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The
Trouble with Tides
by
Miles Mathis
First
posted October 7, 2005
Tidal
theory is one of the biggest messes in contemporary physics. I
will start with tides on the Earth, since they have gotten the
most attention and the most theory. We know the ocean tides are
caused by the Moon, since they follow lunar cycles. But are they
caused by the Moon’s gravity? Let’s look at some numbers.
Let’s compare the Sun’s field to the Moon’s field, at the
Earth.
a_{S}
= force on the Earth by the Sun a_{M}
= force on the Earth by the Moon a_{S}
= GM_{S}/r^{2}
= .006 m/s^{2}
a_{M}
= GM_{M}/r^{2}
= .000033 m/s^{2}
You
can see that the Sun has a much stronger gravitational effect on
the Earth, if we look strictly at field strength. We could have
guessed this without the math, since if the Moon had a stronger
gravitational effect, we would be orbiting it, not the Sun. If
tides are caused by gravity, then it seems like we should be
experiencing Sun tides that utterly swamped our Moon tides. By
the math above, Sun tides would be about 180 times as great as
Moon tides, making the Moon tides invisible. They would follow
the movements of the Sun overhead.
Why aren’t we experiencing Sun tides that are stronger than
Moon tides? According to an article at Wikipedia, which is
following the Standard Model and which is reprinted all over the
web,
Gravitational forces follow the
inverse square law (force is inversely proportional to the square
of the distance), but tidal forces are inversely proportional to
the cube of the distance. The Sun's gravitational pull on Earth
is 179 times bigger than the Moon's, but because of its much
greater distance, the Sun's tidal effect is smaller than the
Moon's (about 46% as strong).
F_{S}
= GmM_{S}/r^{3}
= 2.4 x 10^{11}
N F_{M}
= GmM_{M}/r^{3}
= 5.1 x 10^{11}
N
These equations, as I have simplified them here, don’t
give the right numbers, but we do get 46%. How was this "inverse
cube law" derived? According to a University of Washington
website^{1},
Tidal
forces result from imperfect cancellation of centrifugal and
gravitational forces a distance L away from the center of gravity
of the system and have the form F_{t}
= GmML/R^{3}
Other websites agree. Here
is one that is especially funny, considering everything:
So
the gravitational attraction of the Sun is 178 times greater than
that of the gravitational attraction of the Moon. But how can
this be? We all know the moon is more effective in producing
tides than the Sun. There is a simple explanation for this, and
it is not that we have been lied to! It is only the proportion of
the gravitational force not balanced by centripetal acceleration
in the Earth’s orbital motion that produces the tides.^{2}
Two
major problems here. One, the gravitational force causes the
centripetal acceleration. There can be no lack of balance. As for
the gravitational and centrifugal forces, although they are
caused separately, they cannot cancel, since they both tend to
create tides. In fact, most physics books and websites use a
summation of centrifugal effects and gravitational effects to
create tides on the Moon, as I will show below, since both are
tidally positive. That is to say, gravity would create tides even
without circular motion, and circular motion would create tides
even without gravity. So the two are additive. There is no
possible cancellation, in the way that is assumed above. Besides,
the Earth is not feeling a centrifugal effect from the Moon,
since the Earth is not orbiting the Moon. Even if it were
orbiting a barycenter, it still would not be in circular motion
about the Moon. Therefore the tides on the Earth could not be an
imperfect cancellation of centrifugal forces and gravitational
forces, even if these forces were in opposition. There are no
centrifugal forces on the Earth directly caused by the Moon,
since there is no angular velocity around the Moon.
Secondly, the math above is dishonest. If we look at the
Sun/Earth system, then the center of gravity of the two bodies is
so close to the center of the Sun that it makes no difference.
The Earth has almost no effect on the Sun. Therefore, the
distance L is just the radius R, and the equation is the same
as
F_{t}
= GmMR/R^{3}
That is not an inverse cube law, it is an inverse square
law in poor disguise. But the author seems to realize that, so he
continues to add layers of heavy clothing. He continues:
However, for a spherical object with
an average density d and a radius r we can view its mass as the
product of its density and volume, so that M = d (4/3πr^{3}).
