[Rogers Misunderstanding] Only the Moon Experiences a Net Acceleration in the Earth-Moon Rotating System (Heliocentric System)

[T Pagano]

On Wed, 18 Apr 2018 15:52:46 -0700, Bill Rogers wrote:

> On Wednesday, April 18, 2018 at 6:05:03 PM UTC-4, T Pagano wrote:
> <snip repetition required by Tony's thread hopping>
>> *****************HELIOCENTRIC MODEL ONLY******************************
>>
>> 1. The issue has NEVER been "whether" the Earth accelerates in the
>> Earth-
>> Moon rotating system (in the Heliocentric Model).
>
>
>> 2. The issue I've raised consistently and repeatedly is the direction
>> of the "NET" acceleration vector on the Earth at any time t in the
>> heliocentric model.
>
> The direction of the "net acceleration" of the moon is directly towards
> the earth, and vice versa. That you don't understand that shows you
> don't understand the simplest aspects of Newtonian dynamics. Not
> acceleration, not orbital motion, not radial forces.


*****EARTH AND MOON IN A ROTATING SYSTEM******************

Centri- Gravit- Gravit- Centr-
fugal ational ational fugal

earth moon

<------ + ------> <------ + ------>

[COM is not depicted]


1. (Heliocentric Model) *Gravitational Forces* on the Rotating Earth-
Moon System:
(a) These forces are equal and opposite radial forces directed
towards the COM along the Earth-Moon Line.
(b) If these gravitational forces were unbalanced by the
centrifugal forces they would cause a net acceleration on the Earth and
Moon along the Earth Moon line towards the COM and the Earth and Moon
would spiral into each other.
(c) The Earth and Moon do not spiral into each other.

2. (Heliocentric Model) *Centrifugal Forces* on the Rotating Earth-Moon
System:
(a) These are radial forces opposite to the gravitational forces
on the Earth and the Moon.
(b) The magnitude of Centrifugal Force is dependent upon the
mass and velocity of the moving body and its distance from the COM.
Centrifugal Force = (m)(v^2/r). It's direction is opposite to the
gravitational force.
(c) If there is a net acceleration on the Earth in the direction
of motion the Centrifugal Force increases and the Earth recedes from the
Moon. This Earth is not receding from the Moon.
(b) If there is a net deceleration on the Earth's motion along
the direction of motion then the Centrifugal Force decreases and the
orbital diameter decreases. This is NOT occurring.
(c) The only net acceleration in the Earth-Moon rotating system
is on the Moon. The tidal bulges are actually offset from the Earth-Moon
Line. The tidal bulge on the Moon-side is slightly ahead of the Moon's
orbital path. This offset causes a slight gravitational force on the
moon offset from the Earth-Moon Line causing a slight net acceleration in
the direction of the Moon's motion. This slight increase in acceleration
causes a slight increase in the Moon's orbital velocity; which increases
the Centrifugal Force on the Moon; and causes the moon to recede from the
Earth.


3. (Heliocentric Model) Is there ANY evidence that there is a net
acceleration on the Earth in the Earth-Moon Rotating System? ABSOLUTELY
NO:
(a) The Centrifugal and Gravitational forces on the Earth are
perfectly balanced. According to Newton's First Law this means there is
no net acceleration on the Earth.
(b) The magnitude of the Earth's tangential velocity is constant;
it's angular velocity is constant. The Earth is NOT being accelerated or
decelerating in the direction of motion.
(c) The Earth's orbit is neither receding or spiraling in.
(d) The only attribute changing in the Earth's orbit is the
direction of the fixed magnitude velocity vector.

[gdgu...@gmail.com]

As always. You answer no direct questions and you learn nothing.

Just to be clear, you believe that an object that is traveling in an orbit is *not* undergoing any acceleration? And you believe that Newton would agree?

[Bruce Stephens]

On 20/04/2018 20:58, T Pagano wrote:
> (b) If these gravitational forces were unbalanced by the
> centrifugal forces they would cause a net acceleration on the Earth and
> Moon along the Earth Moon line towards the COM and the Earth and Moon
> would spiral into each other.
> (c) The Earth and Moon do not spiral into each other.

Imagine the Earth is entirely stationary and the moon is orbiting in a
perfect circle (at constant speed) around it. (Assume the Earth is at
the origin and just consider the orbital plane: so 2D.)

You can easily write down the position of the moon as a function of time
(make the orbit of unit radius, and take the simplest units of time):
(cos(t),sin(t)).

You can then calculate the moon's velocity (as a function of time): take
the first derivative (wrt t).

And then you can calculate the acceleration: just differentiate again.

What's the direction of that acceleration?

