Moon's "tidal lock" rotation theory - What is wrong with it?

Originally posted by ThinkingHuman
It says that the elongation is due, not to gravity by itself, but to the difference between gravity and centrifugal force. Those two forces are equal in size at the center but their amounts are, at the closest and furthest points, different from each other - but in opposite directions. Therefore the elongation.

This makes sense but even the line you pointed out does not address what I said, that the earth is not subject to centrifugal force with regards to the Moon. Only the Moon is subject to centrifugal force relative to earth because it is in earth's orbit.

Earth should not experience elongation, at least not because of a difference between gravity and centrifugal forces.

TH: I haven't had time to sit down and compose a lengthy response, between near 16 hour shifts at work and feeling like warmed over dog poo due to this bronchitis (at least I gave it to the military staff - take that! Sounds like a TB ward in the hangar now).

Since I haven't, here's the short form answer: I think I see it from your viewpoint - you see the Earth as being motionless and the Moon falling around it, so there ought not be centripetal force acting on the Earth. (centrifugal forces are illusory)

However, that's not true, exactly. Another thought experiment. Say you have two bodies. One very very massive, say a star, and one comparatively small, say Mercury. From an outside viewpoint, Mercury seems to go around the Sun and the Sun seems to sit there. Now, replay it with two bodies of the same size. What orbits what? Does one sit still and the other go around it? Which is still and which moves? What chooses that?

The truth is, that's not what happens. In any two body orbital system, BOTH orbit each other. There is a "still place" that they both orbit about, the location of that is somewhere on a radial line between the two centers of mass. Celestial mechanics calls this the "barycenter". In the Earth-Moon system, the barycenter of orbit is not the center of the Earth, it's somewhere between, but due to the mass difference, it's still somewhere inside the Earth, IIRC.

nice link with pictures and math

BTW, this is how astronomers "detect planets" around distant stars. They can't resolve them directly with a telescope, but you can spot the stars orbiting around their barycenters as the planets go around. The star is actually moving back and forth a bit as it orbits a point inside itself. This motion causes very tiny Doppler shifts in the spectrum of the star. If you analyze this with some neat math tricks (Fourier, primarily) you can pick out the various influences on the star's motion, and through magic, you can pick out how many planets and about what size they are.

Originally posted by Bedlam

Bedlam, thanks for your reply. I do understand the time constraints, I should have given you more time to reply. Sorry about that.

In any two body orbital system, BOTH orbit each other.

I anticipated that you may consider this to be an explanation for tidal force to apply to earth also. In a previous post I stated "I am aware that earth and Moon rotate around a common center of gravity but this is a 28 day cycle, not a daily one." But the NOAA site you linked also refers to the difference between centrifugal and gravity. (NOAA only talks about centrifugal, not centripetal). So I reconsidered and came up with this.

Earth orbits around the center of gravity of the earth-moon-mass, the radius of the orbit is 4,651 km (which is 1,720 km below the surface of the earth), in about 29.5 days.

The two forces of interest are "X" for centrifugal force around the common center of earth-moon-mass, and "G" for gravity of the Moon that applies on two points on earth, "n" for the spot that is nearest to the moon and "f" for the spot that is furthest from the moon. We observe that the two daily tides are almost equal in size. Then we can write the following formula:
Gn + Xn ≈ Gf - Xf

The two forces add to each other on the near side because the act in the same direction, while they subtract from each other on the far side. This means that on the far side the centrifugal force must be significantly greater than the gravity in order to be equal or almost equal to gravity plus centrifugal force on the near side. The furthest point on the far side has a radius of 4,651 km, which means it travels a total of 29,220 km during its 29.5 day orbit. That comes out to 41.3 km per hour. Imagine you travel a circle with a length of 29,220 km at 41.3 km, I believe the centrifugal force would be all but zero. That force does not seem, to me, to be able to explain the tide on the far side. But I may be wrong, so please correct me if you find where I blew my math.

reply to post by ThinkingHuman


Right now I've got a major case of drug-head. I'm not dodging you, I just feel like hell, so instead of doing math I'm reading the last two books in WOT.

Did that NOAA link not spell out how the far side tide occurs? I didn't read it all the way through, just to the barycenter part.

reply to post by Bedlam


I have not read it in depth enough to have a full understanding but I am working on it. I believe I understand the forces and their direction that apply to it. Now its a matter of doing the math and that is not as easy for me. Btw, in comparison, the moon is travelling at a speed of 3,370 km per hour. The balance between the two forces must be VERY different. I will try to look into that after I understand the math regarding earth.

reply to post by Bedlam

The NOAA site indicates that the centrifugal force applies equally and same direction everywhere on the surface of earth moving in a circle, the radius of which is equal to the distance between the center of the earth-moon-mass, and the center of earth, which amounts to 4652 km. As I pointed out earlier, 29.5 days for one revolution gives me a speed of 41.3 km per hour (25mph).

To get an idea of the centrifugal force I decided to calculate the curvature: Lets make a straight line, 100 meters long, with the end points (A and B) on this circle. How far away does the circle stray from the straight line at its mid-point, M? We can resolve this question with the theorem of pythagoras, a2 + b2 = c2. We start with one triangle ABC (C being the center of the circle). The CM divides that triangle into two equal triangles, AMC and BMC (they have a right angle at M). CA is the radius of 4652 km, AM = 100/2 = 50 m. sqrt( CA2 - AM2 ) = 4651999.99973 m. Therefore the distance from M to the circle is 0.27 mm, or less than one sixteenth of an inch.

Put your car in an empty parking lot, stick a glass of water on your dashboard, measure a straight line of about 110 yards, and crank your wheel such that it will be off the straight line by one sixteenth of an inch after about 55 yards. I predict that you will not notice any centrifugal force. From this force you need to reduce even further, the amount of gravity from the moon that applies at the far point. At that is supposed to move oceans up by several meters?

So, at this point I would lean towards you and the theory being correct. Thanks for your input. I would like numbers to support the theory, though. For NOAA, such a calculation should be easy, easy to display for public benefit, and an issue of significant importance to their work.

But to get back to the main topic, how can a centrifugal force so minute cause a moon into "tidal lock"?

But to get back to the main topic, how can a centrifugal force so minute cause a moon into "tidal lock"?

4.5 billion years of time?

Originally posted by Mogget

But to get back to the main topic, how can a centrifugal force so minute cause a moon into "tidal lock"?

4.5 billion years of time?

What makes you suggest that this is a long time with regards to tidal lock to occur?

We are talking about a scientific theory, so the theory should somehow explain a detail like this, should it not? And I still do not understand how tidal forces can result in tidal lock on a solid body, because no tide will ever cause a rock to be "squeezed", or elongated.

I think the most plausible answer to the locked lunar orbit is that the Moon early in the formation of the Solar System was ejected from the Earth. In this way the moon would get the similar orbital velocity and a still receding motion (annually 3.5 cm)

originally posted by: IvarNielsen
I think the most plausible answer to the locked lunar orbit is that the Moon early in the formation of the Solar System was ejected from the Earth. In this way the moon would get the similar orbital velocity and a still receding motion (annually 3.5 cm)

Ejecting something off a planet doesn't result in a constant receding motion (unless it achieved the escape velocity). The "stuff" would either eventually fall back to Earth, or achieve an orbit at a certain distance from the planet, which is what happened to the Moon.