Then the tidal force has the form F_{t}
= (π/6)GmLd(2r/R)^{3},
so that F_{t} ~ d
(2r/R)^{3}
Beautiful.
You don’t see magic like that everyday. The trick here is in
losing the L variable. The author needed to get rid of that so
that he would not have to cancel it with his R^{3}
in the denominator, bringing it back down to R^{2}.
So he surrounded it by as many variables as he could. This
confuses the reader, who he then hopes will miss the flaw in his
quick claim that F_{t}
~ d (2r/R)^{3}.
But that last equation is flat
wrong. It does not follow from the previous equation, since if
you are computing proportionalities, you cannot create a
proportionality to distance, and then leave one of the straight
distance variables hanging. The last equation is a
proportionality between force, density and radius. But L is part
of the radius. It cannot be separated out like that and then left
behind. I have shown that with the Sun/Earth, L is almost the
same as R. Even with the Moon/Earth, L is very close to R. You
cannot mathematically just ditch it like that.
You can see this more clearly if you go back to the equation F_{t}
= (π/6)GmLd (2r/R)^{3}.
Remember how he got there. If that equation is applied to the
Sun, then L = R and the equation reduces back to F_{t}
= GmM/R^{2}.
It therefore can’t differ from the force that is 180 times the
Moon’s force. It is true that according to the equation F_{t}
= (π/6)GmLd (2r/R)^{3},
the Moon should have a slightly different force on the Earth than
the one above. If we take the barycenter of the Earth/Moon to be
4671 km from the center of the Earth, and take R = 384467, then L
= 379,740 km F_{t}
= (π/6)GmLd (2r/R)^{3}
= 1.9 x 10^{20}
N All of the fancy clothing
didn’t change much of anything. All we did was incorporate the
center of gravity of the Earth/Moon, and that did not give us an
inverse cube law or lower the force from the Sun. In fact, it
only made the problem worse. The Sun tides are now 190 times
those of the Moon, instead of 180.
This has not kept large parts of mainstream physics from
accepting the idea that tidal forces are dependent on density or
angular size in the sky, and that this density dependence can be
given to gravity. The data tells us that the Sun’s force is 46%
of the Moon’s, someone throws together a dishonest equation to
prove it, and it becomes dogma.
A much better explanation
of the inverse cube law is supplied by Wikipedia:
Linearizing
Newton’s law of gravitation around the center of the reference
body yields an approximate inverse cube law. Along the axis
through the centers of the two bodies, this takes the form F_{t}
= 2GmMr/R^{3}
"Linearizing"
means differentiating the equation with respect to R, so that
this new equation represents a change in the field, rather than
the strength of the field. Despite being weaker, the field of the
Moon changes more quickly. This causes a greater difference from
center to far or near edge. Another way to express this without
differentiation is: a = GM[1/R^{2}
– 1/(Rr)^{2}]
Where R is the distance
between objects, and r is the radius of the gravitating object.
They tell us this equation is approximately equal to a = GM2r/R^{3}
Giving us an inverse cube law.
It is clear that the differentiating proves that there
would be
an inverse cube effect in the tideproducing differentials,
supposing that the postulates of this theory are true. I don't
know that I would call it an inverse cube "law", since
it does not apply to the field itself. It applies to the
differential field. It comes from the fact that tides in a static
gravitational field are determined by the rate of change of the
field, not by the strength of the field. What I mean by static is
that this calculation does not take into account the circular
motion of the object in the field. Even objects in straight
freefall would be subject to this tidal inverse cube law, as
Wikipedia and current theory admits. But the Earth is not in
simple freefall around the Sun. It is in orbit. We must therefore
add a centrifugal effect to the static effect of the field. Once
again it appears that this must take the Sun’s effect beyond
the Moon’s effect on the Earth. To
find out, let us actually calculate
a force. We know that the centrifugal force varies in a different
way from the centripetal force. The centripetal force gets weaker
as you go out, since it must be assigned to the gravitational
field. But the centrifugal force, in this case, increases at
greater radii. This is because the far side of the Earth in its
orbit must have a greater orbital velocity than the near side. To
calculate this force we must first find the acceleration of
different parts of the Earth using the equation a = v^{2}/R.
