Cfrgtkky

Reading this question (Is the cupola, on the inside of the ISS, cold or warm to the touch?) prompted me to wonder about an object in orbit's orientation as it orbits its host object. For example, consider a hollow tube that is in orbit around the earth and it is perfectly perpendicular to the earth. Spaceman Spiff is immediately beneath the cylinder, and is able to look through the cylinder and observe a specific star.

As he and the tube orbit around the earth (assuming no external influences other than the earth's gravity), would the tube remain pointed at the star (hence the tube changing from perpendicular orientation in relation to the earth to parallel as it gets to 1/4 around its orbit) and Spiff would continue to see the same star in his view? Or, would the cylinder remain perpendicular to the earth, and Spiff would see a continually changing field of stars during his orbit, until he gets back to his starting position, when the star would come back into view as he completes his orbit?

Most spacecraft, I believe, are designed and controlled very carefaully to maintain a specific orientation with the earth, but I don't know what would happen if there was no spacecraft control mechanism or external forces acting on it.

tl;dr: Spaceman Spiff will be cursing under her breath for having listened to @RusselBorogove and believed all those up votes, and gone to space hoping to enjoy viewing stars through a space telescope with no attitude stabilization or damping!

This is an invited answer, the invitation is by the author of another answer. I'd commented on some problems there but rather than fix the problem, I was invited to write elsewhere.

So here it goes...

If the cylinder's rotation was carefully stopped, then nothing else acted on it, it would hold its orientation and Spiff would see the same star through a complete orbit, while the tube went from Earth-surface-perpendicular to parallel and back.

This isn't correct, not even close!

The Earth's gravity gradient (variation in the 1/r potential over the physical extent of the spacecraft) results in a slightly different force pulling on each point on the spacecraft.

For the math, see this question and especially @Litho's answer. For a thin rod of mass $m$ and length $l$ in circular orbit, with a perpendicular moment of inertia $frac{1}{12}ml^2$, rotating in-plane around the short axis, the torque (to first order) is given by

$$L_G = -frac{GM_Eml^2}{8R_C^3}sin 2theta,$$

and the instantaneous angular acceleration is simply

$$ddot{theta} = -frac{3GM_E}{2R_C^3}sin 2theta.$$

That's a pretty amazing result! With $GM_E$ of about 3.986E+14 m^3/s^2 and an altitude of 400 km, $ddot{theta}$ can be as large as 0.4 degrees per minute^2 at 45 degrees, and that is independent of length!

...long skinny objects in orbit are subject to tidal gradient forces, so if the tube is long enough, it will slowly get pulled into an always-perpendicular-to-Earth rotating orientation.

This isn't correct, not even close!

It will oscillate for a very, very long time, unless your object has build-in dampers to slowly absorb the angular acceleration as it spins at first, or eventually achieves "capture" in a gravity gradient stabilization, where it will oscillate and slowly decay. The details depend on initial conditions.

Natural damping mechanisms include differential atmospheric and photon drag, and induced tidal distortions in the Earth by the telescope's gravity. These are very very small on an orbital timescale!

This is all from Gravity Gradient Stabilization of Earth Satellites by R. E. Fischell:

The maximum angle to which the satellite will swing is of great interest. If this angle is less than 90°, capture of the satellite into gravity gradient attitude stabilization will result. The angle can be calculated rather simply by equating the angular kinetic energy that the satellite must develop (to achieve an angular rate of 1.0 rpo) with the work
done by the gravity gradient torque as the satellite moves out to that maximum angle.

Calculation shows that if the satellite is vertical at the time the boom is deployed, then it will swing out to a peak angle of 35.36 0. If the satellite is initially at an angle greater than 54° off the vertical, it will swing out to 90°; this, therefore, is the limiting angle for capture.

Python script: https://pastebin.com/yVq6WuCu

$theta$ and $dot{theta}$ in inertial frame for a thin rod in a 400 km circular orbit starting at various initial angles with respect to the nadir, with no initial rotation in the inertial frame (i.e. looking at some star):