Thank you very much for sharing this with us. I was wondering whether this behavior may similarly occur for nozzles immersed in pressurized fluid-chambers or if the same effect is observable instead for inlet pressure increments for a fixed exit pressure. If immersed, What will be the conditions for the shock waves to happen within the surrounding fluid?
@mz0095618 күн бұрын
3:02 wow I am impressed you have a map where L3 L4 and L5 are not on the orbit of earth, but above! thank you, this is great
@KUSHALGOKHALE25 күн бұрын
Good work. Appreciate your hard work.
@jeffreywwilsonАй бұрын
So you are assuming that the Earth is in fact ROUND?
@christianfrassl7107Ай бұрын
Great video
@dalelerette206Ай бұрын
Trust is faith graduated to perfect healing. ❤
@dalelerette206Ай бұрын
Trust is faith graduated to perfect healing. ❤
@is47374 ай бұрын
Excellent!
@TheGravityAssistant2 ай бұрын
Thank you and thanks for watching
@kapilsharma-xz7os4 ай бұрын
Oh man thank you for making me understand about CD nozzles in compressible flow case and ..Most importantly, Pe and PA variations.
@TheGravityAssistant2 ай бұрын
Glad the video was helpful! Thanks for watching
@wiictorr4 ай бұрын
OMG THIS VIDEO SAVED ME IN HYPERSONIC AERODYNAMICS!!!!!!! Thanks from Brazil🟢🟡
@TheGravityAssistant2 ай бұрын
Glad it was helpful!
@TAasali4 ай бұрын
Amazing!!😮😮😮
@CarlosAM14 ай бұрын
"What happens next will SHOCK you" ...badum tss
@bjornisaksson13324 ай бұрын
Very informative and nicely animated. I would love to see the same but with the moon<->earth instead, obviously focusing on tidal effects
@spoddie6 ай бұрын
Well done mate. I had naively thought the Lagrange points were points where the gravity was equal, although if I ever thought about it, that won't make sense.
@TheGravityAssistant6 ай бұрын
Thanks for watching and the kind words!
@mz0095618 күн бұрын
yes because of the mesh that was shown at the start. there it cancels out, but only because the fake/imaginary value from the rotation is added sun = earth + rotation
@mrnnhnz7 ай бұрын
Someone has explained it to me, but I still only have a rough grasp of: why spacecraft at a L point dong just go there and stay there. It's seemed to me that this would minimize station-keeping fuel use. But, apparently, orbiting the L points instead is actually more efficient, station-keeping wise. I suppose it's to do with the, in real life, changeable nature of the L points. I wonder if you would perhaps include a visualization of the changing location a L points over long timespans? This video came very close to doing that, but it was mostly about what happens if you perturb a spacecraft at the L point, not so much about the dynamic nature of that point itself over long time frames. Obviously, aside from station-keeping concerns there are other good reasons for orbiting that point instead of going right there, like there might be several spacecraft who want to be there, and they can't all inhabit the same point in space! Good video, enjoyed this.
@mrnnhnz6 ай бұрын
@@TheGravityAssistant Yes, that was helpful. Thanks for taking the trouble to reply.
@b.h13628 ай бұрын
great video, can you make a video on nrho orbits ?
@TheGravityAssistant7 ай бұрын
Thank you! It's on my to-make wish list, but unlikely to happen any time soon unfortunately. My understanding of NRHO orbits at the moment is not deep enough for me to concisely explain them in a coherent way. The things higher up the list are topics I'm in a better position to explain!
@josecarlosribeiro36288 ай бұрын
Congratulations Master for your beautiful presentation and interesting manter! I learnt so much! Thank's a lot and my best regards! Jacareí-Sao Paulo- Brazil.
@paperclips13068 ай бұрын
Does KZfaq also put cookies in my brain because I literally thought of this " ok the math says so but what's the reason for supersonic flow".
