I would like to point out a conceptual mistake in Chronos’ description of orbits. He says that “An orbit can be considered as a balance between gravity and centrifugal force.” That is incorrect. If that were so the resulting force would be zero and therefore the acceleration of the orbiting mass would be zero as well (Newton’s second law). Uniform circular motion is, however, accelerated motion: the velocity vector changes in time. Contradiction.
The correct description of a circular orbit is thus: there is only one force, namely the gravitational attraction F. We have an expression for this force F, the Newtonian law of gravitation. Once we know that the orbit is circular we can use geometry to prove that the acceleration of the orbiting mass is a=v^2/R where v is the tangential velocity and R the orbit radius. Equating F=ma gives the equation of the orbit. Only one force, hence the accelerated (circular) motion. This is the correct description based on Newtonian mechanics in the inertial frame where the source of the gravitational force is at rest. Non-circular orbits are not much harder, but let’s keep it simple.
Chronos’ mistake, which appears in many textbooks, originates in that m v^2/R is also the expression for the centrifugal force felt by a rotating observer in the absence of gravity. Did he mean to
describe the orbit in the frame of reference of the orbiting mass, the only frame where the centrifugal force exists?. Maybe, but that is a non-inertial frame, where Newtonian Mechanics doesn’t apply. Things are a lot harder then. For example one has to argue that free gravitating masses move in precisely the right way for the centrifugal force that exists in their rest frame to cancel the gravitational force. That is, one has to invoke a form of Einstein’s equivalence principle, even though we are only working out a simple nonrelativistic problem.
I am under the strong impression that Chronos really believes that the orbits of a free gravitating mass seen from an inertial frame are a balance between two opposing forces, even though one of them only exists in non-inertial frames. Perhaps he can clarify this point in his own words.
But the difference in centrifugal force from the near side to the far side can be completely attributed to its rotation. If you only look at the contribution from its orbit, the difference is zero.
No, the rotation of the Moon is immaterial. At any given instant, the point on the Moon facing the Earth is moving more slowly (linearly) than its opposite point on the other side of the Moon.
If the moon were not rotating, then each point of the moon has identical instananeous “linear” velocity. If you say that’s not true, then we have something to discuss, to clear up the misunderstanding. But you have to consider a case where the moon is not rotating, to do that.
RM Mentock: I was considering the center to be at the Earth. Given that, the “top” point of the Moon is experiencing a larger centrifugal force than the “bottom” point. While it’s true that the Moon itself must be rotating for this distinction to be meaningful, I’m not sure that that’s relevant.
marcos alvarez, it depends on what reference frame you’re using. If one is working in an inertial reference frame, then you’re correct, one should not be using centrifugal force. In an inertial reference frame, the only force is gravity, which is the centripetal force, and since there is a net force, there is an acceleration.
However, I was not working in the inertial frame; I was working in the (non-inertial) co-rotating frame. In that frame, the Moon is not moving, and remains at rest due to the balance of two forces, the gravitational force and the centrifugal force. Since we’re in a non-inertial frame, we need that centrifugal force.
With regards to the energy versus force methods for explaning the effect, it may just be a reflection of my training, but I find the conservation of energy explanation to be more intuitive. If you prefer to use the forces instead, go ahead.
And as for tidal bulges, for small bulges, the asymmetries cancel out, but for large bulges, they don’t, even in simple two-body systems. In the exreme case, an object will be distorted into a sort of teardrop shape called a Roche lobe, with a pointy cusp pointing towards the other object. This is often the case in binary star systems where one has become a red giant.
John W. Kennedy, thanks for the graphic! I didn’t intend for you to go to that work, I’d produced a graphic of my own a while back and it seems I’ve lost track of it. I’ll see what I can dig up.
I missed that comment. They don’t completely cancel out, see my post above answering jezzaOZ.
I reread the thread, and I realized I hadn’t been as specific as I thought. If you calculate the centrifugal force on the points of the moon using John W. Kennedy’s diagram, you notice that the centrifugal force is proportional to distance. However, part of the centrifugal force on the surface of the moon is due to the rotation of the moon–not its orbit around the Earth. If you calculate that, and subtract it, you find that the result is that the centrifugal force left is constant, no variation with distance from the Earth!
If you draw another diagram where the moon orbits the Earth without rotation, it’s apparent that each point of the moon follows a similar shaped orbit, but around an offset center. That offset is enough to cause the centrifugal force not to vary by distance from the Earth.
