# Why Is The Moon Not Spinning Then?

Discussion in 'Pseudoscience Archive' started by common_sense_seeker, Sep 6, 2008.

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The Standard Model reason of why the Moon is moving away from us is supposedly due to the Earth's rotation inducing the Moon with angular momentum. But if this were the case, surely you would expect the Moon to be rotating slightly, due to it's irregular orbit, if nothing else.

Cheers, AL

3. ### OphioliteValued Senior Member

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It is spinning. It rotates once a month, more or less, as it moves around the Earth.

5. ### Steve100O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔OValued Senior Member

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It does rotate slightly from our viewpoint.

It just goes back and forth.

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I still don't think that the relative amount of rotation of the Earth compared to that of the Moon is enough to induce the amount of angular momentum needed to increase it's orbit by 3cm a year.

I propose that it is simply the size of the gravitational field being produced from the center of the Earth which is reducing due to the effect of CONSERVATION OF ENERGY.

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A full explanation can be found in PSEUDOSCIENCE under 'Gravity Problem Solved'.

AL :soapbox:

9. ### OphioliteValued Senior Member

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Quite correct. That was why I put in the remark about 'more or less'. As a consequence, and I am sure you know this, over time we can see more than 50% of the moon's surface.

10. ### Janus58Valued Senior Member

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1,617
You think wrongly.

The Earth slows its rotation due to interaction with the Moon by some 1.5 milliseconds per century. It is this angular momentum lost by the Earth that is given to to the Moon.

The total kinetic energy of a rotating sphere can be found by

$E = \frac{\omega^2 M r^2}{5}$

Where w is the angular velocity in radian/sec.
M is its mass

Plugging in the values for the Earth for both its present rate of rotation and the amount it would have slowed in one year, and taking the difference, we get a value of:

4.34E+18 joules.

This is the amount of energy the Earth has to transfer to the Moon.

The total energy of the orbiting Moon is found by

$E= \frac{GMmMe}{2a}$

Where G is the gravitational constant
Me and Mm are the masses of the Earth and the Moon
a is the average orbital radius for the Moon.

Plugging in the correct values for the Moon's present orbit and one 3.8 cm further out and taking the difference we get a value of:

3.77E+18 joules

Which is the amount of energy it would take to raise the Moon's orbit by 3.8 cm and is less than the amount of energy the Earth has to transfer in one year.

The difference is due to the fact that some of the energy given up by the Earth is lost due to tidal heating.

So yes, the Earth does give up enough rotational energy to the Moon in order to increase its orbit by 3.8 cm per year.

Last edited: Sep 6, 2008
11. ### Janus58Valued Senior Member

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I think that it fair to point out that it rotates once per sidereal month. During the synodic month (the "month" most commonly referred to.) the Moon actual rotates 1.08 times.

12. ### orcotValued Senior Member

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3,410

the angular velocity of earth is: 0,0000727 rad/s
the earths mass is 5,97E27 gram
and has a radius of 6378100 meters

(0,0000727²*5,97E27*6378100)/5= 2,57E32 not 4,34 E18 joules
are you sure abouth that formula?

13. ### Janus58Valued Senior Member

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First, you need to use kg, not grams for the Earth's mass, which gives an answer of 2.57e29 joules for the KE of the Earth.

Second, the 4.34e18 joules is the KE the Earthloses by slowing its rotation by 15 microseconds in the course of one year. IOW, the KE difference between now and one year later when it will take 15 microseconds longer to complete a rotation.

14. ### OphioliteValued Senior Member

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I was trying to keep it simple. It also osscilates up and down as well as from side to side in a regular, but quite complex pattern. Additionally when you consider its path from the perspective of the sun things get even more complex. You are aware I am sure that the moon does not go round the Earth, but both rotate around their barycentre, which admittedly is inside the Earth, for the moment. I didn't see the need to introduce that for Common_Sense_Seeker/

15. ### XyleneValued Senior Member

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1,398
Isaac Asimov once worked out that the Moon and Earth would part from each other until they rotated around each other with a period of about 50 days--i.e. about seven revolutions per year. Of course, at the same time--because the same process of the loss of angular momentum is occurring between the Sun and the planets, the Earth is withdrawing slowly fromthe Sun, and its orbit is getting longer as well.

16. ### AlphaNumericFully ionizedModerator

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You mean like this http://en.wikipedia.org/wiki/Image:Lunar_libration_with_phase_Oct_2007.gif due to the motion of the Moon during it's orbit?

The reason the same side of the Moon always faces the Earth is due to tidal breaking, something completely explained by Newtonian gravity.

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Thanks for the link, I've never seen the lunar libration in video. It looked like just the kind of motion you would expect from my theory, being due to it's irregular orbit of an uber-condensed Earth's inner core. The explanation I give of a reduced gravitational field is an elegant and simple one. What is the transfer mechanism of your angular momentum? If it is by a particle, which I propose will be proven from the 10th Sept, then it would have to have a large mass, would it not?

Do gravitons have mass?

AL :shrug:

18. ### AlphaNumericFully ionizedModerator

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You don't have a working theory though. You have no quantitative predictions, no equations, no derivations, nothing but a load of BS and ignorance.
Newton's ego, voodoo and the Christian faith.

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I'm not claiming to be an expert. Perhaps you can summarise the models for me?

AL

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The natural decline of the Earth's gravitational field strength is a simple and elegant solution. It shouldn't be dismissed out of hand quite so easily.

AL

21. ### StrangerInAStrangeLaSubQuantum MechanicValued Senior Member

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If the moon rotates, we should see well over 90% of its surface.

22. ### James RJust this guy, you know?Staff Member

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Try this:

Get yourself 2 coins, a large one and a small one. The large one is the Earth, the small one is the Moon. Rest them flat on a table top.

Now, with the Earth held still, move the Moon around the Earth in a circle to simulate the Moon's orbit around the Earth, with the "Heads" side of the coin facing upwards so you can see it.

When the Moon is above the Earth, would a person looking from the Earth's surface towards the Moon see the top or bottom of the "head" on the Moon? What happens when the Moon has moved to the other side of the Earth?

Now, in fact you know that we always see the same side of the real Moon. Using the coins, how can you make sure that the "head" on the Moon coin always faces the same way, as seen from the Earth?

And now the big question: do you think the Moon rotates now, or not?

23. ### one_ravenGod is a Chinese WhisperValued Senior Member

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13,387
I understand what you are saying about relative motion, but my answer would be no.

If you are sitting right on the nose of the man in the moon and looking at earth, you would see the earth rotating, but it would never be out of your sight.
I would conclude the mmon does not rotate.

Likewise, imagine you have a ball tied to a rope at the end of a string.
The ball may appear to be rotating if the observer is orbiting you at the same speed and believes that you are orbiting around the ball - but the fact remains that the ball is orbiting around you.