The Moon moves away from the earth - Is the theory correct ?

In summary: Moon's gravity?In summary, the moon moves away from the Earth 3.8 cm per year. This is because the Earth's gravity falls off with distance, and the Moon's gravity is differentially stronger in different parts of the Earth. The friction between the Earth and the tidal waves is what's causing this.
  • #1
Bjarne
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The moon moves away from the Earth 3,8 cm per year.
In the past the velocity has been gradually increased.

I have read at the internet, that when the theory explaining this cause of this phenomena is correct the moon would for about 85 million years been orbiting 4 meters above the earth.

My own calculation shows this would have happen for about 1, 2 billion years ago (if this theory is correct).

We know that the moon is more than 4 billion years old, so how is it possible to keep believing such dictionary theory?

(sorry if this is not perfect English)
 
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  • #2
I doubt the recession speed was vastly higher in the past, but anyway, if we use 3.8 cm/yr and the current distance of 385,000 km, we get 10 billion years. So it would have had to average more than double that recession velocity to have started off very near the earth.
 
  • #3
I believe (according to the prevailing theory) the moon will not forever move away from us with 3,8 cm per year, but rather slower and slower.

Therefore I also think (that according to the prevailing theory) that the moon in the past was moving away from Earth faster than now, and properly proportional with >>> g = G x M1xM2 /r^2

But I do not fully understand the prevailing theory, and would appreciate to have some simple explanation to way this happen?
 
  • #4
Okay, here goes. Because gravity falls off with distance, there is a differential in the Moon's gravity across the Earth. This differential is call a tidal force. The tidal force raises bulges in the ocean called tidal bulges. If nothing else interfered, these tidal bulges would align with the Moon.

The Earth, however, rotates. As it does so, friction between the Earth and the tidal bulges tries to drag the bulges along with the Earth. As a result, the tidal bulges lead the Moon a little. The moon tries to pull back on the bulges, but this alos means the bulges pull forward on the Moon. This transfers angular momentum from the Earth to the Moon. The Moon tries to speed up in its orbit. But doing so causes it to climb into a higher orbit and the Moon recedes from the Earth.

As far as the recession being faster when the Moon was younger, its not that simple. There are a lot of factors besides the difference in gravitational attraction. The friction between the Earth and the tidal bulges has a huge effect. Reduce the friction and the bulges lead the Moon by less and thus pull forward on the Moon less, causing a lower recession rate.
The continents play a large role in determining this friction. Because of plate tectonics, the continents weren't always in the configuration they are now. In fact, in the past they were clustered together in one landmass centered on the pole. In this configuration, they offered little resistance to the tidal bulges and the friction between Earth and the bulges is greatly reduced causing a much smaller recession rate then otherwise.
 
  • #5
Thank you
Nice and simple
 
  • #6
This has often troubled me! Maybe somebody can enlighten me?

From the Moon's POV, we see the Earth stationary in the sky. So, from the Moon's POV, why doesn't gravitational attraction happen?? Shouldn't the Earth approach the Moon?

If we were living on the Moon and studying the receding Earth, wouldn't we come to a conclusion that involved spinning objects creating a repulsive force or something?

I'm confused by this. Talking about "tidal forces" and stuff confuses me too, because what "force" are we exactly talking about here? Gravity? Electromagnetic? Strong nuclear? Weak nuclear? Argh!

Can somebody please explain lucidly why the Earth, from the vantage point of the Moon, is receding? Does the "background universe" somehow have an effect on this?

EDIT: AH forget it, I just re-read Janus's lucid explanation and it cleared things up in a way I haven't experienced before. Thank you Janus!
 
  • #7
Cryptonic have a point here
The Moon is forcing the tidal waves forward on its way over the oceans, thereby avoiding the gravitational effect from this mass until the waves hit land and the Moon is moving over the waves.