R = 1.4959787 x 10^{11}
m v = 2πR/t t = 365.257d = 31558205s v = 29784.68322
m/s R + r = 1.4960424 x 1011 m R – r = 1.4959149 x 1011
m v_{o}
= outer velocity = 29785.95147 m/s v_{i}
= inner velocity = 29783.41297 m/s a = .00593008 m/s^{2} a_{o}
= .005930332 m/s^{2} a_{i}
= .005929827 m/s^{2} Δa
= 2.53 x 10^{7}
m/s^{2}
So let’s show the basic equation for the math above: Δa
= [v^{2}/R]
– [v_{i}^{2}/(R
– r)] = [4π^{2}R/t^{2}]
– 4π^{2}(Rr)/t^{2}
= ω^{2}R
– ω^{2}(R
– r) where ω is the angular
velocity (= 2π/t) Δa = ω^{2}r
All
that work to get the same number we found at first, way above
[.006]. I wanted to show you that the circular motion equation
generated the same number as the gravitational equation. This is
no accident, of course. Measured from the center of the Earth,
the two numbers would be expected to be the same, since it is the
acceleration due to gravity that keeps the Earth in orbit,
according to gravitational theory. The math above just mirrors
the math of the differentiated equations. It is the same in form,
but not in output, since the centrifugal field varies differently
than the gravitational field, as I have said. Here is the
equation for the static gravitational field of the Sun at the
Earth: Δa = GM[1/R^{2}
– 1/(Rr)^{2}]
= 5.08 x 10^{7}
m/s^{2}
Doing the same math for the gravitational field of the Moon at
the Earth, we find Δa = 1.14 x 10^{6}
m/s^{2}
These two equations yielded the number 46%, remember. But now we
have some more tidal effect to add from the Sun. The total tidal
effect from the Sun is now Δa = 2.53 + 5.08 = 7.61 x 10^{7}
m/s^{2}
This does not take us over the effect from the Moon, but it takes
our number for Sun tides up to 67% of Moon tides.
The
Standard Model, as expressed in Wikipedia and elsewhere, adds the
centrifugal effect using this equation: Δa = ω^{2}mr
ω is the angular velocity, so, according to Kepler’s law, ω^{2}
= GM/R^{3}.
This makes the equation equivalent to the math I used. This term
ω^{2}mr
gives us half of the value of the first term, 2GMmr/R^{3}.
In other words, the tidal effect caused by circular motion is
half the tidal effect caused by the static gravitational field.
On this much we agree, as you can see from my numbers for the Sun
[2.53, 5.08]. But the Standard Model goes on to apply the full
equation to the tidal effect on the Earth from the Moon: Δa =
ω^{2}mr
+ 2GMmr/R^{3}
This equation is equivalent to my math above: Δa = ω^{2}r
+ GM[1/R^{2}
– 1/(Rr)^{2}]
But
neither equation is applicable, since the Earth is not orbiting
the Moon. The first term on the
right side cannot be applied, because if you reexpand it, you
find that it contains the variable R. Like this, remember:
a
= ω^{2}R
– ω^{2}(R
– r)
This R applies to the EarthMoon radius. But if the
Earth is actually orbiting the barycenter, then this radius R
does not apply in the first term of the equation. We must use the
number 4,671 there, not 384,400. To get the correct angular
velocity, we must use the correct radius. The current equations
use 384,400 for R, but the value should be 4,671. That throws off
all the numbers, and prevents them from getting the 46% they
desire. If we correct the math, we find 67%, as I showed above.
But that doesn't match data.
