# Curious about gravitational sluing detail

bowlegged
Let's say a big rock is headed toward a small moon, impelled only by existing momentum and gravitational pull. The rock is not rotating or tumbling at all, leastwise not in relation to the local time/space norm. The rock is sped up by the moon's gravity and it falls ever faster and faster. But the rock just misses crashing into the planetoid, which BTW has no atmosphere, and the diving rock gets slued around the much bigger entity and careens (slingshots) off on a greatly altered trajectory. My only question is, wouldn't such a momentous encounter tend to impart some spinning/tumbling feature to the runaway rock? (And can you prove it?)

Many thanks.

bahamagreen
If the purpose of mentioning the moon is only to increase the approach speed of the rock toward the planet, why not just simplify; you already have a stated initial condition that the rock has a momentum, so that can be as large as you want. The rock can approach the planet without the moon at any speed you choose for the thought experiment.

So, leaving out the moon, we just have a rock and the planet.

"The rock is not rotating or tumbling at all, leastwise not in relation to the local time/space norm."

Let's examine this... I'm not sure what you mean by local time/space.

If a spacecraft is in orbit around the planet such that the orientation of the spacecraft is that its belly always faces the planet (the view out the front window will always show the planet below and space above), then the pilot of this craft will measure that he is not rotating - his craft is not tumbling.
An observer floating a ways off will measure the craft as tumbling, making one rotation per orbit.

Likewise, if the craft is oriented so that its long axis is aligned to its line of sight to a distant star and this long axis is parallel to the orbital path twice per orbit, then the pilot will measure a tumbling rotation, one per orbit, and a distant floating observer would measure that the craft is not tumbling.

All this just to examine what it means for a craft or rock to be said to be tumbling when orbiting, passing, sling-shotting, or crashing with respect to a planet.

I'm assuming that what you mean by local time/space is that the rock or craft is following its local geodesic straight non-rotating path and orientation. From this perspective there is not rotation or tumbling, only non-accelerated linear translation in a local straight line.

That said, from the rock's perspective, since it is not seeing itself as tumbling initially (it has an axis that maintains alignment with the path of its orbit), even as that path begins to curve around the planet and then uncurve, that alignment will maintain to that curvature... and no local measurement of the rock will find a rotation component with respect to its orbital path.

Yet, clearly there has been a rotation with respect to the distant stars. So there are really two questions; what really constitutes a rotation and from which vantage point is it being measured, and (to what I think you are asking) does any rotation during the maneuver stay with the rock as an additional angular rotation - does an initially "non rotating rock" become a "rotating rock" after the sling-shot maneuver?

I think the answer to this second question, from both the rock's and a floating observer's perspectives must be "No".

The rock will see the universe rotate about it during the maneuver but will be unable to measure anything local that would be a rotation with respect to its path.

The floating observer would see the rock begin to slowly rotate so as to keep its same orientation to its orbital path, and this rotation would speed up as the rock approached its closest point to the planet, then the rotation would slow as the rock left the planet, and with sufficient distance from the planet the rotation would approach zero; the whole rotation amount being the inner angle of the asymptotes of the two long arms of the approach and departure paths

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bowlegged
Sure, the mention of a moon was kind of odd. You could say the rock had an encounter with an actual planet. But it's relevant that the encountered body has no atmosphere (or at least that the runaway rock never touches or interacts with such atmosphere, which could tend to complicate the thought experiment).

Yes, you understood me precisely about the initial lack of tumbling/rotation.

Okay, so you believe the answer is "no additional angular momentum results from the slingshotting". Thank you.

bahamagreen
Well, don't just take my word for it... my reasoning does not constitute a proof;
let's see if anyone offers a different sense of what would happen.

MikeGomez
from the rock's perspective, since it is not seeing itself as tumbling initially (it has an axis that maintains alignment with the path of its orbit), even as that path begins to curve around the planet and then uncurve, that alignment will maintain to that curvature...

The floating observer would see the rock begin to slowly rotate so as to keep its same orientation to its orbital path,

Hmm, what would cause the rock to rotate and keep it's same orientation with it's path?

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My understanding is that for the rock to rotate synchronously(i.e., the same side always facing the planet) while on a hyperbolic orbit(as in slingshot), angular momentum of the rock's rotation would have to be continuously altered during the approach.
From zero rotation(in an inertial ref.frame) at the infinity from the planet, accelerating to maximum angular velocity at periapsis, and decelerating to zero again during the last leg of the orbit.

While in principle tidal forces act exactly in such a way, the effect should be too slight to measure after just a single approach, unless extreme circumstances are invoked. Such as extremely strong gravitational field, extremely low moment of inertia or extreme elongation of the orbiting body.

