Bandersnatch
Science Advisor
- 2,743
- 1,564
Hmmm, that stick analogy - isn't the far end lagging behind only due to air resistance?
There is zero torque on an object that is circularly orbiting some other object and is tidally locked to that other object.The constant torque here is the difference between the centripetal forces on the near and far side. There's a constant force on the near side that is stronger than the one on the far side. If there's a constant acceleration, there's a constant force, and a constant torque.
That's a bad example, because once again there is no torque once the object does become tidally locked.You are saying there shouldn't be any shearing expected, but I can't see that. The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.
Wrong. The gravitational force at the point closest to the central body, furthest from the central body, or at any point along the line from the central mass to the center of mass of the body in question contribute absolutely nothing to the torque. That's Physics 101. It's the points off that line that result in gravity gradient torque. To first order, the Newtonian gravity gradient torque on some body located a distance [itex]\vec R[/itex] from some central mass and with a moment of inertia tensor [itex]\boldsymbol I[/itex] about the body's center of mass expressed in inertial coordinates is (without derivation; it's long) [itex]3 \mu/R^5\, \vec R \times (\boldsymbol I \vec R)[/itex], where [itex]\mu=GM[/itex] is the standard gravitational coefficient of the central mass.Before a body achieves tidal lock, there is a torque applied on the body. This torque emerges from the differing forces on the near and far sides, the near side experiencing a stronger force.
I think the far side would catch up and oscillate if the point from where the torque is applied (the near side) was not accelerating, that's right.The far side will catch up with the near side and overcome it. Perfectly elastic body would be oscillating around middle point with far side sometimes being ahead and sometimes behind the near side.
The centripetal force isn't constant?? If it wasn't, the moon would either fly off or fall down. Consider this: the tangential velocity of the moon is constant, if the acceleration from gravitational force gets weaker, the moon would start to orbit farther away, the opposite would be true if the force increases. The force from gravity is constant, the torque is constant, the period of rotation is constant. What am I missing?A constant torque on the near side means that you are spinning faster and faster.
And then - yes, the far side will try to catch up but will never be able to.
And then - no, it does cause a varying period of rotation.
And centripetal force is not constant either in that case.
But if that is the case, than we wouldn't need torques to create a tidal locked body in the first place, would we? Everything would be naturally in tide lock.It might help if you considered the mechanics of planetary condensation from an orbiting cloud.
If the forces and effects you are proposing as inferred from Keplerian mechanics were operating as you surmise, then how could condensation ever occur?
The differing orbital velocities at the near and far sides of the diffuse mass would prevent anything like spherical symmetry from ever forming wouldn't they? Would keep the matter smeared out around the orbit.
But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets
Nope. Imagine a light year long stick in vacuum and start to spin it.Hmmm, that stick analogy - isn't the far end lagging behind only due to air resistance?
I agree, but you are considering the body after it is in tidal lock. The point is that there is a torque for the body to reach tidal lock, and my argument is that the body will reach tidal lock but will suffer shearing.There is zero torque on an object that is circularly orbiting some other object and is tidally locked to that other object.
That's a bad example, because once again there is no torque once the object does become tidally locked.
I agree, that's why we have proposed tidal bulges to produce the torque needed to lock the body.Wrong. The gravitational force at the point closest to the central body, furthest from the central body, or at any point along the line from the central mass to the center of mass of the body in question contribute absolutely nothing to the torque. That's Physics 101. It's the points off that line that result in gravity gradient torque. To first order, the Newtonian gravity gradient torque on some body located a distance [itex]\vec R[/itex] from some central mass and with a moment of inertia tensor [itex]\boldsymbol I[/itex] about the body's center of mass expressed in inertial coordinates is (without derivation; it's long) [itex]3 \mu/R^5\, \vec R \times (\boldsymbol I \vec R)[/itex], where [itex]\mu=GM[/itex] is the standard gravitational coefficient of the central mass.
I agree as well, torque is produced from travelling tidal bulges. Plus, with no gravity, the linear velocity alone would not cause rotation, so we may assume that the linear velocity would tend to keep the far side from turning in alignment with the near side, but that's not fundamental. I'm considering the situation a bit like Newton solved the orbital speed, he let the body travel straight to the point where it would go without gravity, than he worked on the distance from that point to where the object ends up in an orbit.Note that there is zero torque if the body is aligned such that one of its principal axes points toward the central mass. Now consider an object in a circular orbit whose rotational angular velocity is equal to the orbital angular velocity and that is aligned in this manner. It will always be aligned in this manner. This is tidal lock in its simplest form. (Aside: Note that there is zero torque of the body in question has a spherical mass distribution as any set of axes are principal axes for such a body. Bodies with a spherical mass distribution cannot become tidally locked.)