@chibuezendiokwelu638 ай бұрын
Hi I left a comment on your follow up comment looking forward to ur response
@andrzejekandrzejek61328 ай бұрын
Thank you, well done. I fell about 50 years younger. All the best to you:))
@TheGravityAssistant8 ай бұрын
Thanks very much for the kind words. All the best to you too
@chibuezendiokwelu639 ай бұрын
I have a question, u mentioned that pressure at the exit nozzle must match the atmospheric pressure. So how do you reduce the atmospheric pressure. Also the part abt shock was a bit confusing Looking forward to your reply
@TheGravityAssistant9 ай бұрын
Hi, atmospheric pressure at the exit can be reduced by increasing altitude (e.g. a rocket or plane flying higher). Alternatively by conducting a rocket engine test in a vacuum chamber, or with a pressure reducing device such as an inducer. You could instead imagine that we increase the pressure of the fluids going into the engine, as it is the ratio between fluid total pressure and atmospheric pressure that matter. What was confusing about shocks, could you elaborate a bit more please?
@chibuezendiokwelu638 ай бұрын
12:58 since u said shock is an inefficient compression process, and since shock travels in opposite direction to the flow in the nozzle, how does it's increase help match the nozzle exit pressure to the atmospheric pressure p.s- what is supersonic expansion
@TheGravityAssistant8 ай бұрын
Apologies for the slow reply. Can you please restate your questions, I don't really understand what you're asking.
@chibuezendiokwelu638 ай бұрын
@@TheGravityAssistant how does increase in shock help exit pressure to match atmospheric pressure
@Me.is.Malhar9 ай бұрын
amazing video!! i totally thank myself for picking aerospace!!
@TheGravityAssistant9 ай бұрын
Thanks for watching, glad it was helpful
@salesv10 ай бұрын
This was 2 years ago but I can't wait for the next one
@TheGravityAssistant9 ай бұрын
Thanks for the kind words! I have plans to make more videos, when life permits, unfortunately can't say when..
@manjuna112311 ай бұрын
Great video.
@TheGravityAssistant11 ай бұрын
Thanks for watching!
@Borsting8911 ай бұрын
I am halfway trhough the video. This video does a great job explaining. Finally I understand the L4 and L4 points.
@TheGravityAssistant11 ай бұрын
Thanks for watching, very glad to hear that it was helpful!
@andrewscoppetta4944 Жыл бұрын
Thank you… just thank you so much. I finally understand why Lagrange points work
@TheGravityAssistant Жыл бұрын
Very glad to hear that :). Thanks for watching!
@The1andOnlyWog Жыл бұрын
It is criminal that you are so under-subbed! I did my part in trying to rectify that grievous error by subscribing. Lol. Great videos, very informative and easy to follow. Keep up the great work, don't give up, and I'm sure you'll go far on this platform. I can't wait to see how far you go!
@TheGravityAssistant Жыл бұрын
Thanks for the kind words and for watching!
@icojb25 Жыл бұрын
Where is part 2!!!!
@RyanSmith-ow6cm Жыл бұрын
I hope you're able to make more of these videos, this was absolutely fascinating.
@TheGravityAssistant Жыл бұрын
Thanks for watching and the kind words. I am in the process of making more videos, but unsure when I will be able to complete them unfortunately. Work and life is keeping me very busy these days
@mubashirahsan8524 Жыл бұрын
BEST FUCKING VIDEO EVER, SAVED MY PROPULSION MID TERMS
@technocracy90 Жыл бұрын
Where's part 2??
@TheLordGojira Жыл бұрын
The biggest question I have is just “how do you reduce the pressure at nozzle exit?” Does the air coming out of the nozzle play any part in this? I can certainly understand a pressure controlled room or environment, but I don’t think that’s really what’s meant or else it’d be kind of useless.
@TheGravityAssistant Жыл бұрын
What is important is the ratio between the pressure in the combustion chamber (high pressure) and the ambient (low) pressure at the exit. If you want to achieve supersonic flow at the exit, you can increase the high pressure or decrease the ambient pressure (or both). You're right that if we want a rocket nozzle to work at sea level, it doesn't make sense to try and lower the environment pressure - so first stage rocket engines usually have very high combustion chamber pressures to achieve a large pressure ratio. Conversely, 2nd and 3rd stage engines which operate in near-zero pressure environments can have much lower chamber pressures and still achieve supersonic flow / a high pressure ratio. Anything divided by almost zero is very big.