If e.g. the water of the Atlantic Ocean was pressed above Euroasia, the gravitational force between Earth and the Moon would merely result in a negative anomaly above the Atlantic Ocean and a positive anomaly above Euroasia. The variation of the mass attraction between Earth and the Moon would probably neutralize each other. If so, we can forget about any additional gravitational effect caused by a "returning tide wave"

Neither friction between the oceans’ bodies of water and the sea bottom can explain that friction between 2 internal bodies on Earth can have such an effect on the Moon that its acceleration increases, simply because it has not been explained how a dynamic rotational force could be transferred to the Moon through space.

So what does really cause this phenomenon?
How is "mechanic" forces "transferred through space / to the moon?"
Friction gives normally only heat and can not adapt to the mass attraction connection?
It seem to be a "missing link” here ?
 
  • #8
Bjarne said:
Neither friction between the oceans’ bodies of water and the sea bottom can explain that friction between 2 internal bodies on Earth can have such an effect on the Moon that its acceleration increases, simply because it has not been explained how a dynamic rotational force could be transferred to the Moon through space.

So what does really cause this phenomenon?
How is "mechanic" forces "transferred through space / to the moon?"
Friction gives normally only heat and can not adapt to the mass attraction connection?
It seem to be a "missing link” here ?

No, I think the mechanism as explained by Janus is pretty well understood. Perhaps this text from http://en.wikipedia.org/wiki/Orbit_of_the_Moon makes it easier to understand (bold print by me, not in original):

The tidal bulges on Earth are carried ahead of the Earth–Moon axis by a small amount as a result of the Earth's rotation. This is a direct consequence of friction and the dissipation of energy as water moves over the ocean bottom and into or out of bays and estuaries. Each bulge exerts a small amount of gravitational attraction on the Moon, with the bulge closest to the Moon pulling in a direction slightly forward along the Moon's orbit, because the Earth's rotation has carried the bulge forward. The opposing bulge has the opposite effect, but the closer bulge dominates due to its comparative closer distance to the Moon. As a result, some of the Earth's rotational momentum is gradually being transferred to the Moon's orbital momentum, and this causes the Moon to slowly recede from Earth at the rate of approximately 38 millimetres per year.
 
  • #9
according to wikipedia tidal forces are proportional to 1/d^3
so if tidal forces doubled then how much higher would the tides be?

also, assuming that the bulges lag by the same amount, what does this all translate into in terms of net acceleration on the moon? (the tidal bulge on one side of the Earth partially cancels the effect of the bulge on the other side)

and lastly, would the bulges lag by the same amount if friction remained the same?
 
  • #10
so the 2 bulges together form a sort of dipole. so I would expect that the acceleration of the moon, if the mass of the bulges remains the same, would be 1/d^3
 
  • #11
grandpa correctly notes that tidal forces are proportional to the cube of distance. The other conclusions are incorrect.
 
  • #12
ok. apparently that was wrong.

let the moon be at the origin. let the center of the Earth be at (d,0). let bulge one be at (d-r,-x). let bulge two be at (d+r,x). let d>>r>>x. let x be constant. let L1=distance from moon to bulge one. let L2=distance from moon to bulge two.

to get a unit vector pointing in the direction of bulge one we simply divide each component of the coordinates of bulge one by L1. likewise for bulge two.

net force from bulge one on the moon is 1/L1^2. multiplying by the unit vector and taking only the x component we get x/L1^3. for bulge two we get x/L2^3. the sum of these 2 numbers (one of the x's is negative) is the net acceleration of the moon in the direction of its orbit.

this is proportional to L1^3-L2^3/(L1^3*L2^3)
 
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  • #13
L1^3-L2^3/(L1^3*L2^3)

to solve this we need to approximate.. (or at least I do)
let L1=d+a (where d>>a)
let L2=d-a

the numerator reduces to exactly 6d^2a+2a^3 (the d^3 terms cancel out)
the denominator I didnt work out exactly but its obvious that its largest term is d^6.

since d>>a the a^3 term can be ignored. so I get net force/acceleration (for a given set of bulges) is proportional to 1/d^4tidal force is proportional to 1/d^3. assuming that twice the tidal force will result in tides twice as high (and twice as massive) and that the bulges continue to lag by the same amount then the net acceleration of the moon is 1/d^7.

then it becomes a matter of orbital mechanics. its obvious that if we double the acceleration of the moon in its present orbit that we would double the rate at which it is moving away. but what if it were in a different orbit and we kept the acceleration the same?
 