And since none of these apply to the scenario in question, the rock will retain the initial angular momentum of its rotation throughout the encounter.

But even if we allow for the tidal forces to alter the rotation of the rock, there would be no net change in angular momentum between the initial and final points on the orbit, as the tidal forces would act to speed up the rotation during half of the orbit, and slow it down during the other half. Only the final orientation would change(i.e., the rock would keep facing the planet).

Staff Emeritus
Gold Member
In order to change the rotation of this rock a torque must be applied. You did not mention the shape of the rock, that is critical. A perfectly spherical rock with uniform density will not feel any torque, so will pass with no change in rotation. On the other hand a rock with non uniform shape or non concentric variations in density will experience a torque as the pass occurs. Thus the rotation will change.

bahamagreen
MikeGomez:Hmm, what would cause the rock to rotate and keep it's same orientation with it's path?

The rock is in free fall... but that's a good point. If you toss up a rock into a parabolic arc I wonder if it too "rotates" in order to preserve its initial axial alignment with its path... now I'm thinking that is doesn't.
I think throwing something heavy like an initially non-rotating bowling ball would demonstrate that it is too massive to rotate that much in the short time it was moving...?

Bandersnatch:My understanding is that for the rock to rotate synchronously(i.e., the same side always facing the planet) while on a hyperbolic orbit(as in slingshot), angular momentum of the rock's rotation would have to be continuously altered during the approach.

I agree, but I also think the stipulation that the same face stays toward the planet is different from maintaining an axial alignment with the orbital path.
If the incoming and outgoing paths approach being a 270 degree angle, then if the rock aligns to the path the rotation approaches 90 degrees, but if the rock maintains a face toward to planet the rotation is 270 degrees.
But in both cases, the angular momentum would be continuously varying because of the acceleration and distance relations, such conditions in both cases being the same, yet needing different amounts of momentum change for the respective rotations...?
I'm thinking the alignment to path is "natural", but the alignment of face to planet would have to be done by adding and then subtracting additional angular momentum, somehow.

Integral:You did not mention the shape of the rock, that is critical.

Yes, I assume the rock is simplified for the purpose of this thought experiment... Considering the rock to be a point mass kind of blows the whole question about rotation and angular momentum, but perhaps considering cases of non-symmetric extension of mass and density of the rock would help sort out what the effects are acting on the "small round rock" case...?

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This is actually a very interesting problem and I don't think we are at the answer yet...

And I'm wondering why it was moved out of the Relativity section. The whole question concerns a moving mass in a gravitational field, the mass's orientation with respect to its geodesic through curved space, and whether an apparent torque applied to an orbiter should be attributed to the field curvature, or to the geodesic, or to the mass...?

I'm thinking the alignment to path is "natural", but the alignment of face to planet would have to be done by adding and then subtracting additional angular momentum, somehow.
The "natural" thing for a moving body is to obey Newton's laws of motion. So, unless there's a torque acting on the body, it will not change its angular velocity.
The only force acting on the rock is the gravity of the planet, which produces torque acting to align one of its axes(for non-spherical, non-point-like rock) with the direction towards the centre of the planet.

Maintaining prograde alingment without any other forces acting on the system would be "unnatural", in the sense of violating the laws of nature.

My only question is, wouldn't such a momentous encounter tend to impart some spinning/tumbling feature to the runaway rock?
Only minimal, but if it kept orbiting the moon for very long, it would start spinning due to tidal forces:
http://en.wikipedia.org/wiki/Tidal_locking

2021 Award
Gravity effects the centre of mass of the rock.

But now consider two rocks, separated, but traveling on different parallel initial paths. Each will travel an independent path. They would have very slightly different final paths after the encounter. So if the rock was now modeled as a long cylinder with two halves, each centred on one of those two points, it would change its astronomical orientation during the pass.

bowlegged
I've been 'away' in that I wasn't notified of the last 6 or 7 posts -- could have something to do with moving this thread elsewhere. ('strange that Thread Tools says I'm still subscribed)

I really appreciate Integral's reply. Sure, let's make the rock I gigantic femur-shaped thing, or barbells. If this indeed has nothing to do with general relativity then my question is pretty much answered: the time/space distortions would not impart angular momentum (even if the runaway femur-rock were slingshot by a veritable Black Hole?).

As for synchronized orbit (always facing the larger entity), my understanding is that can occur only after eons in the same orbit. 'Course I might err.

Mentor
If the rock has a spherical symmetry in its mass distribution, it will not start to rotate.
If the rock has no spherical symmetry, it can gain angular momentum, and continue to rotate after it passed the planet. This is easier to show in the limit of large separations, but it stays true in this case as well.

