Many thanks to Ming Yan for the snapshots!

Tutorial 1

Tutorial 2

Then we talked about a real Comet Tempel-Tuttle (notice a typo in Tempel's name on blackboard), which almost crosses the path of Earth every 33.83 years and was at least once as close as 0.02 AU to Earth in the 14th century. Who knows, maybe it will hit the Earth in the distant future, so we want to know the vectors of velocity of both objects at impact. We found that it is easy to calculate the angle θ of the entounter point, assuming for simplicity that the inclination is 180 degrees instead of 162.7 degrees (a counterrotating comet, in Earth's orbital plane, which BTW is called ecliptic). T2c (jpg), T2d (jpg), T2e (jpg). We got the a rough estimate of the speeds: 40 km/s (C/55) and 30 km/s (Earth), direction: almost head-on.

Left as an exercise for you was to: (i) estimate the mass of the comet assuming a reasonable density like, say, 1200 kg/m³, radius 1.8 km, and from the very large, ~70 km/s, impact speed to calculate E_imp in units of Tsar Bomba energy (50 Mt TNT = 2e17 J), (ii) then to speculate on the global effects of the comet T-T impact, given that 2e17 J detonated in New York would shatter windows in Toronto, 550 km away. Notice there was a misprint in the exponent of the bomb's E, it is 2e17 not 2e14 J.

Solution: E_imp = 1200*(1800.**3)*4*(70000.**2/2) J = 6.86e+22 J. (Notice I assumed spherical shape but took π/3 ~ 1 in this estimate). According to Python, this is 342922 Tsar Bombs.

That would be a really lousy day. Almost as bad as one 66.04 Myr ago, when 3/4 of all plant and many animal species including dinosaurs were wiped out in the aftermath Chicxulub asteroid collision (E_imp ~70 million Mt TNT or about 1400000 Tsar Bombs).
Read about it! The impactor was bigger & denser than comet Tempel-Tuttle. A ~5.5 km radius asteroid with significant amout of iridium inside, which left a global Ir-rich layer worldwide, hit the Yucatan peninsula at the speed of only ~20 km/s. It most likely came from the prograde-orbiting asteroid belt, and hence did not collide with the Earth head-on. If it did, 10x more energy would be released, we the mammals might have been done in as well and, the worst of all, ASTC25 could not be offered before 10⁸ years from now.

Tutorial 3

Problem 1.

Problem 2.

☆ Problem 3.

☆ Problem 4.

Answer:
Forward or backward but not toward the center of the Earth, since in this case a small radial component of velocity (< 1/1000 of orbital speed of almost 7.9 km/s) adds very little (less than 0.5/1000000) to the length of the vector of velocity. Energy, a, and P are changed very very little, at most in the seventh digit, possibly not enough to avoid the trash bag striking the space station after one P. When you throw the bag forward/backward, you add/subtract a few m/s from the orbital speed arithmetically - thus changing already the 3rd or 4th digit of the speed of the bag, w.r.t. the space station value. Suppose this changes P by P/1000 (a more precise estimate can be made), then the trash bag arrives at its launch point about dP = P/1000 late or early, w.r.t. station, i.e. 90 min/1000 ~5.4 s apart. In that time interval the Space Station moves by ~7.9 km/s * 5.4 s ~ 40 km, and that will be the minimum spacing of the bodies after one P, growing in subsequent orbital periods. Collision is still possible, after many periods, but requires extremenly bad luck.
See the blackboard: T3a (jpg).

☆ Problem 5.

Tutorial 4

Problem 1.

Problem 2.

Tutorial 5

Ideas for T7.

Problem 1. Stubborn donkey paradox.

Problem 2. Critical push parameter.

Problem 3. Galactic Roche lobe.

Problem 4. Galactic sphere of influence, again.

Tutorial 7

Problem 1. R3B problem and the linear algebra of vectors

Problem 2. R3B problem and the curious, universal, triangles

Solution:

It is easiest to show the equilateral triangles are in fact equilibria or points of zero total force, and then retrace all the steps to generalize that implication to equivalence. See the blackboard cited above. The sum of accelerations f does equal zero, based on the result of problem 1.

Problem 3. Sometimes called colinear, sometimes collinear, but always unstable!