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  • #14
the angular momentum of the moon is mvd.

torque is force times d. it produces change in angular momentum just as force produces change in momentum.

centrifugal force is mv^2 / d

gravity equals 1/d^2

the cetrifugal force must equal the gravitational force
v^2/d=1/d^2
v=1/√d
therefore the velocity of the moon at distance d from Earth is 1/√d
therefore the angular momentum of the moon at distance d is √d
therefore the derivative of √d gives the amount of potential angular momentum stored at each distance d.
d/dd*√d=1/(2*√d)

the tidal force acting on the moon (to accelerate it) at distance d is 1/d^7
therefore the torque (rate of change of angular momentum) acting on the moon at distance d is 1/d^6

the infinitesimal time spent at each infinitesimal distance is potential angular momentum per distance divided by torque.
this equals [1/(2*√d)]/[ 1/d^6].
which equals d^6/(2*√d)≡d^5.5
therefore t(d)=d^6.5

d(t)=t^(1/6.5)

continued in post 18
 
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  • #15


The topic referred is called tidal acceleration or tidal friction.

Wikipedia said:
tidal acceleration would continue until the rotational period of the Earth matched the orbital period of the Moon. At that time, the Moon would always be overhead of a single fixed place on Earth. Such a situation already exists in the Pluto-Charon system. However, the slowdown of the Earth's rotation is not occurring fast enough for the rotation to lengthen to a month before other effects make this irrelevant: About 2.1 billion years from now, the continual increase of the Sun's radiation will cause the Earth's oceans to boil away, removing the bulk of the tidal friction and acceleration.

I believe the key here is that tidal acceleration would continue until the rotational period of the Earth matched the orbital period of the Moon.


Reference:
http://en.wikipedia.org/wiki/Tidal_friction" [Broken]
 
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  • #16
Agree with orion. It's called tidal locking - see Mercury. The earth-moon relationship is not nearly ancient enough to achieve this state given their relative masses.
 
  • #17
I understand that the "level difference" of the tidal bulge gives larger mass attraction and therefore also larger accelration.
But I do not understand how friction on Earth can be "connected" with a mass attraction connection. (All the calculations of granpa I do not understand)

How can friction between the oceans’ bodies of water and the sea bottom - explain that friction between 2 internal bodies on Earth can have such an effect on the Moon that its acceleration increases, - it has not been explained how a dynamic rotational force could be transferred to the Moon through space.
Can someone explain that, in simple words.
 
  • #18
continued from posts 14, 13, and 12

I have finished my calculations. I just needed to get some sleep and get a fresh perspective on it.
the infinitesimal time spent at each infinitesimal distance is change in angular momentum per unit (distance) divided by torque (rate of change of angular momentum).
dt=[1/(2*√d)]/[ 1/d^6]*dd
therefore dt≡d^5.5*dd. let's define an arbitrary unit of distance equal to 38 mm. just call them units. the moon is 10 billion units from Earth today. the recession rate of the moon today is 1 unit/year so if dd=one unit then dt=one year.

given dt=(10^-55)*d^5.5*dd (measured in units and years)
then t(d)=(1.53846*10^-56)d^6.5

t(10,000,000,000)=153,846,000 which is totally wrong.
 
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  • #19
ok that doesn't seem to work either. unless I made a bonehead mistake somewhere I can only assume that my assumptions must be wrong. there were only 2. one was that the tides lagged by the same amount and the other was that the height of the tides was proportional to the tidal force. tides certainly didnt lag LESS in the past so maybe the height of the tides isn't proportional to tidal force.

if I assume that the height of the tides is proportional to the square root of the tidal force then the result is that it would take the moon 2 billion years to reach its current position. still not 4.5 billion but much better than 150 million. but I see no reason why the tides would follow such a law.

but that doesn't seem to be the case:

http://en.wikipedia.org/wiki/Tide#Forces
the tidal force depends not on the strength of the gravitational field of the Moon, but on its gradient (which falls off approximately as the inverse cube of the distance to the originating gravitational body; see NASA).[18] The gravitational force exerted on the Earth by the Sun is on average 179 times stronger than that exerted on the Earth by the Moon, but because the Sun is on average 389 times farther from the Earth, the gradient of its field is weaker. The tidal force produced by the Sun is therefore only 46% as large as that produced by the Moon. (According to NASA the tidal force of the Moon is 2.21 times larger than that of the Sun.

