Careful, careful, the BA is listening.

I think you are just misunderstanding the textbooks, in this case. Your theory doesn't seem to match the evidence.

Geo3gh DoctorDon GrapesOfWrath

Let me try to state or restate the evidence [img]/phpBB/images/smiles/icon_frown.gif[/img]

1) In 1969 the laser reflector was placed on the Moon. As you stated the Earth and Moon, are moving apart at a rate of "3.82±0.07 cm/year."
2) It has been known for hundreds of years that the Moon *appears* to slow down by 1 revolution every 18.6 years.
3) The above two, are components of the same vector.
4) The leap second. Even before the atomic clock the Earth's *appearent* slowing was detectable. Since the atomic clock (1956 ish) the amount has been fixed at around 0.68 second per year.

Other than a physics book and logic I think that's about it.

Geo3gh said:
"Don and I have both brought out evidence that the moon's orbital radius is indeed increasing (on the order of 3cm/year) and that the length of the day is increasing as well."

GrapesOfWrath said:
"I think you are just misunderstanding the textbooks, in this case. Your theory doesn't seem to match the evidence."

I believe that most all the astronomy texts say or infer that the Earth is *slowing* down, or as just stated "the length of the day is increasing." I don't believe that I can misunderstand that unless the word slowing has some new definition. I don't believe there is evidence that the Earth, Moon, or the Earth-Moon system are slowing.

Example:
If there is a stop light at the end of a freeway on ramp, and if you stop at it, your speed is 0 and your acceleration is 0.
The light turns green you accelerate (speed up) to 70 mph and set the cruise control. Here you aren't accelerating, but you are cruising at 70 mph.
Brake lights ahead!
You put your foot on the brake pedal.
Are you speeding up, no.
Are you cruising, no.
Are you stopped, no.
You are accelerating, this time you're "slowing" down. 60-50-40-30-20-10
OK! OK! Reset the cruise control.
Are you stopped no.
Are you speeding up no.
Are you cruising ????
Are you "slowing" down, no.
You are simply going more slowwwwly.
"Slowing" is a negative acceleration.

Going slow is a relative term whereby you need to compare one fixed speed to another. If you were going 70, but now only 10 you're going slower or slow. If you're going 10 and the car pool lane is still at 70 again you're slower or slow. The amount in this case is a difference of 60. That difference of 60 will not change unless an acceleration +/- takes place in one or the other of the speeds.

The Leap Second has been about 0.68 seconds every year for some 40+ years, which says to me it is not an acceleration only a difference in speed. I could be wrong there could be an acceleration, but the amount is far to small to measure.
I say that the leap second data, recession of nodes data, and the laws of physics prove that the Earth and Moon are not "slowing".

Yes, the observational data Earth-Moon in relation to the stars shows them to be moving slow, but nothing shows them to be "slowing".
All the laws stated in physics textbooks seem to show that no slowing of the Moon or Earth can be possible. What the laws show is that the Moon and Earth are slowing the Sun, but in turn the Sun is speeding up the Earth and Moon.

Gary

<font size=-1>[ This Message was edited by: Gary Redmond on 2002-03-22 07:56 ]</font>

You're confusing "slowing" with "slower"
The evidence has been reference above, in this thread. Believe it or not.
You are right that the leap seconds are a result of a difference in speed, between our current rate and the standard rate which was referenced to Earth rotation speed of the year 1900. But the evidence shows that the earth slows only about 2 milliseconds a day per century. That is much, much smaller, but it is still large enough to measure.

That means that the earth is taking 2 milliseconds longer to rotate than it did in 1900. So, every 500 days or so (it is not a steady rotation, mostly because of redistribution of the mass of the atmosphere and in the Earth's interior), we need a leap second. That is close to your .68 second per year.

They don't.

That is only your interpretation.

Pardon me for jumping in here, Gary, but I need a clarification on this . . . you seem to accept that the Moon is farther away from Earth this year than last year, and will be even farther away next year, but you say that its orbital velocity is not slowing?

Am I correct in that, or am I missing something?


_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2002-03-22 08:27 ]</font>

Hi,

> > What does the Earth's motion around the sun have to do with this?"
>
> *Speed*. If a body such as the Moon travels a greater distance than the Sun in the same time of one year, which goes faster?

