Unexpected findings in need of an explanation

[Sylvain9595]

it's still not the right geometric quantity to address the question you're asking in this thread.

I admit you are right, I get it now. I was too focused on the retrograde motion for some reason...

 

[PeterDonis]

suppose we try to model the planetary motions in order to try to understand what determines the apparent size of the retrograde loop in the sky

I've tried constructing an approximate model by using small segments of circles to approximate the elliptical planetary orbits near perihelion and aphelion. Unfortunately that model isn't working out well--even for small segments, the error using circles instead of the actual ellipses is too large. However, working through that process showed me that "geometry" in and of itself can't be the entire answer. Sure, you can draw geometric pictures of the two oppositions and see how the angles compare, but in order to draw those pictures, you have to already know the dynamics--where the Earth and Mars actually are in their orbits at the start and end of Mars's retrograde loop. And that is a problem of dynamics, not just geometry.

As a try at describing the relevant dynamics, let's consider two approximations, which don't exactly capture the true conditions of the 2003 and 2012 oppositions, but will at least illustrate a key difference between them. The shortest possible Earth-Mars distance at opposition would occur if Earth were exactly at aphelion and Mars were exactly at perihelion (the 2003 opposition was fairly close to this). The longest possible Earth-Mars distance at opposition would occur if Earth were exactly at perihelion and Mars were exactly at aphelion (the 2012 opposition was not quite as close to this).

Under those conditions, the problem should be time symmetric about the time of opposition: i.e., Mars's apparent retrograde motion should start some time $\Delta t$ before the time of opposition, and should end exactly the same time $\Delta t$ after the time of opposition. So we only need to consider one half of the problem, and then just reflect it in mirror image to get the other half. We'll focus on the first half.

Of course we're not actually interested in $\Delta t$, we're interested in $\Delta \theta$, the angle between the Earth-Mars line at the start of Mars's apparent retrograde motion, and the Earth-Mars line at opposition (which for both of the cases described above is just the major axis of both the Earth and Mars orbit ellipses, i.e., the line passing through the Sun, Earth, and Mars). What determines that angle is where on their respective orbits Earth and Mars are when Mars starts its apparent retrograde motion. But time is also a factor in when that is.

A comment about the shapes of the orbits: the segments of each planet's orbit that we are considering have the same shape. In other words, the curvatures of both orbits along segments centered on the major axis, at perihelion and aphelion, are the same. So in both cases described above, we are looking at the same two curves; the only things that vary are how far apart they are, and how fast each planet is moving along its curve.

Also, because Mars's orbit is more eccentric than Earth's, the curvature of its orbit at perihelion and aphelion is greater. That means that more of its motion between the start of its apparent retrograde motion and opposition is parallel to the major axis, and less is perpendicular to the major axis.

Now, a key point: since the shapes of the curves are the same in both cases, and since what we're interested in is the Earth-Mars line, we can actually plot both cases on a single diagram. We take one Earth curve, and two Mars curves, at the two different distances (smallest and largest). We mark the two points on the Earth curve where the apparent retrograde motion of Mars starts for the two cases, and the corresponding two points on the Mars curve. Then we just look at the respective angles that the Earth-Mars lines make with the corresponding lines through Earth and parallel to the major axis of the orbits.

If we also draw lines through the two Mars positions parallel to the major axis of the orbits, we can call the Earth lines $E_S$ and $E_L$ (for Earth lines at smallest and largest Earth-Mars distances at opposition), and the corresponding Mars lines $M_S$ and $M_L$. If we then call the perpendicular distances between the Earth and Mars lines $D_S$ and $D_L$, and the Earth-Mars distances at opposition $O_S$ and $O_L$, then the tangents of the two angles we want to compare are (approximately) the ratios $D_S / O_S$ and $D_L / O_L$. So the question is, which of these two ratios is larger? Or, if we want to combine everything into a single formula, is the quantity $D_S O_L / D_L O_S$ greater than or less than $1$?

Let's look at how the distances compare. We know $O_L > O_S$, indeed $O_L$ is larger by about a factor of two. Call the actual factor $f_O$, i.e., $O_L = f_O O_S$. We also know that $D_L > D_S$ by a significant factor, because the Earth is moving faster in its orbit in the $L$ case while Mars is moving slower. Indeed, to a first approximation, if we call the times from the start of retrograde motion to opposition $t_S$ and $t_L$, and the orbital speeds of Earth and Mars in the two cases $v_{ES}$, $v_{EL}$, $v_{MS}$, and $v_{ML}$, then we expect that $D_S = \left( v_{ES} - v_{MS} \right) t_S$ and $D_L = \left( v_{EL} - v_{ML} \right) t_L$. Or, adopting obvious notation, $D_S = \Delta v_S t_S$ and $D_L = \Delta v_L t_L$. And we can define the velocity factor $f_v$ by $\Delta v_L = f_v \Delta v_S$, and the time factor $f_t$ by $t_L = f_t t_S$, so $D_L = f_V \Delta v_S f_t t_S$.