The theoretical amplitude of oceanic tides caused by the Moon is about 54 cm at the highest point,which corresponds to the amplitude that would be reached if the ocean possessed a uniform depth, there were no landmasses, and the Earth were not rotating. The Sun similarly causes tides, of which the THEORETICAL amplitude is about 25 cm (46% of that of the Moon)it may simply be that modelling the tides as 2 point masses at the poles of a sphere just isn't good enough. I'm trying to add a ring of negative mass at the equator but the math isn't working out.

http://www.physics.buffalo.edu/~sen/documents/field_by_charged_ring.pdf [Broken]
 
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  • #20
tidal locking...


The system that is being described is called tidal locking, every planet orbiting every star and also with at least one moon in the entire Universe experiences this natural phenomena. The conceptual mathematical model is shown in reference 3.

The primary star Sol and all the planets in the entire solar star system cause a tidal acceleration on Terra, the tidal pull from the Lunar moon is the strongest, followed by the Sun which has half the effect of the Lunar moon, it is very much larger than the Lunar moon but farther away, and tidal acceleration falls off strongly with distance, because the tidal acceleration to first order falls off as [tex]\frac{1}{r^3}[/tex]. The rest of the planets' effects are infinitesimal in comparison.

Tidal acceleration at point A:
[tex]a_T = Gm \left( \frac{1}{r_L^2} - \frac{1}{(r_L + R)^2} \right)[/tex]

Tidal acceleration at point B:
[tex]a_T = Gm \left( \frac{1}{(r_L - R)^2} - \frac{1}{r_L^2} \right)[/tex]

The time for a body to become tidally locked:
[tex]t_{l} \approx \frac{w a^6 I Q}{3 G m_p^2 k_2 R^5}[/tex]

Reference:
http://en.wikipedia.org/wiki/Tidal_friction" [Broken]
http://en.wikipedia.org/wiki/Tidal_locking" [Broken]
http://staff.washington.edu/aganse/europa/tides/tides.html" [Broken]
http://en.wikipedia.org/wiki/Earth" [Broken]
http://en.wikipedia.org/wiki/Moon" [Broken]
 
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  • #21
http://en.wikipedia.org/wiki/Tidal_friction

A tidal bulge (called an equilibrium tide) does not really exist on Earth because the continents break up the tide when they pass under the Moon. Oceanic tides actually rotate around each ocean basin as vast gyres around several amphidromic points where no tide exists. The Moon pulls on each individual undulation as Earth rotates—some undulations are ahead of the Moon, others are behind it, while still others are on either side. The equilibrium tide in the shape of a prolate spheroid that actually does exist for the Moon to pull on is the net result of integrating the actual undulations over all the world's oceans. Earth's net equilibrium tide has an amplitude of only 3.23 cm, which is totally swamped by oceanic tides that can exceed one metre.

There is geological and paleontological evidence that the Earth rotated faster and that the Moon was closer to the Earth in the remote past. Tidal rhythmites are alternating layers of sand and silt laid down offshore from estuaries having great tidal flows. Daily, monthly and seasonal cycles can be found in the deposits. This geological record is consistent with these conditions 620 million years ago: the day was 21.9±0.4 hours, and there were 13.1±0.1 synodic months/year and 400±7 solar days/year. The length of the year has remained virtually unchanged during this period because no evidence exists that the constant of gravitation has changed. The average recession rate of the Moon between then and now has been 2.17±0.31 cm/year, which is about half the present rate.[2]

then what was the recession rate 620 million years ago?
 
