What worries me is that they’re stealing a little bit of Jupiter’s momentum every time. If they’re not careful it’ll fall towards the sun and we’ll have a Jupiter landing on our heads.
As the other commenter has stated, it’s not about the gravity - because you have to ultimately counter that gravity to escape. But it is about the kinetic energy of the planet as it orbits.
While it’s true that your spacecraft would have to ‘counter that gravity’ to escape, that’s from the frame of reference of the planet. From the frame of reference of whatever distant object you want your craft to ‘accelerate’ towards, your craft will appear to have gained momentum. If it were a zero sum game- there would be no “gravitational slingshot” effect (aka gravity assist maneuver).
The way your spacecraft ‘steals’ kinetic energy from the planet it orbits is by using the “gravity” of the planet. The two objects never come into physical contact with one another, the mass of the ship and the mass of the planet effect each others path through space-time- although very slightly. That is to say they seem to ‘pull’ on each other- what we call gravity.
The Earth and the Moon likewise ‘steal’ energy from each other through ‘tidal’ interactions. This causes the Earth to rotate more slowly and the moon to recede from our planet- this is all due to ‘gravity’.
Black holes also have kinetic energy that you can ‘steal’, to boost yourself toward a third celestial body just like planets do.
funkajunk@lemmy.world
on 31 Aug 16:03
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Spacecraft can use the gravitional energy from Jupiter to perform a “slingshot” maneuver, gaining significant momentum and reaching the outer solar system with less fuel.
if i remeber correctly, it just slings most of fast moving things around (roughly equally in all direction), and only slow moving things actually hit it.
slung out of the system.
that seems a bit too strong for jupiter, that seems more like suns behaviour
Jupiter's pull is so great, compared to earth, that the ones that do get past or then pulled more towards the sun.
this seems correct.
but i have not actually done any courses on celestial mechanics, and mostly basing on yt videos that i watch, so you maybe are correct on this one.
This is a sling. Bottom right corner is a catapult and uses stored energy (the elastic - sorry if wrong wording.) To me at least the idea of a sling that pivots around the user’s hand is much closer to the action of using a planet’s gravitational pull to sling a satellite craft onwards.
You wouldn’t want to do it with a satellite and I’ve only ever heard of it being called a slingshot (which usually refers to the device pictured in the meme, likely increasing clarity) or a gravity assist.
Satellites are objects orbiting a planet or some other smaller object. One could argue that it is a satellite on the way though too.
I find the idea of an object like the moon doing a gravity assist around Jupiter to get to us very amusing.
mr_azerty@lemmy.world
on 01 Sep 07:52
nextcollapse
What bothers me is I often read they are using the planet’s gravity to gain speed. Whatever speed an objet may gain while entering orbit should be lost when exiting it, right ?
So I guess it’s the cinetic energy of the planet that is actually fuelling the spacecraft, isn’t it ?
tiramichu@sh.itjust.works
on 01 Sep 09:26
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Yes. If the planet was stationary in space, it wouldn’t work. Approach from ‘behind’ the planet and you get a boost, approach from the ‘front’ and you hit the brakes.
Whatever speed an objet may gain while entering orbit should be lost when exiting it, right ?
That is true from the frame of reference of the planet. From the frame of a 3rd distant object that you want to accelerate towards, it appears you have gained momentum.
So I guess it’s the cinetic energy of the planet that is actually fuelling the spacecraft, isn’t it ?
Yes, but the mechanism for ‘extracting’ the kinetic energy from the planet is by using ‘gravity’, hence the name, “Gravitational Slingshot”.
You are gaining (or losing) energy based on if you are traveling in the same direction at the planet or not.
If you are coming from behind (travelling in the same direction) you an falling into the gravity well for longer. Thus gaining more energy. The extra energy is based on the speed of the planet through space.
Conversely if you an coming from the front, you fall for a shorter period. You lose energy at you climb up the gravity well.
But as you fall towards a planet (any gravity well); you pick up speed, if the planet is moving away from you, you fall for longer before you catch up. As you climb back up, you don’t spend all of the energy you gained on the way down. That difference is the Slingshot effect.
