I have always wondered about this with regard to Newton's Third law:
For every action there is an equal and opposite reaction.
I ask because Earth is an third body scenario between the sun and Jupiter. Jupiter has enough gravity to occasionally pull the Earth slightly (not significantly) out of orbit from the Sun, but Earth's orbit to the Sun is self-correcting due to the difference in mass between the Sun and Jupiter. Quick web searching reveals Jupiter's pull on Earth is only approximately 0.005% of the Sun's after accounting for both mass and distance, but that number rises to 0.011% after accounting for syzygy with the moon.
> So if a craft departs from rendezvous with another craft it must do so by pushing away from that other craft.
More commonly we push away propellant, but pushing on the other craft is an option.
> If the rendezvous was in orbit does that mean departing from rendezvous pushes both crafts out of orbit?
Assuming we're pushing on the other craft, it means both crafts will change orbits, but in opposite directions. If we're talking about an instantaneous push in the direction of travel one craft will move to an orbit where it is closer to earth one half rotation later, and the other craft will move to an orbit where it is farther from earth one half rotation later.
Both, either, or neither craft could exit orbit like this, but one would be exiting orbit by crashing into the thing its orbiting around (e.g. earth) and the other would exit by reaching escape velocity and flying off into the distance.
> I ask because Earth is an third body scenario between the sun and Jupiter. Jupiter has enough gravity to occasionally pull the Earth slightly (not significantly) out of orbit from the Sun, but Earth's orbit to the Sun is self-correcting due to the difference in mass between the Sun and Jupiter. Quick web searching reveals Jupiter's pull on Earth is only approximately 0.005% of the Sun's after accounting for both mass and distance, but that number rises to 0.011% after accounting for syzygy with the moon.
Yes, Jupiter constantly (not occasionally) perturbs earths orbit, and technically earth constantly perturbs jupiters orbit (though the influence in that direction is completely negligible), but as you note its not enough for either to reach escape velocity or crash into the sun, and it appears to more or less even out over time.
the "push" is generated by escaping gas from burning fuel. it's possible that if the gas impacts the other craft it could affect that craft's orbit but I can't imagine that 60 years after people walked on the moon you and I are the first to think about that, so I presume its been accounted for
As Larry Niven wrote in The Smoke Ring and The Integral Trees...
"West takes you In, In takes you East, East takes you Out, Out takes you West, North and South bring you back again."
For anyone who is remotely interested in this, a considerable chunk of the Gemini program was all about solving some of the practicalities involved with Rendezvous, and it is quite interesting even hearing some of the astronauts come to grips with some of the physics while orbiting in space trying the various types of rendezvous and docking maneuvers that were attempted.
Buzz Aldrin (who was the second person to walk on the moon) wrote his MIT thesis on methods for astronauts to handle the complexities of orbital dynamics when performing rendezvous maneuvers in orbit:
Spend years reading and doing calculations before you understand orbital rendezvous. Or spend a couple days learning to dock in kerbal space program, with the real-sized earth.
I have a hard time imagining physics. For example take a train moving 100 kmh to the north which wants to reverse direction to the south. It has to break and then accelerate again, a very costly operation. Except when the tracks make a turn? But how can a northward momentum change to a southward momentum?
The same confusion I have when trying to imagine satellites going around Earth or slingshot maneuvers. Would an X-Wing turn in space differently than in the atmosphere of Hoth? Would it in space just rotate, but keep its forward (now backwards) momentum instead of turning like a fighter jet?
> The same confusion I have when trying to imagine satellites going around Earth or slingshot maneuvers.
I can't recommend KSP enough. It's a "silly" game with "on rails physics" (so not exactly 100% accurate wrt general relativity stuff) but it's got a very nice interface and it will make you "get" orbital mechanics by dragging stuff around. You'll get an intuition for it after a few hours of gameplay / yt video tutorials. Really cool game.
This is how I now “get” orbital mechanics better than I ever did trying to study it. Play is the best education.
> For example take a train moving 100 kmh to the north which wants to reverse direction to the south. It has to break and then accelerate again, a very costly operation. Except when the tracks make a turn? But how can a northward momentum change to a southward momentum?
Your train is decelerating, and then accelerating southwards. It really is.
If you were on a train that was travelling in a straight line northwards and the driver applied the brakes, it would decelerate, which really is acceleration with a negative value (and I can hear that in my old high school physics teacher's voice, hope you're doing well, Mr Siwek). You would feel yourself being thrown forwards if the acceleration was strong enough because your momentum wants to keep you moving north.