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  • #22
if I assume that the height of the tides is proportional to the square root of the tidal force then the result is that it would take the moon, by my calculations, 2 billion years to reach its current position. still not 4.5 billion but much better than the 150 million that I get when I assume that the height of the tides is proportional to tidal force. but I see no reason why the tides would follow such a law.

its easy to see that on a perfectly spherical Earth that if gravity is 10% weaker at one point on its surface then the ocean will have to be 10% deeper at that point to produce the same amount of pressure at the bottom. it doesn't have to be 10% deeper but if it isn't then it won't be stable.
any ideas on why the height of the tides would be proportional to the square root of the tidal force? the simplest thing to try would be to add viscosity to the water. if that doesn't work then presumably it would hove to be because of the effect of the continents.

A tidal bulge (called an equilibrium tide) does not really exist on Earth because the continents break up the tide when they pass under the Moon. Oceanic tides actually rotate around each ocean basin as vast gyres around several amphidromic points where no tide exists. The Moon pulls on each individual undulation as Earth rotates—some undulations are ahead of the Moon, others are behind it, while still others are on either side. The equilibrium tide in the shape of a prolate spheroid that actually does exist for the Moon to pull on is the net result of integrating the actual undulations over all the world's oceans. Earth's net equilibrium tide has an amplitude of only 3.23 cm, which is totally swamped by oceanic tides that can exceed one metre.The theoretical amplitude of oceanic tides caused by the Moon is about 54 cm at the highest point

http://en.wikipedia.org/wiki/Tidal_friction

it would seem that ocean tides are proportional to tidal force but the equilibrium tide is proportional to the square root of the tidal force.

do the tides form some kind of standing wave?
 
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  • #23
apparently tides are much more complicated than they seem.

http://en.wikipedia.org/wiki/Tide#Phase_and_amplitude

they are called 'kelvin waves'.

http://en.wikipedia.org/wiki/Kelvin_wave

also one thing I read seemed to imply that their speed of propagation is less than (about half) the speed at which the moon moves overhead (approximately the speed of rotation of the earth).

http://www.ocean.washington.edu/people/faculty/susanh/402/lecture17.pdf [Broken]

they even form kelvin waves in the hudson bay (page 20):I think that makes it clear why Earth's net equilibrium tide has an amplitude of only 3.23 cm,
 
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  • #24
http://www.coas.oregonstate.edu/research/po/research/tide/index.html [Broken]

check out Hudson bay. its unreal. how can it do that?

but the Mediterranean doesnt.this is obviously some kind of standing wave modified by Coriolis force.
 
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  • #25
Newton said that if we assume the moon to be avoiding gravity then it must travel away from the surface of Earth at very specific velocities.I was wondering what speed the moon moves at now,and what the speed of it must be to avoid the Earth,when it is four meters above the ground,and are they different?
 
  • #26
I wonder if the frequency could be a factor?

the frequency of the tides is proportional to 1/d√d
d=distance to the moon.

for an inductor:
When there is a sinusoidal alternating current (AC) through an inductor, a sinusoidal voltage is induced. The amplitude of the voltage is proportional to the product of the amplitude (IP) of the current and the frequency ( f ) of the current.

inductance is the electrical equivalent of mass.

so the height of the tides (current)=tidal force (voltage) divided by frequency

height of tides=d√d/d^3=d√d

which works perfectly. the moon would take 2 billion years to reach it current position.
 
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  • #27
I think I finally got it (famous last words). it was just a matter of factoring in the effect of frequency.

what is the acceleration of the moon in the direction of its orbit given tidal bulges of mass=1 that lag behind the moon by distance x?
let the moon be at the origin. let the center of the Earth be at (d,0). let bulge one be at (d-r,-x). let bulge two be at (d+r,x). let d>>r>>x. let x be constant. let L1=distance from moon to bulge one. let L2=distance from moon to bulge two.

to get a unit vector pointing in the direction of bulge one we simply divide each component of the coordinates of bulge one by L1. likewise for bulge two.

net force from bulge one on the moon is 1/L1^2. multiplying by the unit vector and taking only the x component we get x/L1^3. for bulge two we get x/L2^3. the sum of these 2 numbers (one of the x's is negative) is the net acceleration of the moon in the direction of its orbit.