It also works in reverse, if the planet is moving towards you. You catch up quicker, thus gain less speed. And spend overall more energy than you gained when you climb back out. Slowing down in the process.
kuberoot@discuss.tchncs.de
on 01 Sep 12:36
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I’m confused, but this doesn’t make sense to me.
It shouldn’t matter whether you’re moving in the same direction or not for this, because ultimately it’s all relative - if you set the planet as the frame of reference, the direction you come in from doesn’t matter - just the velocity and angle.
What I can see working is calculating the in and out angles - if the exit velocity is at a sharper angle relative to the planets velocity than the entrance angle, then your exit velocity “gains” more of the planet’s velocity than the entrance velocity “loses”.
If you were completely stationary, from the planet’s point of reference, you’re moving with the velocity of the planet. If you then did half an orbit, exiting in the other direction (theoretically), from the planet’s point of reference you have the same speed, just in the other direction - but from the sun’s point of reference, you’re now moving at the planet’s speed on top of the planet’s own speed, thus gaining double the velocity of the planet.
The issue is, of course, I have no idea if I’m making sense, or missing the point.
agroqirax@lemmy.world
on 01 Sep 16:53
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kuberoot@discuss.tchncs.de
on 01 Sep 17:54
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Ayy, I’m not crazy, that sounds like exactly what I described… The only question is, is the explanation of “you spend longer falling” is bs, or if it makes sense if you conceptualize it differently?
But you also have to choose the most convenient observer to help you get it.
I would say the easiest way to “get it” would be to consider it from the Suns observation point of view. Choosing the planet or spacecraft just means that you have to consider a lot more relative motion.
Situation D: Same setup, you catch the planet quicker because it is now traveling toward you.
Situation E: You have only gained “11” speed rather than the 29 when the planet was moving away from you.
Situation F: You still spend 20 speed to climb out of the gravity well; for a total speed loss of 9.
This is obviously simplified and the numbers are meaningless. But the concept stands.
Depending of the incoming and outgoing angles; the energy changes are more or less…
Hope this illustrates it a little better.
kuberoot@discuss.tchncs.de
on 02 Sep 09:16
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Well, relative motions are more intuitive to me - they make sense, and I can use calculations for them.
In the first example, you presented 101 speed - this means only 1 speed relative to the planet, and that’s all that’s getting redirected (in the planet’s frame of reference your enter and exit velocity should be the same, since that’s how orbits work). The number is just too small, but your velocity would be planet velocity + 1 on a different vector, which will be less than 101 total.
If we estimate the angle on the picture is about 60 degrees from the velocity vector, and the speed to be 100+v1, the speed from the planet’s frame of reference is v1 - so, the exit velocity will have components of (100+v1cos(60°)) and (v1sin(60°)), so the final speed relative to the sun should be
sqrt((100+x*cos(60°))^2+(x*sin(60°))^2)
Wolfram alpha suggests this simplifies to sqrt(x(x+100)+10000), and comparing the equation by appending <x+100 gives the solution of x>0
This means, if my math is correct, with an entry angle of 0° and exit angle of 60°, you always lose speed.
I could try replacing the angle with a variable and setting a constraint of x>0 and see if the free version of wolfram alpha would spit out something, but just replacing the 60 with y is spitting out some convincing solutions, since in those x is never greater than 0.
PS: Thanks for taking the time to explain, the fact that you went out of your way to draw the diagrams is not lost on me. I unfortunately still don’t see it, but I really do appreciate the effort!
Taking the planet as the reference point. Complicates the situation a lot, but here we go.
If you contrast “A” and “D”. The initial velocity in “A” is 1, whereas “D” is 201. The acceleration due to gravity in “A” SEEMS LOWER (this is why external observer is way easier) on the way in and in “D” it seems higher. In “A” you are literally falling for much longer (gaining much more speed); than in “D”.
In “C” and “F” the situations are also different, I over simplified a bit too much. In “C” you would spend more energy than in “F”; since the acceleration due to gravity would seem higher, but not that much more. I should have made the exit angle 90°, to make them exactly equivalent…
The calculations are significantly more complex from the point of view of either the planet or space craft.
