Neville Jones wrote:
Regner,
Your analysis of the cause of the tides is invalid, because
gravitational attraction between parts of extended objects being
exactly equivalent to all mass of each object being concentrated into
mathematical points is valid for these objects being rigid, not
flexible. You cannot assume the World to be rigid on the one hand and
then subtract the theoretical force on a rigid World from both sides of
a flexible World in order to 'explain' the tides.
There is no such assumption in my analysis.
Close to a body, you have to take into account the mass distribution in
that bode, in order to
get an accurate value of the gravitational force from that object.
To calculate an accurate value of the gravitational acceleration by the
Earth on you, we would
need to know the detailed mass-distribution inside the Earth -
considering the Earth a point-object
in this case, would be a bad approximation.
The gravitational force on you by the Moon, however, is accurately
given by approximating the
Moon by a point-mass.
Likewise, the difference in gravitational force by the Moon on
opposite sides of the Earth, are
accurately given by approximating the Moon by a point-mass.
If the Earth was completely rigid, the tidal forces from the Moon
would still be the same - they
have nothing to do with the composition of the Earth - only the
distance from the Moon and
the diameter of the Earth enters here.
The response of the Earth to this tidal force, obviously depends on
the composition of the Earth,
but I know too little about that to be able to calculate anything
beyond orders-of-magnitude
estimates. Notice, however, that my post didn't say anything about the
amplitude or phase of the
tides on Earth - I only gave it's cause, and the cause of tides is what
was misunderstood by other
posters here.
The acceleration of the centre-of-mass of the Earth behaves exactly
as if all the Earth's mass was
in that point. That can be fairly easily proven mathematically (and is
what we observe). The same
obviously goes for the Moon. So the orbits of the Earth and the Moon
around their common bary-
centre, can be accurately described in the point-mass approximation.
If you subtract that motion and multiply by the mass, you can get the
tidal forces on opposite
sides of the Earth. There is nothing inconsistent here.
Where would these forces on the oceans be acting? At the centre of mass
of the ocean, or on individual water molecules, or at the centre of
mass of the World?
Gravity acts between each and every particle with
mass.
The pull would vary as the inverse-square of the distance, which would
result in very serious tidal waves arising as the Moon-side ocean was
pulled more than the other was 'left behind'.
Well, this is what we experience as tides. If you
want to call that a "...very serious tidal wave...",
be my guest. Remember that Earth's own gravity is still pulling every
part of the Earth towards
it's own centre-of-mass (Earth is self-gravitating).
The gravitational acceleration by the Moon at
the surface of the Earth, is about a millionth of that due
to the Earth in the same spot. The tidal acceleration is another factor
of 30 smaller. It is not that hard to
actually put some numbers in, and check ones assumptions.
Incidentally, the centrifugal force produced
by Earth's daily rotation is about 30,000 times larger than
the tidal force from the Moon, and produces the (stationary) 21.3km
equatorial bulge. Assuming the
displacement to be proportional to force (as with an ideal
spring/elastic band), the Lunar tidal force
would result in a 7mm difference between high- and low-tide in the
Earth's crust. The observed effect
is a few centimeters, and obviously depends on the composition of the
local crust - orders-of-magnitude
estimates can be pretty powerful (this one wasn't far off)...
Rather than being [I moved these a bit to
align with my equations below/Regner]
<-
| -> as depicted, the force would
be proportional to
x*x^(-1) (x*x+2xR+R*R)^(-1) (x*x+4xR+4R*R)^(-1)
It is the same thing, Neville. You give the total
gravitational force by the Moon on
various parts of the Earth in those expressions. Now let me rephrase in
terms of the
distance, d, between the centre-of-mass of the Earth and the Moon
(d=x+R) to make
it symmetric. Then we get:
(d-R)^(-2) d^(-2)
(d+R)^(-2)
which is the same as
(d*d-2dR+R*R)^(-1) d^(-2) (d*d+2dR+R*R)^(-1)
Just substitute d=x+R and you'll see it is the same. Now since, R
<< d, we can
Taylor expand in (R/d). First we prepare our _expression_ by taking the
leading
term, d^(-2), outside, to get
[d*d(1-2R/d+(R/d)^2]^(-1) d^(-2) [d*d(1+2R/d+(R/d)^2)]^(-1)
Since (R/d) is very small, we can to the first approximation, neglect
(R/d)^2 so that
[d*d(1-2R/d)]^(-1) d^(-2) [d*d(1+2R/d]^(-1)
Again, since (R/d) is very small, we can approximate 1/(1+2R/d) by
1-2R/d - that
is the Taylor expansion, and it gives us
1/d^2 + 2R/d^3 1/d^2 1/d^2 - 2R/d^3
If we now subtract the motion of the centre-of-mass of the Earth,
1/d^2, then we are
left with the tidal forces:
+ 2R/d^3 0 -
2R/d^3
-2/d^3 is the derivative of 1/d^2 with respect to d, and then we
multiply by the radius
of the Earth, R, to get the tidal force - exactly as I wrote in my
previous post.