[T Pagano]

1. According to the standard explanation the acceleration vector is
perpendicular to the velocity vector.

2. However the problem with the standard explanation is that the
magnitude of that radial acceleration vector is certainly not 0:
a = v^2/r. So this net radial acceleration vector (if unbalanced) would
certainly do more than change the direction of the moon's velocity vector.

3. [In Stephens example] If there is a "net" (non-zero) radial
acceleration vector on the Moon directed towards the center of mass of
the rotating system (located at the center of the earth in Stephens'
example) then there must also be a net force vector, a net velocity
vector and a net displacement of the Moon towards the Earth. There would
be no stable circular orbit; the orbital radius would decrease.

4. What does the standard explanation ignore? The Centrifugal Force
(while considered fictitious in the standard explanation) is NOT
dispensable in explaining a stable circular orbit. As such there is NO
"NET" radial acceleration as claimed.

5. A stable circular orbit is maintained ONLY if the radial centrifugal
acceleration balances the radial centripetal acceleration on the moon.
(a) If the radial centrifugal acceleration > the radial
centripetal then the Moon recedes and
(b) if the radial centrifugal acceleration < the radial
centrifugal then we have an orbit with a decreasing orbital radius.
(c) Since the standard explanation claims there is a net
centripetal acceleration on the moon there would not be a stable circular
orbit.


6. What has been the actual state of our Moon's orbit?
(a) It is generally agreed that the moon's orbital radius is
INCREASING at a rate of 3.8cm/year.
(b) That is, the centrifugal force experienced by the moon is
greater than the centripetal force.
(c) The net radial acceleration on the moon is pointing away
from the COM.


7. How is this unbalanced condition explained?
(a) A tangential force must be applied in the direction of the
Moon's orbit accelerating the moon and causing the magnitude of its
velocity to increase.
(b) This increasing velocity (and increase in angular velocity)
is causing the centrifugal force to be greater than the centripetal
force. The radial centrifugal acceleration is greater than the radial
centripetal force causing the radius of rotation to increase (receding
orbit).

8. Where does the tangential force come from to cause the Moon to
accelerate in the direction of motion?
(a) The tidal bulges are not exactly aligned with the Earth-Moon
line; the earth's-moon-side high tide bulge is slightly ahead of the
Earth-Moon line in the direction of the Moon's orbit.
(b) Due to this offset the mass of the high tide bulge applies a
slight gravitational force to the moon offset from the earth-moon line in
the direction of rotation.
(c) A component of this force is tangential to the Moon's
direction of rotation causing the moon to accelerate slightly in the
direction of orbital motion.

[Bruce Stephens]

On 21/04/2018 17:07, T Pagano wrote:
> 1. According to the standard explanation the acceleration vector is
> perpendicular to the velocity vector.
>
> 2. However the problem with the standard explanation is that the
> magnitude of that radial acceleration vector is certainly not 0:
> a = v^2/r. So this net radial acceleration vector (if unbalanced) would
> certainly do more than change the direction of the moon's velocity vector.

If something is moving and accelerating always (and only) in a direction
perpendicular to its direction motion, then its velocity will only
change in direction (not magnitude).

Why are you avoiding the (nowadays trivial) maths?

Ignore the physics for the moment and consider a point in a plane whose
coordinates are (cos(t),sin(t)) at time t. (Use radians when thinking of
the trigonometric functions just to avoid some ugly constants.)

What kind of path does that point have?

Is its velocity as I suggested (the derivative of the position with
respect to t)?

If so, what, concretely, is that? I claim it has constant magnitude
(i.e., the point is moving with constant speed); is that true?

Is its acceleration as I suggested (the derivative of the velocity with
respect to t)?

If so, what is it? Does it, too, have constant magnitude, for example?

[Bob Casanova]

On Fri, 20 Apr 2018 21:54:53 +0100, the following appeared
in talk.origins, posted by Bruce Stephens
<bruce.r....@gmail.com>:

You do realize, don't you, that you're using terms
unfamiliar to (or at least misunderstood by) Tony, right?
Terms like, taken in order of appearance:

orbiting
function
orbit
cos
sin
velocity
derivative
acceleration
differentiate

Tony has shown repeatedly that he understands none of those.
--

Bob C.

"The most exciting phrase to hear in science,
the one that heralds new discoveries, is not
'Eureka!' but 'That's funny...'"

- Isaac Asimov

[Ernest Major]

On 21/04/2018 18:11, Bruce Stephens wrote:
> On 21/04/2018 17:07, T Pagano wrote:
>> 1.  According to the standard explanation the acceleration vector is
>> perpendicular to the velocity vector.
>>
>> 2.  However the problem with the standard explanation is that the
>> magnitude of that radial acceleration vector is certainly not 0:
>> a = v^2/r.   So this net radial acceleration vector (if unbalanced) would
>> certainly do more than change the direction of the moon's velocity
>> vector.
>
> If something is moving and accelerating always (and only) in a direction
> perpendicular to its direction motion, then its velocity will only
> change in direction (not magnitude).

