over that time interval. That is a perfectly valid geometric quantity, but it is not what is meant by the amplitude of retrograde motion in standard astronomical usage. It mixes longitudinal motion (which defines retrogradation) with latitudinal motion (which does not).

This distinction matters conceptually:

• A planet can exhibit a large change in declination without being retrograde at all.
• The onset and end of retrograde motion are defined by the longitudinal (or RA) angular velocity passing through zero, not by extrema or sign changes in declination.
• The north–south motion reflected in declination arises primarily from orbital inclination and viewing geometry, not from the Earth overtaking the planet, which is the physical cause of retrograde motion.

For this reason, authoritative treatments—both historical and modern—describe retrograde loops and their amplitudes along the ecliptic (or in right ascension), not by total great-circle distance traveled on the sky.

So while your calculation correctly describes how far Mars moved in total on the celestial sphere during those intervals, it is measuring a different quantity than retrograde amplitude. The RA-only values are not incomplete; they are intentional, because they isolate the longitudinal reversal that defines the phenomenon.

I agree with you that the differing behavior of declination during retrograde periods is an interesting and subtle topic in its own right—but it is indeed a separate discussion from how retrograde amplitude is defined and measured.