Thinking about trying to solve a real set of equations is a bit much; there are other concerns; like the fact that gravity drops off at 1/d^2^; so distance between the objects matters, the integration over distance of the equations is beyond me (I haven’t had to do that since uni, 20yrs ago). But the concepts are not too complicated; and for me at least the external observer makes it so much less complicated.
kuberoot@discuss.tchncs.de
on 02 Sep 22:25
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So, the issue is, as far as I know the calculations are dead simple - you “enter” and “exit” the planet’s influence at the same distance from the planet, which means your potential gravitational energy didn’t change, so from the orbital mechanics point of view, from the planet’s frame of reference, your velocity should stay the same.
As you “fall” in the orbit around the planet, you’re converting potential gravitational energy to kinetic energy, but as you “climb” you convert it back into potential gravitational energy, ending with the same amount of each kind of energy. The only change is that the velocity is redirected.
With that in mind, it’s why, from my knowledge, the equations are really simple, with the only complications being trigonometry (to resolve the angles) and pythagoras (squaring, adding and getting the square root make the result unintuitive).
Going back to your graph, if I were to do the math, according to my theory:
In A, let’s say you go in with 200 speed, and 0° angle (for simplicity). That means relative to the planet you have 100 speed.
In B, you gain some speed by converting potential energy to kinetic. We can’t say how much you gained, because we’re missing any real measure of distance and mass, but the neat thing is - it doesn’t matter, because:
In C, you turn that kinetic energy back into potential energy, and end up with the same speed you entered at, at the same distance. This means you now again have 100 speed relative to the planet, but aimed at a 60° angle. We can now add the velocity vectors of the planet and the velocity relative to the planet to get the velocity relative to the sun, using the planet’s velocity as one axis, getting a vector of [100+100*cos(60°); 100*sin(60°)], or [150; 86.6025], with magnitude of 173.2051, which is less than the 200 we went in with.
If you want an intuitive example of what I’m referring to, consider a planet approaching you as you are stationary relative to the sun. If we assume ideal, presumably impossible, entry and exit angles of 0° and 180°, leaving the planet’s gravity field moving in the exact opposite direction than what you entered, you’ll note you’ll be gaining speed on exit either way, despite not moving towards the planet on the approach and “catching up”.
The graph doesn’t really show anything other than illustrate your thoughts - but there’s absolutely nothing backing that as being true :/
Either way, it does feel like we’re going around in circles, and I don’t want to be taking up your time unnecessarily. If you have something to disprove my math (maybe my understanding of orbital dynamics is wrong, and it’s not that simple), that’d be a starting point to try to figure out what’s wrong; if you’re interested, I could try to make diagrams, though I feel like they might kind of look the same, just with different numbers based on calculations.
I guess one last thing I can offer is a video somebody replied to me with elsewhere in the thread, explaining this idea: youtube.com/shorts/kD8PFhj_a8s
As others have said, you are stealing kinetic energy from the planet to go faster. Or giving kinetic energy back to the planet to go slower.
So, relatively, you slow down and the planet speeds up or the planet slows down and you speed up.
kuberoot@discuss.tchncs.de
on 01 Sep 17:51
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Right, but as I explained, it’s the how that doesn’t make sense to me - the explanation that you “fall for longer” doesn’t make sense, since 1. with how orbits work, it takes the same energy and time to “fall” as it does to ascend, and 2. at these scales you can use the planet as an inertial frame of reference, so the angle of approach doesn’t matter for how “long” you “fall”, it’ll be the same regardless of whether you’re moving towards or away from the planet.
You mentioned “from the perspective of the planet” before, and I think perhaps that’s the key, from the planet’s perspective you fall and rise with equal velocities and equal accelerations, but crucially the planet is moving relative to other things and curves your orbit, so whilst you might might have the same falling and rising speeds relative to it, they’re not in the same direction, so your velocity has changed, and from an external perspective you’ve gained velocity from it.
