Long rod in orbit

[GTOM]

Lets suppose we could place a long rod in orbit, at the distance of Moon.
Which would be the stabile position of the rod, perdendicular to the orbital path, or parallel to it?
I think parallel, to minimize tidal forces between the two opposite parts of the rod.

 

[FactChecker]

As a casual amateur, here are my two cents. Consider some extreme cases:
1) If the rod has the exact curvature of a circular orbit, parallel would be stable because every particle of the rod would have the same orbital speed, even if it was separated.
2) If the "rod" was separate pieces and vertical, the pieces would separate due to their different orbital velocities.
3) If the rod is solid, rather than separate pieces, and vertical, the ends at different orbital speeds start to rotate to a parallel position. Whether they would overshoot and oscillate or become a stable parallel bar is a harder question.

CORRECTION: This is wrong. I forgot to consider what happens if the rod is perturbed. See @Bandersnatch 's posts #5 and #10

 

[Hornbein]

As a casual amateur, here are my two cents. Consider some extreme cases:
1) If the rod has the exact curvature of a circular orbit, parallel would be stable because every particle of the rod would have the same orbital speed, even if it was separated.
2) If the "rod" was separate pieces and vertical, the pieces would separate due to their different orbital velocities.
3) If the rod is solid, rather than separate pieces, and vertical, the ends at different orbital speeds start to rotate to a parallel position. Whether they would overshoot and oscillate or become a stable parallel bar is a harder question.

My rule of thumb is, if it can oscillate then it will.

 

[FactChecker]

My rule of thumb is, if it can oscillate then it will.

Good point! I guess there is no damping force, so that seems right.

 

[Bandersnatch]

A rod would try to orient itself radially, due to tidal forces, in the same way as tidally-deformed ellipsoids of planets and moons orient themselves with their long axes radially, towards the centre of the gravitational field. If there's oscillation, it's around that position.
Orbit-parallel/tangent is unstable to perturbations.

 

[FactChecker]

A rod would try to orient itself radially, due to tidal forces, in the same way as tidally-deformed ellipsoids of planets and moons orient themselves with their long axes radially, towards the centre of the gravitational field. If there's oscillation, it's around that position.
Orbit-parallel/tangent is unstable to perturbations.

I think I see why. (Please stop me if my amateur ideas are off the track.)
Suppose it starts parallel with orbital speed $V_o$, and there is a small perturbation which raises one end up. That end is at a higher altitude but still has speed $V_o$. Orbital speed decreases with altitude, so that end is going faster than its new orbital speed. Therefore, that end tends to "escape" orbit and go up more. That is unstable.
My original idea did not consider a perturbation from parallel.

ADDED: This is unnecessarily complicated. See @Bandersnatch 's post #10

 

[GTOM]

A rod would try to orient itself radially, due to tidal forces, in the same way as tidally-deformed ellipsoids of planets and moons orient themselves with their long axes radially, towards the centre of the gravitational field. If there's oscillation, it's around that position.
Orbit-parallel/tangent is unstable to perturbations.

And what makes that stabile? At the end farer from planet, bigger speed than lower end, but weaker gravity field. Wouldnt that rotate further until it oscillate around parallel position?

 

[Baluncore]

Lets suppose we could place a long rod in orbit, at the distance of Moon.

How long is the rod?
Is the rod straight or curved?
When it was placed in orbit, was it rotating once every month, so it always remains parallel with the Earth's surface, or was it aligned with a fixed star?

 

[GTOM]

A rod would try to orient itself radially, due to tidal forces, in the same way as tidally-deformed ellipsoids of planets and moons orient themselves with their long axes radially, towards the centre of the gravitational field. If there's oscillation, it's around that position.
Orbit-parallel/tangent is unstable to perturbations.

And what makes that stabile? At the end farer from planet, bigger speed than lower end, but weaker gravity field. Wouldnt that rotate further until it oscillate around parallel position?

How long is the rod?
Is the rod straight or curved?
When it was placed in orbit, was it rotating once every month, so it always remains parallel with the Earth's surface, or was it aligned with a fixed star?