If a spacecraft is in orbit around the planet such that the orientation of the spacecraft is that its belly always faces the planet (the view out the front window will always show the planet below and space above), then the pilot of this craft will measure that he is not rotating - his craft is not tumbling.
An observer floating a ways off will measure the craft as tumbling, making one rotation per orbit.
That is not right, if you measure rotation with a gyrometer (or any other device used to measure rotations) you will get rotation in the first case and no rotation in the second case. You get a wrong impression if you look at the planet - look at the sky or your instruments to see if you are rotating.

bahamagreen said:
Yet, clearly there has been a rotation with respect to the distant stars. So there are really two questions; what really constitutes a rotation and from which vantage point is it being measured
There is an observer-independent way to check if an object rotates.

bahamagreen said:
The rock will see the universe rotate about it during the maneuver
It will not, unless it starts to rotate (=change the alignment of its axes relative to the stars and other objects far away). If it starts to rotate, you can measure it internally in the rock with gyroscopes, you don't need any external reference points.

bahamagreen said:
but will be unable to measure anything local that would be a rotation with respect to its path.
What is "Rotation with respect to its path" supposed to mean?

bahamagreen said:
The floating observer would see the rock begin to slowly rotate so as to keep its same orientation to its orbital path,
This is wrong. Nearly all satellites are counterexamples.

Bandersnatch said:
My understanding is that for the rock to rotate synchronously(i.e., the same side always facing the planet) while on a hyperbolic orbit(as in slingshot), angular momentum of the rock's rotation would have to be continuously altered during the approach.
From zero rotation(in an inertial ref.frame) at the infinity from the planet, accelerating to maximum angular velocity at periapsis, and decelerating to zero again during the last leg of the orbit.
The result does not have to be zero.

Baluncore said:
Gravity effects the centre of mass of the rock.
Gravity effects all parts of the rock.
But now consider two rocks, separated, but traveling on different parallel initial paths. Each will travel an independent path. They would have very slightly different final paths after the encounter. So if the rock was now modeled as a long cylinder with two halves, each centred on one of those two points, it would change its astronomical orientation during the pass.
It would also change its angular velocity (in general, the change can be zero as well).

bahamagreen
mfb, That is not right, if you measure rotation with a gyrometer (or any other device used to measure rotations) you will get rotation in the first case and no rotation in the second case. You get a wrong impression if you look at the planet - look at the sky or your instruments to see if you are rotating...

I guess we are touching on the idea of absolute rotation and its detection.

Does a craft whose long axis aligned with its direction of motion stay that way when not influenced by a gravitational field? I think yes, and a free floating object within the craft initially at rest and not rotating with respect to the craft would remain in that condition...

But, if the craft approaches a large mass, the path of the craft will turn... the path itself rotates. So which becomes the basis for an absolute rotation, the curved path of free fall or the "flat" geometric orientation of the craft to the stars?

The question is does the craft turn as well as the path (and the floating object inside it) with respect to the mass so as to keep their orientations on the curved path, or do the craft and floating object inside maintain the same orientation wrt the stars and rotate wrt the line of the path...?

Sounds like you are saying that the absolute rotation will always be wrt the stars... does not GR have something to add to this in terms of the curved path of the free fallers, including any gyrometers in the craft?

Mentor
Does a craft whose long axis aligned with its direction of motion stay that way when not influenced by a gravitational field?
If it is not rotating, and its direction of motion is not changing.
I think yes, and a free floating object within the craft initially at rest and not rotating with respect to the craft would remain in that condition...
Right.
But, if the craft approaches a large mass, the path of the craft will turn... the path itself rotates.
Are you thinking about general relativity here? For planets, those effects are tiny compared to the regular, Newtonian part of gravity.
The question is does the craft turn as well as the path (and the floating object inside it) with respect to the mass so as to keep their orientations on the curved path, or do the craft and floating object inside maintain the same orientation wrt the stars and rotate wrt the line of the path...?
If they have a spherical symmetry (so standard 1/r gravity does not apply a torque), the spaceship keeps aligned with the stars and with its gyroscopes.

GR can lead to a difference between gyroscopes and stars, indeed. But those are effects of micro- or even nanoarcseconds, whereas the spaceship experiences a deflection of many degrees.

2021 Award
Consider two particles, traveling initially on different but parallel paths. Their paths are then deflected by the gravitational attraction of another body. One particle travels slightly further, the other a shorter distance, faster.
Does their mutual orientation relative to distant stars remain unchanged ?
Do they remain the same distance apart after deflection as they were before ?
(I am deliberately ignoring the mutual attraction of the particles here).