But what possible relevance does this have? Speed and distance are relative to some arbitrary frame and origin. I'm talking about the earth-moon tidal system, so I am considering a rotating frame of reference, centered on the barycenter of the Earth-moon system, rotating with the angular velocity of the moon's (current) revolution. I am ignoring the effect of the sun, as it is very small at this distance.

> > I don't care how the moon came to be where it is, I am modelling the
> > system as it is now and extrapolating that motion into the far future.
>
> OK, now do it backward and tell me what you find.

Interesting question. I haven't tried that.
Still isn't relevant for this particular discussion, though. I'll take a look at it and get back to you.

> > "Center of gravity is an ill-defined concept. I assume you mean center of mass."
>
> Absolutely not! The center of mass (or momentum) is the point around which a mass or system rotates.

Um, first, the center of mass is not the same thing as the center of momentum. The center of mass is a point, while the center of momentum is a frame in which the net momentum of a system is zero. Second, a mass or system does not necessarily rotate around its center of mass. That is not the definition of a center of mass. Take that 4x4 you mentioned by the end and spin it around your head: it's rotating, but not around its center of mass, nor around the center of mass of you and the board together.

The definition of the center of mass is the integral of the density times the displacement over the integral of the density (i.e. the total mass). While it is true that in the absence of external forces (like your hand), a body will rotate around its center of mass, that is not part of the definition of center of mass. Rather, it is a consequence of the definition.

Similarly, the net torque on a system is the integral of the cross product of the displacement of each mass element and the force vector on each mass element. Clearly, there must be a non-zero torque on the earth-moon system, because that integral will be negative and large over the bulge ahead of and nearer the moon, and positive and small over the bulge on the far side of the earth. (While "r" takes on equivalent values in both directions, the gravitational force goes as 1/d^2, where d is the distance to the moon from the volume element. So the mass closer to the moon feels a stronger gravitational force. Since the magnitude of r cross F goes as the sine of the angle between the vectors, the sign of the torque will clearly be opposite between the two bulges, since r flips direction but F does not. Finally, if we define a right handed coordinate system such that the Earth's current rotational velocity is positive, then the bulge nearer the moon will feel a negative torque by the right hand rule, and the far bulge will therefore feel a positive torque, and the net torque will be negative. QED.)

Since Torque = dL/dt = I dOmega/dt (where I is the moment of Inertia tensor for the earth, in this case, I_{ij} = Volume integral of density times the difference between the displacement squared times a delta function in ij and the quantity x_i times x_j). Those integrals are clearly going to be positive for the coordinate system defined above, so dOmega/dt, the change in the angular velocity of the earth, has to be negative: i.e., it's slowing down. QED.

> In my book the center of gravity is described as follows. "The moment about any axis produced by the weight of the body."

That sentence is poorly worded. It is not clear if the word "produced" refers to the axis or the moment. It's also scientifically not very useful because "weight" is always defined in terms of an external gravitational field, so whatever this "center of gravity" is, it's not an intrinsic property of the object, as it will change given the external circumstances. The center of mass will only change in so far as it is a vector, and thus is defined relative to some arbitrary origin.

> But in large and distant bodies it is appearent that even though the mass is the
> same the gravitational and inertial workings are different...Gravity works on
> weight, inertia works on mass.

Okay, I think I see why we're talking past each other. You seem to think that a graviational force will not accelerate a mass in the same way as, say, an electromagnetic force (which is what force it would be if you pushed on it with your hand). If you think that, you're out of synch with all of physics. The equivalence of gravitational and inertial mass, while not proven analytically, has been measured to be consistant to some phenomenal degree of accuracy (I forget how many, but it's many many decimal places) and there's never been any evidence that they're different (your claims to the contrary I find frankly baffling). This is what allows us to say "F=ma=GMm/r^2" and is the basis for my saying that dOmega/dt must be negative. If you don't accept that, then we can't move any further in this conversation, because everything I'm saying is based on this premise.

You still haven't explained how you think the moon can move away from the earth without slowing down. In the light of Kepler's Third Law, this would seem impossible, regardless of what the earth is doing.

Don

HUH! [img]/phpBB/images/smiles/icon_confused.gif[/img] This not only nonsensical, it is irrelevant (what does the Saros cycle have to do with the topic under discusion?).

Please excuse the extensive quotations I felt the context better.