@TheLordGojira Жыл бұрын
@@TheGravityAssistant I see, that makes intuitive sense in retrospect. If the starting total pressure is higher, to the point that the reduced static pressure is close to atmospheric conditions while holding onto a lot of dynamic pressure, it would just keep moving through similar conditions as though not much had changed. If the static pressure is similar but the temperature is much higher, does that mean the compressed gas is actually released in the exhaust at densities lower than STP? I’ll have to try to wrap my head around that… I’ve been trying to look into a means of numerically estimating the motion of airflow based on pressure differentials and figuring out how far an open air supersonic flow will continue traveling at supersonic speeds before reaching equilibrium with normal air, and I’m not quite sure what-in open air away from the turbine-stops a flow from forming normal shock outside of just powering through it with high pressure…? And how the duration of that air burst impacts this. Do you have any suggestions on where I should look to that? I’m not really sure how to mathematically define the behavior of a compressible thermodynamic gas to move from a region of high pressure to low pressure in the first place. >.<
@TheGravityAssistant Жыл бұрын
@@TheLordGojira To answer your first question, the ideal gas law PV = nRT can be rearranged to P = rho * R * T, where rho is density and R = R_universal / M_molecule. From that we can see that yes, gasses at higher temperature for a given pressure will have lower density. It also makes sense when you think about a candle or a camp-fire, their combustion is happening at 1 atmosphere of pressure, and the heated gasses float away because of their reduced density (and the effect of buoyancy). I'm not sure I totally understand your second question. I'm going answer the question 'how does a supersonic jet stream break down and dissipate', and hope that helps :). When the supersonic stream exits a nozzle, it is at the same static pressure as the atmosphere, but it is has a much high velocity than the ambient air, so when it moves past there will be a large shear force. This shear force acts to slow down the jet, (and accelerate the ambient air a little bit). Viscous shear and turbulence dissipates the supersonic jet's high dynamic pressure. Have a look at figure 12 and 13 of this paper: iopscience.iop.org/article/10.1088/1742-6596/1240/1/012019/pdf And look at the bottom of the exhaust stream from this photo: space.stackexchange.com/questions/29758/temperature-and-pressure-of-rocket-exhaust Similarly, the temperature will eventually equalise once the hot exhaust gases have radiated or convected their excess heat away to the rest of the atmosphere.
@sandeepdwarka Жыл бұрын
I'm out of mind! Accidentally watched this video, now I'm forced to subscribe this channel. ❤❤❤ More info expected on DSN communication. Best wishes, keep educating.
@TheGravityAssistant Жыл бұрын
That's for the kind words and thanks for watching!
@jamesmaddison4546 Жыл бұрын
Make more videos please? pleeaassseee?🙏🙏🙏
@TheGravityAssistant Жыл бұрын
Thanks for watching! I have more videos planned and some draft scripts written. The problem is finding the time to make them :(
@samuelsnowdon2271 Жыл бұрын
Great for ksp FAR
@GemstoneActual Жыл бұрын
Where's part 2?
@jamesmaddison4546 Жыл бұрын
Read the other comments and youll have your answer....
@mastershooter64 Жыл бұрын
gonna try this in ksp now
@TheGravityAssistant Жыл бұрын
Unfortunately it doesn't work in KSP (at least it hasn't work when I have tried). I suspect it is because of the basic aerodynamics model in KSP. AGA really needs a good supersonic / hypersonic model in order to work.
@nikelinq2899 Жыл бұрын
@@TheGravityAssistant what about with Ferram Aerospace Research installed?
@TheGravityAssistant Жыл бұрын
@@nikelinq2899 I'm not familiar with how FAR models hypersonic flight. If you feel like giving it a go, I'd be very interested to hear the outcome!
@mastershooter64 Жыл бұрын
9:35 yooo dudee whaaatt?? awesome!! nice!
@Iceflkn Жыл бұрын
A couple questions: 1. How much gravity do LaGrange Points have? Is there a direct relationship to the strength of that points gravity when compared to the celestial bodies that created it? 2. Are LaGrange Points taken into account when trying to predict the orbits of the planets? 3. Do Lagrange Points create gravitational lensing? Efforts are currently underway to use the suns gravity as a telescope, thanks to the gravitational lensing the sun creates. If these LaGrange Points have the necessary gravity, could they be used the same way as, "Gravity Telescopes", so to speak?