Imagine you start stationary relative to the sun, with Jupiter barrelling towards you (not on a collision course!). From Jupiter’s perspective you fall towards it, and so from the suns perspective you gain velocity opposite jupiters orbit, but you’re not directly head on so it twists your course (let’s say 90 degrees to keep things simple) then as you leave Jupiter it indeed decelerates you relative, but crucially you’re in a different direction now, (from jupiters perspective) you’re pointed right towards the sun, so as you pull away Jupiter is decelerating you in the sun direction (aka accelerates you away from the sun). So you were both accelerated in the anti-Jupiter-orbit direction and then again in the anti-sun direction. Added together those give you a vector which is non-zero, so you’ve gained speed from Jupiter.
If your orbit didn’t curve (eg if you could pass straight through the middle of Jupiter without colliding) I think perhaps it’d cancel out its own effects on your velocity, though I’d need to check to be certain…
kuberoot@discuss.tchncs.de
on 02 Sep 08:41
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I’m sorry, but this comment thread genuinely makes me feel like I’m going insane. You seem to have explained exactly the same thing as me, with the same example, and none of it includes the “fall for longer before you catch up” bit.
As for the orbit not curving, yeah, I think you’re right - the obvious case is if you’re sitting stationary on the planet’s orbit, but the curious case is if you’re approaching from the sun, with the planet’s velocity plus velocity away from the sun. If I’m not mistaken, in that case you’d end up with the same velocity (minus what you might have lost to the sun’s gravity), but on the other side of the planet’s gravity well, which means you still gained energy.
I guess the original claim works if you imagine it along a specific axis only (1 dimensionally) in that perspective you either fall quickly then leave slowly or fall slowly and leave quickly, matching up to a change in velocity along that axis.
ITT, a bunch of people who simultaneously admit that they don’t really know for sure arguing with the people explaining it to them.
It’s ok to not know things. It’s okay to be confused. It’s much better to ask for clarification or do your own research than to tell people who do know that they are wrong.
threaded - newest
LMFAO
Long Mission Fling Assisted Operation
Wow
or the Way Of Woosh, yes indeed, you’re spot-on with your acronyms
What I see: level up flush
Fellow Balatro enjoyer
Context?
Jupiter is used as a gravity assist to launch spacecraft further. This maneuver is known as a slingshot.
What worries me is that they’re stealing a little bit of Jupiter’s momentum every time. If they’re not careful it’ll fall towards the sun and we’ll have a Jupiter landing on our heads.
</s>
It’s alright, we send someone out every few years to give it a bit of a wind up, like a grandfather clock.
<img alt="" src="https://sh.itjust.works/pictrs/image/6ef736be-957e-41d8-95e8-a215e32d4cd7.webp">
Ah ok. I remember we also did this with the Moon for one if the Apollo missions
sg1 taught us that, its either jupiter, or another gas giant or a black hole.
Not a physicist but i don’t think a black hole would work…
Why wouldn’t it? A gravity well is a gravity well. As long as you remain outside the event horizon it should work in the same way.
As the other commenter has stated, it’s not about the gravity - because you have to ultimately counter that gravity to escape. But it is about the kinetic energy of the planet as it orbits.
While it’s true that your spacecraft would have to ‘counter that gravity’ to escape, that’s from the frame of reference of the planet. From the frame of reference of whatever distant object you want your craft to ‘accelerate’ towards, your craft will appear to have gained momentum. If it were a zero sum game- there would be no “gravitational slingshot” effect (aka gravity assist maneuver).
The way your spacecraft ‘steals’ kinetic energy from the planet it orbits is by using the “gravity” of the planet. The two objects never come into physical contact with one another, the mass of the ship and the mass of the planet effect each others path through space-time- although very slightly. That is to say they seem to ‘pull’ on each other- what we call gravity.
The Earth and the Moon likewise ‘steal’ energy from each other through ‘tidal’ interactions. This causes the Earth to rotate more slowly and the moon to recede from our planet- this is all due to ‘gravity’.
Black holes also have kinetic energy that you can ‘steal’, to boost yourself toward a third celestial body just like planets do.
Spacecraft can use the gravitional energy from Jupiter to perform a “slingshot” maneuver, gaining significant momentum and reaching the outer solar system with less fuel.