I have had readers answer:
"But we are subtracting the R out of the equation, so it
doesn't matter what it is." It matters because the value of
v and ω are chosen to match that radius. The Earth wouldn't have
the same angular momentum around a barycenter that it would have
around the Moon, would it? The current equations use a value for
ω that implies the Earth is orbiting the Moon; when, at best,
the Earth is only orbiting a barycenter.
This means that
all the standard model math fails. The mainstream has been
publishing false equations. I assume they know they are doing
this, since the holes in the equations are so big. Using the
wrong radius is a huge error, one that is difficult to explain
away as an oversight. It would have to be an oversight of many
decades, involving thousands of specialists. I believe it is a
purposeful fudge.
[Note added August 2007: Confronted with
parts of this paper in late 2005, Wikipedia deleted all its tidal
theory math, its tidal theory page, and ordered a rewrite with
lots of new illustrations. It appears they are perfecting their
propanganda rather than admitting that their math and theory
doesn't work. This change affected many other websites as well,
since Wiki is linked to a large percentage of online encyclopedic
entries. Large parts of tidal theory have gone into hiding since
the publication of this paper. One place that is still hanging
the dirty laundry out in the open is the department of
Oceanography at Texas A&M (ocean.tamu.edu), as I was informed
by a reader. All the tidal math there falls to this critique,
since it is equivalent to the math that was up at Wiki. It is a
pretty variant, but it includes the inverse cube law above, and
uses the wrong radius. It is very clever at hiding all the
problems, since it hides some variables and refuses to define or
assign others. It also hides the barycenter problem, revealed
just below.]
Another major problem with tidal theory
concerns its use and misuse of the barycenter. The barycenter is
the center of gravity of the Earth/Moon system, which both bodies
are said to orbit. Feynman was one of the most famous to suggest
that the Earth has a nonnegligible tide created by orbiting this
barycenter. Is this true? Let's do the full math.
R =
4671 km v = 2πR/t t = 27.32d = 2360448s v = 12.43 m/s R
+ r = 11042 km R – r = 1707 m v_{o}
= outer velocity = 29.39 m/s v_{i}
= inner velocity = 4.54 m/s a = 3.31 x 10^{5}
m/s^{2} a_{o}
= 7.82 x 10^{5}
m/s^{2} a_{i}
= 1.2 x 10^{5}
m/s^{2}
Δa_{o}
= 4.51 x 10^{5}
m/s^{2}
Δa_{i}
= 4.51 x 10^{5}
m/s^{2}
We certainly do find a significant effect from the Earth
orbiting its own barycenter. In fact, it swamps all other
effects. It is 40 times as great as the gravitational effect from
the Moon and almost 60 times the total effects from the Sun.
However, Feynman was wrong in one very important way. The effect
doesn’t just raise a tide on the far side of the Earth from the
Moon; it raises an equal tide toward the Moon. Feynman obviously
didn’t know what to do with that negative radius. But as you
can see from my diagram, it produces a positive tide. You must
follow the steps of the math I did previously, and if you do it
exactly, you find that you must subtract a_{i}
from a, to achieve the proper differential. As vectors, they are
pointing in opposite directions, so you subtract a negative,
which is the same as addition. Δa_{i}
= a – a_{i}
The
barycenter falsifies the entire standard analysis, since it would
swamp all effects from the Sun and Moon. You cannot include
effects from the barycenter, since they cannot be made to fit the
given data. And you cannot fail
to include effects from the barycenter, since current gravity
theory demands a barycenter. This is called a failed
theory.
Some have tried to squirm out of this by telling
me that since the barycenter numbers are equal forward and back,
the tide is a constant and the other tides can just be stacked on
top of it. But this is illogical. The numbers show the barycenter
tides equal front and back: they do NOT show an equal tide all
the way around. The barycenter is a discrete point, and there is
a vector pointing from the center of the Earth to the barycenter.
This gives the barycenter tide position and direction on the
surface of the Earth, just like any other tide. In other words,
it would create heaps. That is what my illustration shows. The
math and the illustration do not show an equal heaping in all
directions. All points on the oceans would not swell equally, and
at the same time. Therefore, if we had a barycenter tide, it
would be quite obvious. It would create huge swells and it would
swamp all other tides. We wouldn't need a list of 388 tidal
harmonics, we would need only 1.