this is proportional to L1^3-L2^3/(L1^3*L2^3)

to solve this we need to approximate.. (or at least I do)
let L1=d+a (where d>>a)
let L2=d-a

the numerator reduces to exactly 6d^2a+2a^3 (the d^3 terms cancel out)
the denominator I didnt work out exactly but its obvious that its largest term is d^6.

since d>>a the a^3 term can be ignored. so I get net force/acceleration (for a given set of bulges) is proportional to 1/d^4

but what is the size of the bulges as a function of the moon distance from the earth?
tidal force is proportional to 1/d^3.

the frequency of the tides is proportional to 1/d√d
d=distance to the moon.

for an inductor:
When there is a sinusoidal alternating current (AC) through an inductor, a sinusoidal voltage is induced. The amplitude of the voltage is proportional to the product of the amplitude (IP) of the current and the frequency ( f ) of the current.

inductance is the electrical equivalent of mass.

so the height of the tides (current)=tidal force (voltage) divided by frequency

height of tides(size of the bulges)=d√d/d^3=1/d√d

which multiplied by the 1/d^4 derived above gives the net acceleration of the moon in the direction of its orbit at distance d. now we want to know the infinitesimal time the moon spends at each infinitesimal distance from the earth.

the angular momentum of the moon is mvd.

torque is force times d. it produces change in angular momentum just as force produces change in momentum.

centrifugal force is mv^2 / d

gravity equals 1/d^2

the centrifugal force must equal the gravitational force
v^2/d=1/d^2
v=1/√d
therefore the velocity of the moon at distance d from Earth is 1/√d
therefore the angular momentum of the moon at distance d is √d
therefore the derivative of √d gives the amount that angular momentum when the moon moves across distance d.
d/dd*√d=1/(2*√d)

the tidal force acting on the moon (to accelerate it) at distance d is 1/d^5.5
therefore the torque (rate of change of angular momentum per time) acting on the moon at distance d is 1/d^4.5

the infinitesimal time spent at each infinitesimal distance is change of angular momentum per distance divided by torque (rate of change of angular momentum per time).
dt= [1/(2*√d)]/[ 1/d^4.5]*C*dd
dt=d^4.5/(2*√d)≡d^4*C*dd

current rate of recession of the moon is approximately 30 mm/ year.
1 year=(380,000,000,000)^4*C*30 mm
C=1.6*10^-48

solving for t(d)
t(d)=C*d^5/5

t(380,000,000,000 mm)=2.5 billion years.

why 30 mm/year? according to wikipedia:
geological record is consistent with these conditions 620 million years ago: the day was 21.9±0.4 hours, and there were 13.1±0.1 Synodic months/year and 400±7 solar days/year

from that I calculate that the average rate of recession for the last 600 million years is 30 mm/year
 
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  • #28
if the tides are indeed standing waves then a simplistic analogy would be to think of the nodes as capacitors and the antinodes as inductors. the circuit would have to have very little resistance. tidal forces would be an ac voltage.

to have the frequency response required by the equations below it would have to be operating well above its resonant frequency. I hope that makes sense. because if it doesn't then the equations would have to be wrong.
 
  • #29
Bjarne said:
The moon moves away from the Earth 3,8 cm per year.
In the past the velocity has been gradually increased.

I have read at the internet, that when the theory explaining this cause of this phenomena is correct the moon would for about 85 million years been orbiting 4 meters above the earth.

My own calculation shows this would have happen for about 1, 2 billion years ago (if this theory is correct).

We know that the moon is more than 4 billion years old, so how is it possible to keep believing such dictionary theory?