NASA uses Jupiter’s gravity well to slingshot craft into farther orbits. Edit: Farther trajectories may have been the better phrasing.
I can’t help but infer a joke about NASAs budget.
Imagine bouncy spring soundeffect for full enjoyment
It’s also a shield for earth.
but jupiter also slings a lot towards earth too
Not really. It does pull some in, but not directly at earth, and the majority is either “eaten” by Jupiter or slung out of the system.
Jupiter’s pull is so great, compared to earth, that the ones that do get past or then pulled more towards the sun.
At least that is how my professor described it.
if i remeber correctly, it just slings most of fast moving things around (roughly equally in all direction), and only slow moving things actually hit it.
that seems a bit too strong for jupiter, that seems more like suns behaviour
this seems correct.
but i have not actually done any courses on celestial mechanics, and mostly basing on yt videos that i watch, so you maybe are correct on this one.
.
Ironic that Goliath was defeated by a sling and now our solar system’s Goliath is a sling
<img alt="" src="https://lemmy.world/pictrs/image/44140612-a02f-4e15-92b6-dfae8749e8d7.jpeg">
This is a sling. Bottom right corner is a catapult and uses stored energy (the elastic - sorry if wrong wording.) To me at least the idea of a sling that pivots around the user’s hand is much closer to the action of using a planet’s gravitational pull to sling a
satellitecraft onwards.en.m.wikipedia.org/wiki/Sling_(weapon)
You wouldn’t want to do it with a satellite and I’ve only ever heard of it being called a slingshot (which usually refers to the device pictured in the meme, likely increasing clarity) or a gravity assist.
The motion does seem to be more accurate though.
Good point! A satellite is more likely to orbit a planet monitoring things I guess? It’s been a while since I’ve played Kerbal Space Program.
Satellites are objects orbiting a planet or some other smaller object. One could argue that it is a satellite on the way though too.
I find the idea of an object like the moon doing a gravity assist around Jupiter to get to us very amusing.
What bothers me is I often read they are using the planet’s gravity to gain speed. Whatever speed an objet may gain while entering orbit should be lost when exiting it, right ? So I guess it’s the cinetic energy of the planet that is actually fuelling the spacecraft, isn’t it ?
Yes. If the planet was stationary in space, it wouldn’t work. Approach from ‘behind’ the planet and you get a boost, approach from the ‘front’ and you hit the brakes.
That is true from the frame of reference of the planet. From the frame of a 3rd distant object that you want to accelerate towards, it appears you have gained momentum.
Yes, but the mechanism for ‘extracting’ the kinetic energy from the planet is by using ‘gravity’, hence the name, “Gravitational Slingshot”.
You are gaining (or losing) energy based on if you are traveling in the same direction at the planet or not.
If you are coming from behind (travelling in the same direction) you an falling into the gravity well for longer. Thus gaining more energy. The extra energy is based on the speed of the planet through space.
Conversely if you an coming from the front, you fall for a shorter period. You lose energy at you climb up the gravity well.
So you gain speed if you circle rotation-wise and lose it if you circle counter-rotation wise?
Is that how they did it in 2010?
No, it’s hard to explain without diagrams.
But as you fall towards a planet (any gravity well); you pick up speed, if the planet is moving away from you, you fall for longer before you catch up. As you climb back up, you don’t spend all of the energy you gained on the way down. That difference is the Slingshot effect.
It also works in reverse, if the planet is moving towards you. You catch up quicker, thus gain less speed. And spend overall more energy than you gained when you climb back out. Slowing down in the process.
I’m confused, but this doesn’t make sense to me.
It shouldn’t matter whether you’re moving in the same direction or not for this, because ultimately it’s all relative - if you set the planet as the frame of reference, the direction you come in from doesn’t matter - just the velocity and angle.
What I can see working is calculating the in and out angles - if the exit velocity is at a sharper angle relative to the planets velocity than the entrance angle, then your exit velocity “gains” more of the planet’s velocity than the entrance velocity “loses”.