To add to the confusion,
the National Oceanic and Atmospheric Administration (NOAA)
explains tides on its website^{3}
by disregarding fields and differentials altogether. In various
places it glosses (very poorly) the effects I have calculated
here. It mentions the barycenter effect, the tidal bulges, and so
on, if only to mention all the physics thought to be involved.
But after outlining and diagramming every possible cause and
effect, it resorts to tides as a straight force from the Moon.
The author states that the force is not great enough to overcome
the gravitational pull of the Earth on the ocean water directly
beneath the Moon, but at the tangents the Moon’s effect is
unresisted by Earth’s gravity. Gravity does not pertain at a
perpendicular. So the Moon tends to draw all the waters of the
Earth from the tangents to the sublunar point. The author accepts
that water must also “heap” at the antipodal point, but he
does not say how an attractive force at the tangents would heap
water at the antipodal point. In fact it would not. A force of
this sort would tend to decrease the total amount of seawater in
the far half of the ocean and heap it all in the near half. There
would be no far tide, just a large far depression.
It is
clear that there is no mainstream view of what causes ocean
tides. Several views are held by different mainstream
organizations. The NOAA is in the dark ages, CENPA is publishing
some very dishonest math, and other experts are all over the map.
Feynman, who no one would call marginal, weighed in on the
barycenter explanation, but got it wrong. NASA and JPL appear to
accept the inverse cube law, but avoid the issue on their
websites. They give only PR glosses for a mainstream audience.
The Standard Model, or its
upper levels, appears to be currently founded on the inverse cube
"law," but it is inconsistently applied. It is applied
in conjunction with the centrifugal effects in regard to tidal
effects on the Moon, but it is not when explaining tides on the
Earth caused by Sun and Moon. This is because the current model
needs to keep the Sun’s effects low, so that they do not
conflict with data. In order to explain spring tides and neap
tides, the Sun’s effects must be squeezed to fit data, and 67%
is just too high; 45% is about right.
Regarding the barycenter answer, it has not been accepted even
though Feynman liked it. To disregard Feynman, physics has to
have a pretty good reason, and I have shown you the reason. If we
apply the correct math to the barycenter theory, then we find
that if it is true it swamps everything. It gives us two high
tides that vary 1/1.22, which we could readily accept. Except
that these tides are so huge that all the other variations are
lost. The neap tides and spring tides are easily measurable: the
data can’t be explained if the effect from the Sun is 72 times
smaller than the main effect. Lunar perigee and apogee variations
also would become negligible if they were compared to a
barycenter tide. All effects from variation in the lunar orbit
would be lost. All this is a
terrible problem. If you accept the postulates of current
gravitational theory, then you are led inexorably to the
barycenter tide. But you cannot accept it because it conflicts
strongly with all data. It cannot be absorbed by even the most
creative theory or math. But if you throw out the barycenter
tide, then you have to throw out all the bathwater too. We used
the same physics and the same differential equations to find the
barycenter numbers that we did to find the Solar and Lunar
numbers. If the physics and equations are wrong, they must be
wrong all the way down. We cannot just go back halfway, taking
the numbers we like. We have to throw out all the numbers and
start over. Even more, we have to throw out all the assumptions.
No matter how we followed the assumptions, we arrived at numbers
that did not work. This is called a failed theory.
Ask yourself why Feynman didn't insist on the barycenter
solution. He was in a position to insist, and he was the type to
insist if he knew he was right. He didn’t insist because it was
only a suggestion, one he couldn’t ultimately make fit the
data. He saw correctly that it was the logical answer given
gravity, but once it didn’t fit the data, he gave it up. He was
always an empiricist, and never let his theory get before the
facts. Now, ask yourself why Feynman did not follow up with
another theory. Surely he could see that gravity insisted on the
barycenter tide. If the barycenter tide does not work, then there
is something seriously wrong with gravity. The only way the
barycenter tide could be false is if it is itself being swamped
by another force field. What could this field be? Feynman
obviously hadn’t a clue.
