Does Gravity / curved space cause rotation?

Meaning, if a spaceship is heading not directly toward Earth, but slightly off to one side, and when finally being close to the Earth it falls into earth orbit, does the spaceship continue to point in the same directions as it was when approaching the earth, or does it now rotate at a frequency that is equal to its orbital frequency?

I would expect the spaceship to be pointing in the opposite direction after completing one half of an earth orbit.

  • $\begingroup$ Are you asking if an extended object tilts (and then rotates) when headed in the general direction of another massive object? $\endgroup$ – Nikolaj-K Feb 6 '14 at 13:41
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    $\begingroup$ I doubt this is what you meant but have you seen anything about Gravitoelectromagnetism? It has some relevance to your question. $\endgroup$ – Brandon Enright Feb 6 '14 at 18:47
  • $\begingroup$ You did not react to any answer. Is it that they all missed your point? What else were you expecting? $\endgroup$ – babou Feb 12 '14 at 17:17

I assume there is only Earth and the ship in your system.

For one thing the ship will not fall into orbit just like that. While falling, it gathers kinetic energy which will allow it to leave Earth again towards outer space, after its trajectory has been deflected by Earth gravity. The amount of deflection depends on its speed and how close it comes to Earth. The shape of the trajectory is that of (half) a hyperbola

enter image description here (source: Wikimedia Commons)

Assuming the ship has a long axis (i.e. it is not a homogenous sphere, or concentric homogenous spherical shells), it will be subjected to a rotation because of tidal forces.

Tidal forces are simple to understand. The ship is supposed to be a reasonably rigid and strong solid (no need for a General Products hull though). The trajectory of the center of mass of the ship is the same as if the whole ship mass were concentrated there.

However, that is not the case. Some parts of the ship are closer to Earth than the center of mass, while others are further away. So the parts that are closer to earth tend to be deflected more than the parts that are further from earth. Or to see it another way, they will be more attracted by Earth. This creates a torque that makes the ship rotate around its center of mass, and tries to align its long axis with he center of gravitational attraction.

How fast this rotation takes place is dependent on actual figures.

Note that the ship follows a curved hyperbolic trajectory. So, even assuming that, at some point, the long axis is right on (the tangent to) the trajectory, thus balancing the gravitational effects between both ends of the ship, this will no longer be the case a bit later as the tangent changes it orientation. The forward part will tend to be a bit outside the trajectory while the rear part will tend to be inside. This will then be accentuated by the tidal effect so that the forward part will tend to point more and more outside of the trajectory and away from the Earth which is inside, while the rear part will tend to point closer to the Earth center of gravity.

However, after the ship passes closest to the Earth, the tidal force pulling more the rear part than the forward part will tend to realign the ship on its trajectory, with the forward part in front.

I think the total effect depends on actual figures. I suspect there may be a residual rotation speed rotating the ship as it departs from Earth, but I have not done any calculation to be sure of it. This would imply a minute, very very ... very minute, change in Earth rotation, because of angular momentum preservation.

Side remark: If Earth is replaced by a neutron star, the intense gravity differential between fore and aft parts may also tear the ship to pieces.


"When finally being close to the earth it falls into earth orbit" Won't work. If the body was gravitationally unbound incoming, you must dump binding energy (do a burn or collide) or it is equally unbound outgoing (OK - Shapiro delay and such. Close enough).

If you want gravitation theory to be predictive, it operates in spacetime. What you perceive as curves in 3-space are minimum action straight lines in 4-space. Linear and angular momenta are conserved (close enough).

The Earth's gravitational field is divergent. If an orbiting body has a longest principle moment of inertia, that axis at equilibrium will point to the Earth's center of mass as it orbits. ISS FUBAR flying tangent to the surface is madness, hence its forever failing reaction wheels. Except for a small central volume, there is no "weightlessness" in ISS FUBAR.