In our calculations, we can just as well use the exact expressions,
before the Taylor
expansion and before throwing away the 2nd order term in (R/d). The
difference is
very small, however, and the form I wrote, conveys a lot more meaning
to a physicist.
where x is the distance from the centre of mass of the Moon to the
closest tip of the ocean nearest to the Moon and R is the radius of the
World, and x >> R in the heliocentric model.
I'm a bit confused. Why does x and R change if
the Universe is Geocentric?!?
Ocean swell would also vary with the Moon's orbit.
As they do.
The
analysis is thus far too simplistic.
Yes - and I never did say that I provided a
thorough analysis of the tidal interactions
between the Earth and the other bodies in the Solar system. I gave a
simple analysis
aimed at being understood by laymen, and which explains the major
forces that
govern tides on Earth.
As
with other aspects of the so-called 'solar system', such forces would
lead very quickly to disorder and destruction.
Nope. Why would they? See the actual numbers in green, above.
This reply got much longer than I had intended - sorry.
Regner
Neville.
There has been a couple of references to tides, lately,
and much
confusion
on the subject. I'll do what I can to clear it up a bit.
Please take the time to read the whole post before commenting.
The force of gravity moves objects
The Moon and the Earth orbit each other due to their gravitational
interaction.
It can be shown that you can simplify this problem:
Every single bit of the Moon interacting gravitationally with every
single
bit of the Earth.
is exactly equivalent to
All the mass of the Moon being located at the centre-of-mass of the
Moon,
and all the mass of the Earth being located at the centre-of-mass of
the
Earth.
From such an analysis we find that the two objects will orbit their
common
centre-of-mass in elliptical orbits.
Tidal forces deform objects
But the Earth and the Moon are extended objects so what effect does that
have? Well, the Moon is still close enough to the Earth, that the
gravity
from the Moon will change over the diameter of the Earth:
Moon Near Centre Far
side of Earth side
O <-------- <-------
<------
Okay, if we now subtract the part that actually moves the Earth, that is
we subtract <------- which is the same as adding ------->
to all three
vectors (arrows) above, and we get the tidal forces:
Moon Near Centre Far
side of Earth side
O <- | ->
So looking at the Moon-Earth system from the outside we see that the
Moon-side of Earth experience a larger gravitational pull from the Moon
than the centre of the Earth, and the opposite side of Earth experience
experience even less gravitational pull from the Moon. Seeing it from
the
Earth point of view, we have a Moon-ward force on the Moon-side of Earth
and an opposite force on the opposite side of Earth. This is what causes
the tidal bulge, and this is why there are (approximately) two high-
and two
low-tides a day. The oceans, being a liquid, obviously responds much
stronger
than the bedrock, but the Earth's crust is flexing too, as has been
measured.
The Earth rotates under this tidal bulge, which makes it a dynamical
phenomenon.
The gravitational force is proportional to 1/r² where r
is the distance
between the Earth and the Moon. The tidal force is the difference in
gravitational force over a certain distance, which means it can be
expressed
as the r-derivative of the gravitational force. The tidal force
is therefore
proportional to 1/r³. That means the tidal force falls off more
rapidly with
distance than the gravitational force. This is the reason that the gravitational
force of the Sun dominates the orbit of the Earth, but the Moon's tidal
forces
on Earth are larger than the Sun's.
The Sun also contributes a significant tide (about half that of the
Moon)
and when the two are aligned at new Moon or full Moon, we get more
powerful tides called spring tides. At either half Moon we get neap
tides.
The ellipticity (eccentricity) of the Moon-Earth and the Sun-Earth
orbits
also causes variations in the strength of tides on Earth.
The Moon experience (about 22 times) stronger tides from the Earth
because of the Earth/Moon mass ratio of 81 and the size ratio of 3.7.
The Lunar tide from the Sun is 3.7 times weaker than here on Earth, due
to the 3.7 times smaller diameter of the Moon.
The tides would of course be exactly the same in a geocentric Universe.
Allen's concerns in the "Acceleration thread"
To account for tidal influences, the centre-of-mass of the
test-mass of
the accelerometer, should coincide with the centre-of-mass
of the free-
falling laboratory. Tides would have NO effect on such a set-up.
But then the tidal effects on such a lab are very small in the first
place:
Less than a billionth per meter of the gravitational force keeping the
lab
AND the test-mass in orbit.
That's all for now (sorry for such a long post), and I
hope
it clarified a
few things.
- Regner
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