Tidal theory, like so much
other contemporary theory, has become a farce. Newton proposed
the Moon’s gravitational field as the cause of tides and no one
has seen fit to correct him, even though we have data now that
makes his theory ridiculous to keep. If Newton had known a mass
and distance for the Sun and Moon, he would never have proposed
the theory he did (I hope).
Now, I admit that tidal theory has become very advanced in some
ways. New models can predict the effects of tides with greater
accuracy. And tidal theory is quite successful in showing how the
given forces can create the tides we see. But it has made no
progess since Newton in explaining the genesis of the fields
themselves. As I have shown here, the foundational theory of
tides is little more than a bad joke. The Standard Model tries to
keep all this out of sight, and it is amazingly successful in
doing so. Most tidal analysis does not mention the relative
strengths of the fields of the Sun and Moon, since it immediately
explodes the theory. Only places like Wikipedia are foolish
enough to hang the dirty laundry in the open air. Most books and
websites are long on computer graphics, historical glosses, and
advanced mathematics, and very short on foundational theory. We
can now see why.
Tidal Effects on the Moon
Now
let us look at tides on the Moon. I will start over with my
analysis, pretending once again that the reader knows nothing
about tides; but in this section I will hit some topics that we
missed in the first section. The Standard Model, as glossed in
textbooks at all levels, explains tides by showing that real
bodies do not behave like point particles. Because they have real
extension, different parts of the body must be feeling different
forces. If we take the Moon as an example, we can compare three
points on or in the Moon. We take the point nearest the Earth,
the point at the center of the Moon, and the point farthest away.
The point at the center feels a force from the Earth that is just
sufficient to make it orbit. That is why, in fact, it is
orbiting. It feels no tides of any kind. The point nearest the
Earth requires less force to make it orbit than the point at the
center, but it actually feels more force. The point farthest from
the Earth requires more force to make it orbit, but it is feeling
less force than the point at the center. The point nearest
therefore feels a resultant force toward the Earth and the point
furthest feels a resultant force away from the Earth. This causes
a tide that maximizes at the near and far points.
So far so good. The Standard Model applied to the Moon follows
what we have already found regarding the Earth. But before we
analyze it again, let’s look at something interesting. Notice
how theorists who claim to believe in General Relativity always
revert to Newton when it comes time to explain forces in
gravitational fields. In the chapters on General Relativity, we
are told that an orbiting body is feeling no forces. It is simply
following curved space, the “line” of least resistance. We
are shown the ballbearing on the piece of rubber, and the tiny
marble orbiting it with no centripetal force. All quite
ingenious, except that it does not explain the genesis of the
forces at a distance used in tidal theory. How can an orbiter
that is feeling no forces achieve tides? Even more to the point,
how can an orbiter that is traveling in the curved space of its
primary recurve that space in order to transmit a tidal force to
the primary? Is the gravitational field between the Moon and
Earth curving convex or concave, relative to the Moon? I would
think it must be one or the other. It cannot be curving both ways
at once. If anyone answers
“gravitons,” then I think we can throw out the curved space
idea as superfluous. If we have gravitons mediating the force,
then the Moon is feeling a force. In which case we don’t need
curvature to explain anything.
But the current theory
isn’t even that advanced, regarding tides. Graviton or no
graviton, the theory reverts to Newton for the explanation. To
cover all its bases, the theory gives the situation a sort of
double cause. The first cause is given to the gravitational
field. Nearer parts of the body will accelerate toward the Earth
faster than farther parts, regardless of their weight or mass.
Remember that acceleration in a gravitational field has nothing
to do with mass. All objects fall at the same rate. Acceleration
is dependent only on radius. So the analysis should always be
talking about accelerations, not forces. This part of the theory
is at least logical, given Newton’s equations. It is true that
the static gravitational field would create tides as claimed,
near and far. But it would create these tides even if there were
no circular motion and no orbit. An object in freefall would
experience this sort of tide, as the Standard Model admits.
