# I need some support, please, for a gravity assist analysis

• I
• LesRhorer
In summary, the two obtuse individuals are promoting themselves as something they are not, and they are not physicists. They are not backing up their claims with evidence. A few minutes of video or voice chat would be very helpful. Thanks.
LesRhorer said:
Adding a third body introduces a second force which destroys the symmetry.
Not if the force from the third body is approximately uniform gravity, which accelerates the first two bodies nearly the same way.

What destroys the symmetry is switching from the CoM-frame of the the two bodies to the CoM-frame of all three bodies. But either frame is valid, and you can use both in the same same analysis if you transform the velocities correctly between them.

PeroK
PeroK said:
The classic example is an elastic collision between two balls of equal mass. In the rest frame of the first mass, the second mass loses all its KE. And vice versa. In COM frame, both balls rebound with equal and opposite velocities.
Perfectly correct, of course. The point is the total kinetic energy of the system remains constant. If we fly a jet over the two balls, the relative energy of the system in the FoR of the jet is HUGE, but at the moment of the collision the energy of the two ball system WRT the jet does not change, even though the velocities of both balls do.

In a slightly more apt analogy for this case, instead of two balls of equal mass, we can take a freight train and bounce the ball off it. To a person on the train, the ball arrives and leaves at the same speed, just in opposite directions. If the ball's initial velocity is V, then the total acceleration is 2V. To a person at thetrain station, who sees the ball initially travelling at 1/2V, the ball will come back at 1-1/2 V. In that frame of reference, the total acceleration is 2V. To the jet flying overhead, the initial velocity of the ball might be, say, 100V. After the collision, depending upon the direction the jet is flying, the new velocity is either 98V or 102V, resulting in an overall acceleration of 2V. I trust everyone can see a very distinct pattern, here, The point is while velocites are always relative to the particular frame of reference, accelerations are not. We can fiddle with relative velocities all we want, but it does not change the amount f energy involved in some interaction. The changes in Voyager's velocites were not just a matter of changing reference frames. Their speed increased in every reference frame, including that of Jupiter and Saturn.
PeroK said:
For the gravity assist, the local kinematics are largely independent of any other mass.
Well, yeas and no. Mostly yes. Jupiter and Saturn do have orbital velocities rather close to the incoming velocities of Voyager I and II. to that end, the velocities cannot be ignored, and they are 100% produced by the influence of the mass of the Sun. That said, over a relatively short time period we can take the velocity to be constant and the trajectory flat, rather than curved. During the time the craft is within a few planetary diameters, it is close enough for a first approximation. The salient point, however, is that even though the local field strength is much larger than the remote field strength, the sum total of all the contributions of the local gravitational filed is zero. Suddenly the very tiny, but still non-zero contribution of the second order field gets important to the overall result.
PeroK said:
However, when we switch to the rest frame of another mass, such as the Sun, we are looking for a collision that is favourable in terms of imparting KE to the space probe. That's the trick: to organise the collision so that the probe gains maximum KE relative to the Sun.
Not just relative to the frame of reference of the Sun. For a speed increase maneuver, the change in velocity compared to that of an unaccelerated system is greater than the speed change relative to the Sun. The planet is accelerating away from its original locus after all. My point, however, is if the Sun has the mass of a pea, or is a few parsecs away, then no speed-up at all occurs. The Sun's presence and influence is absolutely required.
PeroK said:
The misconception is that the Sun's gravity is involved in the local kinematics. It is only involved to a marginal degree. This has been fully explained in posts by @A.T. and @Filip Larsen
Again yes and no. At any point along the path of the spacecraft which can be deemed local to the planet, the Sun's contribution is indeed small compared to the planet's, at some points perhaps by many orders of magnitude. There is a catch, however. While the planet successfully and inexorably pulls on the craft to speed it up while inbound, it pulls just as inexorably on the craft slowing it down when outbound. As I showed above, for any span whose endpoints are equidistant on either side of perigee, the total contribution to the speed / energy is zero Throughout that same span, the sum of the Sun's differential contributions is not zero. This is why the amount of speed / energy gain is independent of the mass of the planet. It is not, however, independent of the mass of the Sun nor the distance from the Sun. Planets closer to the Sun can produce much higher gravity boosts than the very much larger gas giants.

It is also absolutely critical the planet is orbital. The entire effect is driven by the fact we are dealing with the angular momentum of the planetary system and the torque applied by the Sun.

