3
$\begingroup$

Consider two nonelastic spherical bodies with uniformly distributed density, a small such body in a circular orbit around the bigger one.

And consider the smaller body's rotation is matched (as if "tidally locked") to its orbit so the same hemisphere always faces the larger body.

Now, here's my question: a particle on the daylight surface of the orbiting body is orbiting the "sun" with a smaller radius than a particle resting on the "night" side. If the two particles weren't attached to the body, the daylight particle would have a faster solar orbit and the nighttime particle a slower orbit.

Would this not impart a retrograde rotational force on the orbiting body? If this is an already understood concept, what is it called? I read up on tidal locking and tidal acceleration on Wikipedia, and this dynamic wasn't mentioned.

I think it's kind of interesting because I never thought about tidal forces doing anything with inelastic objects.

Improve this question
$\endgroup$
1
  • $\begingroup$ A rotational force will do work to increase/decrease the rotational kinetic energy of the orbiting body. Where is said energy coming from in your analysis? $\endgroup$
    – Willie Wong
    Commented Oct 7, 2011 at 12:12

3 Answers 3

Reset to default
1
$\begingroup$

It's a good question.

A similar situation is a satellite, in a circular orbit about a planet, consisting of two bodies linked by a tether, aligned on a radius to the planet's center.

The tether is under tension because the lower body is in a stronger gravitational field than the center of mass, the upper body is in a weaker field than the center of mass.

This tension exactly offsets the higher gravitational field, so the net acceleration on the lower body is equal to that at the center of mass. (Except for the centripetal acceleration due to the once-per-orbit rotation of the composite.)

Similar logic applies to the upper body.

Edit: In terms of rotational force, it is true that if the tether were cut, the lower body would be pulled into a lower (slightly eliptical) orbit with a lesser orbital period than the upper body and would thus drift ahead of it. In fact, the point at which the tether is cut becomes the apogee of the lower body's orbit, and the perigee of the upper body's orbit. However, when the tether is not cut, the lower body is prevented (by the upward pull) from entering that lower orbit, and similarly for the upper body.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Mike, that doesn't address any rotational force imparted to the tethered pair. $\endgroup$
    – Doug
    Commented Oct 7, 2011 at 8:14
  • $\begingroup$ @Doug: Edited to try to clarify that. $\endgroup$
    – Mike Dunlavey
    Commented Oct 7, 2011 at 12:26
  • $\begingroup$ @Doug: Something that does happen in the earth-moon system is that the earth is not tidally locked with the moon. This causes a momentum transfer from the earth to the moon, lifting its orbit (and slowing the earth's rotation). $\endgroup$
    – Mike Dunlavey
    Commented Oct 7, 2011 at 12:46
  • $\begingroup$ @Doug: My understanding of how that works is the moon sees a rotating earth, with a tidal wave on that earth which is not actually pointing toward and away from the moon, but offset forward by friction with the earth's rotation. So in effect it is orbitting two bodies, one nearer and offset forward, the other farther and offset back. Since the gravitational pull of the nearer body is stronger, the moon feels a stronger force from it, thus accelerating it in its orbit. $\endgroup$
    – Mike Dunlavey
    Commented Oct 7, 2011 at 13:07
0
$\begingroup$

Good question indeed.

As you noted, there will be a "tidal force" acting on the body. This "force" tends to deform the body (and the "sun" as well).

As a first-order approximation the bodies will become ellipsoid. However the exact shape will probably be complex, especially if we don't neglect the fact that the "sun" is deformed as well, and its gravitational force is not as if it was a point mass.

Now, if after the deformation the bodies spin so that they constantly "face each other" - this situation is stable, and there will be no "retrograde rotation".

It's easy to see that there's no momentum on any of the bodies as a whole (both bodies are symmetric WRT the line connecting their centers of mass).

OTOH both bodies will already be deformed such that the overall force applied to each body part will make it rotate about the center of the rotation (of both bodies) with the same angular velocity.

Improve this answer
$\endgroup$
-1
$\begingroup$

Maybe the solution to my misunderstanding of orbital dynamics is that if we disconnected the daytime and nighttime particles, and took away the rest of the satellite, the particles wouldn't continue in circular orbits; one would be too fast and the other, too slow so they would be at their elliptical "peri" and "apo" positions of elliptical orbits.

My assumption was that the particles would be naturally pulled into their circular orbits by some force that doesn't really exist.

Does anyone agree? There is no such rotational force?

Improve this answer
$\endgroup$
3
  • $\begingroup$ Your problem is a bit strange. First You demand us to envision two bodys (Inelastic, what evewr You mean with that) then a thid one called sun pops up. You should clarify this first. A sketch of that system would be very helpfull. And read about elasticity and on "inelastic" ! You have wrong pictures about this. $\endgroup$
    – Georg
    Commented Oct 7, 2011 at 11:38
  • $\begingroup$ I started calling the larger body "sun" and its satellite's orbit a "solar" orbit. That also allowed me to describe the "daylight" and "nighttime" hemispheres. By inelastic I mean these bodies don't have molten cores and oceans that respond to the tidal forces; consider the spheres rigid so that we can isolate and describe a single dynamic between them. $\endgroup$
    – Doug
    Commented Oct 7, 2011 at 11:50
  • $\begingroup$ Thats the problem: rigid means elastic! Tidal lock comes from inelasic properties! Any real astronomical body is more or less inelastic, because ideal elasticity does not exist in real bodys. $\endgroup$
    – Georg
    Commented Oct 7, 2011 at 12:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.