Solution:

The blackboard shows the form of perturbation equations near a saddle point (upper right corner of left panel, two ODEs for ξ and η). Constants a, b are both positive real numbers and represent a saddle potential of arbitrary curvatures. Notice the positive sign in front of one, and negative in front of the other! The substitution of time variation in the form e^λt, while keeping the amplitude factors as ξ and η with tildas (waves above the symbols) results in a linear matrix equation shown in the center. Determinant of the shown matrix must be zero if solutions of perturbation equations are non-zero. That condition gives a bi-quadratic equation solved for λ² on the right, which for any a,b > 0 gives one of λ² positive (meaning that e^λt grows in time exponentially, without bound). That concludes the proof that every saddle point in a rotating system is an unstable equilibrium.
Colinear L-points are saddle points of potential, therefore they are always unstable.
FYI, it can be shown that the triangular Lagrange points L4 and L5 are unstable too, if the mass parameter μ is larger than about 1/27, otherwise they are stable (there are small oscillations around them with two frequencies, with amplitudes that neither grow nor are damped).

Tutorial 8

Problem 1. Average time between collisions in replenished debris disks.

Estimate the mean time of collision with another disk particle (for a typical grain of dust at 100 AU from the star), taking into account the vertical motin of grains in a disk. They have non-zero vertical average speed, because they are on slightly inclined orbits, as evident from the observed z/r aspect ratio of the disk equal, say, z/r ~ 0.05.

Estimate collision speed of grains at 100 AU and argue that it leads to catastrophic shattering of them into finer dust.

Show that t_coll is much smaller than the age of the star equal 20 Myr, therefore the similar disks need to be constantly replenished, rather than primordial. Hence the name "replenished dust disks" for the whole category of objects. Alternative names are "β Pictoris dust disks" or "Vega-type phenomenon" or exoplanetary disks, or Kuiper belts around stars.

SOLUTION:
There are many ways to compute the estimate, mainly
(i) considering a box in which number density of particles is n (number per m³) and the cross-sectional area of one dust grain is S₁, or
(ii) assuming that an average particle moved through the disk vertically back and forth once per orbital period. The results are close to each other in either method.

Let's use (i) like we did in the tutorial T8. Mean collisional time is inverse of the mean collisional rate = n v 4S₁, where a factor 4 comes from the fact that effective cross section for collision is 4 times the area of one grain (glancing collision happens when centers are spaced at 2 times the radius of each grain, so the collision happens when a center is passing through area 4S₁ around the other particle.

What's v? It is the mean inter-particle speed that (neglecting eccentricity of particle orbits) we can assume to be the vertical speed of a typical particle on a slightly inclined orbit: v ~ v_K (z/r), i.e. the circular orbital speed (or 2π r divided by orbital period P) times the aspect ratio of the disk. This gives v ~ 2π z/P and thus t_coll = P/(8π n z S₁). At this point it would be good to do units check! They're ok. As far as the values are concerned, at 100 AU from the sun the circular speed is 3 km/s and perhaps 3.5 km/s at that distance from &beta Pic, times z/r ~ 0.05 which gives typical encouter speed 170 m/s. This definitely is enough to crush minerals to small sub-micron sized dust. Mean collisional time is thus the survival time of silicate particles.
Meanwhile, this is how we evaluate the optical thickness perpendicular to disk by imagining a photon piercing through the disk: τ_⊥ ≈ n z S₁. This is because n z is the number of particles per unit area of disk, and τ is the total cross section of these particles per unit area. As a result,

t_coll ≈ P/(8τ_⊥). Units are ok.

snapshot T8d.jpg  

Evaluation of orbital period P at 100 AU from β Pic yields P = 1000 yr/1.75^0.5, and
t_coll ≈ P/(8*3e-3) ~ 10⁴ yr. This is a very short time compared with 20 Myr, the age of the system. Therefore, particles must be continuously replenished (by asteroid collisions and cometary dust release). The same happens in other circumstellar disks!