If you were on a train that was travelling around a U-shaped bit of track looping from northbound to southbound, then you'd be thrown towards the outside of the curve. Guess what? The train is not moving north so fast, and your momentum is trying to keep you moving north.
The difference here is that if you brake the train to a stop and throw it in reverse then you're dissipating energy as heat to stop it, and then applying more energy from the drivetrain to get it moving again, but if you go round a U-shaped track the energy going north is now energy going east. You have not added or removed energy, just pointed it a different direction.
Turning around a track definitely dissipates some heat energy through increased friction with the rails. Imagine taking a semicircle turn and making it tighter and tighter. At the limit, the train is basically hitting a solid wall and rebounding in the other direction, which would certainly transfer some energy.
The energy question is this: going from a 100kmh-due-north momentum to a 100kmh-due-south momentum via slowing, stopping, and accelerating again clearly takes energy. You can also switch the momentum vector by driving in a semicircle. Turning around a semicircle takes some energy, but how much - and where does it come from? Does it depend on how tight the circle is - or does that just spread it out over a wider time/distance? If you had an electric train with zero loss from battery to wheels, and you needed to get it from going north to going south, what would be the most efficient way to do it?
There is no "required" energy to change direction, even for a zero-radius change, think of a bouncing ball:
This only applies in perfectly elastic systems, where the bodies can convert kinetic energy to potential energy and back with perfect restitution. Which, thanks to the second law of thermodynamics, doesn't exist in reality. It's only a question of how much energy is lost. (Unless, of course, you include the medium into which the energy dissipates as heat into the system itself. But such a model is not useful in almost all practical scenarios.)
A bouncing ball is elastic. There is some loss in the process of storing the energy from the movement into the ball and then releasing it into the opposite direction. Good example though!
> Turning around a track definitely dissipates some heat energy through increased friction with the rails.
No it doesn't, but we're talking about identical spherical frictionless trains in a vacuum.
You are also talking about a track with infinite mass because otherwise the reason train can change direction is because it's pushing the track northwards
What's going on here is that your momentum changes whenever you experience a force. Your energy changes whenever you experience a force towards or from the direction that you are traveling.
The force from the rails at all points is at right angles to the direction of motion. So your energy doesn't change. Your momentum is constantly changing. And you're doing it by shoving the Earth the other way. But the Earth is big enough that nobody notices.
Now to the orbital example. In the Newtonian approximation, an orbit works similarly. In a circular orbit, you're exchanging momentum with the planet, but your energy remains the same. The closer the orbit, the more speed you need to maintain this against a stronger gravity, and the faster you have to move.
In an elliptical orbit, you're constantly exchanging momentum with the planet, but now you're also exchanging between gravitational potential energy, and kinetic energy. You speed up as you fall in, and slow down as you move out. Which means that you are moving below orbital speed at the far end of your orbit, and above when you are close.
Now to this paradox. Slowing down causes you to shift which elliptical orbit you are in, to one which is overall faster. Therefore slowing down puts you ahead in half an orbit, and then you'll never stop being ahead.
In your train example, the rails exert a force on the train as it turns. In orbit, the planets are constantly exerting a force on the satellite.
What's happening is that you exchange forward momentum for angular momentum. When the track straightens out again, you trade the angular momentum for forward momentum again. The train pays for this in friction losses; the orbital maneuver costs some fuel for steering.
When you brake you generate a ton of heat.
Doing a U-turn generates less heat, but still quite a bit. The train will have to slow down depending on the radius of the curve, and even then the turn will slow it down some more.
But yeah, less heat generation means kinetic energy is conserved.
Cars have to slow down when they turn because it’s too much to ask of the tires to accelerate (throttle) and turn, since turning is in itself acceleration.
Caveat: when the tires are already at the limit of adhesion (e.g. on an F1 car). In a road car, you are not normally turning at 1g and probably can’t accelerate at 1g so you can turn and accelerate when you have enough margin.
It’s just the average driver doesn’t realize how much margin is available.
A train has momentum in the direction of the track. If the track makes a 180° turn the train will lose some momentum to increased friction with the track during the turn, but essentially the momentum still follows the track.