@TheGravityAssistant Жыл бұрын
Hi, thanks for watching! To answer your questions: 1 - Lagrange points don't 'have' any gravity of their own, as they are just points in space with no mass. The two bodies (e.g. the Sun and the Earth) have mass and so generate gravitational fields. The Lagrange points are just empty points in space where the two gravity fields of the bodies interact in a way that allows a third body with a small mass to orbit around the main body in a way that would not 'normally' be possible if there was no second body. 2 - No - similar to point 1, the Lagrange points do not attract / repel / interact with anything as they are just empty points in space that are a result of two interacting gravity fields. It is the behaviour of the gravitational fields around a Lagrange point that causes the peculiar motion of the third body. The gravity of each of the planets however, must be taken into account when predicting the orbits of the other planets, as the gravitational fields of each planet propagate infinitely and interact and change the motion of each of the planets. 3 - No, because Lagrange points don't have any mass (and therefore gravity) of their own, they will not cause lensing.
@rat_king- Жыл бұрын
tanh(x)for fluid supersonic. cosh(x) for object driven in flow..... why?
@TheGravityAssistant Жыл бұрын
I'm sorry, I don't understand your question. Could you please clarify what you mean?
@carultch Жыл бұрын
@@TheGravityAssistant I'm guessing that Rat King is referring to the graph at 7:27 looking like a tanh(x) curve, and the graph at 10:38 looking like a cosh(x) curve, and wondering why those two functions describe the situation..
@loretacancela5297 Жыл бұрын
Hi! I'm currently designing a Nozzle for a CubeSat, I used MOC to obtain the design of the wall. I then passed it to ANSYS and i have an underexpansion effect which was expected if my ambient pressure is 0. So, I want to know how much I can expand the exit area to get the pressures to balance. I want to model this just like your video where I can have two charts and see the change in Mach and Pressure changing the velocity and exit area, what program do you use or what can you recommend for this. Thank you!
@TheGravityAssistant Жыл бұрын
Hi, thanks for watching. I hope that I am understanding your first question correctly - it is not possible/practical in reality to match the exit static pressure of the nozzle with the vacuum pressure (~0) in space. It is possible to calculate a maximum theoretical exit velocity from a nozzle where all the thermal energy is converted to kinetic energy and the exit temperature is 0 K (and therefore the exit static pressure is 0), but this would require a very long (and heavy) nozzle with a very big expansion ratio. Additionally a real gas would condense into liquid or solid at a temperature >0 K and so the maximum ideal gas expansion could not be achieved. Additionally + 1, viscous losses with the long nozzle wall would remove some amount of energy from the flow, further preventing us reaching this theoretical idea. In practice, all space nozzles are under-expanded, and the designer has to make a trade-off between factors such as performance (Isp), mass, packaging (length and diameter) etc. etc. Regarding the plots of Mach / pressure / temperature - I made them in matlab using the geometry of a simple converging-diverging nozzle, and the isentropic flow relations (please see the link below). If you know the cross section area at each position along the nozzle, you can use equation 9 (the Mach-area relation) to calculate the Mach number at each position. Once you know the Mach number at each position along the nozzle, you can use equations 6 and 7 to calculate the temperature and pressure at each location. www.grc.nasa.gov/www/k-12/airplane/isentrop.html Hope that helps!
@bengrindell7693 Жыл бұрын
Outstanding video!!!!!
@TheGravityAssistant Жыл бұрын
Thank you! Thanks for watching
@andre-ww5dd Жыл бұрын
Why this video has so little views?
@ariebos7872 Жыл бұрын
Great explanation with very explanatory graphics. Many thanks. A question though: what are design parameters for a nozzle to create certain effects? E.G., if I wanted te create maximum shockwaves, regardless wat happens to the airspeed at the exit, what would that mean for the design of the nozzle?