Problem 2. Radiation to gravitation ratio, β

SOLUTION:

We start by computing the rate of energy absorbed by the dust grain per unit time. That can be converted to rate of momentum addition to the grain per unit time, by dividing it by c, the speed of light in vacuum. The convertion rule is just the property of photons, and all other ultrarelativistic particles. Since in physics d(momentum)/dt = F_rad, we can write down radiation pressure force. It's the flux of energy of stellar radiation at distance r, L_*/(4π r²), times the cross-section area A of the dust grain. The gravity force with correct physical units of newtons is of course F_grav = GM_* m / r², so
β = (L/M)_* (A/m) /(4πGc).
Here, A/m is the cross section area-to-mass ratio of the dust grain, equal to A/m = 3/(4ρ α), with α (alpha) being your chosen designation of the dust grain radius (R would perhaps have been clearer, after all!).

snapshot T8a.jpg   snapshot T8b.jpg

Because both radiation flux and gravity of the star fall off with the same power of distance, the β ratio does not depend on r, which we were supposed to prove.
Numerical evaluation of the ratio gives something like

β ≈ 2 μm / α

Blow-out happens when β > 1. We conclude that the blow-out particle radius around β Pic is α = 2 μm (2 micrometers or microns). Incidentally, if you think we haven't encountered enough β's already, particle streams detected in the Solar System that flow outward to leave it forever under radiation pressure of solar radiation, have been named β-particles (α-particles being those on stable elliptic orbits) much before the disks like β Pic were discovered.

Is the calculation correct, even though we: (i) assumed that the full unobscured flux of stellar radiation is seen by a particle, and (ii) we used geometrical optics rather than wave optics, far more precise for particles of diameter similar or smaller than the wavelength of absorbed light?

Yes, it is. Condition (i) is satisfied well in the very optically thin disk of β Pic. And because the peak of energy of the starlight is in the sub-micron range, while the diameter of the most relevant dust sizes is a few microns, (ii) is also ok. We were just lucky, in general one should use precise solutions of Maxwell's equations, in place of the assumption that the effective absorption cross-section is constant and equal to π α².

This is the subject of the so-called Mie theory of scattering and absorption from spherical particles, based on work of Gustav Mie. Particles of size ~ wavelegth are known to sometimes scatter as if they were bigger than they really are (which is called resonant scattering of radiation), whereas particles much smaller than that are always pretending to be much smaller than they really are (we say that they decouple from EM radiation).

There is a remarkable agreement of this theoretical value of 2 μm with the determination of the typical size of grains surrounding the star β Pictoris, using modelled optical properties (efficiency of scattering in different direcitions, thermal radiation efficiency) of dust in 3-d models of dust distributions, which are compared with astronomical observations after projecting the radiation produced in the model onto the 2-d plane of the sky. Indeed, this is the smallest type of grains observed in scattered light because normally the law of distribution of sizes of grains is a steep power law, predicting huge numbers of the smallest grains, which dominate the total scattering area, although not the total mass (mainly contributed by the largest particles). So it is also in other exo-Kuiper belts (circumstellar dust disks), only the small size cutoff differs a little. In the solar system, we tend to observe grains down to sub-micron size, on account of o few times smaller (L/M)_* ratio, causing a few times smaller β and a smaller blow-out size: about 0.5 microns, for a typical porous silicate grain density = 2x water density.

Tutorial 9

Problem 1. Orbital adventures of a particle.

The disk is truncated (density of gas goes to zero) at its outer radius. The particle, upon arriving there, changes to an orbit that is elliptic, with periastron just inside the truncation radius. Calculate the equilibrium eccentricity by requiring that at periastron the solid particle does as fast as the gas, and not receive a push or pull from the gas.

Compare the result with the problem described as tutorial problem in lecture notes L16-18 and comment.

Suppose that the gas disk in β Pic ends at about 200 AU. How far will some solid silicate grains be found from the star, if their radiation pressure coefficient is 0 < β < 1, for instance β = 0.3 or 0.4?

Problem 2. Imagine it is 2034.

67 minutes is almost certainly the minimum orbital period of a near-surface satellite. Deciphering the communications will surely take years, as all the world's computer centers, and all the software including the newest ChatGPT ver. 8.0, did not make any progress within the 3 years of the discovery.