Lets suppose km long rod, straight, aligned with Sun.

 

[Bandersnatch]

And what makes that stabile? At the end farer from planet, bigger speed than lower end, but weaker gravity field. Wouldnt that rotate further until it oscillate around parallel position?

Velocities are a distraction. That's just initial conditions. You want to focus on dynamics. The only force acting here is gravity.

(image credit R.Verrault, w/modifications)
The centre of mass C experiences just the right gravitational acceleration g_MC to keep the rod in orbit.
Due to the inverse square dependence on distance, the mass element m at the near end will experience stronger gravity g_M than the centre of mass, while the far end will experience weaker gravity, g'_M.
The rod is rigid, so all of its parts are kept in orbit by the same centripetal acceleration g_MC as the centre of mass.
From the CM point of view, there is a pair of tidal forces pulling on the ends of the rod along the radial direction r_MC.
You can decompose each of the forces on the ends of the rod to get a pair of components. One acting along the length of the rod - which tries to stretch it - the other providing torque. The latter disappears only when the rod is fully aligned radially towards the central body.
Much like with a pendulum, any overshooting of the equilibrium position gives rise to a restoring force with opposite direction.
Providing there is some way to dissipate energy, the rod will eventually come to a rest, pointing towards the central body.

This is how tidal locking works.

 

[Baluncore]

And what makes that stabile? At the end farer from planet, bigger speed than lower end, but weaker gravity field. Wouldnt that rotate further until it oscillate around parallel position?

It may seem counterintuitive, but the further end would orbit at a lower speed, while the lower end would orbit at a higher speed. That is because the lower orbit requires a higher speed to counter the greater gravitational attraction.

That means a vertical rod would start to rotate backwards, which is the opposite of the rotation needed to become stable. Stability would require rotating forwards, once per month, in a lunar radius orbit.

 

[GTOM]

It may seem counterintuitive, but the further end would orbit at a lower speed, while the lower end would orbit at a higher speed. That is because the lower orbit requires a higher speed to counter the greater gravitational attraction.

That means a vertical rod would start to rotate backwards, which is the opposite of the rotation needed to become stable. Stability would require rotating forwards, once per month, in a lunar radius orbit.

So what it will do, ultimately?

 

[Baluncore]

So what it will do, ultimately?

It would increase its backward spin gradually over time.

If you started it instead, with a forward rotation having a period greater than one month, then that rotation would gradually slow to a one-month period, when it would finally become tidally locked, showing only one side to the Earth, with phases, like the Moon.

Eventually, it might take up a stable orbit shepherded by, thrown out of orbit by, or impacting with the Moon.

 

[Filip Larsen]

Gravity gradient has even been used as a passive attitude stabilization mechanism, where the orbiters minor inertial axis are stable in the radial direction. For an oblong (pencil shaped) rigid object the minor axis is typically along the geometrically long axis, e.g. in direction of the mass boom in the linked case. The Space Shuttle likewise had a stable orientation with its nose pointing at Earth.

A quasi rigid object, i.e. a rigid object that internally convert mechanical stress into heat, will allow change of rotational energy such that the in-orbit radial orientation is stable. For instance, if the linkage of mass boom in the above case is flexible attached (i.e. think dampened spring) then this would significantly help stabilize the orientation. Note that spinning satellites often employ nutation dampers to null out non-pure rotation ("wobbling"), but that is a somewhat different purpose than gravity gradient stabilization.

 

[bob012345]

Lets suppose we could place a long rod in orbit, at the distance of Moon.
Which would be the stabile position of the rod, perdendicular to the orbital path, or parallel to it?
I think parallel, to minimize tidal forces between the two opposite parts of the rod.

I think the bigger issue is if the orbit of the rod itself is stable. Of the five LaGrange points in the Earth-Moon system, two are stable, L4 and L5. Factoring the sun, they become regions of stability. All this was worked out in the 1970’s when there was interest over hypothetical space colonies. Back in the day I joined the L5 Society which advocated building O’Neill style colonies at L5.


https://nss.org/what-is-l5/