(Note: the planet's mass and particularly density are not completely irrelevant. They do determine the escape velocity profiles and the altitude of perigee, both of which affect the efficiency of the transfer,)

LesRhorer said:
It is also absolutely critical the planet is orbital.
You are using a very narrow definition of "gravity assist", while most people here have a more general understanding of the concept.

A.T. said:
Not if the force from the third body is approximately uniform gravity, which accelerates the first two bodies nearly the same way.
False. This is not an effect of the tidal potential. it is an effect of the fact (and magnitude) the system has a angular momentum
A.T. said:
What destroys the symmetry is switching from the CoM-frame of the the two bodies to the CoM-frame of all three
Again, that is false. The path is not symmetrical to the CoM of the two bodies. Inbound the speed is close to escape velocity of the planet. Outbound it can be close to twice the escape velocity of the planet, not to mention more than the escape velocity from the Sun. What is more, the trajectory can be twisted through a fairly large angle relative to the planet's orbit. What would ordinarily be a hyperbolic orbit around the planet gets flattened and turned into an immelman by the interaction with the Sun
A.T. said:
bodies. But either frame is valid, and you can use both in the same same analysis if you transform the velocities correctly between them.
Of course, but that means converting from a non-inertial reference frame to an inertial one, and accounting not only for radial acceleration of the spacecraft y the planet and the Sun, but also the cross product between the spacecraft's velocity vector and the angular velocity of the system. This is also not symmetrical to either the CoM of the planet or the Sun.

Even if this were not correct, however, my main point still stands. It does not happen without the presence of the Sun,

LesRhorer said:
Perfectly correct, of course. The point is the total kinetic energy of the system remains constant. If we fly a jet over the two balls, the relative energy of the system in the FoR of the jet is HUGE, but at the moment of the collision the energy of the two ball system WRT the jet does not change, even though the velocities of both balls do.
That's energy conservation, which holds in each reference frame. That's not energy invariance. Note that below you use the invariance of acceleration (valid) and conclude results about energy invariance (which is not valid).

A ball accelerating by the invariant ##1 m/s## in some direction has a different change in KE in each reference frame. That change may be positive or negative and have any value.

This is a fundamental misunderstanding on your part of basic kinematics.

LesRhorer said:
The point is while velocites are always relative to the particular frame of reference, accelerations are not. We can fiddle with relative velocities all we want, but it does not change the amount f energy involved in some interaction.
The above confuses acceleration with energy.
LesRhorer said:
The changes in Voyager's velocites were not just a matter of changing reference frames. Their speed increased in every reference frame, including that of Jupiter and Saturn.
This is completely wrong. In one particular frame, the result of the gravity assist maneuver was to reduce Voyager's speed to zero. There is always a frame where the initial velocity is zero. And there is always a frame where the final velocity is zero.
LesRhorer said:
The Sun's presence and influence is absolutely required.
Not for the local kinematics which provide the gravity assist. The gravity assist is to be measured in the Sun's rest frame, as the whole point is to help escape the solar system.

Filip Larsen and jbriggs444
PS that velocity and acceleration are vector qualities is something else you should note. Change in velocity (vector) is invariant (across all inertial reference frames), but change in speed (magnitude of velocity) is definitely not invariant.

PPS and KE, of course, is a function of speed, rather than velocity.

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Juanda said:
So it seems Work / Energy transfer is also frame dependent. Let me check some calculations because, as you pointed out, my argument relied on that. If work is frame dependent, then I'm longer sure the 3rd body is necessary in order to be able to "steal" energy from one body and put it on the other.
I'll try solving an elastic collision on two different FoR (at whatever point and at the CoM) to better understand what's going on and that might give me some insight related to this orbital problem.

You are right. In an elastic collision, the energy transfer (work) between the bodies does depend on the speed of the reference frame. Therefore, the argument I proposed is not valid.
I can't still fully understand it. Something is not clicking in my head about the whole situation.
But the link provided by @A.T. definitely helped. Especially the picture with the train and the explanation in the link.
A.T. said:

The third body (the sun) is not necessary to add any force to the other two bodies (Jupiter and the satellite). Its importance is just in giving us a reference frame that we could consider inertial since the CoM of the 3 bodies will be basically static and inside the sun. With respect to that reference frame, the satellite will see its speed increase after the maneuver.

Thank you all for your patience.

jbriggs444 and Dale
PeroK said:
That's energy conservation, which holds in each reference frame. That's not energy invariance.
I fully recognize the difference.
PeroK said:
Note that below you use the invariance of acceleration (valid) and conclude results about energy invariance (which is not valid).
Not quite.I recognize they are two different things. They have to be, of course, since the kinetic energy is not proportional to the velocity.
PeroK said:
A ball accelerating by the invariant ##1 m/s## in some direction has a different change in KE in each reference frame. That change may be positive or negative and have any value.
Of course. I never said differently. My point (and its is not the main one) is the only way to effect a change of energy in a gravitationally constrained system is to apply a force due to gravity, We can change the overall kinetic energy of the system to anything we want by changing the FoR but that does not enable us to claim we have extracted or added any energy to the system internally.

