# Thread: Why Is The Moon Not Spinning Then?

1. Ken, again you have shot yourself in the foot (Same as with the Foucault pendulum)!

Yes I watched YouTube video (link below)* of the Russian blond's hammer throw - the slow motion frames just before the ball hit the ground. The chain and handle grip are always stretched out from the ball in straight line by the centrifugal force of the spinning ball.

This straight line of chain does revolve around the spinning ball; even despite the air drag which would eventually stop the ball's spin with the chain trailing behind if the ball did not hit the Earth. The last slow motion part of the video begins with the chain out in front of the ball, but as the ball continues its spin the chain swings, first away for the recording camera, then continues to spin around to trail behind the ball and finally as it hits the ground, the chain is pointing towards the camera.

Thus the camera only records about 3/4 of one complete rotation or spin of the ball; however the slow motion sequence is only the tiny end fraction of the ball's total flight. I would estimate that during the ball's flight it made more than dozen complete 360 degree spins, but of course it too is spinning about some point only very near the center of the ball due to the relatively slight mass of the chain.

What is it you have against the very useful term "spin" to want to abolish it or at least replace it with "rotating about a barycenter" in two body cases?

Why do you not at least admit a few of your errors? For example, start by admit the sun has much more influnce on the moon than the Earth does (8.45 times more). After you have done that I will suggest the next error to admit.
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PS later by edit: That the only poster to agree with you is Dwayne, should also give you reason to think again. He has his own POV which is always quite disconnected from reality.

2. Originally Posted by Billy T
Ken, again you have shot yourself in the foot (Same as with the Foucault pendulum)!
Billy-bob, do you have any actual proof that a Foucault pendulum would work on the moon? If so, please do post it!

Originally Posted by Billy T
Yes I watched YouTube video (link below)* of the Russian blond's hammer throw - the slow motion frames just before the ball hit the ground. The chain and handle grip are always stretched out from the ball in straight line by the centrifugal force of the spinning ball.

This straight line of chain does revolve around the spinning ball; even despite the air drag which would eventually stop the ball's spin with the chain trailing behind if the ball did not hit the Earth. The last slow motion part of the video begins with the chain out in front of the ball, but as the ball continues its spin the chain swings, first away for the recording camera, then continues to spin around to trail behind the ball and finally as it hits the ground, the chain is pointing towards the camera.

Thus the camera only records about 3/4 of one complete rotation or spin of the ball; however the slow motion sequence is only the tiny end fraction of the ball's total flight. I would estimate that during the ball's flight it made more than dozen complete 360 degree spins, but of course it too is spinning about some point only very near the center of the ball due to the relatively slight mass of the chain.

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Since you're such an observant chap, did you notice that the Russian gal was RAPIDLY spinning her hammer in the counter-clockwise direction prior to release, but that very slow rotation of the ball you noticed was instead in the opposite CLOCKWISE direction?????

Billy-bob, putting aside the fact that ball was spinning at high rps prior to her release, and that the ball was clearly spinning much more slowly in the OPPOSITE direction AFTER the release (I say from cartwheeling prior to the release), can you explain why the ball shifted from a counter-clockwise rotation to a clockwise rotation prior to landing?

Originally Posted by Billy T
What is it you have against the very useful term "spin" to want to abolish it or at least replace it with "rotating about a barycenter" in two body cases?
Nothing wrong with the word "spin," so please re-post any comment of mine that gave you that idea????

As for "rotating about a barycenter," while many people do improperly use "rotate" when describing that, it's better usage to say "REVOLVING about a barycenter."

As in, "the earth *rotates* on its polar axis as it *revolves* around the sun."

If you prefer saying that "the earth *spins* on its polar axis as it *spins* around the sun," then that's OK by me too.

Originally Posted by Billy T
Why do you not at least admit a few of your errors? For example, start by admit the sun has much more influnce on the moon than the Earth does (8.45 times more). After you have done that I will suggest the next error to admit.
Define "more influence?" The moon's gravity affects the Earth's tidal bulges roughly twice as much as our sun's gravity affects our oceans' tidal bulges.