DoctorDon said:
While it is true that in the absence of external forces (like your hand), a body will rotate around its center of mass, that is not part of the definition of center of mass. Rather, it is a consequence of the definition.

That is what I meant to say thanks.
If the 4 x 4 were out in space and spun it would spin around the center of mass. At the same time all the particles in it having momentum would spin around that point thus the center of momentum. Did I say that wrong? Perhaps I need the word angular.

I said:
> In my book the center of gravity is described as follows. "The moment about any axis produced by the weight of the body."
DoctorDon said:
That sentence is poorly worded. It is not clear if the word "produced" refers to the axis or the moment. It's also scientifically not very useful because "weight" is always defined in terms of an external gravitational field, so whatever this "center of gravity" is, it's not an intrinsic property of the object, as it will change given the external circumstances. The center of mass will only change in so far as it is a vector, and thus is defined relative to some arbitrary origin.

Thank you again.
You are 100% correct except for when you say: "It's also scientifically not very useful."
This is the most important part, mass is as you say intrinsic to inertia, weight is intrinsic to gravity. "And never the twain shall meet." Inertia keeps the Earth and Moon apart; gravity ties them together.

DoctorDon said:
The equivalence of gravitational and inertial mass, while not proven analytically, has been measured to be consistent to some phenomenal degree of accuracy (I forget how many, but it's many many decimal places) and there's never been any evidence that they're different (your claims to the contrary I find frankly baffling). This is what allows us to say "F=ma=GMm/r^2" and is the basis for my saying that dOmega/dt must be negative.

Thanks again yes the masses are equivalent.
In your formula "F=ma=GMm/r^2" the m in the ma is a singular mass.
In your formula "F=ma=GMm/r^2" the Mm in the GMm/r^2 is a product of two singular masses.

The book said:
"The moment about any axis produced by the weight of the body."
WEIGHT = the magnitude of the gravitational force exerted on a body by the nearest astronomical body.
I believe the weight of the body mentioned means the sum of the weights of all the particles of the body. Here is the rub. Because all those weights are directed/accelerated toward the center of gravity of the other body they create internal torques, but here again as in any mass (system of particles) the sum of all the internal forces equal zero. Therefore any external gravitational mass cannot cause a torque within another mass. A singular gravitational force is not an unbalanced force side to side, nor front to back, only along the line connecting the gravitational centers, not the centers of mass. Therefore I say that the Moon's gravitational pull cannot exert a torque on the Earth's inertial mass. The Earth cannot become locked face on to the Moon other than by being at the correct altitude such as a geosync satellite.

DoctorDon said:
"You still haven't explained how you think the moon can move away from the earth without slowing down. In the light of Kepler's Third Law, this would seem impossible, regardless of what the earth is doing."
I hope you're ready for this.
I say and believe that the Moon is in what might be called a polar conical trajectory not an equatorial orbit. Over half of that trajectory lies in the Northern hemisphere. As the Sun and Earth travel around the Galaxy they are moving away from the Moon. It is the length or height of the cone that is changing not the radius. That is the recession of nodes and regression of the Moon. Likewise the Moon when it joined the Earth it retarded the Earth's motion around the Sun. Therefore the Earth is on a conical trajectory behind the Sun, hence the Leap Second.
So what is going on? Because the Sun is 325,600 times more massive than the Earth-Moon it attracts them, the Earth-Moon will nearly catch the Sun as it goes Red [img]/phpBB/images/smiles/icon_smile.gif[/img] The Moon will nearly catch the Earth about that time [img]/phpBB/images/smiles/icon_smile.gif[/img]
The Earth and Moon are *actually* speeding up not slowing down.

Have you heard it all yet?


Gary

PS Kaptain K the Saros cycle is not the same as the recession of nodes.

Bzzzt! Wrong!

Here is a diagram of forces a mass far to the right exerts on a rod-shaped object:

...# --->
..# -->............................(MASS)
.# ->

The upper and lower arrows should be tilted toward the middle at the same angle, but the forces are not identical. The rod's end closer to the mass is attracted more strongly. Net force will bring the nearer end nearer still and the farther end farther until the rod is lined up with the mass. That's the crux of tidal torque, in a nutshell.

The rod is an extreme example -- and in fact elongated satellites, which orbit high enough to be completely free of atmospheric drag, become tidally locked (long axid always points toward Earth) in the matter of months. But given enough time any two objects orbiting each other which are not perfect spheres will eventually line up with thier long axes toward each other. Pluto and Charon are one known example of such pair having achieved stability.

