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I know this question has actually been answered before in a couple of places such as Here

Unfortunately I am dealing with an irascible and impatient person who simply will not bother to read any more in depth text from any of the innumerable sources out there. He also won't believe me, irrespective of my background in physics. He has the simple formula for the Delta-V in a so-called "slingshot maneuver", but he insists the velocity given for the plant can be any velocity, not matter how it obtained. Obviously, this is complete nonsense. In a simple 2 body problem, where the primary is not in orbit around some other object, the velocity leaving the primary will be quite the same as it was when inbound, He actually agrees, but then he says, "In the frame of reference of the primary. I cannot make him understand a change in velocity is (for any situation) the same for every single frame of reference in the universe.

Can someone here do a better job than me at explaining

  1. Any change in velocity of any object is the same for every reference frame.

  2. For a free large body - like in deep space - a small object passing close by will experience a zero Delta-V, completely irrespective of the constant velocity of the large body with respect to any other object. Clarification Delta-V here is defined as the change in the magnitude of the velocity function of an object after a maneuver as compared to prior to the maneuver. Obviously, the direction of the vehicle changes, but at every given specific altitude on either side of the planet, the speed is exactly the same. The path is perfectly symmetrical. This is NOT true if and only if the large body is significantly influenced by a 3rd body. In the case of a very large 3rd body (like the Sun) in residence, the path of the vehicle is not symmetrical at all, about the large body or otherwise. It will pick up or drop some momentum. So, in point f fact, will the large body. It loses or gains angular momentum. If the initial orbital angular momentum of the large body is zero, the slingshot produces no change in energy for the vehicle on its return path. None.

  3. A gravity assist / braking is a 3 body problem. Expansion In every elastic collision and every gravitational interaction between two objects the paths are perfectly symmetrical. No exceptions. The only way an elastic collision between two objects can result in a net change in energy or momentum is if a 3rd object is involved. In the case of the slingshot maneuver, the additional momentum is "stolen" from the planet's angular momentum. Of course, if the planet were spinning, and the vehicle actually collided with it, the planet could give up some angular momentum that way, but it is not being physically impacted. Barring such a thing, the only angular momentum that can be available is orbital momentum around the third body. Note: the situation can be analyzed by creating two separate reference frames, one involving one pair of objects and the other involving the other pairing, but it still requires three bodies. See: The fundamental concepts of the gravity-assist manoeuvre

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    $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Jun 10, 2023 at 4:01

2 Answers 2

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I think I agree with your friend though it’s kinda hard to tell what exactly you/him believe.

Slingshot is a 2 body maneuver.

In particular if you were in outer space trying to travel to a another galaxy and there was a massive meteor (or a lone fast star if you need something sufficiently massive) going in the same direction as you faster than you, you could steal some of it’s momentum/energy in a slingshot maneuver to get where you’re going faster. No other nearby planets/suns are required.

As you’ve noted, in the frame of reference of the planet, the slingshotting object’s speed is more or less constant. However, the direction can change, so the velocity can change so from a different reference frame it can gain or lose speed (and so energy).

If the planet is traveling at speed $u$ relative to some other coordinate system and the space probe is at a velocity $v$, then in the planet’s frame you’ll shift from traveling at velocity $v-u$ to $u-v$, so back in the initial reference frame, you’ll change from traveling at velocity $+v$ to $2u-v$. Since $|2u-v|$ can be different from $|v|$, you can gain or lose energy from a slingshot maneuver.

Another way to think about it is that you’re bouncing off the planet. If a car hits an obstacle and that obstacle goes flying, it’s not surprising that the before velocity and the after velocity of the obstacle changes through the collision even though from the reference point of the car, the obstacle just changed from going towards it to going away from it. This is exactly the same thing at a bigger scale.

Here’s neat site that illustrates slingshots and ends with how they can be modeled like 2 body elastic collisions: https://d-arora.github.io/Doing-Physics-With-Matlab/mpDocs/mec_slingshot.htm

Going through your points.

  1. Yes. If by change in velocity you mean the literal change in the velocity vector, then yes, that’s independent of your inertial reference frame (though not necessarily true in non-inertial ones). Note that this is different from delta v.

  2. No, it depends on the reference frame. From the planet’s frame, you can’t gain speed due to symmetry. However you can change your direction i.e. velocity vector. If you use a different frame of reference, then that lets you gain energy. You strangely seem to insist that using a different frame of reference involves a different object, even though there are no significant relevant forces between whatever third object (i.e. the sun) and our two objects over the time of the maneuver.

  3. A slingshot maneuver involves two objects and an outside reference frame. That outside reference frame has no necessary force interactions with the two bodies in question. Normally, when counting the bodies of an interaction, you don’t care about the reference frame (i.e. when studying elastic collisions, you don’t call it a 3 body problem when you have object 1, object 2, and the earth’s reference frame). Therefore, even though the sun is necessary for the actual interaction and is usually necessary for studying the global dynamics of orbital flight, the slingshot maneuver itself only involves two relevant bodies, not including the sun. The link you give is a very good one and in the introduction it explains how people often mistakenly consider the slingshot maneuver as a 3 body problem even though they try to emphasize that the gravitational effects of the sun on the two bodies is insignificant and an analysis of just the forces in the planet satellite system (i.e. 2 bodies) is sufficient to explain the essential aspects of the maneuver.