The Earth's gravity likewise is the stronger force that has, over the eons, caused the moon's two tidal bulges to solidify pointing towards us and not towards the sun.

---snip---
Mathematical Explanation of Tides
[ ]
Since R is small compared to D (about 1/60 of it) this is nearly equal to G * Mmoon * (2 * D * R) / D4, or (a constant) / D3. Tidal acceleration is therefore approximately proportional to the CUBE of the distance of the attracting body. If the Moon were only half as high above the Earth, the tidal accelerations would be EIGHT TIMES as great as they are now!

As it happens, both the Moon and the Sun cause such tides in our oceans. The Moon is about 400 times closer than the Sun, so it causes a tidal acceleration equal to 4003 or 64 million times that of an identical mass. But the Moon's mass is only about 1/27,000,000 that of the Sun. The result is that the Moon causes a tidal acceleration that is about 64,000,000/27,000,000 or 64/27, or a little more than double of that of the Sun. That's why the tides due to the Moon are larger than those due to the Sun.

At different times of a month, the Sun and Moon can cause tides that add to each other (called Spring Tides) (at a Full Moon or a New Moon, when their effects are lined up.) However, at First Quarter or Third Quarter Moon, the tides cause by the two sort of cancel each other out (with the Moon always winning!) and so there are lower tides then, called Neap Tides.

3. Is my post even here?

I said, a small change, like a moon having a moon, results in new variables so that we may not even exist.

4. Ken: Another way to show how silly your POV is follows:

For convenience assume the moon Earth are in “deep space” far from any other masses but all else is unchanged. I.e. the moon turns around once in exactly one orbit period, which is 28 days. Everyone but you and Dwayne describes this as “spinning” on its polar axis with period of 28 days, but you two say: “No that is not it is not spinning, it is tide-locked orbiting the barycenter without any spin.”

Now suppose a third body rapidly passed by, making a gravitational impulse (acting on the moon’s center of mass of course and applying no net torque* to the moon) which throws the moon into a new, much more elliptic, orbit about the barycenter with a period of 50 MONTHS. Now at apogee the moon is no longer keeping the same face turned towards the Earth, but turns 360 around MANY times so we see all side of the moon. (We already can see more than half of the moon as it is not now in a purely circular orbit.)

Do you at least agree it is “spinning” in its new orbit, even though it angular momentum about it polar axis is exactly what it was before the third body approached? If you do, why do you say it is NOT now spinning about its polar axis when the rotation about that axis (rate = 360degrees / 28 days) is exactly the same before as after the third body passed?

You are in the silly position of saying that the moon is spinning around it polar axis if the Moon day is 28.00001 or more Earth days long or if it is shorter than 27.999999… Earth days long; but if it happens to be rotating at exactly 28 days, then it is not spinning – it is orbiting about the barycenter!

Dwayne delights in holding silly positions, but you seem capable of more understanding. We all are trying to help you understand.
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*That is certainly possible, and the case I am assuming for simplicity, but may not be true in general.

5. Originally Posted by Reiku
Is my post even here?

I said, a small change, like a moon having a moon, results in new variables so that we may not even exist.
Hi - I don't think I saw any earlier post by you. However, I think your comment concerns how a moon(s) may influence a planet's ability to evolve life?

Earth's moon is very unusual in being so large relative to the Earth, and there are indeed theories that our moon's strong gravity was helpful in allowing life to form here.

E.g., the moon was likely very much closer to the Earth when life began to develop, thus creating massive tides and waves in our planet's primordial soup. While these early extreme conditions wouldn't be very friendly to later more advanced life forms – even so, these rough conditions would actively stir the soup, which may have been very beneficial in the beginning

As for a moon having its own moon, I won't say NO, but I don't recall that being observed in our solar system?