If you were completely stationary, from the planet’s point of reference, you’re moving with the velocity of the planet. If you then did half an orbit, exiting in the other direction (theoretically), from the planet’s point of reference you have the same speed, just in the other direction - but from the sun’s point of reference, you’re now moving at the planet’s speed on top of the planet’s own speed, thus gaining double the velocity of the planet.
The issue is, of course, I have no idea if I’m making sense, or missing the point.
This explains it quite nicely: youtube.com/shorts/kD8PFhj_a8s
Ayy, I’m not crazy, that sounds like exactly what I described… The only question is, is the explanation of “you spend longer falling” is bs, or if it makes sense if you conceptualize it differently?
It is difficult to conceptualise.
But you also have to choose the most convenient observer to help you get it.
I would say the easiest way to “get it” would be to consider it from the Suns observation point of view. Choosing the planet or spacecraft just means that you have to consider a lot more relative motion.
<img alt="" src="https://lemmy.nz/pictrs/image/dbc5ebb7-6160-4971-a191-bde1d9fb3c86.png">
<img alt="" src="https://lemmy.nz/pictrs/image/002c390c-c4b5-41b5-ac57-e05fc28dc138.png">
This is obviously simplified and the numbers are meaningless. But the concept stands.
Depending of the incoming and outgoing angles; the energy changes are more or less…
Hope this illustrates it a little better.
Well, relative motions are more intuitive to me - they make sense, and I can use calculations for them.
In the first example, you presented 101 speed - this means only 1 speed relative to the planet, and that’s all that’s getting redirected (in the planet’s frame of reference your enter and exit velocity should be the same, since that’s how orbits work). The number is just too small, but your velocity would be planet velocity + 1 on a different vector, which will be less than 101 total.
If we estimate the angle on the picture is about 60 degrees from the velocity vector, and the speed to be 100+v1, the speed from the planet’s frame of reference is v1 - so, the exit velocity will have components of (100+v1cos(60°)) and (v1sin(60°)), so the final speed relative to the sun should be
Wolfram alpha suggests this simplifies to sqrt(x(x+100)+10000), and comparing the equation by appending
<x+100gives the solution of x>0This means, if my math is correct, with an entry angle of 0° and exit angle of 60°, you always lose speed.
I could try replacing the angle with a variable and setting a constraint of x>0 and see if the free version of wolfram alpha would spit out something, but just replacing the 60 with y is spitting out some convincing solutions, since in those x is never greater than 0.
PS: Thanks for taking the time to explain, the fact that you went out of your way to draw the diagrams is not lost on me. I unfortunately still don’t see it, but I really do appreciate the effort!
Taking the planet as the reference point. Complicates the situation a lot, but here we go.
If you contrast “A” and “D”. The initial velocity in “A” is 1, whereas “D” is 201. The acceleration due to gravity in “A” SEEMS LOWER (this is why external observer is way easier) on the way in and in “D” it seems higher. In “A” you are literally falling for much longer (gaining much more speed); than in “D”.
In “C” and “F” the situations are also different, I over simplified a bit too much. In “C” you would spend more energy than in “F”; since the acceleration due to gravity would seem higher, but not that much more. I should have made the exit angle 90°, to make them exactly equivalent…
The calculations are significantly more complex from the point of view of either the planet or space craft.
Thinking about trying to solve a real set of equations is a bit much; there are other concerns; like the fact that gravity drops off at 1/d^2^; so distance between the objects matters, the integration over distance of the equations is beyond me (I haven’t had to do that since uni, 20yrs ago). But the concepts are not too complicated; and for me at least the external observer makes it so much less complicated.
So, the issue is, as far as I know the calculations are dead simple - you “enter” and “exit” the planet’s influence at the same distance from the planet, which means your potential gravitational energy didn’t change, so from the orbital mechanics point of view, from the planet’s frame of reference, your velocity should stay the same.
As you “fall” in the orbit around the planet, you’re converting potential gravitational energy to kinetic energy, but as you “climb” you convert it back into potential gravitational energy, ending with the same amount of each kind of energy. The only change is that the velocity is redirected.
With that in mind, it’s why, from my knowledge, the equations are really simple, with the only complications being trigonometry (to resolve the angles) and pythagoras (squaring, adding and getting the square root make the result unintuitive).