How do you think Moon (and every other moon in the Solar system) became locked face on? Or for that matter Pluto and Charon are locked to each other?

<font size=-1>[ This Message was edited by: Ilya on 2002-03-24 18:04 ]</font>

<font size=-1>[ This Message was edited by: Ilya on 2002-03-24 18:05 ]</font>

I am deleting all the other stuff because I believe the crux of the problem is right here. If we can settle this, then we can move on to the other stuff, if you still feel like it.

First of all, a torque is not a force. It's a *torque*. R *cross* F. It has units of Newton-meters, not Newtons (or kg-(m/s)^2 rather than kg-m/s^2, if you prefer). Therefore, your second premise is inapplicable, and your syllogism fails.

If you still don't see this, please actually *try* to do the integral and see that it's non-zero. I don't see any point in arguing this when you haven't even done the calculation. To make it even simpler, forget about the integral. Consider a long rod (length = r) with a masses on each end of mass m. (The mass of the rod is negligible. There is a mass a distance R away (r << R), and the axis of the rod is tilted with an angle theta with respect to the line between the center of the rod and the distant mass.

By your argument, this system should be static. You say that "internal torques" will sum to zero. If there is no torque on the system, the rod will not move. Ever hear of the Cavendish experiment? Well, never mind the fact that the rod will move, and this movement has been measured (you can use it to measure the value of G - heck, I did that experiment myself in college), we can actually do the calculation. Since it's harder to write the notation in this format than it is to actually do the calculation, I will leave it to you, but you will find that the torque is non-zero. Simply write the two F vectors and the two r vectors, take the cross products and sum them. You *will* get a non-zero answer.

If you don't see how your heuristic argument is flawed, please do the math and see what the answer is. Once you see that there is a net non-zero torque, come back and we'll talk about the rest of it.

Don

[quote]
On 2002-03-24 18:03, Ilya wrote:
Ilya says

Bzzzt! Wrong!

Here is a diagram of forces a mass far to the right exerts on a rod-shaped object:

...# --->
..# -->............................(MASS)
.# ->

Thanks but, it sounds as if you are suspending your rod from the middle (center of mass) as DoctorDon and Cavendish did their dumbbells. If you take your car to have the tires balanced what do they do? They add "weight"s until the tire spins freely around its mechanical center. Now suppose the wheel is bent or there is a large chunk of tread missing. They can still balance the thing but it is going to go bump. OK, what did they do they suspended the wheel from a predetermined point then rotated it in a gravitational field.
The question is from where are the Earth and Moon suspended? I say the Earth and Moon, are suspended from their centers of weight or centers of gravity. If they were not, they would fly off in straight lines. All of your reasoning or book learning seems to suspend the Moon and Earth from their centers of angular momentum rather than centers of gravity.

...# --->
c of g ......
..# --> ''''''......(MASS c of g)
.# ->



Quote:
-----------------------------------------------------------------------
How do you think Moon (and every other moon in the Solar system) became locked face on? Or for that matter Pluto and Charon are locked to each other?
-----------------------------------------------------------------------

Perhaps the question should be, how do moons become moons?
Moons are satellites and satellites need 3 things.
1) Altitude.
2) Speed.
3) Tangency - which is the proper direction for the speed at the altitude.

How do moons or how did our Moon achieve tangency?

How do you draw a line tangent to a circle?

You touch the circle.

A moon must touch its parent, to become tangent, and to become locked face on.

Why moons stay locked on, is another question?
Does the Sun have a center of gravity? Or 8-9 or?
Could the Earth have two or more centers of gravity?
Could the Moon have two or more centers of gravity?
Ah ha, could multiple centers of gravity cause torque? Tee Hee.
Guess what, it takes two forces to cause Torque. It takes one force to hold the body, and one to turn it.

DoctorDon says
If you don't see how your heuristic argument is flawed, please do the math and see what the answer is.

The math is from 30 years ago for me. I have since found the premise on which it is based to be in error. I'm now into what might be called new math. Here is a multiple part story problem for you.