  • $\begingroup$ It's a bit technical (perhaps not for the OP, but for me). Does this mean that the only way to be in stable circular orbit for a, say, tube-shaped ship is for it to point downwards or upwards and have an (initial and thereafter) rotation equal to the orbiting frequency (if it doesn't use reaction wheels)? Or did I not get it? $\endgroup$ – Keep these mind Feb 6 '14 at 16:09
  • $\begingroup$ An orbit is any bound path, overall an ellipse for two bodies. That is independent of (small) body orientation. "Long Duration Exposure Facility" When NASA lost all control for 5+ years, its long axis pointed downward, <BR> wanderingspace.net/wp-content/uploads/2008/11/ldef.png <BR> see.msfc.nasa.gov/mod/modimage/mod.jpg $\endgroup$ – Uncle Al Feb 6 '14 at 16:49
  • $\begingroup$ Hi Uncle Al, can you clarify what you mean by "ISS FUBAR"? $\endgroup$ – Brandon Enright Feb 7 '14 at 17:06
  • $\begingroup$ @BrandonEnright I guess ISS could be Intergalactic Space Ship, or International Space Station. Translating FUBAR is forbidden by site rules. -> Uncle Al : Are you actually saying that the ISS is maintained into an unstable equilibrium with active mechanisms ? - - - Regarding the OP's original question: If the ship comes from outer space and goes around earth in a hyperbolic trajectory, its rotation induced by tidal force may not have time to reach equilibrium. $\endgroup$ – babou Feb 7 '14 at 20:15
  • $\begingroup$ en.wikipedia.org/wiki/Reaction_wheel A fifth leg on NASA's dog. What does ISS FUBAR accomplish? If a body loops without orbiting, one doubts there is time for tidal equilibrium. Absent "stuff," an orbiting body maintains its orientation versus the fixed stars, thus Foucault's pendulum. (FUBAR is an evolved SNAFU). $\endgroup$ – Uncle Al Feb 8 '14 at 0:26

I have some troubles with the other answers. As a sanity check I provide this answer.

I assume that the spaceship is not too big.

  1. If the spaceship enters its orbit initially not rotating (perhaps due to some burn to get into orbit), then gravity will not make it rotate. If it is initially pointing forward relative to its local orbital path, then after half an orbit it will point backward relative to its local orbital path there. (It will keep pointing at the same faraway fixed point, notwithstanding the direction of its actual movement.)

  2. If it is initially rotating, then it will keep that rotation.

  3. If it is initially rotating (pitch-downwards) with the orbital frequency, then it will also keep that rotation. Only then it will maintain its pitch relative to the Earth's surface. Only in this case (described by John Rennie, v1) will it appear as though gravity does the rotating, but 1 and 2 make clear that it is not gravity that explains the rotation, but rather the initial angular momentum of the spaceship that is preserved.

The key thing is that the angular momentum of the spaceship, whatever it is, will be preserved. (I think this is what Uncle Al, v1, means, but I'm not sure.) And even if it is affected, say because of the weight distribution of the spaceship, or due to relativity considerations, it will not be affected such that a stable rotation would be achieved instantly.

The rotation (in the pitch coordinate) of the spaceship may be adjusted by a reaction wheel. This will adjust the rotation, but keep angular momentum preserved.

  • $\begingroup$ I do not think this is correct because of tidal effect. Possibly my answer (written after yours) will explain intuitively why. If you find it unclear or have reasons to disagree, please let me know. $\endgroup$ – babou Feb 7 '14 at 20:32
  • $\begingroup$ @babou Sure, I had hoped to avoid that by stating that the spaceship is "not too big". Also, I think, the tidal effects won't cause instantaneous continuous rotation. However, I was pleasantly surprised by your remark regarding regarding angular momentum transferring between the two bodies. I never thought of that. Thanks. $\endgroup$ – Keep these mind Feb 7 '14 at 20:59
  • $\begingroup$ Thanks. Actually it is very hard to understand the detail what goes on through pure qualitative reasonning. I suspect that are several ways things can occur, depending on the various parameters of the trajectory and the ship mass strcture. There is another analysis that will have the front rather than the back closer to the planet if the ship is very very long with some mass in front and most of the mass in the back, so that the center of mass is close to the rear end. But in all cases, there will be instabilities that make the ship rotate as it passes the planet. $\endgroup$ – babou Feb 11 '14 at 1:41

The same side of the moon always faces the earth due to Tidal Locking, so I think that yes gravity does cause an object to rotate.

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    $\begingroup$ I was about to answer the same. This is likely to happen if the spaceship can be deformed by gravity for instance if it contains some air. However this will take a lot of time, it cannot happen in half revolution as the OP suggests. $\endgroup$ – DarioP Feb 6 '14 at 20:04

Here's how I picture this, w/out any calculations. Imagine the spaceship consisting of two massive objects, "front" and "rear" ones, connected with a long massless rod. They follow the same trajectory before and after interaction with the planet, with the "rear" one always a little behind the "front" one. The forces transmitted through the rod are always aligned with the motion: when "front" is closer to the planet and tries to accelerate faster it pulls the "rear", and vice versa. Therefore after the spaceship continues on its (supposedly hyperbolic) orbit the rod would rotate by the same angle the spaceship direction turned.

  • $\begingroup$ Your reasonning would work if the front and rear masses where independent, not linked by a rod. However, two independent masses following the same trajectory will have their distance vary in proportion to their speed. However the existence of the rod will not allow this variation. Hence the behavior of the system cannot be analyzed as if it had two independent masses following the same trajectory. And if the front mass is being pulled back while the rear mass is pulled forward, how can you expect them to follow the same trajectory? Your reasonning does not hold physically. $\endgroup$ – babou Feb 9 '14 at 2:09

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