Conservation of Energy/Momentum means that there never is a change in total energy/momentum. However, you can easily interchange momentum/energy between two bodies. The way they use angular momentum to analyze the motion isn’t revealing any fundamental facts about interaction with the sun, but is just using conservation laws in different reference frames to simplify calculations.

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    $\begingroup$ Very incorrect, I am afraid. Any two body system is gravitationally symmetrical. The work done on the vehicle by the large body when it accelerates the vehicle towards itself is precisely equal to the work done on the vehicle as it leaves. Thus, the vehicle's path wull always be perfectly symmetrical if the only source of momentum is the gravitational attraction between the two objects. $\endgroup$ Jun 10, 2023 at 5:59
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    $\begingroup$ Which part of what I said is wrong? Are you arguing that you can’t change direction when flying past a planet? That’s clearly false. Given you can change direction, you’re changing your velocity vector which means that from a different inertial reference frame (e.g. in the initial inertial frame of the object it goes from still to moving), energy is transferred from the planet to the object. Note that work being done is a reference frame dependent concept. $\endgroup$
    – Eric
    Jun 10, 2023 at 6:07
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    $\begingroup$ Err, you can use the sun to slingshot to change your direction but since we normally care about things relative to the sun, it wouldn’t gain you any velocity in our solar system’s reference frame (and in other reference frames, i.e. galactic ones, we’re moving in the same direction as the sun so it wouldn’t help). But you could totally slingshot around Proxima Centauri if you were trying to reach Canis Major if you really wanted to. $\endgroup$
    – Eric
    Jun 10, 2023 at 12:53
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    $\begingroup$ The point here is that we’re not using the sun due to it’s gravitational force or anything weird like that. We just like using the sun as a fixed reference point. $\endgroup$
    – Eric
    Jun 10, 2023 at 12:56
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    $\begingroup$ It’s just that the planets have any velocity relative to the reference frame that you care about. If the sun has zero mass and the planets just happened to be in the right location still, you could still do a slingshot. If the planet was traveling in any direction (I.e. directly towards the sun), you can still do a slingshot - the angular momentum around the sun is irrelevant. All that matters is that it has some nontrivial velocity in the reference frame you care about. $\endgroup$
    – Eric
    Jun 11, 2023 at 15:39
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Eric's answer is correct (and I have upvoted it), but I think addressing your specific statements in a separate answer may be useful.

1: This is true but I don't think it means what you think it means.

If I walk north at 10 meters per second relative to the street, and then accelerate until I am walking south at 10 meters per second, my change in velocity is 20 meters per second southwards, and my change in speed is zero. If you're on a train going 100 meters per second south relative to the street, my velocity has changed from 110 meters per second to 90 meters per second northwards, my change in velocity is 20 meters per second southwards, and my change in speed is 20 meters per second.

2: This is false.

If you mean change in speed, not change in velocity, it is true only in the frame of the planet, as your interlocutor says. To conserve momentum and kinetic energy in the frame that measures a change in speed, the minute change in velocity of the much larger body must be considered.

Elastic collisions are very closely analogous. It may be useful to consider the case of a person on the street releasing a tennis ball into the path of an oncoming truck: the truck driver sees the ball approach at some speed, bounce off, and recede at the same speed. The person on the street sees a stationary tennis ball collide with the moving truck and launch away at twice the speed of the truck, and needs to account for the minute change in the much heavier truck's velocity to conserve kinetic energy and momentum.

3: This is false.

No third body is required for a change in speed from a two-body interaction, just any reference frame that isn't comoving with the planet.

The orbits of bodies around the sun are important to the practical problem of if, where, when, and at what angle the change in velocity from the planet-probe interaction will take place, but they are irrelevant (except for very precise calculations) to the actual interaction.

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  • $\begingroup$ No, I know EXACTLY what it means. It means, for one thing, assuming an additional linear velocity for the large body in addition to its orbital velocity does not mean the Delta-V of the vehicle changes. Just because Jupiter is traveling at 200,000 Km/Hr relative to some distant star (which it is compared to virtually any star other than Sol) does not mean the Delta-V of the slingshot around Jupiter is different relative to that qalaxy. $\endgroup$ Jun 10, 2023 at 7:18
  • $\begingroup$ The point is, a slingshot maneuver around a large body in deep space will result will result in absolutely ZERO change in the vehicle's energy when it has left, and ZERO change in the magnitude of its momentum when it has left, even if the the body is traveling at a million miles per hour toward (or away from) our solar system. The ONLY thing that can result in a higher speed for a vehicle in a slingshot maneuver is passing by an object in orbit. Not only that, but for a given vehicle altitude at perigee, the closer to the Sun the planet is, the greater the Delta-V, Capiche? $\endgroup$ Jun 10, 2023 at 7:29
  • $\begingroup$ Sorry, I meant "star", not "galaxy" for the last word in my comment above. $\endgroup$ Jun 10, 2023 at 7:35
  • $\begingroup$ One little error in my comment above. Strictly speaking, I should not have said "...a slingshot maneuver is passing by an object in orbit." We usually say an object traveling in excess of escape velocity is not in orbit. Give that usage, the large body does not have to be in a restricted orbit. A gravity assist will work fine on an object traveling above escape velocity,. $\endgroup$ Jun 10, 2023 at 7:56
  • $\begingroup$ I am in a chat with Eric. Is it possible to add you and perhaps others to the chat? $\endgroup$ Jun 11, 2023 at 9:43

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