Given
The Moon is 3.911*10^8 meters from earth.
C=pi*d
The Moon goes around the Earth in (lets call it) 28 days.
1) How fast "v" does the Moon travel?
The Moon had to strike the Earth on a tangent pass to get into orbit.
2) Moving away from Earth against gravity, how fast "V" was it going when it hit?
Hint: V=v*e^2
The Moon's radius = 1 738 000 meters.
The Earth's radius = 6 366 270 meters.
3) If the Earth and Moon touched their center to center distance D would be what?
The Moons mass = 7.352*10^22 kg.
G = 6.6726*10^-11.
4) Find g If g = (Gm)/D^2
Using the above answers and the ballistics formula for altitude A = (V^2) / (2*g)
5) Find A
The answer "A" tells you where the Moon was at the end of its first month in orbit around the Earth.

So if you know where it was its first month, and you know where it is today you can figure out when our Earth mother received her baby.

I can help with this if you need help.
Hint, the recession of nodes is one revolution every 18.6 years.


Gary

Touch it? Are you saying that the moon had to have actually touched the Earth, or it could not have slowed down to where it was locked to where it kept one face to the Earth?

Um . . . the Earth and Moon are not suspended at all, they're in free-fall . . .

__________________
SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

I think I see part of the problem here. What you say here is true for spheres. If both the Moon and the Earth were spheres of uniform, rigid mass, then there would be no torque on either body, and there would be no exchange of angular momentum from the Earth to the Moon.

But the Earth and Moon are not uniform, not rigid, and are not spheres. They are close to spherical, but not quite there. They are deformed from their revolutions about their axes, and from the differential pull to the near and far sides. And with oceans and continents, etc., the Earth is obviously not of materials.

Hmmm. Maybe another approach is in order.

Gary, I hope you agree that we have tides. Tides are caused by the diferential pull of the moon (and the sun) on the near and far sides of the Earth. Do you agree with me up to here?

If yes, let's go on. The Earth is spinning ,and this moves the deformity (the tidal bulge) so that it is not in direct line with the Earth-Moon line of centers. (I mean an imaginary line connecting the two centers of mass.) The tidal bulge leads ahead of the moon. That is, when the high tide comes, the moon passes it's highest point overhead later (on the order of an hour, I think). Still with me? If not, where do you think my model is wrong?

Now, here's the kicker. The Moon pulls on the bulge's high points different than the low points. The forces do NOT sum to zero, and this non-zero force retards slightly the Earth's revolution. The angular momentum needs to be conserved. Since the Earth loses momentum, it has to go somewhere. Where it goes is to the Moon's orbit. It gains momentum, which means that it needs to be in a higher orbit.

Also, you still haven't answered the evidentiary question. I and others have shown evidence that the Moon's orbital radius is increasing, and the length of the Earth's period of revolution is increasing. You can't just hand-wave it away and say that it doesn't exist.

Please explain why the length of the day has increased by 2 hours since the Devonian.


_________________
Jeff Schwarz
__________________________________________________
Oh freddled gruntbuggly thy micturations are to me
As plurdled gabbleblotchits on a lurgid bee.

<font size=-1>[ This Message was edited by: Geo3gh on 2002-03-26 16:51 ]</font>

<font size=-1>[ This Message was edited by: Geo3gh on 2002-03-26 18:00 ]</font>

Actually, Jeff, you got it right except for one thing. Earth's rotation carries the tidal bulge ahead of the Moon... the bulge leads the Moon, which is why the torque slows the Earth's rotation (by pulling back on the tidal bulge), and speeds up the Moon (by tugging it forward along its orbit). (These descriptions are, of course, quick and dirty, but this is the basic idea.)

Details, details. [img]/phpBB/images/smiles/icon_smile.gif[/img]

Thanks for the correction. I'll go edit the post now.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

At last. I was starting to wonder if anyone ever listened to what the others were saying.
Thank you! Thank you! Thank you!

If you are interested the reason is simply this. A change in momentum is equal to the force applied times the time that it is applied. If we send a satellite to another planet it must have a rocket to force it into orbit, it needs that little extra push.
Moons don't have rockets so how can they achieve orbit/tangency I say they must touch the planets surface. The next thing is lunar locking. If you have a small rubber ball give it a spin as you drop it to the floor, 90% of the time it bounces back up at a slight angle, but with its rotation stopped or locked to the floor. So when moons touch their parent they stop any previous rotation and take on that of the parent.

Gary

Can you do the math I suggested? I hope so because it's more fun when you get the answer yourself.