The problem I see, is that once these binary moons become tidally locked to each other (as Pluto and Charon are locked together), then their primary planet will eventually halt their revolution around their common barycenter, and eventually that would cause them to close together and eventually meld into one moon. Once that happened the evidence of a binary moon system would be lost.

The same thing is projected to happen to our Earth and our moon – as the theory goes, eventually the moon will tidally lock the Earth, and at that point they will both revolve around their barycenter, roughly once every 50 to 60 days – i.e., the Earth will have the same type of day the moon now has. Then, the moon would gradually close back in and meld into the Earth.

Of course, none of that will ever happen since the Earth and our moon will be consumed by our sun as it goes red-giant long before the moon tidally locks the Earth.

Ken
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6. Originally Posted by Ken Dine
Billy-bob, do you have any actual proof that a Foucault pendulum would work on the moon? If so, please do post it!
I do not think one has been observed on the moon for 28 days. I based my statement that it would appear to turn 360 once in 28 days to an observer sitting on the moon without moving on the fact that the Foucault pendulum keeps in the same plane (or one parallel to it) fixed in space. That is its "claim to fame." - Why Foucault invented it. Even you admit that the moon turns relative to planes fixed in space. You just want (for some silly reason - see my post 124) to claim this is not spinning on it axis IFF the spin rate is exactly the same as the orbit period.
Originally Posted by Ken Dine
Since you're such an observant chap, did you notice that the Russian gal was RAPIDLY spinning her hammer in the counter-clockwise direction prior to release, but that very slow rotation of the ball you noticed was instead in the opposite CLOCKWISE direction?????
No, not originally, did I notice that, but on second viewing I agree that is what the camera records.

However two different cameras were used. One close to the Russian lady and viewing her swing the hammer around from BELOW THE PLANE IT WAS SWINGING IN. The second camera was high in the stands with a telephoto lens and looking DOWN ON THE PLANE the ball and its chain were spinning in. That one camera recorded clockwise and the other counter clockwise is EXACTLY what they should if the spin was always in the same direction.

As Ophiolite advised you: When deep in an embarrassing hole stop digging!

You do not seem to know that clockwise and counter-clockwise are RELATIVE to the viewer terms, not absolute terms independent of the viewer's position. If you have a big clock, turn it face down on the floor and then you will note that your perfectly good clock now has the hands going around counter clockwise! One camera had a view from below and the other from above so of course one recorded the VERY SAME SPIN as clockwise and the other as counter clockwise. I am losing count, but think that is error six for you in this thread.

Originally Posted by Ken Dine
...can you explain why the ball shifted from a counter-clockwise rotation to a clockwise rotation prior to landing?
I just did. - You are just adding to your list of silly statements, all based on appearances, without any understand of the facts.

You seem like a nice guy. Please stop shooting yourself in the foot and stop digging deeper in your embarrassing hole. That must hurt with both feet bleeding now.

You do seem to be trying to come to the correct POV, which I explained earlier (in post 96) to you that it is the gradient of gravity which falls off as the cube of the separation, which causes the moon to produce stronger tides than the sun despite the quadratic fall off of gravity, which causes the sun to control the motion of the moon thru space except for the very tiny (1/4 of one percent radial wobble).

Here is part of post 96 again:
Originally Posted by Billy T
...No you are wrong here too. The tides are caused by the GRADIENT of gravity not by gravity itself. The gradient falls off as the CUBE of the distance, not the square. This is why the moon is much more important than the sun in producing tides on Earth.
Just do the calculations with mass of sun at 1 AU from moon and mass of Earth at one moon /Earth separation. When calculating gravity force on moon the denominator distance is squared and the sun is much more important.
When calculating the gradients the inverse CUBE makes the moon's gradient larger at the Earth than the sun's is. So the moon does dominate the tides. Do it and then you will understand why moon dominates the tides on Earth but the sun, not the Earth, controls the moon's orbit around the sun. Earth only causes an "hardly noticicable" percentage variation or "wobble" in the distance between the moon and the sun.
You do not understand what causes the tides. I.e. that it is not gravity but the gradient of gravity.
Thus it would be interesting to hear you try to explain why there are TWO tides each day, (one on each side of the Earth), instead of just one bulge on the side of Earth towards the moon. The correct answer has is due to fact the moon's gravity is stronger on the side near to the moon than on the side most distant from the moon - I.e. depends upon the GRADIENT of the moon's gravity, not on the gravity itself....