MikeGomez
Does their mutual orientation relative to distant stars remain unchanged ?

Yes.

Do they remain the same distance apart after deflection as they were before ?

No.

With the extremely minor exceptions that have been mentioned above, objects under the influence of gravity change path without changing their rotation. Are you having difficulty with that concept?

2021 Award
MikeGomez said:
Are you having difficulty with that concept?
No. It is a trivial concept.
The thing I am having trouble with is the boundaries of the model in which that concept remains true, and the transition situation where such particles make up a bigger object.

Mentor
Does their mutual orientation relative to distant stars remain unchanged ?
No. The reason can be found in Newtonian gravity, GR effects are negligible unless we consider black holes.
You can imagine an extreme example where one object is far away and does not get deflected significantly, while the other one makes nearly a 180°-turn.
Single rigid objects are not two objects moving independent, however - they have to follow a single trajectory with a fixed separation between parts of them.

Do they remain the same distance apart after deflection as they were before ?
No, that's why we cannot use them as a model.

Does their mutual orientation relative to distant stars remain unchanged ?
Yes.

Do they remain the same distance apart after deflection as they were before ?
No.

That is a weird combination of answers. If the distance changes, then accelerations must have been different, so the orientation can have changed too.

Yes & Yes if tidal forces are negligible
No & No if tidal forces are not negligible

MikeGomez
@A.T.

Oh. I guess I didn't understand post #16 the same as you. I thought that he meant there were two different gravitational attraction strengths for two different objects flying by, or they fly by at different distances from the surface so one feels a stronger force than the other.

bowlegged
...GR effects are negligible unless we consider black holes...
So some falling asymmetrical object would pick up non-negligible spin if it were slingshot (dramatically) around a black hole -- you think?

bowlegged
Or...
The slingshot object would pick up spin but physicists attribute that to "tidal forces" and exclude it from being a "GR effect". That might be splitting hairs but I could see their point of view I suppose.

Mentor
So some falling asymmetrical object would pick up non-negligible spin if it were slingshot (dramatically) around a black hole -- you think?
You don't need a black hole for that. I don't know the size of the effect.

Tidal forces are a fully classical effect, the field is simply not homogeneous.

With planets (so no black holes), you get (with reasonable precision) exactly the same results with Newtonian gravity and with GR. That is usually described as "GR effects are negligible" - everything in GR that is not present in Newtonian gravity is negligible.

Staff Emeritus
There's no need to invoke general relativity here. All that is needed is Newtonian gravity. Assuming that the object is orders of magnitude smaller than the planet, the gravity gradient torque exerted by the planet on the object is, to first order, ##\frac{GM}{r^3}\,\hat r \times (I \hat r)## where ##r## is the distance between the object's center of mass and the center of the planet, ##\hat r## is the unit vector pointing from the object's center of mass to the center of the planet (or vice versa; changing the sign yields the same result), and ##I## is the object's inertia tensor. Note that ##\hat r## and ##I## must be expressed in the same frame of reference for this to make sense.

The OP asked specifically about a planet. I'll have more to say on this later. For now, assume the planet is a rogue planet, i.e., it's not orbiting a star. After the encounter is over, the object's linear velocity will have changed but it's speed will be the same as it was before the encounter began. That's because the approach and departure are symmetrical with respect to translation.

There is no such symmetry for rotation. The object's orientation and angular velocity prior to and after the encounter are not conserved quantities.

That the OP specifically asked about a planet makes this even more complicated. The encounter can now transfer linear and angular momentum to the object.

I thought that he meant there were two different gravitational attraction strengths for two different objects flying
That is how I understand it too. And I think in this case the orientation can change, hence surprised by your "yes" to the first question, indicating that orientation cannot change, .

bahamagreen
If I may, let me ask a little more about the possible things to consider about the curved path itself (to see if GR has any relevance or may be correctly eliminated as a concern).

Let's shift for a moment to another situation to clarify what is happening with the path itself; the bending of light passing close by the Sun... I'm choosing this to remove intertial mass from the problem and just look at the relationship between path curving and rotation - in this case the rotation of the direction of movement instead of rotation of a mass's orientation to its path line...

In this case the path is curved and the direction vector of the light rotates in alignment with the rotation of the path.

Yet, the Sun's gravity does not subtract a lateral component of the light's forward speed along its path, it remains c. That is, if c was the vector component sum of the light's forward velocity and the lateral acceleration from the Sun, then in order for the component vector sum to show a measured c, the actual velocity of the light wrt it forward path would be less than c.