(sorry if this is not perfect English)

please excuse me and correct me if i am wrong. but are you implying that the moon some 85 million years ago. was orbiting some 4 meters above the earth?? if you are this theory would be nothing less than insane. If you really know anything about physics whatsoever. you would understand that if the moon was just 4 meters above the Earth the moon and the Earth's gravitational pull on each other alone would cause the moon and the Earth to increse in rotational speed astronomically and ultimatly would cause the two to rip each other apart. When you have an object spinning it creates its own gravitational pull.. so explain this to me? the closer they get the faster they spin. no life, and no water would be able to stay and reside on the planet. not to mention the affects of the sun on the Earth and the moon as well
 
  • #30
Please read the next few posts after the first one...while recognizing the thread is more than a year old...and dropping the attitude.
 
  • #31
Bjarne said:
How can friction between the oceans’ bodies of water and the sea bottom - explain that friction between 2 internal bodies on Earth can have such an effect on the Moon that its acceleration increases, - it has not been explained how a dynamic rotational force could be transferred to the Moon through space.
Can someone explain that, in simple words.

Easy. The friction causes the tidal bulges to lag behind the rotating Earth. Thus there's a non-conservative process occurring and the Moon is accelerated slightly by the bulge that's a bit ahead of rotationally. The gravity of the bulge, offset from perfect symmetry by friction, is what transfers energy between the Earth and the Moon. A perfectly frictionless interaction between the two would mean no net acceleration and no recession of the Moon.
 
  • #32
Bjarne said:
The moon moves away from the Earth 3,8 cm per year.
In the past the velocity has been gradually increased.

I have read at the internet, that when the theory explaining this cause of this phenomena is correct the moon would for about 85 million years been orbiting 4 meters above the earth.

My own calculation shows this would have happen for about 1, 2 billion years ago (if this theory is correct).

We know that the moon is more than 4 billion years old, so how is it possible to keep believing such dictionary theory?

(sorry if this is not perfect English)

If the tides lost energy to friction as quickly as they do at present, then the Moon wouldn't last so long in its orbit. However there's nothing theoretical or observational saying the friction should remain the same and detailed modelling of how the tides change with the rotation rate and positions of the continents has shown the frictional loss was much, much lower in the past. Only in the last ~500 million years or so has the loss been so high. Tidal rhythmites are periodic sedimentary deposits laid down by daily tides. Fossil examples are known all the way back to ~3.2 billion years ago and they show the Moon wasn't much removed from its present orbit - at the closest it was about 38 Earth radii away (presently it's at ~60.)

When the Moon was formed or captured, the tidal forces were sufficient to cause large scale motions of the semi-fluid silicate mantle, causing very rapid energy dissipation and flinging the Moon out to a more distant orbit very rapidly. Further out the body tides, as such tidal deformations are called, became minor and the energy-loss rate become low, until the system hit its present configuration.
 
  • #33
Your are ignoring fluid friction, which increases the braking effect. Waves are the peak of the gravitational iceberg.
 
  • #34
Chronos said:
Your are ignoring fluid friction, which increases the braking effect. Waves are the peak of the gravitational iceberg.

Who's ignoring fluid friction? Which post are you replying to?
 

1. How does the moon moving away from the earth affect tides?

The moon's gravitational pull is what causes tides on Earth. As the moon moves further away, its gravitational pull becomes weaker, resulting in lower tides. However, the effect is very small and would not significantly impact our daily lives.

2. Is the moon moving away from the earth a new phenomenon?

No, the moon has been moving away from the earth since its formation. This is due to tidal forces caused by the Earth's rotation. However, the rate at which it is moving away is very slow and has only become noticeable with modern technology.

3. Will the moon eventually leave the Earth's orbit?

No, the moon will not leave the Earth's orbit. The moon is actually moving away at a very slow rate of about 3.8 centimeters per year. It will eventually reach a point of equilibrium where it will stay in a stable orbit around the Earth.

4. What impact does the moon moving away have on Earth's rotation?

The moon's gravitational pull on Earth actually causes a slight slowing of its rotation. As the moon moves further away, this effect will become even smaller. However, it would take billions of years for the Earth's rotation to significantly slow down.

5. How does the moon moving away from the earth affect the length of a day?

As mentioned before, the moon's gravitational pull causes a slight slowing of Earth's rotation. This results in a lengthening of the day by about 1.5 milliseconds every century. However, this change is so small that it would not be noticeable in our lifetime.

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