Which of the following can be inferred from the above information:
(i) The escape speed from the planet is higher than from the Earth. Astronautics on that planet encountered larger obstacles than on Earth. Perhaps chemical fuel does not suffice to fly into space from the planet in ET1/2031.
(ii) The type and most likely composition of the planet.
(iii) The size of the planet.
(iv) Surface gravity acceleration in units of g=9.81 m/s², and therefore suitability for human habitation.
(v) None of the above. The information is insufficient.

(Radius of the Earth = 6371 km, mass = 5.97e24 kg, and G = 6.673e-11 in S.I. Other numbers you can google, if you need them).

Some of you already figured out the answer after the tutorial, but I don't want to spoil the fun for the rest of you. I'll just add a question for everybody.

Good scientists never rest on their laurels. Based on what they've learned, they formulate new questions for Nature, or in this case, E.T. technology. They ask questions in the form of an experiment or observation.
What further measurement that can be accomplished in less than 1 hour can we propose to answer *all* of the above questions, not just some? Hint: No super-technology of the future is needed, spacecraft or imaging with space telescopes. Instruments from the end of last century would be quite sufficient.

Tutorial 10

Problem 1. Planetary physics of mountains.

A. Are these mountains standing on top of underlying basaltic rocks (density 3.3 g/cc)? The extra pressure of Himalayas would have to be born by the strength of these underlying rocks. Estimate the average pressure produced by a pyramid-like mountain, and separately a pressure due to a linear mountain ridge, both of height H = 10 km above the nearby basaltic sea floor.

Compare the estimated pressure with the properties of rocks. Uniaxial, ultimate compressive strength (i.e., maximum stress) of rock samples is given in literature as 58-300 MPa for basalt, and 170-250 MPa for granite. These values of limit stresses are larger than the measurements of maximum yield stress (or plasticity limit): 20-60 MPa. The indicated ranges come from different samples and different rock sample sizes, the amount of nonuniformities, volume of mineral- or water-filled cracks, and so on. When rocks become plastic, they inelastically and irreversibly bend. The smallest limiting stress is however the maximum shear stress of rocks: 1.7-4 MPa (see the paper cited below).

Can Himalayas be supported by underlying rock strength? If yes, which way will the pendulum deflect away from the vertical as you apprach the mountains?

B. Have you heard the expression "the tip of an iceberg"? An order of magnitude more vertical extent, volume, and mass of ice, is found underwater than seen above the water line. Find out that ratio, if the density of ice is 0.92, while that of salty cold water is 1.02 (both in g/cm³, or g/cc). Assume uniform densities, and a similar shape of the ice mountain above and below the water surface.

Apply the same principle, in geology called isostasy, to the Himalayas, assuming that rock strength plays a minor role at best. Predict what depth D below the sea floor is reached by the granitic mountain rocks. Densities are those listed in part A. As you approach Himalayas, which way a resting pendulum will deviate from the vertical (defined as a line pointing toward the Earth's center).

Finally, consider that temperature rises as you descend below the surface, in case of Earth quite rapidly: at about 100 km of depth under continents, 60 km below ocean floor, and in some locations even much less than that, T = 1800 K is reached, rocks lose elastic strength and become a malleable, very slowly flowing and mixing magma. This region is called asthenosphere (Ancient Greek 'asthenos' = lacking strength). Granitic/continental "iceberg" cannot extend lower than the asthenosphere-lithosphere boundary. What height H of the highest mountains theoretically possible on Earth follows from the principle of isostasy?

Notes:

1. Mountains on Earth are additionally affected by erosion, which is another limiting factor. Ice formation in cracks is the main process that fragments rocks at high altitudes and sends them tumbling down. H₂O is common in the universe, therefore ice-driven erosion may happen on exoplanets as well. If there is not piling-up affect of continental plate collisions (which on Earth float on asthenosphere, and move sideways driven by magma currents in it), then erosion will be happy to gradually level any mountains. But if the pileup of plates pushes rocks up quicker than erosion levels them, mountains grow. Then the H-limit that you are going to derive comes into play.

2. Cf. this document with measurements of rock properties. W.E.Lewis, Master Thesis, 1946. Uses psi (pound per square inch) units, 1 psi = 6.895 kPa.