@TheGravityAssistant Жыл бұрын
Thanks for watching and the kind words. That's an interesting question and I had to take some time to think about it. On a fundamental level, the most important variable that we can control is the pressure ratio between the total pressure in the chamber and the pressure of the environment at the exit. Please see the image I linked at the end of this message. This following paragraph assumes that we have a constant nozzle geometry, like shown in the linked image. At point 7 we have perfectly expanded flow at the exit and no shocks. As we move upwards to point 6, the environment pressure is a little bit higher than the final nozzle pressure, and so some 'weak' oblique shocks are needed to match the exit pressure to the environment pressure. As we increase the environment pressure more, the weak oblique shocks eventually turn into a 'strong' normal shock at point 5. This shock occurring at point 5 is the 'strongest' shock that a converging-diverging nozzle (of a given geometry) can create. If we increase the environment pressure more, the normal shock has to move into the nozzle and it gets weaker because the subsonic expansion now contributes to some of the pressure recovery. If we want to create a stronger normal shock at the nozzle exit, we need to increase the mach number of the flow at the nozzle exit, and so we need a bigger nozzle exit. However, with the same chamber-to-environment pressure ratio as before, this nozzle will now be over-expanded and the normal shock will actually occur inside the nozzle (situation 4). And so we will need to increase the chamber pressure to move the shock back to the nozzle exit and return to situation 5. So to summarise, if our goal is to make the strongest shock possible, we want a huge expansion ratio nozzle and a (quite high) pressure ratio between the chamber pressure and environment pressure that allows us to achieve situation 5. aerospaceengineeringblog.com/wp-content/uploads/2016/06/NozzleConDiv.jpg This was a longer message than I expected! I hope it is helpful.
@ariebos7872 Жыл бұрын
Thank you so much for your elaborated answer. Is there I way I can email you directly?
@TheGravityAssistant Жыл бұрын
@@ariebos7872 No worries. I can be reached at [email protected], and I will reply when I can
@24pavlo Жыл бұрын
Is there part 2?
@TheGravityAssistant Жыл бұрын
There will be!
@24pavlo Жыл бұрын
@@TheGravityAssistant So there is part 1, part 3, but no part 2?
@TheGravityAssistant Жыл бұрын
@@24pavlo Correct, there will be a part 2 on oblique shock waves, but it is still in progress.
@Leader-th4dn Жыл бұрын
As an aerospace engineer, I think your videos are fantastic! The animation is awesome and the detailed explanation of the theory is pretty clear. 😇 I can't wait to see more videos.(I love hypersonic topics🤩)
@TheGravityAssistant Жыл бұрын
Thanks for the kind words! I'll do my best to put out some new videos in the near future, work has been hectic the last few months
@LauridsDenmarkCool Жыл бұрын
Great informative video, but at 5:07, the formula you use is T^2/2pi, where I believe I should have been (T/2pi)^2 or T^2/4pi^2. can anyone confirm this? thank you
@saimadhav7695 Жыл бұрын
Great video. Can I know what software or tool you used to illustrate the flow field and post shock properties?
@TheGravityAssistant Жыл бұрын
Thanks for watching. The flow field is solved and animated in MATLAB.
@cleon_teunissen Жыл бұрын
In this video you put a lot of emphasis on the rotating coordinate system. I believe you are not doing your audience any favors with that. Please allow me to explain. Before I get to my main point, let me first acknowledge the history of the Lagrange points problem. Today we live in the age of computers that can do high precision numerical analysis, allowing us to create visualizations such as the potential surface representation. Before the age of electronic computers the choice of way to bring the Lagrange point problem to a solution was determined by computational efficiency. Using a rotating coordinate system was more efficient, so that is what physicists used. Choice of computation method and how to physically understand a phenomenon do not necessarily coincide. Very often they do, but not always. Example: there is the branch of electronic engineering that deals with the equipment for generating, or receiving, or processing electric oscillations. These circuit boards have resistors, capacitors and inductors. The electric oscillations in those circuit board can be modeled with sines and cosines, but when using sine-cosine notation it is tedious to keep track of phase. As we know, sines and cosines can be represented with exponential notation, by using complex number notation. To make the calculations more efficient electronics engineers introduce imaginary current and imaginary voltage, allowing them to move the calculation to complex number space. The electronics engineers are aware that the imaginary current and imaginary voltage have no physical counterpart; they are purely computational tools. To *explain* to students what is happening in the circuit board the teacher uses only the physical current and the physical voltage. In Celestial mechanics: The intuitive understanding of celestial mechanics is in term of the quantities momentum, angular momentum, and kinetic energy. Example: Halley's comet: The major axis of Halley's comet's orbit is about four times greater than the minor axis. Let's start from the aphelion of Halley's comet. Moving at its slowest velocity of all its orbit the comet starts falling to the Sun. The comet has only a bit of radial velocity, but it is enough to not hit the Sun. Moving down