OK, I think I see what you’re getting at, RM, and yes, you can analyze it that way, but it seems to be more compicated conceptually for giving the same result. As for the symmetry of the bulges, what I should properly have said is that the low-order terms are symmetrical, but the higher order terms are not. Technically, those higher order terms are always there, but for small bulges, the symmetrical portion is much more significant than the asymmetric part.
Saying they give the same result is kinda like saying that the asymmetries cancel out. They don’t.
What if the moon rotated seven times faster than it does now? In your analysis, would you back out seven revolutions, or six? If you back out the rotation of the moon, then your claim is not true. Your analysis may have some heuristic value, but it is certainly wrong. You should have consulted Cecil on this.
Since it looks like this thread is active I thought I’d post a related question that has been on my mind from time to time.
Is the abundance of mare on the near side of the Moon and the absence of mare on the far side realted to the uneven distribution of mass in the Lunar body? It’s hard to believe it could be a coincidence.
If the Lunar highlands (non-mare) are similar to continental formations on Earth and the mare similar to oceanic basalts then this would indicate the center of mass is on the sub-Earth side of the Moon.
Just curious 'cause I’ve never read any sort of reference to this sort of thing.
More or less. There are a lot of webpages out there that explain the tidal locking of the moon by saying that this displacement of the moon’s center of mass is towards the Earth, and that is why it is “locked.” The center of mass is displaced a large amount (about 2 kilometers) but this website says it’s towards a point midway between Mare Serenitatis and Mare Crisium–probably not where you expected? (Moon map)
First, this cannot even be calculated without specifying how fast the Moon is rotating. Second, the centrifugal force due to rotation at the innermost point (assuming that everything’s in the same plane, which is close enough for the present) is toward the Earth, reducing (even negating, if the Moon is rotating a great deal faster than it in fact is) the already weaker centrifugal force due to the orbit at that point. Third, the centrifugal force due to rotation on the outermost point is away from the Earth, increasing the already stronger centrifugal force due to the orbit.
Innermost point:[ul][li]Orbital centrifugal force is away from the Earth, but is less than at the center.[]Rotational centrifugal force is toward the Earth.[]Gravity toward the Earth is more than at the center.[/ul][/li]
Center:[ul][li]Orbital centrifugal force exactly balances gravity.[]Rotational centrifugal force is zero.[]Gravity toward the Earth exactly balances orbital centrifugal force.[/ul][/li]
Outermost point:[ul][li]Orbital centrifugal force is away from the Earth, but is more than at the center.[]Rotational centrifugal force is away from the Earth.[]Gravity toward the Earth is less than at the center.[/ul][/li]
All three factors, orbital centrifugal force, rotational centrifugal force, and gravity, draw the innermost point in and drive the outermost point out.
It is not “already weaker”. That’s my point. Without the effect of the moon’s rotation, there is no variation in centrifugal force across the body of the moon.
Trace the path of the points on the moon as it orbits. If the moon does not rotate, but translates, each point will describe curves that are identical but displaced from each other by a traslation. The reaultant vectors are identical. So, if you back out the rotation, there is no variation of the centrifugal force across the body of the moon.
We know how fast it’s rotating now, but the original point of the question was how it got that way.
Yes, but now you are describing a situation that never existed, will never arise under natural conditions (because the locked configuration is stable), and, even if it were created, would be unstable (because, unless the Moon were infinitely rigid, purely gravitational tidal effects would drag the Moon around until the locked configuration was reached).
All I’m talking about is the analysis, not whether the moon lags, leads, or is locked. The point is, it is rotating, and that rotation has nothing to do with its orbit. It could be rotating ten times as fast and the orbit would not change significantly.
You have to back out the effects of that rotation whether it is rotating ten times as fast, just as fast, or not at all. If it were not at all, it would be easy.
That rotation also produces a bulge on the left and right sides of the moon–the sides from the point of view from the Earth. And that bulge is the same as the bulge produced by the rotation on the front and back sides.
RM, could you do me a favor and explain what it is you’re saying? Your points up to now have been a little too fragmented for me to understand, but I have a great deal of respect for your points in general. Chronos’s tides explanation using a co-rotating reference frame and centrifugal force seemed satisfactory to me, and I can’t figure out what you’re point is about the lunar rotation about its axis.