SOLUTION

Let's call the height of the mountain $H$, and the area of its fotprint $A$. It doesn't matter what precise shape the base/footprint is. Archimedes showed that volume of a pyramid with arbitrary footprint is $V = A H / 3$ (you can easily prove it yourself by integration or simply knowing that area under a curve $y=(h/H)^2$ from $h=0$ to $h=H$ equals 1/3. Volume of a ridge with the same base area is, however, $V = A H /2$. (Pyramid can be divided in a large number of horizontal slices with area growing like a square of the distance from the top, and a ridge can be divided in slices with area growing linearly with the same distance. Hence the integration gives 1/3 and 1/2, correspondingly.)

A. In case of the pyramid shape of a mountain, the average pressure (weight divided by area) is $ p = \varrho_{gran} H g/3 $, while a triangular ridge exerts a larger average pressure, $p = \varrho_{gran} H g/2$, where $\varrho_{gran} =2700 $ kg/m$^2$, $g = 9.81$ m/s$^2$, and $H = 10^4$ m. The pressures equal 88 Mpa and 132 MPa. These values both exceed the plasticity limit of an underlying basaltic rock, and even more so the shear stresses/strength of bedrock basalts. Therefore the answer is no, the Himalayas cannot just be standing on the underlying horizontal layer of rocks. These mountains would slowly make a deeper and deeper impression in such hypothetical horizontal rocks, and gradually sink to a certain lower height by deforming them.

If you approach Himalayas, in this incorrect model the additional mass of granites would deflect the motionless pendulum away from the line toward the center of the Earth towards the Himalayas. This was expected by early explorers, but did not actually happen, much to their surprise.

B. Iceberg equation equates the weight of the whole iceberg (including the under-water part, let's call its volume $V_{tot}$) with the weight of denser water that it displaces (volume submerged under water is $V_{subm}$). Weight is volume times density, so $$ V_{tot} \, \varrho_{ice} = V_{subm} \, \varrho_{water}\,, $$ from which we calculate that the ratio of underwater part to total volume of an iceberg is equal to $\varrho_{ice}/\varrho_{water} = 0.92/1.02 = 0.902$. The rest, i.e. only 10% of the whole iceberg, is sticking above the water surface, justifying the saying "it's just the tip of the iceberg".

Let's now apply that idea (called isostacy) to granitic mountains floating in denser, deformable basaltic rocks: the densities are 2.7 and 3.3 (units don't really matter for the ratio), and so the part of volume of the continental rocks (the lighter ones) located under the ocean floor is 27/33 = 9/11 of the total volume. Only 2/11 = 1/5.5 of the total volume is forming continental mountain ranges above the seafloor. As long as the shape of the mountain is the same over and under ocean's floor (like a mirror reflection, although with different vertical extents), the depth of a mountain formation (whether pyramid or ridge-shaped) equals $H=10$ km times 9/2, hence 45 km. That depth is still within the lithospere and above asthenosphere boundary where T=1800 K is reached and rocks, independent of type, completely melt. The conclusion is that Himalayas are about the highest mountains possible on our planet. Much higher mountains could not be supported by their required deeper roots, because granite and basalt at a depth larger than 50-60 km melt nad freely mix in magma.

The plumb line (pendulum) should not deflect towards Himalayas, and maybe even deflect away from Himalayas a bit, as you approach them, sice they're a very large rock (mostly under-surface), but underdense w.r.t. to the bedrock.

Isostacy can be used in planetary geology as well as Earth geology.

Problem 2. Netflix movie with Lagrange point L1 in a lead role.

SOLUTION OUTLINE

The 'observed' transit lightcurve from movie "Clara" can be seriously analysed to see what could be true and what not in the discussed scenario. First, the depth of the transit is 1.00-0.9962=0.0038 or possibly 0.0037.
A blackboard snapshot T10-1 shows that, neglecting the limb brightening effect, the area ratio of planet to star is equal $\simeq 0.0037$, so the radius ratio equal to $\sqrt{0.0037} = 0.0608$. The star's radius was given as 1.135 million km (larger than our sun), which gives the physical radius of the planet equal $R_p \simeq 69000$ km, which is 1 Jupiter radius. So far so good, that's quite possible, the planet could be like Jupiter.