Current theory gives a second
mechanism, and this mechanism requires an orbital velocity.
Nearer parts of the Moon orbit in a slightly smaller
circumference than farther parts. They travel this circumference
in the same time as the rest of the Moon. Therefore they have a
slower orbital velocity. With more acceleration and less orbital
velocity, the near tide is increased. Likewise, farther parts of
the Moon have less acceleration and more orbital velocity, once
again increasing the proposed tide. This analysis is once again
(mostly) true, but this second cause has nothing to do with
gravity. It is an outcome of all circular motion, whether you
have a gravitational field or not. Whirl any dimensionally
consistent object and the circular motion will create tides in
the object just like these, if you apply the forces in the same
way. You will say that a spin
dryer or a ride at the fair does not cause elongation like this,
forcing the object to spread out along the radial line. Just the
opposite: a fair ride or spin dryer causes the object to flatten
out along the orbital line. But this is because all the force is
applied from the back of the object. If we applied all the
gravitational force from the back of the Moon, the Moon would
also flatten in this way. No tide could be created at the back,
since the force would constrain it, just like the wall of the
dryer. The rest of the Moon would feel no force and would be
totally unconstrained, except by its inner structure. Feeling no
centripetal force, it would naturally follow the vector of the
tangential velocity until it too was constrained by the force
from behind or by internal structure. This is exactly what
happens in the dryer or at the fair.
But if Big Uncle Joe swings you by the arms, you do not flatten
out like a towel in the dryer. You flatten out along the radius.
The force is applied from the front, and you spread out in a line
away from the center, with your legs flying behind you.
To begin to create the Moon analogy, you would have to have three
ropes along the same radial line. Lie on the ground with your
head pointed toward Big Uncle Joe. He ties one rope to your
hands, one rope to your waist, and one rope to your feet. As he
swings you, Uncle Joe wants to keep your hands, waist and feet in
the same line, so that your legs don’t lead, and your hands or
waist either. From his point of view, he wants to see just your
screaming face. Your waist and feet should be in a direct line
behind your head. If your
hands, waist and feet all weigh about the same, then it is
obvious that the rope about your feet will have more tension on
it. Your feet are going faster to keep up, and they weigh the
same, therefore more force. By the same token, the rope on your
hands has less force. Of course Big Uncle Joe is not a magician:
he can’t swing you at three different rates at the same time.
We can imagine the forces being different along each rope, but
Uncle Joe is causing your tangential velocity at the same time,
through the same ropes. He can’t impart three different torques
down the same line. So this example cannot be created. It is a
useful visualization however. A
similar example could be created in the spin ride at the fair, or
in a centrifuge. Take three equal weights and tie them to three
separate points along the same radial line, at different
distances from the center. Get the centrifuge up to speed and
measure the forces. The greatest force will be on the one
furthest from the center. To see how an object that was a sum of
the three weights would react as a whole, subtract the force on
the inner weight from the force on each one. The force on the
inner one is now zero, the force on the middle one is bigger
(middle minus inner) and the force on the outer one is biggest
(outer minus inner). The outer weight wants to move away from the
center a lot, the middle one less and the inner one even less.
This is immediately clear with the outer one. If all three ties
were cut simultaneously, then the outer one would move away from
the center faster than the middle one. The distance between the
two weights would increase. Therefore the outer weight seems to
feel a force away from the middle weight. The same is true
comparing the inner and middle, although it is a bit less
intuitive for some. If the ties were cut the inner weight would
move away from the center more slowly than the middle weight.
Once again the two weights would increase the distance between
them, so that there is a sort of force away from the middle
weight once again. If the middle weight is the baseline, then
both the outer weight and the inner weight will tend to move away
from it. This will make the object deform along the radial line,
just like the tides on the Moon.