Going back to your graph, if I were to do the math, according to my theory:
If you want an intuitive example of what I’m referring to, consider a planet approaching you as you are stationary relative to the sun. If we assume ideal, presumably impossible, entry and exit angles of 0° and 180°, leaving the planet’s gravity field moving in the exact opposite direction than what you entered, you’ll note you’ll be gaining speed on exit either way, despite not moving towards the planet on the approach and “catching up”.
The graph doesn’t really show anything other than illustrate your thoughts - but there’s absolutely nothing backing that as being true :/
Either way, it does feel like we’re going around in circles, and I don’t want to be taking up your time unnecessarily. If you have something to disprove my math (maybe my understanding of orbital dynamics is wrong, and it’s not that simple), that’d be a starting point to try to figure out what’s wrong; if you’re interested, I could try to make diagrams, though I feel like they might kind of look the same, just with different numbers based on calculations.
I guess one last thing I can offer is a video somebody replied to me with elsewhere in the thread, explaining this idea: youtube.com/shorts/kD8PFhj_a8s
youre taking advantage of the planets immense kinetic energy.
As others have said, you are stealing kinetic energy from the planet to go faster. Or giving kinetic energy back to the planet to go slower.
So, relatively, you slow down and the planet speeds up or the planet slows down and you speed up.
Right, but as I explained, it’s the how that doesn’t make sense to me - the explanation that you “fall for longer” doesn’t make sense, since 1. with how orbits work, it takes the same energy and time to “fall” as it does to ascend, and 2. at these scales you can use the planet as an inertial frame of reference, so the angle of approach doesn’t matter for how “long” you “fall”, it’ll be the same regardless of whether you’re moving towards or away from the planet.
You mentioned “from the perspective of the planet” before, and I think perhaps that’s the key, from the planet’s perspective you fall and rise with equal velocities and equal accelerations, but crucially the planet is moving relative to other things and curves your orbit, so whilst you might might have the same falling and rising speeds relative to it, they’re not in the same direction, so your velocity has changed, and from an external perspective you’ve gained velocity from it.
Imagine you start stationary relative to the sun, with Jupiter barrelling towards you (not on a collision course!). From Jupiter’s perspective you fall towards it, and so from the suns perspective you gain velocity opposite jupiters orbit, but you’re not directly head on so it twists your course (let’s say 90 degrees to keep things simple) then as you leave Jupiter it indeed decelerates you relative, but crucially you’re in a different direction now, (from jupiters perspective) you’re pointed right towards the sun, so as you pull away Jupiter is decelerating you in the sun direction (aka accelerates you away from the sun). So you were both accelerated in the anti-Jupiter-orbit direction and then again in the anti-sun direction. Added together those give you a vector which is non-zero, so you’ve gained speed from Jupiter.
If your orbit didn’t curve (eg if you could pass straight through the middle of Jupiter without colliding) I think perhaps it’d cancel out its own effects on your velocity, though I’d need to check to be certain…
I’m sorry, but this comment thread genuinely makes me feel like I’m going insane. You seem to have explained exactly the same thing as me, with the same example, and none of it includes the “fall for longer before you catch up” bit.
As for the orbit not curving, yeah, I think you’re right - the obvious case is if you’re sitting stationary on the planet’s orbit, but the curious case is if you’re approaching from the sun, with the planet’s velocity plus velocity away from the sun. If I’m not mistaken, in that case you’d end up with the same velocity (minus what you might have lost to the sun’s gravity), but on the other side of the planet’s gravity well, which means you still gained energy.
I guess the original claim works if you imagine it along a specific axis only (1 dimensionally) in that perspective you either fall quickly then leave slowly or fall slowly and leave quickly, matching up to a change in velocity along that axis.
But yeah, I wouldn’t have explained it that way.
ITT, a bunch of people who simultaneously admit that they don’t really know for sure arguing with the people explaining it to them.
It’s ok to not know things. It’s okay to be confused. It’s much better to ask for clarification or do your own research than to tell people who do know that they are wrong.
Why are we like this?