The trouble with the movie scenario starts with the speed $v_K$ and distance $a$ of the Jupiter-like planet. Knowing that $M_* = 1.62 M_\odot$, these parameters are: $v_K = 602$ km/s, and $a = 3.96\cdot 10^{-3}$ AU. Which means a very short period of motion and an impossibly small star-planet distance equal $a = 0.594$ mln km, smaller than the radius of the sun and of the star. No wonder that the surface temperature based on assumed luminosity $L_* = 5.43 L_\odot$, albedo $A=0.34$ equal to Jupiter's albedo, is also impossibly high, $ T_{surface} = 6118\, K$. Notice that I put a wrong number on the blackboard, 3366 K, it is really 6118 K! It's higher than sun's effective temperature (and the star's temperature) because we use a formula that doesn't care if the radius is outside or inside the star. But it's interesting to note that there are hot planets out there. A Jupiter-type planet called KELT-9b has an estimated temperature of 4643 K.

So we now know that the movie hasn't been consulted too well, anyway not by a current of former ASTC25 student. But is the duration of transit the only thing that is wrong? Let's see.

(i) The movie claims that the planet and its artificial moon or station located in L1 point are near the Habitable Zone (HZ) boundary, in fact the station is just inside it, and the planet outside it. Suppose the timing of the transit was mistakenly as 4000 seconds, and try time which is 20 times longer: ($\Delta t = 20*4000s = 22.2$ hr). It would mean a 20x slower $v_k = 30.1$ km/s, and the star-planet distance 400 times larger than previously calculated, or $a=1.58$ AU). Estimate of T would drop to 6118 K / 20 = 305 K, which is a good temperature for liquid water to exist (although we know there are nuances: this is an *average* temp. without the greenhouse effect, valid for albedo equal to Jupiter's albedo). We should actually assume a slightly longer time of the transit, so that the planet is placed just outside the HZ, say, at T=270 K. That would require 6118K/270K = 22.7 times longer transit (lasting 25.2 hrs). OK!
(ii) But that is not all. There is a separate problem with the shape of the perturbed lightcurve profile, happening in every observation at the same phase of the transit (so it cannot be due to a exo-moon, which would produce signatures at different phases of, or outside the planetary transit). Namely, a hypothetical object precariously hanging around L1 point, where trajectory must occasionally be corrected by intelligent beings, would produce TWO events, if it can be seen at all: one near the beginning and one near the end of occultation. See the figure. In the movie, there is only one event and the other is missing.
Cf. a sketch T10-2 (png)
(iii) What about the observed *brightening* event? An object blocking light of the star should cause dimming of the observed light, not brightening, as in the movie.
(iv) The above sketch is helpful in establishing the criterion of visibility of an object placed at L1. The line joining planet and star centers has inclination $arctan(R_*/a) \approx R_*/a$. At the distance of $r_L = a (\mu/3)^{1/3}$, the L1 point is at locating shifted sideways w.r.t. planet's center by approximately $r_L R_*/a = R_* (\mu/3)^{1/3}$, toward star's center. Notice that the shift projected on the sky does not depent at all on star-planet distance! The planet has Jupiter's radius, so it cannot have mass very different from Jupiter (it's a gas giant). Let's assume $m_{pl} = m_{Jup}$, then the mass ratio is $\mu = 10^{-3}/1.62$ and therefore a shift toward the center equals $0.059 R_* = 67000$ km. From the depth of occultation we got $R_{pl} = 69000$ km, which is a similar but 3% larger. L1 point would not be visible, but almost visible behind the planet. A moon-like space station would have non-zero extent, so it might be detectable during the transit, as a symmetric dimmming of light, though I would have to think how to disentangle the signature of the L1 object from limb darkening effect of the star. It would not be easy.

If the object is not exactly at the L1 point, then it may happen that it is visible behind the planet only just after the beginnig of the transit or just before its end - this would agree with observations in the movie. That would however not be reasonable. A civilization advanced enough to build such a thing, would know to keep its artificial object very close to the unstable L1 point. The trajectory corrections are much cheaper energy-wise there. The futher one allows the object to drift away from L1 point, the faster it moves away (exponentially in time!) and the more troublesome is the orbital correction to force it back to L1. So why would the aliens allow a significant drift (as the movie suggests)?
(iv) It is possible but very unlikely that a civilization would build a whole artificial home just 6% closer to the star (to be just inside HZ). The concept of HZ is rather fuzzy and approximate, it's not a sharp physical boundary. There are a few possible ways to raise the temperature on a planet by several degrees Celsius or more. Such an engineering is called terraforming. It's way cheaper and feasible than moving to L1 with all belongings, from the planet's moon. (Why a moon? The original abode of alien civilization would be an a moon & not on the Jupiter-type planet itself, as it lacks a solid or liquid surface, which is certainly making life easier - pun intended).