I haven't seen tide lock explained by heat the way you state, could give me a reference? And the body isn't aligned when forces are acting on the body to achieve tidal lock. And truly rigid bodies can't be tidal locked, I agree.What if the object has a non-spherical mass distribution and isn't aligned in this manner? Now you will get a non-zero gravity gradient torque, and that torque will be in the direction of tidal lock conditions. If the body truly is a rigid body what you'll get is something akin to harmonic motion. Truly rigid bodies cannot become tidally locked. A non-rigid body will have some plasticity or viscosity to it, and this in turn means that some of that change in rotational energy that would nominally result from that gravity gradient torque instead heats the body (which is eventually radiated away). These dissipative forces, along with a non-spherical mass distribution, are a necessary conditions for tidal lock to occur.
You are correct, these would complicate a lot the problem, but I am trying to simplify things to understand the basis of the forces that create tidal lock. Also notice that the semi major axis of the moon is not from the near to the far sides, but from the leading to the trailing sides... actually I found differing statements regarding this online, so I am not completely certain of this, but everywhere you look you will find that the near side is shallower than the rest of the moon. But that will just complicate things up, consider any other body with tidal bulges front and back and the forces that work on it before it is in tidal lock, and notice that these forces, specially the torque, should cause shearing. Everywhere you have, you have to expect shearing, and to keep a body in a constant orbit, you have to apply a constant force, if you turn off the centripetal force, the body flies off at the tangent. So we have a constant centripetal force, a constant linear velocity and a constant torque that must cause shearing, am I wrong? Or is gravity not a constant force/acceleration? If you turn off the tangential velocity, won't the body fall down? Doesn't that imply a constant force/acceleration?What if the orbit isn't circular, or the central mass doesn't have a spherical mass distribution, or there are other objects around (we're no longer in the two body problem world)? Things become a bit more complicated here. Now there are time-varying tidal forces on the body. The Moon, for example, has small but observable time varying body tides. The observability of these time varying body tides is one of the legacies of the Apollo program. The Apollo astronauts left retroreflectors on the Moon. The McDonald Observatory has collected a large time series of measurements of the distances between the observatory and these retroreflectors. These data yield a very accurate picture of the Moon's orbit and also of the Moon itself, even it's interior.
Yes, I am. There's a beautiful animation of it on the wikipedia page on the moon. And I don't disagree with you, I am not contesting that bodies are in tidal lock in real life, with no shearing, I am trying to point out that the forces (constant centripetal and torque) that act on the body before it achieves tidal lock should cause shearing, shouldn't they? You are saying that there's no constant torque, because if there was the period of rotation would vary, but there must be a constant torque if the centripetal force is constant - gravity is a constant. Are you saying that gravity is not a constant centripetal force/acceleration?altergnostic, are you familiar with the moon's libration? The moon rocks back and forth exactly like a big, solid pendulum would on your desk. When it rocks to the left, the torque pulls to the right. When it rocks to the right, the torque pulls to the left. When centered, there is no torque.
Perhaps if you drew a diagram showing the forces acting on a body in tidal lock, it would help here.
The relativistic effects that matter across a light-year (about 10^{16} meters) are completely irrelevant at the diameter of the moon (less than 10^{7} meters. There are simply no relativistic phenomena involved in understanding the tidal lock of these systems (yes, you CAN solve them using the methods of the GR gravitational theory... but the first step in that solution is to apply the weak-field approximation and that gives you the same results as Newtonian gravity).Nope. Imagine a light year long stick in vacuum and start to spin it.
In this context, is "combined Ricci and Weyl effects" anything but a long-winded way of talking about classical Newtonian gravity, which is completely sufficient to explain the phenomenon of near-spherical planets?But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets
I was not talking about relativistic effects, i was talking about deformation due to torque. I think the point of our misunderstanding is that you consider that the torque is not constant, but if the force from gravity is constant, so is the torque. And this force is always greater on the near side. Think of it this way: the law of inertia states the if there are no forces on a body, it will remain in its rest state, either travelling at a constant linear velocity or standing still (which are the same thing). To maintain a stable curved path, like an orbit, the centripetal acceleration must be constant. This constant force generates a constant torque, like in a mery go round. If you don't hold on to something, you will fly off at the tangent. This must mean a constant centripetal force, and if the body is not falling towards the central mass, but orbiting, there is a constant force holding it in its curved path. This constant force is gravity, and it generates a constant torque. Why do you believe the torque ceases?The relativistic effects that matter across a light-year (about 10^{16} meters) are completely irrelevant at the diameter of the moon (less than 10^{7} meters. There are simply no relativistic phenomena involved in understanding the tidal lock of these systems (yes, you CAN solve them using the methods of the GR gravitational theory... but the first step in that solution is to apply the weak-field approximation and that gives you the same results as Newtonian gravity).
Bandersnatch is correct. If no torque is applied to the non-relativistic rotating rod it will be straight (although if has been elastically deformed by an applied torque it will take a moment for the deformation to relax and the rod to regain that undeformed straight shape). The only reason for the outer edge to lag would be drag, atmospheric or otherwise.
You're argument is wrong, and has gone on far too long.I agree, but you are considering the body after it is in tidal lock. The point is that there is a torque for the body to reach tidal lock, and my argument is that the body will reach tidal lock but will suffer shearing.
This is old, old stuff, some going back to George Darwin and A.E.H. Love. More recently, but still old stuff,I haven't seen tide lock explained by heat the way you state, could give me a reference?