7. Well Billy T , are you trying to say that the moon does not orbit the earth, but instead orbits the Sun in a seperate orbit from earths prime orbit around the sun.

If the moon has a spin rotation as appearantly suggested in discussion, then the moon must gain the momentum for that spin accross 638 miles of it trajectory before a cancelation of that angular trajectory which occurs every 15 minutes. So in short terms the moon acutally spins for 15 minutes before it cancels that spin motion and begins another 15 minutes of spin in the opposite direction.

DwayneD.L.Rabon

8. "Why is the moon not spinning then?"

The moon is not spinning (we don't see both sides) because it turns once on its axis for each orbit, so we see the same side, because its axis of rotation is aligned with ours, more or less .

This thread has spun a tale or two but it started to spin out with what apparent and actual spinning are. Or someone thinks they've found a hole in Newtonian mechanics again, but spin is a distinct kind of motion - we all know what it is and we can all see the moon doing it.

The question should be: "If the moon isn't spinning, why don't we see both sides?"

P.S.
Oh crap, back in the "the moon orbits around the sun, not the earth" ballpark.
The earth-moon system is coupled to the sun's gravitational influence, primarily. The moon is primarily coupled to the earth's. Any satellites between them can be coupled to either, primarily, or in a lagrange point, so coupled equally to both (earth and moon). We have observational craft orbiting earth, the moon, the sun, I'm sure there are one or two at these lagrange points.

9. Originally Posted by DwayneD.L.Rabon
Well Billy T , are you trying to say that the moon does not orbit the earth, but instead orbits the Sun in a seperate orbit from earths prime orbit around the sun.

If the moon has a spin rotation as appearantly suggested in discussion, then the moon must gain the momentum for that spin accross 638 miles of it trajectory before a cancelation of that angular trajectory which occurs every 15 minutes. So in short terms the moon acutally spins for 15 minutes before it cancels that spin motion and begins another 15 minutes of spin in the opposite direction.

DwayneD.L.Rabon
Hard to follow you, but yes Every body orbiting the sun not in a perfectly circular orbit does have changing (linear) momentum (More near the sun than at apogee) but its angular momentum is constant.

10. The maxium stable orbit for moon orbit of the earth is about 331,559 miles from earths center after this point its orbit will become erratic, where it would become oblong in orbit like a comet, as if it where to strike earth or be lost to space.

The limit of this orbit is determine by the sun. the exstent of the current orbit is due to interaction with the sun, such as the rate of motion at which the moon gets farther from the, estimated to be 3 cm per year.

In mention of the moons trajectory changes relavant to spin, the moon travels to the gravitional inductance of the sun in its motion around the earth of which it meets the center point every 15 minutes (at times tis may be as short as 7 minutes) in that time the moon has traveled 638 miles (20 miles on earth surface or 0.3 degrees) when the moon meets this inductance point the inductance force is is canceled and the moon gains next motion of inductance of the sun which is traveling in opposite motion causing the moon to turn backward (simular to sine frequency).
Variations in the time of the moon to travel to the suns inductance point cause accumulation of motion in one direction or the other to be greater at different times, therorically if enough momentum is gained in one direction the moon could actually make a complete turn.

The overall motion is simular to walking. As well this same motion may the resulting cause of the number of subtle earthquakes that occur every second or minute.

DwayneD.L.Rabon

11. Originally Posted by Ken Dine

#2 - your 4th graphic, if viewed from a planet at the center of its orbit, would be tumbling around two axes. It thus would NOT be tidally locked ...UNLESS, there were other moons close by that were causing it to tumble and keeping it from becoming tidally locked.