Conclusion: If we agree to fix the terrible timing mistake by a factor of 22.7, then the only serious problem seems to be the dimming/brightening conundrum, and a less certain objection to a single, as opposed to double, dimming events during each eclipse/transit. Anything else you've noticed?

Other problem ideas

Solar eclipse of 2024

Info: On April 7 2024 the Moon was went through periapse (perigee) at $d_m = 258850$ km distance from (the center of) Earth. The Moon has radius of $R_m = 1737.5$ km. The Sun was then at 1.001293 AU from Earth, i.e. $d_s = 149791367$ km. Its diameter is $2 R_s = 1392684$ km. The Earth (radius $R_e = 6371$ km), is tilted to the ecliptic by 23.5 degrees, and rotates with period about $P_{rot} =$ 23 hr 56 min.

SOLUTION

Let's work in geocentric and observer-centric frames. The observer is on the surface of the Earth, at distance $\xi R_e$ away from its rotational axis. Since we are interested in $\phi = 45$ degree N latitude, $\xi \approx \cos \phi \approx 0.707$.
As it is nearly the time of Spring equinox, I will assume that the axis is tiled at angle $\theta_e = 23.5^o$, and lies in the plane perpendicular to the Sun-Moon line.

The Sun is much further than the Moon. So let's introduce a concept of 'reduced sun' (or equivalent sun that would look exactly like the true sun), placed at the distance of the Moon, having a reduced radius $R^m_s= R_s (d_m/d_s)$ and a reduced speed along ecliptic of $v^m_s =$ 30 km/s $(d_m/d_s)$.

The situation is this: The observer on the moving surface of Earth, the Moon, and the reduced sun, are all moving in the same direction, but the observer is the only one moving at an angle to the other two. Alright, lets get those two other speeds first. I think the Moon moves at its perihelion speed 1082 m/s. Calculation based on Earth+Moon mass and distance gives 1062 m/s. Whatever. The reduced sun moves only 70 m/s or so. The observer moves quite fast, at speed $\xi R_e \cdot 2\pi/P_{rot}\sim 328$ m/s, at an angle.

Derivation 1:
Imagine the three objects (m,s,obs) move on rails with different speeds. The Moon is the fastest. The sun moves in the same direction at only 70 m/s and the Moon catches up easily, the difference of speeds being about 1010 m/s. Observer moves in geocentric frame at speed 465 m/s if on equator, and at a more realistic place, at about 300 m/s, at 23.5 degrees to the 1010 m/s speed. If the observer moved at 1010 m/s parallel to the moon-sun motion, then the eclipse would not progress (would last almost forever, except for curvature of trajectorie s that we neglect). ok. $$\Delta v = v_m - v^m_s - v_{obs}$$ and that's a vector subtracttion, that makes the shadow move on a perpendicular plane at speed $\sqrt{1010^2 + 300^3 -2\cdot 300\cdot 1010\cdot \cos(24^o)}$ m/s = 740 m/s. The agreement with actual speeds of the shadow in Mexico is decent. Due to subtraction of rotation, the angle on the target grows to $\gamma$, given by equation $1010\, \sin 24^o = 719\, \sin \gamma$. This gives $\gamma\approx 40^o$ if I remember correctly. That's not a bad approximation to the path of eclipse on the map, except in the parts of Earth where its surface is far from perpendicular to the light rays. Likely due to this, near Toronto the eclipse moved almost 1000 m/s.

The time of totality (considered happenig near the noon) can be obtained from difference of angular diameters of sun and moon as seen from observer's point of view, divided by the $\Delta v$ and multiplied by the observer-moon distance $d_m - \xi R_e$.

(...)

The result was about 4.2 minutes of total eclipse duration, and almost 200 km diameter of the shadow, moving 700-1000 m/s across the surface, depending on time and latitude.