Suppose we have a vehicle in orbit around a free body in deep space with an inbound velocity of 3/4 escape velocity at some point. We can set a FoR so the initial velocity of the vehicle is zero in that frame of reference. When the vehicle swings round the body it will have a velocity in the external FoR of 1-1/2 times escape velocity of the body. This does not mean the vehicle will fly of into space just because it has a greater energy and momentum WRT the external FoR. He claims it will. It does not mean the body, by itself is producing a gravity assist of 1-1/2 times the escape velocity of the free body. He claims it does. He claims merely asserting the body "has a velocity of X" - which is of course utter nonsense - means the body will produce an increase of speed of X.
PeroK said:
The above confuses acceleration with energy.
Not at all. For any mechanical system, the change in energy of a massive body is equal to the integral of the scalar product of the force vector and the differential of the position vector. The magnitude of said force vector is always proportional to the acceleration of the body in question,unless of course there is some other force being applied to the body in addition to the one in question. For a two body gravitational system, this inexorably ties the acceleration of both components to the increase in kinetic energy of the system internally. We can change the kinetic energy of the system or any part of it by merely changing the FoR, but that will never change the relative velocities of the two bodies WRT to each other.

The relationship does not apply in an inelastic system, of course.
PeroK said:
This is completely wrong. In one particular frame, the result of the gravity assist maneuver was to reduce Voyager's speed to zero. There is always a frame where the initial velocity is zero. And there is always a frame where the final velocity is zero.
Of course. This is not relevant to any of my points. The fact Voyager's velocity in the first frame is zero does not mean Voyager stopped dead in its tracks relative to Jupiter, More importantly, it does not mean somehow Jupiter increased Voyager's speed by a factor of three, rather than two. The amount of energy ( a total of about 500 Gigajoules) is dependent upon the forces applied and the distance over which those forces were applied. Without ever leaving Low Earth Orbit, we could have easily come up wth a FoR in which Voyager's KE was 600 Gigajoules.This would not have magically caused Voyager to head out into deep space.
PeroK said:
Not for the local kinematics which provide the gravity assist. The gravity assist is to be measured in the Sun's rest frame, as the whole point is to help escape the solar system.
Once again, yes of course. The point is it does not happen at all if there is no Sun (or other third body)

Does anyone need more of an example why "help me win an argument" is a bad idea?

russ_watters, Tom.G, jbriggs444 and 2 others
A.T. said:
You are using a very narrow definition of "gravity assist", while most people here have a more general understanding of the concept.
Such as? The point of a gravity assist is to enable a spacecraft to achieve very high velocities without using much fuel. Attempting to do this in a 2 body system results in no increase in velocity at all.

PeroK, weirdoguy and jbriggs444
LesRhorer said:
We can change the overall kinetic energy of the system to anything we want by changing the FoR but that does not enable us to claim we have extracted or added any energy to the system internally.
I think I get that point you are trying to make here:

- Given a single massive body, any energy gain by the probe is fully frame dependent.

- Only given a system of at least two massive bodies in relative motion one can extract some energy from that system in a frame invariant sense, by reducing the velocity difference between them.

LesRhorer said:
Such as?
This concept is not restricted to bodies in orbit. You could also gain speed in a desired direction by interacting with a massive body moving independently through interstellar space.

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Gleb1964 and Dale
Dale said:
This is not correct. There is no requirement that the velocity be right at the escape velocity. In fact, it should be above it.
'Not much. The most efficient transfer occurs when the orbit is a hyperbola. Starting out any faster wastes fuel, although of course it may be necessary depending on the alignment of the planets. Nothing is perfect, after all
Dale said:
That is also false. A hyperbolic orbit is always faster than escape validity.
I think you mean "velocity"