Putting all of the above together, we have

$$
\frac{D_S O_L}{D_L O_S} = \frac{f_O}{f_v f_t}
$$

From the numbers already given in this thread, we have

$$
f_O = 1.81
$$

$$
f_t = 1.33
$$

I couldn't find where the orbital speeds had been posted in this thread, so I got them from Wikipedia: for Earth, $30.3$ km/s at perihelion and $29.3$ at aphelion; for Mars, $26.5$ at perihelion and $22.0$ at aphelion. That gives

$$
f_v = \frac{30.3 - 22.0}{29.3 - 26.5} = 2.96
$$

So we have

$$
\frac{D_S O_L}{D_L O_S} = \frac{f_O}{f_v f_t} = \frac{1.81}{2.96 \cdot 1.33} = 0.46
$$

This is less than $1$, so it is telling us that the $L$ angle is larger than the $S$ angle by a factor of a little more than $2$. Of course this is only an approximate number even for the idealized cases described above, which are not quite the same as the actual oppositions we've discussed in this thread; but it serves to illustrate what the angle depends on and why we should not be surprised to find the $L$ angle being significantly larger than the $S$ angle: because the "shrinking" effect of the greater distance (which is what $f_O$ is capturing) is more than outweighed by the greater speed difference ($f_v$) and the longer time ($f_t$), both of which act to increase the angle.

 

[Sylvain9595]

@PeterDonis,

I still don't understand why the speed would change the amplitude.

Under those conditions, the problem should be time symmetric about the time of opposition: i.e., Mars's apparent retrograde motion should start some time before the time of opposition, and should end exactly the same time after the time of opposition.

There seem to be a problem here : Mars's apparent retrograde motion can only start at the opposition (but the hole loop should indeed start before the opposition).

Here is a little animation with circular orbits : I can't even trace a loop whith the Earth-Mars distance only 1,5 times larger in configuration n°2 :

codepen.io/Boanerges-the-solid/pen/QwEKLEW

 

[WernerQH]

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)

Minutes in right ascension (RA) are minutes of time, not minutes of arc. One hour in RA corresponds to 15 degrees, but only for motion along the celestial equator. Because the ecliptic is inclined with respect to the equator, the "hour" may be shorter than 15 degrees in constellations like Gemini or Sagittarius, and longer in Pisces or Virgo. Since the planets tend to move along the ecliptic, it is more resonable to consider the ecliptic longitude. I found your finding as puzzling as you did and checked some almanacs I still have on my bookshelf: the retrograde arc was about 19 degrees in 1997 (at a distance of 99 million kilometers) and about 13 degrees in 2020 (at 62 million kilometers). So we seem to have reached an agreement on the facts.

I still don't understand why the speed would change the amplitude.

As to the why, @mfb in his post #5 has already given an explanation. The small speed difference of 2.8 km/s at a distance of only 55.8 million kilometers translates to an angular speed of about 0.25 degrees per day, whereas the difference of 8.3 km/s at 100.8 million kilometers produces 0.41 degrees per day. Combine that with the shorter duration of the nearer opposition (61 days versus 81 days) you see that the planet will not travel so far during its retrograde motion.

 

[Sylvain9595]

As to the why, @mfb in his post #5 has already given an explanation. The small speed difference of 2.8 km/s at a distance of only 55.8 million kilometers translates to an angular speed of about 0.25 degrees per day, whereas the difference of 8.3 km/s at 100.8 million kilometers produces 0.41 degrees per day. Combine that with the shorter duration of the nearer opposition (61 days versus 81 days) you see that the planet will not travel so far during its retrograde motion.

I have to admit you're right!

I've created a better animation:

Configuration 1 (2003): Earth-Mars distance = 8, Earth's speed = 1.14 times the speed of Mars

Configuration 2 (2012): Earth-Mars distance = 14.4 (1.8 times greater), Earth's speed = 1.33 times the speed of Mars

Link to the animation: codepen.io/Boanerges-the-solid/pen/myEryvL

Thanks everyone for your replies, it was really very kind of you!

 

[DaveC426913]

That's a darned impressive animation @Sylvain9595! Did you write that yourself?

For those that don't follow the provided link, here is a snapshot:

Readers will notice that Configuration 1 has Mars and Earth in close proximity, yet it results in a very short arc of retrograde motion.

It's quite counter-intuitive to the passing-cars-on-a-highway analogy.

 

[Sylvain9595]

That's a darned impressive animation.

You can thank the Claude AI for that, I just gave her the instructions ;)

 

[WernerQH]

The ellipticity of the orbits is important. It just occurred to me that an asteroid with a very elliptic orbit, but a perihelion distance not much bigger than 1 AU, wouldn't produce a loop when it comes close to Earth. It would overtake Earth at a speed bigger than 30 km/s. (It would need that much kinetic energy to return to its aphelion.)

 

[PeterDonis]

Mars's apparent retrograde motion can only start at the opposition

No, it starts before opposition. That's obvious from the ephemeris data on JPL Horizons; the reversal of both right ascension and declination starts before opposition, and ends after it. Those actual cases are not quite time symmetric about opposition, because they don't meet the idealized conditions of my idealized example. But they still show retrograde motion before opposition.