I think this is correct (but let me know). If so, then it looks like the rotation part of the turning/bending ensures that the forward component of light speed is always c and does not change speed with changing radii like an orbiter in an eliptical orbit.

I guess this is just another way to ask if from the rock's perspective its path is a "straight line" in spite of clearly curving around the planet in a similar in which light bending around the Sun is traveling a "straight line".

The way the problem is posed, the answers I'm seeing have the rock maintain alignment with the stars, which means that from the rock's perspective it is rotating wrt its path... and rotating in a specific way - the leading edge of the rock will rotate way from the planet in the direction of the planet's normal line perpendicular to the plane tangent to the planet, where the origin of the normal line and rock are the endpoints of that line, that line's origin at the plane's point of contact where the planet's surface is closest to the rock at that moment. And the back end of the rock will rotate down the opposite way (the rock would tumble "backwards".
But it would tumble backward in both the approach and departure legs on each side of the point of closest approach. Yet the geometry of the situation is symmetric as are the forces for both legs, so why is the rotation non-symmetric?

Is there something in GR that has the nature of the Equivalence Principle with regard to free fallers?

I'm sure my arguments are botched up, but is there nothing else happening here?

Mentor
Light is a bad example, as it cannot rotate like massive objects can do.

That is, if c was the vector component sum of the light's forward velocity and the lateral acceleration from the Sun
You cannot add velocities and accelerations, the units do not match.

I guess this is just another way to ask if from the rock's perspective its path is a "straight line" in spite of clearly curving around the planet in a similar in which light bending around the Sun is traveling a "straight line".
That depends on your mathematical framework, it does not influence physics, especially not in the way you would like.

MikeGomez
That is how I understand it too. And I think in this case the orientation can change, hence surprised by your "yes" to the first question, indicating that orientation cannot change, .

I thought it would help to draw some pictures. Please excuse the horrible programmer art.

Red line is path/trajectory of rock.
Yellow arrow is rock’s orientation relative to the stars. This never changes.

In fig. 1 the rock is deflected by gravitation of planet, but its orientation to the stars (its rotation) remains the same.

In fig. 2 a second identical rock flies by the planet, but at a closer distance, so it is deflected more. They started out with parallel velocities, and ended up going in very different directions, but they did not rotate to align themselves to the path.

 Asumming negligible tidal and relativistic effects.

Did I get it right?

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Mentor
Yellow arrow is rock’s orientation relative to the stars. This never changes.
It can change if the rock does not have a spherical symmetry, as has been mentioned before.

MikeGomez
It can change if the rock does not have a spherical symmetry

Yes, I know, if the rock is not perfectly spherically symmetric. Thats why I went back and edited the post and added "negligible tidal forces".

Besides, the amount of change in rotation of a non-symmetric rock would be extremely small. The poster in post #2 thought that the object would align with the path of deflection. That means in fig. 1 the object (whether symmetric or not) would have rotated several degrees. I was trying to correct that. Now I guess the point is hopelessly lost, but thanks anyway.

as has been mentioned before.

Honestly. Did you really think that was necessary? For once I wish we could just stick to physics.
Why do you do that?

but they did not rotate to align themselves to the path.
I think we talk about different orientations here. Baluncore was asking about the "mutual orientation" of the two test masses. I assume he means the orientation of the line connecting the two test masses. You are talking about the individual orientations of each test mass.

If the distance changes then the orientation of the connecting line can change too.

bowlegged
There's no need to invoke general relativity here.
...
For now, assume the planet is a rogue planet, i.e., it's not orbiting a star. After the encounter is over, the object's linear velocity will have changed but it's speed will be the same as it was before the encounter began. That's because the approach and departure are symmetrical with respect to translation.

There is no such symmetry for rotation. The object's orientation and angular velocity prior to and after the encounter are not conserved quantities.

That the OP specifically asked about a planet makes this even more complicated. The encounter can now transfer linear and angular momentum to the object.

You think a symmetrical object would pick up any spin if slingshot dramatically around the periphery of a black hole?

And... you lost me at "but it's speed will be the same as it was before". It's well known that NASA employs the maneuver to boost a craft's speed.

Mentor
And... you lost me at "but it's speed will be the same as it was before". It's well known that NASA employs the maneuver to boost a craft's speed.
In the reference frame of the planet, the speed is unchanged. In another reference frame (like the sun's), where the planet is moving, the speed can change.

MikeGomez said:
Besides, the amount of change in rotation of a non-symmetric rock would be extremely small.
I'm not so sure about that, it would be interesting to find an upper limit.

Did you really think that was necessary?
If the same wrong things get repeated? I think it is useful.

Staff Emeritus