The shape of an orbit which tends to zero at infinity is a hyperbola. A hyperbola is assymptotic to X = 0 and Y = 0 at Y = infinity and X = infinity, respectively. If the craft does not slow all the way to zero in the limit as its distance tends toward infinity, then its orbit is not properly speaking a hyperbola. Forgetting relativity, in the limit of infinite velocity, the orbit becomes a flat line.
Dale said:
Only in the CoM frame. A third body is not needed. By the principle of relativity we can use any inertial frame we like.
Doing so does not increase the velocity of V WRT P. It does not increase F as V approaches P and it does not decrease F as V passes perigee. It will not turn an elliptical orbit into a hyperbolic or trans-hyperbolic one. Only in a planetary system can asymmetrical orbits be produced. Just because there are an infinite number of frames in which the relative speed of the inbound path is lower than the relative speed of the outbound path in some FoR external to the barycenter of the system in no way means the large body has increased the KE of V.it only means there is no such things as an absolute inertial frame of reference, and that all linear motion is relative to any arbitrarily chosen FoR.

wrobel said:
There are a lot of systems where both momentum and energy conserved but it would be strange to call them all elastic collisions.
I agree, although it s really merely a semantic issue, and not terribly important as long as everyone in a discussion uses the same definitions. To be sure, if both energy and momentum are conserved, the interaction is elastic. I think most people would expect a collision to involve mechanical impact. I know for certain my insurance agent would not like saying a near miss is a collision.

Drakkith said:
But my simulation in Universe Sandbox changed both the speed and direction... the asteroid gained nearly 6 km/s in speed.
In what direction? In order to know the answer one must know the velocity vector of the asteroid compared to the orbital velocity of the Earth. It looks to me like the simulation assumes the earth is not in orbit around the Sun. In this case, it is a simple 2 body problem, and the answer depends only upon the frame of reference. The Earth itself then imparts no net energy to the asteroid. At something like 16 Km/sec relative velocity, the asteroid should likely wind up in a reasonably stable orbit around Earth, unless of course that 10 Km/s initial velocity was something like heading straight toward the Earth. I presume not. The Earth, however, is in fact in orbit with an orbital velocity of nearly 30 Km/s.

If the path of the asteroid was substantially parallel to that motion, the asteroid would fly off into the inner solar system at up to nearly 20 Km/s (if I am adding correctly in my head). If the path of the asteroid was substantially anti-parallel to the Earth's path, then the asteroid would likely crash into the earth, having given up a large fraction of its momentum to the Earth. If the path is substantially perpendicular to the Earth's path, then the asteroid may not pick up or lose much speed at all, but rather will reflect either Northwards or Southwards out of the plane of the ecliptic. That is what being in orbit does for you.

Note in either of the first two cases, the asteroid will wind up going roughly in the opposite direction from which t approached. In the latter two cases, it will wind up traveling not quite 90 degrees to its original path. There is a totally radical difference between a flyby of a free body in deep space and a planet in orbit around a star. One is rather uninteresting, even if one fiddles with frames of reference. The other is perfectly astounding, whether taken from the FoR of the planet or of the star.

LesRhorer said:
An orbital flyby of Mercury results in a huge increase in speed, (over 40 Km/s) despite the fact the planet is only about 5 times as massive as our moon. A flyby of Neptune only results in a fairly small overall acceleration (about 5 Km/s), despite it having more than 300 times the mass of Mercury. That really should tell one all one needs to know anout whether the Sun's influence in a maneuver like that performed by the Voyager spacecraft, Cassini, and New Horizons. The Messenger spacecraft used assists at Earth, Venus, and on three separate flybys of Mercury in braking maneuvers to slow itself down. These maneuvers all resulted in actual changes in speed and direction compared to the Sun. They are not an artifact of changing the reference frame.
It's not about the Sun's influence, it's about the speed of the objects relative to some reference frame. Mercury is moving at 40+ km/s relative to the Sun (or a reference frame 'pinned' to the Sun, or the barycenter, or whatever you want to say). The only reason the Sun is important here is because we are tying a reference frame to it and things in the solar system are moving with respect to it.
LesRhorer said:
Furthermore, no interaction between two bodies is ever capable of achieving the aforementioned results. All two body interactions result in perfectly symmetrical trajectories. Thus, simple changes in the direction of a craft along the plane of its initial trajectory with no net increase in the energy of the craft with respect to the barycenter of the two body system qualifies.
My simulation disagrees with you.
Juanda said:
@Drakkith do you agree with this? Maybe there is something wrong with the simulation? The results you provided no longer make sense to me. I assume it's being solved numerically. Maybe the acquaricy is being an issue.
You can find the various integrators that Universe Sandbox uses here: https://universesandbox.fandom.com/wiki/N-Body_Simulation#Models
LesRhorer said:
In what direction?
I provided the details in my previous post, including the velocity vectors before and after the interaction relative to the coordinate system.
LesRhorer said:
It looks to me like the simulation assumes the earth is not in orbit around the Sun.
That is correct.
LesRhorer said:
In this case, it is a simple 2 body problem, and the answer depends only upon the frame of reference. The Earth itself then imparts no net energy to the asteroid.
Yes, reference frames matter. In the Earth's frame, or the asteroid's frame, the initial and final speeds are the same (but not the velocities). In another reference frame things are different and energy can be gained/lost by each body. As my simulation shows.