So, we can add up the
effects on the Moon just like on the Earth. We can use the
equation Δa_{E}
= ω^{2}r
+ GM[1/R^{2}
– 1/(Rr)^{2}]
= .000012 + .000024 = 3.6 x 10^{5}
m/s^{2}
Δa_{S}
= ω^{2}r
+ GM[1/R^{2}
– 1/(Rr)^{2}]
= 6.9 x 10^{8}
+ 1.4 x 10^{7}
= 2.1 x 10^{7}
m/s^{2}
The
solar tide on the Moon should be 171 times smaller than the
terrestrial effect. More importantly, the visible tide on the
Moon should be symmetrical front and back. Is this what we find?
Not at all. The Moon rotates relative to the Sun, so we would not
expect to find a solar effect on the Moon, beyond a tiny constant
shift in the crust opposite the direction of this rotation. The
rotation of the Moon on its axis relative to the Sun does not
cause a further tide from the Sun, or add to the tidal effect,
but it acts to shift the tides we have already calculated, just
as the rotation of the Earth shifts the ocean tides, causing them
to travel. I am not aware of any experiments on the Moon to
measure lateral shift of the crust in the direction opposite
rotation, to verify the relative strength of the solar tide,
although this would be a very useful experiment. However,
concerning the terrestrial tide, we have ample visual data. This
data is not a confirmation, to say the least. The schematic of
the Moon (above) will show you that the Moon exhibits no real
tide at the back. Even more astonishingly, it has a negative
tide at the front, the crust being almost obliterated in places.
How does current theory explain this? It can't explain it using
gravity or circular motion. This is how it is explained in the
Encyclopedia
of the Solar System:^{4}
"The conventional explanation for the center of
figure/center of mass offset is that the farside highland
lowdensity crust is thicker. It is massive enough and
sufficiently irregular in thickness to account for the effect."
More gobbledygook, in other words. If the farside crust is
lowdensity, this would only add to the problem. To create a
greater tide we need more mass over there, not less.
Before
I move on to solve all these problems, I have one more thing to
say about the orbit of the Moon. In all these analyses, both mine
and those of the Standard Model, it has been assumed that outer
parts of the Moon can travel faster than the inner parts. The
diagram requires it and so we have just taken it as a given. We
do not even ask how it is physically possible for different parts
to have different tangential velocities and different orbital
velocities. The gravitational field cannot be creating them,
since it cannot exert a force tangentially. The field creates
only radial forces. We need either a mechanical cause of the
variance, or we need to show that all orbiters exhibit shearing
along the direction of orbit. Orbiters in tidal lock should
exhibit strong symptoms of shearing, since the forward part of
the object is always in lower orbit and the back part is always
in higher orbit. The back part of the Moon should shear in the
reverse direction of orbit and the front part should shear in the
forward direction. But the data is negative, and we are given no
cause for the negative data. The only mechanical cause would be
some sort of absolute rigidity of the Moon radially. But this is
not true empirically. With current theory, the lack of data is a
complete mystery. General
Relativity can explain it, since according to that theory, the
Moon is feeling no forces. A Moon feeling no forces would not be
showing any signs of shearing. But you can hardly use GR to
explain the tides we don’t see and use Newton to explain the
tides we do see. The Moon is either feeling forces or it
isn’t. The problem was a big
one for Newton, even in his own time, since he is the one who
postulated that the tangential part of the velocity in orbit was
caused by the orbiter's "innate motion". That is to
say, the tangential vector is one the object has prior to or
independent of the gravitational field. But of course the object
could not have a variable innate motion. It cannot speed up outer
parts and slow down inner parts just to suit diagrams.
It is now not just Newton’s problem. Current theory has
inherited it and failed to explain it, or even try to explain
it.
You can now read my newest paper on tides, More
Trouble with Tides. You can also read the first two parts of
my Solution to
Tides.
^{1}www.npl.washington.edu/AV/altvw63.html.
This site is managed by CENPA, the Center for Experimental
Nuclear Physics and
Astrophysics. ^{2}www.sanho.co.za/tides/tide_theory.PDF ^{3}www.coops.nos.noaa.gov/restles1.html#Intro ^{4}Compiled
by NASA and JPL, 1999. p. 252.
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