How do your graphics relate to the issue of whether (or not) our own moon is still *rotating* around its polar axis?
Quite simply, The only difference between the graphics are the axial tilt of the satellite. The sidereal rotation and period of revolution is the same. In the first graphic (Which BTW is the fourth graphic from a different viewpoint), you would say that the satellite tumbles around two axes, but in the 3rd, which closely resembles the orbit of the Moon, you would say that it doesn't rotate at all. Yet you have still failed to answer as where exactly between 90° and 0° axial tilt does the satellite stop all rotation, while to someone outside the system the rotation of the satellite doesn't change, just the angle at which it rotates.

Also, let's go back the the situation where you say the satellite is tumbling around two axes. This situation is very like that of Uranus. It too, essentially lays on its side as it orbits the Sun with its polar axis stationary with respect to the fixed stars. If you were to watch Uranus from the Sun, its apparent motion would be like that of graphic 4, with the exception that it would rotate around its polar axis more quickly.

A rotating body, however, does not naturally rotate around two axes, unless there is precession. Precession is caused by a torque applied to the axis of a rotating body. This torque acts at a right angle to the precession.
In this case it would have to act at a right angle to the orbital plane. However, there is nothing that can cause such a torque, and without a constant torque, there can be no precession.

Ergo, no matter what it looks like from the Sun, Uranus doesn't tumble around two axes (and neither does the satellite in the 4th graphic.) Instead, its polar axis stays fixed in one direction while it rotates around it.

And what holds for an axis of a planet of satellite holds to a face of a satellite. If a satellite keeps one face stationary with respect to the stars, it does not rotate, even though it presents all sides to the planet as it orbits.

The upshot is that, despite your earlier protest, there is a preferred frame for rotation. This is easily demonstrated. Take a bicycle wheel that is not rotating with respect to the fixed stars. You can grab it by the axle and easily turn it in any direction. Now spin it on its axle, and try to twist it, it will fight you.(and will precess as discussed above). So even if the fixed stars are not the preferred frame for rotation, they are very close to it. And thus, rotation is properly judged with respect to the fixed stars in all instances.

If our moon is still rotating around its internal polar axis one (1) time per complete orbit as some here claim, then please do name the force that has (for at least 3 billion years) kept our moon rotating around its polar axis against the torque of tidally locking?
You are arguing a tautology. Essentially, you are saying that since the moon doesn't rotate as it orbits it keeps the same face towards the Earth, and since it keeps the same face towards the Earth, it doesn't rotate.

The reverse argument is that because the Moon rotates once per orbit it keeps one face always towards the Earth.

Tidal locking locks the Moon's period of rotation to 1 rotation to one orbit. Though it doesn't keep the rotational speed matched to the orbital speed. The radians per sec swept out by the Moon's orbits speeds up and slows down from perigee to apogee, the radian per sec due to rotation however stays constant. This is what causes the libration of longitude, at different parts of the orbit the rotation speeds ahead of or lags behind the orbit. This indicates independence of rotation from orbit.

Other than the Earth and moon's mutual pull of gravity on each other (and to a lessor extent of the sun & Jupiter, etc), the only remaining force I see that's affecting the moon's motion would be the moon's own orbital motion around the Earth-moon barycenter, and the force of that orbital motion is not constant since the moon's orbital motion is speeding up as energy is transferred from the Earth through tidal braking.

Pretty graphics, though.

Ken
The Moon's orbital motion is slowing down not speeding up. Higher orbits are slower orbits and the Moon is receding from the Earth. It is gaining energy, it is just that the majority of that energy is gained as gravitational potential.

13. That's why it doesn't rotate the way other moons do.

14. Originally Posted by Janus58
The Moon's orbital motion is slowing down not speeding up. Higher orbits are slower orbits and the Moon is receding from the Earth. It is gaining energy, it is just that the majority of that energy is gained as gravitational potential.
$\omega r = v$

Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.