This reminds me of how the propulsive efficiency of a rocket engine depends on the reference frame.

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Dale and jbriggs444
LesRhorer said:
The shape of an orbit which tends to zero at infinity is a hyperbola.
The (degenerate) case between an elliptic and hyperbolic trajectory is actually called a parabolic trajectory, that is, the two-body trajectory where the relative speed at infinity is zero. So I guess now that when you have said "hyperbolic" earlier you actually meant what is normally labelled parabolic, i.e. a trajectory with near zero hyperbolic excess speed? That would at least explain why those parts of your posts were so hard for me to decipher.

Dale
LesRhorer said:
Such as? The point of a gravity assist is to enable a spacecraft to achieve very high velocities without using much fuel. Attempting to do this in a 2 body system results in no increase in velocity at all.
This is simply false. From the wikipedia article on Elastic collisions:
https://en.wikipedia.org/wiki/Elastic_collision#Two-dimensional_collision_with_two_moving_objects

$$v_x=\frac{\cos (\phi ) (u (m-M) \cos (\theta -\phi )+2 M U \cos (\phi ))}{m+M}-u \sin (\phi ) \sin (\theta -\phi )$$ $$v_y=\frac{\sin (\phi ) (u (m-M) \cos (\theta -\phi )+2 M U \cos (\phi ))}{m+M}+u \cos (\phi ) \sin (\theta -\phi )$$ $$V_x=\frac{\cos (\phi ) (U (M-m) \cos (\phi )+2 m u \cos (\theta -\phi ))}{m+M}+U \sin ^2(\phi )$$ $$V_y=\frac{\sin (\phi ) (U (M-m) \cos (\phi )+2 m u \cos (\theta -\phi ))}{m+M}-U \sin (\phi ) \cos (\phi )$$
Where I have cleaned up the notation a bit using capital letters for the large mass and lower case letters for the small mass, and I have chosen the ##x## axis to be the direction of the initial velocity (##U##) of the large mass.

You can confirm that both momentum and energy are conserved for this system.

Note that this is a 2-body collision. There is no third body. From this expression you can obtain the increase in velocity, which you falsely claim is zero, by directly evaluating $$\Delta v=\sqrt{v_x^2+v_y^2} - u = \sqrt{\frac{m^2 u^2-2 m M u^2 \cos (2 \theta -2 \phi )+2 M u U (m-M) \cos (\theta -2 \phi )+2 M u U \cos (\theta ) (m-M)+M^2 u^2+2 M^2 U^2 \cos (2 \phi )+2 M^2 U^2}{(m+M)^2}}-u$$

We can further simplify this by doing a Taylor series expansion for small ##m## to obtain $$\Delta v \approx 2U-2(U-u)\frac{m}{M}$$

So, contrary to your statement, there can be a large ##\Delta v## for a 2 body elastic collision provided that ##U## is large.

LesRhorer said:
The shape of an orbit which tends to zero at infinity is a hyperbola.
This is false, it is a parabola.

LesRhorer said:
If the craft does not slow all the way to zero in the limit as its distance tends toward infinity, then its orbit is not properly speaking a hyperbola
This is also false. Hyperbolic trajectories are precisely the ones that do not slow all the way to zero in the limit as distance tends toward infinity.

See: https://en.wikipedia.org/wiki/Hyperbolic_trajectory "a body traveling along this trajectory will coast towards infinity, settling to a final excess velocity relative to the central body. ... gravitational slingshots, can be described within the planet's sphere of influence using hyperbolic trajectories."

LesRhorer said:
Just because there are an infinite number of frames in which the relative speed of the inbound path is lower than the relative speed of the outbound path in some FoR external to the barycenter of the system in no way means the large body has increased the KE of V.
In all of those frames the large body has indeed increased the KE of the small one. You cannot increase the speed of a mass without increasing its KE.

LesRhorer said:
there is no such things as an absolute inertial frame of reference, and that all linear motion is relative to any arbitrarily chosen FoR.
While this is true, it contradicts your previous statements. That you don't recognize that is unfortunate, but clearly not something you are interested in correcting.

At this point, this thread is done. Your initial question has been answered, we will not be providing you any material to support your argument as your argument is wrong on many, many points.

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