A fighter jet (or X-Wing in orbit) kind of generates its own "track" with the guiding forces of the wings. You can still do a 180° turn and keep a significant part of your momentum. Though the guiding effects are a lot softer, so your losses are a lot worse
A satellite (or an X-Wing in orbit) has no rails that can go in arbitrary directions. Any momentum is in "orbit direction", but orbits work in weirder ways. If you make your orbit highly elliptical then at the highest point you will have traded nearly all your kinetic energy for potential energy and can make a 180° turn pretty cheaply (because it's only a small change in speed)
A very related physics issue that boggles my mind is when you roll a disk, like a wheel. You can roll the disk north, and it'll lean, curve, and end up going south. What force changed the direction of the wheel?
I understand it, intellectually. It's pushing sideways against the surface as it leans and spins, but it just doesn't feel right. I have no intuition for it.
If you are talking about the gyroscopic precession effect that happens when you push on a spinning disc, this is the best video I've seen so far that explains it in an intuitive way: youtube.com/watch?v=n5bKzBZ7XuM
You too can change direction easier if there is an object (like a pole or something) you can push/pull against. Try it, maybe it will help your intuition.
Run towards a pole and then try to come back around it, once without touching it and once using it to swing around. That's the role the curved tracks play. You exchange momentum with the object, and in the end with the Earth.
play KSP, it will click after a few days.
I feel like none of the answers have addressed you train example correctly. The momentum is exchanged with the Earth. So the Earth+train still have the same total momentum. The energy is mostly conserved (ignoring the friction that's needed to stay on the track). You can do the same by running past a lamp post and extending a hand to grab it - you'll change direction.
They should really teach physics using KSP.
I tried to teach a group of HS students about orbital mechanics as a high school physics teacher using KSP. It was... difficult. Not impossible. But I agree it's an excellent learning tool.
Right, the UI/UX is a lot to just get to the rocket part. KSP is probably the best game that forces that into your head with a classic simulation that's fun, but I gotta say something like Rocket League was better at building my intuition for rocket behaviors.
Yeah, it's amazing. With enough docking and maneuvering practice I developed some kind of intuition for moving in space. I could maneuver without meticulously planning the burns.
Still can't leave Eve though...
I'm a professional astrodynamicist and I owe my base level understanding of orbital mechanics to KSP. It's a fantastic resource for learning the basics of Keplerian motion.
Also, obligatory XKCD: https://xkcd.com/1356/
Arguably aerodynamics is confusing on a whole other level to mere orbital dynamics. :D
I washed my hands of aerodynamics after I got my first job in satellite navigation. Messy stuff, that Navier-Stokes business
No, it's simple. Just make sure the airplane falls nose-first if it ever stops (speed<stall).
I wish ksp 2 hadn't been a boondoggle
I haven't kept up with it, but hopefully Kitten Space Agency will be able to take up the torch.
I doubt SpaceX could put a satellite in orbit with KSP physics. Just the absence of realistic thermal conduction would prevent it. The outer skin temperature typically peaks around 300–600 °C during the densest part of the atmosphere. If you calculate those forces wrong the rocket has a bad day. Best case it is over engineered and has a reduced payload. They might as well do their calculations with pi equal to 3.
What does thermal conduction affect? Is it mostly practical spacecraft construction, or actually related to orbital mechanics?
The FAR mod is touted as being realistic; I haven't played it though.
https://xkcd.com/2205/ comes to mind with your pi approximation.
Nobody is saying KSP physics is perfect.
Until I played KSP, I had no idea how hard orbit was compared with just going up into space (and generally the greater population thinks the same -- they think that sending New Shephard upto 100km is about the same as sending a Dragon into orbit). I had no idea how you move in orbit, how getting from low earth equitorial orbit to Jupiter takes less energy than getting from the same ship to a polar orbit (and even then that the only real way to change your orbit like that is to go out beyond the moon and back), etc.
Forwards is up, up is back, back is down, down is forwards.
It's much easier to reason about when you frame it closer to reality: you're not on a circular path, you're continuously falling, and because you're moving forward, you're continuously missing the earth, with its pull decreasing with distance.
How related to this is the helicopter 90-degree phase lag thing?
Or in the words of Larry Niven (The integral trees)
East takes you Out
Out takes you West
West takes you In
In takes you East
Down is where the enemy gate is.
So precise, he piss on a plate and never splash.
This brings back fond memories of Heinlein's juvenile sci fi series.
A similar thing is true when cornering a race car when measuring time through the corner.
Slow is smooth, smooth is fast.
How so?
"Faster" (higher speed) = wider cornering radius = more distance = slower
But you exit going faster, which means you make up time on the straight after the corner.
It's a balance, many of these cars can accelerate and decelerate very hard so the time to get back to the full speed for the next section is fairly short reducing the effect of slowing down. The effect of taking a too wide racing line though means a large multiple in the distance travelled.