Here is a little animation with circular orbits

If your animation (meaning here your original one, not the second one) doesn't show the same behavior as the actual data, your animation is wrong. Actual data trumps models. That's why I had to throw away the idealized model I mentioned in post #32, which used circular orbits as well--it didn't show the same behavior as the data.

 

[256bits]

My opinion is "How did those guys years ago figure thus all out without calculating computers and access to only crude instruments?"
I feel pretty dumb compared to them.

 

[PeterDonis]

I still don't understand why the speed would change the amplitude.

Because, as my post #32 shows, it affects the distance that each planet travels perpendicular to the line of opposition (the major axis of both orbits in the idealized case I gave there) during the time of apparent retrograde motion, and that in turn affects the angle Mars covers, as seen from Earth, during apparent retrograde motion. The faster Earth is moving compared to Mars, the further behind Mars it will be when apparent retrograde motion starts, and the further ahead of Mars it will be when apparent retrograde motion ends. That increases the angle Mars appears to cover in Earth's sky during retrograde motion.

 

[Sylvain9595]

Because, as my post #32 shows, it affects the distance that each planet travels perpendicular to the line of opposition (the major axis of both orbits in the idealized case I gave there) during the time of apparent retrograde motion, and that in turn affects the angle Mars covers, as seen from Earth, during apparent retrograde motion. The faster Earth is moving compared to Mars, the further behind Mars it will be when apparent retrograde motion starts, and the further ahead of Mars it will be when apparent retrograde motion ends. That increases the angle Mars appears to cover in Earth's sky during retrograde motion.

It is very clear now, I guess I needed a visualisation. Thanks again.

 

[DaveC426913]

It is very clear now, I guess I needed a visualisation. Thanks again.

Very clear for me too. And fascinating.

 

[PeterDonis]

So we have

$$
\frac{D_S O_L}{D_L O_S} = \frac{f_O}{f_v f_t} = \frac{1.81}{2.96 \cdot 1.33} = 0.46
$$

This is less than $1$

Just to belabor this a little more:

We expect that the three factors in the above formula are actually not independent.

We know that $f_v$ increases as $f_O$ increases, because a larger Earth-Mars distance means Earth is closer to perihelion and hence moving faster, and Mars is closer to aphelion and hence moving slower. So heuristically, we would expect some kind of relationship of the form $f_v = k_v f_O$.

It's not as obvious why $f_t$ also increases as $f_O$ increases, but looking at the animations that @Sylvain9595 made might help with that. I think the logic I gave in post #41 is also relevant. In any case, I think that heuristically we would also expect a relationship of the form $f_t = k_t f_O$.

Substituting these two relationships into the formula above gives

$$
\frac{D_S O_L}{D_L O_S} = \frac{f_O}{f_v f_t} = \frac{1}{k_v k_t f_O}
$$

In other words, if we are comparing two retrograde loops at smaller ($S$) and larger ($L$) Earth-Mars distances, we should expect the ratio of the $S$ angle to the $L$ angle to decrease as the distance difference between the $S$ and $L$ cases (which is what $f_O$ captures) increases.

This in itself does not show that the ratio must be less than $1$, i.e., in itself it does not show that the $L$ angle is larger; for that we need to plug in actual numbers for $k_v$ and $k_t$ and see what their product is. We've done that in this thread for the Earth-Mars case and seen that the ratio is indeed less than $1$. I'm not sure if there's a way to find actual analytical expressions for $k_v$ and $k_t$ to show that their product must always be such that the ratio in the formula above is less than $1$.

 

[Sylvain9595]

Wait a second, we did solve the problem I outlined in my first message, but now I've run into an even bigger one!
Here's the actual observation provided by Dave:

It's clear that Mars is observed approaching us in the first part of the loop and then moving away from us in the second part, whereas my animation shows exactly the opposite! In other words, the loops in my animation are "upside-down" compared to the actual observation.

To match the observation, the yellow dot centered on the right-hand loop should be closer to Earth—not farther away (in this image, the trajectories have been reversed compared to the other animation)!
The loops should appear the other way around, just like on the ancient tychonic model :

I should mention that I was inspired by this animation to create my own:

Witch I found here: https://people.highline.edu/iglozman/classes/astronotes/retrograde.htm?utm

 

[DaveC426913]

It's clear that Mars is observed approaching us in the first part of the loop and then moving away from us in the second part, whereas my animation shows exactly the opposite! In other words, the loops in my animation are "upside-down" compared to the actual observation.

Are you sure my image isn't upside down? One of the oppositions was in the southern hemisphere. I had to rotate the image 180 degrees to align it. I was not concerned with "true" orientation; I was only interested in relative magnitude.

In retrospect, I think that's not what you're concerned about.

You are concerned your simulation doesn't represent radial distance properly (the Y-axis). I didn't expect it to. It's a projection, is it not? The Y-axis in your simulations doesn't have to represent anything at all.