Curious about gravitational sluing detail

[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?

 

[A.T.]

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.

 

[mfb]

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.

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.

 

[D H]

Besides, the amount of change in rotation of a non-symmetric rock would be extremely small.

Not true. It's not all that hard to get an angular acceleration on the order of 10^-6 radians/second² at closest approach in the scenario described in the OP. That doesn't sound like much, but keep in mind that the Earth's rotation rate is on the order of 10^-4 radians/second, and that the Moon's is on the order of 10^-6 radians/second. An angular acceleration of 10^-6 radians/second² is pretty large given the slow rotation rate of a typical space rock.