# Equilibrium position of a very long object in orbit

Let's imagine we have delivered a very long/tall tower into orbit. It's a rigid construction, a couple hundred kilometers long, at altitude where air friction is entirely negligible.

If we place it in circular orbit - its center of mass moving at just the right speed for its altitude - but we rotate it to some random angle to the surface, and leave it like that, giving it spin of 1 revolution per day so that - neglecting tidal forces - it would naturally remain in the same orientation to the surface.

It won't stay like that. The linear velocities of all its points relative to Earth are about the same (the bottom end's minimally lower, upper's minimally higher due to that 1RPD spin, but these should be negligible differences).

Now, the end that happens to be at lower altitude, moves slower than orbital velocity proper for that altitude and will tend to be pulled downwards. The other end, exceeding circular orbit's velocity for its altitude, will be pulled upwards. Our tower starts spinning.

What's the stable equilibrium position at which it would remain still - always facing Earth at the same angle?

• See the answer and comments to this question space.stackexchange.com/questions/17191/… The answer says, in part "An orbiting spacecraft of finite size experiences 'gravity gradient torques'. These torques, if left uncorrected, tend to align the long axis of the spacecraft so that it points toward the center of the earth. " Commented Aug 17, 2016 at 13:44
• This is noticeable and useful (or inconvenient) even for objects as little as 1 meter long. Commented Aug 17, 2016 at 20:57

"Gravity Gradient Stabilization" tends to align long masses radially with respect to the Earth. It's an application of tidal forces. It's particularly effective at holding the mass in the radial orientation; it doesn't provide much damping, so a satellite that rotates into a radial orientation solely due to gravity gradient torques will overshoot and oscillate around that orientation for a while.

It's also not the only effect. Magnetic torques, even inadvertent ones, can be enough to have a noticeable effect. Aerodynamic torques tend to be large only if the object is asymmetric, but can be large if there are e.g. solar panels that display the center of area from the center of mass.

It takes money, time and effort to control the attitude of the ISS. If they could have put it into a stable orientation that meets all the requirements, instead of having to actively control it, I'm sure they would have. So that provides some evidence that this is a pretty complex area for large, functional bodies.

On the other hand, the question is about a tower "a couple hundred kilometers long". If we rule out aerodynamic effects (the question says "air friction is entirely negligible", but that's hard to imagine in practice: Even small amounts of aerodynamic drag generates a lot of torque and bending moment when applied over hundreds of kilometers, but never-the-less let's take it as given) and there are no active disturbing torques from large inadvertent magnetic fields, arrivals/departures of momentum, etc, then gravity gradient stabilization should be the strongest remaining effect: It'll stabilize in the radial, up-down direction.

• There is a seven-part technical discussion starting at coursera.org/learn/spacecraft-dynamics-kinetics/lecture/wncQ8/… . In circular orbit, with longest (inertial) axis radial, next longest pointing along the orbit, and shortest axis perpendicular to the plane of the orbit, the position is stable. With small displacements it will oscillate around the stabile position until damped by internal friction or from external torques. There is another stabile configuration but after a not-so-small perturbation is can lose its lock and tumble.
– MBM
Commented Apr 1, 2018 at 20:07
• "... it doesn't provide much damping, so a satellite that rotates into a radial orientation solely due to gravity gradient torques will overshoot and oscillate around that orientation for a while." Would a long rigid rod in orbit around a rigid spherical Earth ever damp out, mathematically? In the real world don't they add separate dampers for this?
– uhoh
Commented Jan 15, 2019 at 13:14
• It’s fundamentally the tide-locking that worked for the moon. You need some form of internal non-rigidity & related friction losses. Dampers help by providing much more of those. A “couple hundred kilometers” tower isn’t going to be rigid, and likely will have plenty of flexibility losses. But I haven’t designed one... Commented Jan 15, 2019 at 23:54
• @BobJacobsen for some reason I didn't get a notification from your comment. I just happened back here now tracking down some damper-less explanations that needed to be corrected. Yes there might be some losses, but wow with those degrees of freedom who knows what kind of resonances and Tacoma Narrows Bridge analogous behaviors are possible! Btw any comments on this answer?
– uhoh
Commented Jan 22, 2019 at 13:24