the gradient of the Sun's gravitational field the comet is constantly being accelerated. As the comet approaches its perihelion the gravitational acceleration becomes ever larger, and that acceleration vector is shifting to perpendicular to the instantaneous velocity. At perihelion there is a local maximum in how much the orbit curves. The ascending journey back to the aphelion is the time inverse of the descend to perihelion. I think we will all agree that the above is the way to understand the orbit of Halley's comet intuitively. Another expression would be 'visceral understanding'. Conversely, any attempt to present the mechanics of Halley's orbit in terms of a rotating coordinate system would be absurd. In my opinion this generalizes to all of celestial mechanics. Why does an asteroid orbiting near L4/L5 of Jupiter never escape? Well: that is due to celestial orbital mechanics, which arises from the inverse square law of gravity and inertia. The inverse square law of gravity gives rise to a Kepler orbit, and a Kepler orbit loops back onto itself. Let's say you are a celestial object, and you are orbiting right at the L4 point of a primary and secondary. Some perturbation (from yet another planet, far away, in that same system) makes you slide away from the L4 point. Well, there is no escape; the orbital mechanics brings you right back to where you started from; a Kepler orbit loops back onto itself. My point is: there should not be a tacit assumption that there is a 1-on-1 relationship between efficient computation and physical understanding. They often coincide, but sometimes they don't. Some years ago I created a physics simulation for orbiting motion along the L4/L5 points. I programmed the calculation for motion with respect to the inertial coordinate system. Using inertial coordinate system or rotating coordinate system, the computer performs the computation in real time either way, so for programming the computer there is no need to use a rotating coordinate system. For programming the numerical analysis: using the inertial coordinate system is simpler. As to the display of that simulation: that is a two-panel display. One panel for the inertial point of view, the other for the rotating coordinate system point of view. The computer performs that coordinate transformation in just a couple of processor cycles, so of course I added that coordinate transformation.
@TheGravityAssistant Жыл бұрын
Thanks for your thoughtful comment, and apologies for my very slow reply. I wanted to take some time to consider your post. Firstly, I agree that in the case of an elliptical orbit such as that of Halley’s comet, it would not make sense to use a rotating reference frame, as the angular velocity of the orbiting body is constantly changing. We could have a variable angular velocity rotating frame, but then that introduces yet another abstract force… However, Lagrange Point analysis assumes a circular orbit of the secondary body (Circular Restricted Three Body Problem). Clearly this assumption will only be valid for planets / moons with a nearly circular orbit about the primary body, and in fact the Lagrange Points become meaningless if the eccentricity of the orbit strays too far from 0. However, this assumption is good enough for many planets in our Solar system. Personally, I don’t find your explanation of L4 stability satisfying. If the stability is simply a result of the inverse-square nature of gravity and a closed Keplerian orbit, why should L4 and L5 be stable while L1, L2, and L3 are not? Why should L4 and L5 become unstable if the secondary body becomes too large? I agree that rotation frame alone is not a total solution, but IMO it can provide an intuitive explanation for something like 80% of the problem. I don’t mean to be rude, but did you watch the entire video? (I know it’s long 😊) At 20:23, 27:45, 32:24, 33:50 I have the side-by-side rotational frame and inertial frame animations that you describe from your previous work. It’s my opinion that humans are not very good an intuiting circular motion in an inertial reference frame, as it’s not something we have day-to-day experience with (additionally, visually keeping track of a quickly rotating object and also analysing its motion at the same time gets tiring quickly). Generally, humans are IN the reference frame that is undergoing rotation. And so, even though the centrifugal force and Coriolis force are not ‘real’, we understand them and their effects in a rotating reference frame more directly. Some daily examples of the centrifugal force being a car or bus going around a corner and ‘throwing’ us outwards, or a roller coaster ‘pushing’ us downwards at the bottom of a loop. In the case of the Coriolis force, kids throwing a ball on a merry-go-round learn that it curves in funny ways, and then in high school we learn about how it causes cyclones and hurricanes due to the Earth spinning. Even things like retrograde motion of the planets are due to us observing them from a rotating reference frame, and the debate around that topic took quite some time to resolve 😉! In any case, the Coriolis and centrifugal forces are basically IF this, THEN that rules which can be visualised quickly. Also, the rotating reference frame allows convenient calculation of the potential surface, which provides a nice graphical illustration of the points and their stability. Feel free to point me towards a reference, but I have never come across such a neat illustration of Lagrange point stability from an inertial point of view. In my mind, the inertial reference frame just provides a useful ‘anchor back to reality’ which is the last 20% of the puzzle in this analogy.