Cars can usually brake and turn harder than they can accelerate.
You also tend to spend more time on the straight after the corner, than in the corner itself
So you mostly optimise for corner exit speed, especially if the car has particularly slow acceleration and a long straight comes after the corner.
For F1 I was under the impression exit speed wasn’t as important as minimizing arc length of the turn.
Assuming there is a long enough straight before the next corner
Yeah depends on the corner but the general thumb-suck approximation is sound.
If there's banking, it can change things.
The distance has no effect.. Its all about speed, you want to take the line that lets you get through the corner while maintaining the highest speed. If you are going faster and spend as little time as possible breaking and accelerating you will gain time. Also a higher exit speed means you will be going faster for the entire straight after the corner making a very big time difference.
Your car, depending on how much grip it has + other variables, will have a theoretical minimum diameter circle it can drive around at various speeds. The higher the speed the bigger the circle. Finding your racing line is just a matter of fitting the biggest circular arc inside the space available in the corner.
Ideally you want to break in a straight line before the corner and reach the speed your car can drive the circle at at just the moment you enter it.
Theres more nuance when it comes to compound corners, FR vs FF cars, oversteer understeer, hills bumps etc. But the basic theory is simply fitting circles.
You are generally not taking a perfect circle corner. You can/should be slowing down as you enter the corner and then speeding up even before you exit. In this way you can shorten the distance traveled while getting a higher exit speed - sometimes higher than the largest possible circle corner. Optimizing this for the car/track/conditions is what makes for a great driver.
The distance is irrelevant.. It is true that depending on the car you may gain time breaking and accelerating while turning.
But that is a more subtle and advanced concept though (like dealing with elevation changes).. People should understand the big circle first.
In the context of winning a race you need to get the subtle and advanced concepts right or you will be in last place. If you are just driving on the street it doesn't matter.
Most times Ive seen anyone playing a racing game they seem to be totally clueless.. They dont even comprehend the big circle. They always go into corners way too fast, break super hard and then crawl out.
Its so common it surprises me racing games have always been so popular.
What I have also noticed is that over time racing games have changed their physics to be totally wacky in order to meet the general public's wacky expectations.. (eg. mario kart or GTA5) I cant play those games cus the physics are so strange.
I was referring to real world races where we cannot ignore physics.
Racing games are very different. They tend to have adaptive AI - you are more likely to win with the naive approach you describe than the physically perfect route. The physically perfect result will get your through the race several minutes faster, but the AI opponents become impossible to beat. Thus the ideal path is the worst thing you can learn. (I haven't played games in years, but IIRC the games you mention don't pretend to be about racing, I wonder how ones that pretend to be a real race compare)
?? I guess you havent played modern racing games. No-one races against AI, its all against other people. Games like Assetto Corsa and iRacing have very good physics models. Real race drivers use them to train and are often seen online.
The circle thing is aimed at most people here. If your average person implemented that they would dramatically improve their times.. All the other stuff (of which of course there is a lot) would result in relatively marginal improvements.
Life and business are often the same.
I thought I understood all this until I played Kerbal Space Program.
Obligatory: https://xkcd.com/1356/
could it be possible to flag a thread when you need to pay or register to read an article ??
very annoyoing, the subject looks good, open tab and rohhhhhhhh... paid or register.
Paste link, good to go. https://archive.is/qrP0p
There is the small, tiny issue of people commenting just based on title, without even reading the article.
In this case I expected it just links to https://www.youtube.com/watch?v=bcvnfQlz1x4 and didn't even notice in links to Wired.
Allowing paywalls vs not been discussed for a long time. Latest comment from dang about it seems to be this:
> The answer is that paywalls are allowed when there are workarounds (such as archive links) which allow ordinary readers to read the article without paying or subscribing, while hardwalled domains (i.e., without such workarounds) are banned. - https://news.ycombinator.com/item?id=43876575
https://hn.algolia.com/?dateRange=all&page=0&prefix=true&que...
This is a consequence of the virial theorem of mechanics.
https://en.wikipedia.org/wiki/Virial_theorem
This theorem also lets you conclude that as a nondegenerate star becomes more tightly bound (smaller, for a given mass) it must also become hotter.
(Why did someone downvote this?)
I'll posit it's because the virial theorem is applicable to the time average of a stable system and is therefore a specific case. The time average is not a requirement for this discussion (i.e. dynamics) and invoking that theorem is unnecessary. Ergo it is not a consequence.