Most likely the velocity is fixed and radius is getting higher and $\omega$ is decreasing.

I don't know though...but I do know what you said makes no real sense.

15. If the Moon had a moon, would that be spinning?

16. Yes.

17. Originally Posted by CheskiChips
$\omega r = v$

Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.

Most likely the velocity is fixed and radius is getting higher and $\omega$ is decreasing.

I don't know though...but I do know what you said makes no real sense.
Orbital speed does decrease and the period of the orbit does get longer, yes.

The orbital speed of any satellite can be found by:

$V_o = \sqrt{\frac{GM}{r}}$

where r is the radius of the orbit.
G the gravitational constant
M the mass of the Planet

As r increases, Vo decreases.

The period of the orbit can be found by

$T = 2\pi sqrt{\frac{r^3}{GM}}$

Again, note how the period increases as r increases.

Lastly, the energy of a satellite can be found by the sum of its kinetic energy and potential energy or:

$E= \frac{mv^2}{2}- \frac{GMm }{r}$

OR

$E = -\frac{GMm}{2a}$

Where m is the mass of the satellite and a is the mean orbital distance.

18. Originally Posted by CheskiChips
$\omega r = v$

Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.

Most likely the velocity is fixed and radius is getting higher and $\omega$ is decreasing.

I don't know though...but I do know what you said makes no real sense.

I agree. Janus58 just quoting the same equation isn't a sufficient explanation.

19. Are you challenging the correctness of the equations? If so, why? On what grounds?
If not, what could possibly be more sufficient than equation, which explains exactly what is happening?

20. Originally Posted by CheskiChips
$\omega r = v$
Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.
Most likely the velocity is fixed and radius is getting higher and $\omega$ is decreasing.
I don't know though...but I do know what you said makes no real sense.
Your “most likely” gave me a good laugh. Thanks. To give you one, I will say: “Most likely, the sun will rise again tomorrow.”
Neither is a probability thing. To start where surely you can follow:

F = ma
And for circular orbiting mass, m, which is r from mass M the F is GMm/r^2.
As m appears on both sides of the equation I will drop it and chose set of units where GM = 1 so I have:

a = 1/r^2 but the acceleration a is either v^2/r or r$\omega$^2 and the second is more convient now so:

a = 1/r^2 = r$\omega$^2 and now not bothering with the left most "a part" of the equation anymore and multiplying both sides of the rest by r^2 we have:

1 = (r^3)$\omega$^2 which you may not recognize is one of Keppler’s three laws. (The third , I think by memory.) Namely that the period, P, squared is proportional to the cube of the radius r.
It is actually more general than just this circular orbit case; because the period in an elliptical orbit depends only on the longest straight line segment with end points on the elliptical orbit, which I will call “L”.
(Normally this length is called the “major axis,” but as both “m” and “a” were already used, I was desperate for different letter, so I called that length “L”)

So the more general statement of that law for elliptical orbits is:
P^2 ~ L^3 or if willing to use the strange units in which GM = 1, then P^2 = L^3.
I like these units as who the hell remembers the mass of the planets, the sun etc. or G? (Except expert workers in the field, like Janus58.) But I digress. Back to:

1 = r^3$\omega$^2 = r (v^2) which tells you, for example, that if r doubles, then the tangential velocity goes down to the square root of what it was. But when r doubles, so does the circumference and even it the tangential velocity did not decrease, it would take longer to get all the way around the orbit. I.e. the orbital period, P, is increasing or $\omega$ is decreasing. {Jeeezzz! that Janus 58 is a smart guy - he just knew this without all this work. hehehe.}

Switching to the general elliptical case (which includes circles) When L doubles, then L^3 goes up 8 times and the square of the period must too. Thus the period more than doubles. – it becomes 2(square root 2) times greater, (“most likely”) just as “most likely” the sun will rise again tomorrow, but who really knows?

Originally Posted by common_sense_seeker
I agree. Janus58 just quoting the same equation isn't a sufficient explanation.
Was the F = ma start simple enough for you?

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