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

1. Oct 3, 2008

### Bjarne

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)

2. Oct 3, 2008

### Staff: Mentor

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. Oct 3, 2008

### Bjarne

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. Oct 3, 2008

### Janus

Staff Emeritus
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. Oct 4, 2008

### Bjarne

Thank you
Nice and simple

6. Oct 8, 2008

### Cryptonic

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. Oct 10, 2008

### Bjarne

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. Oct 10, 2008

### Oberst Villa

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. Oct 10, 2008

### granpa

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. Oct 10, 2008

### granpa

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. Oct 10, 2008

### Chronos

grandpa correctly notes that tidal forces are proportional to the cube of distance. The other conclusions are incorrect.

12. Oct 11, 2008

### granpa

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)

Last edited: Oct 11, 2008
13. Oct 11, 2008

### granpa

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

tidal 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?

Last edited: Oct 11, 2008
14. Oct 11, 2008

### granpa

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

Last edited: Oct 12, 2008
15. Oct 11, 2008

### Orion1

The topic referred is called tidal acceleration or tidal friction.

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:
Tidal friction - Wikipedia

16. Oct 12, 2008

### Chronos

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. Oct 12, 2008

### Bjarne

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. Oct 12, 2008

### granpa

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. lets 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.

Last edited: Oct 12, 2008
19. Oct 12, 2008

### granpa

ok that doesnt 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 isnt 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 doesnt 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 isnt good enough. I'm trying to add a ring of negative mass at the equator but the math isnt working out.

http://www.physics.buffalo.edu/~sen/documents/field_by_charged_ring.pdf

Last edited: Oct 13, 2008
20. Oct 13, 2008

### Orion1

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 $$\frac{1}{r^3}$$. The rest of the planets' effects are infinitesimal in comparison.

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

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

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

Reference:
Tidal friction - Wikipedia
Tidal locking - Wikipedia
Calculating tidal accelerations on Earth due to other solar system planets - washington.edu
Earth - Wikipedia
Moon - Wikipedia

Last edited: Oct 13, 2008