Reading Tony gets one confused about physics, but that's not true. It's
true in the case of a circular orbit, and there might be other examples,
but, for example, the magnitude of the moon's velocity is not constant -
the moon has an elliptical orbit and moves faster at perigee than at apogee.

It's also too advanced for Tony, as he presents himself here. We first
need him to understand Newton's first law of motion, and indicate
whether he accepts or rejects it. (His current argument seems to involve
rejection of Newton's first law, but it is hard to tell whether Tony
rejects something or just doesn't understand the implications.)

>
> Why are you avoiding the (nowadays trivial) maths?
>
> Ignore the physics for the moment and consider a point in a plane whose
> coordinates are (cos(t),sin(t)) at time t. (Use radians when thinking of
> the trigonometric functions just to avoid some ugly constants.)
>
> What kind of path does that point have?
>
> Is its velocity as I suggested (the derivative of the position with
> respect to t)?
>
> If so, what, concretely, is that? I claim it has constant magnitude
> (i.e., the point is moving with constant speed); is that true?
>
> Is its acceleration as I suggested (the derivative of the velocity with
> respect to t)?
>
> If so, what is it? Does it, too, have constant magnitude, for example?
>

--
alias Ernest Major

[Don Cates]

On 2018-04-21 12:46 PM, Ernest Major wrote:
> On 21/04/2018 18:11, Bruce Stephens wrote:
>> On 21/04/2018 17:07, T Pagano wrote:
>>> 1.  According to the standard explanation the acceleration vector is
>>> perpendicular to the velocity vector.
>>>
>>> 2.  However the problem with the standard explanation is that the
>>> magnitude of that radial acceleration vector is certainly not 0:
>>> a = v^2/r.   So this net radial acceleration vector (if unbalanced)
>>> would
>>> certainly do more than change the direction of the moon's velocity
>>> vector.
>>
>> If something is moving and accelerating always (and only) in a
>> direction perpendicular to its direction motion, then its velocity
>> will only change in direction (not magnitude).
>
> Reading Tony gets one confused about physics, but that's not true.

Yes it is true. His criteria that it's "accelerating always (and only)
in a direction perpendicular to its direction motion" defines a circular
orbit. Something in an elliptical orbit has an acceleration that is not
always perpendicular to its direction of motion.

It's
> true in the case of a circular orbit, and there might be other examples,
> but, for example, the magnitude of the moon's velocity is not constant -
> the moon has an elliptical orbit and moves faster at perigee than at
> apogee.
>
> It's also too advanced for Tony, as he presents himself here. We first
> need him to understand Newton's first law of motion, and indicate
> whether he accepts or rejects it. (His current argument seems to involve
> rejection of Newton's first law, but it is hard to tell whether Tony
> rejects something or just doesn't understand the implications.)
>>
>> Why are you avoiding the (nowadays trivial) maths?
>>
>> Ignore the physics for the moment and consider a point in a plane
>> whose coordinates are (cos(t),sin(t)) at time t. (Use radians when
>> thinking of the trigonometric functions just to avoid some ugly
>> constants.)
>>
>> What kind of path does that point have?
>>
>> Is its velocity as I suggested (the derivative of the position with
>> respect to t)?
>>
>> If so, what, concretely, is that? I claim it has constant magnitude
>> (i.e., the point is moving with constant speed); is that true?
>>
>> Is its acceleration as I suggested (the derivative of the velocity
>> with respect to t)?
>>
>> If so, what is it? Does it, too, have constant magnitude, for example?
>>
>
>


--

--
Don Cates ("he's a cunning rascal" PN)

[Bruce Stephens]

On 21/04/2018 18:46, Ernest Major wrote:
> Reading Tony gets one confused about physics, but that's not true. It's
> true in the case of a circular orbit, and there might be other examples,
> but, for example, the magnitude of the moon's velocity is not constant -
> the moon has an elliptical orbit and moves faster at perigee than at
> apogee.

It's quite possible I'm confused (whether by Tony or for other reasons).
But for something in an elliptical orbit the acceleration isn't always
perpendicular to the motion is it?

[Ernest Major]

On 21/04/2018 19:03, Don Cates wrote:
> On 2018-04-21 12:46 PM, Ernest Major wrote:
>> On 21/04/2018 18:11, Bruce Stephens wrote:
>>> On 21/04/2018 17:07, T Pagano wrote:
>>>> 1.  According to the standard explanation the acceleration vector is
>>>> perpendicular to the velocity vector.
>>>>
>>>> 2.  However the problem with the standard explanation is that the
>>>> magnitude of that radial acceleration vector is certainly not 0:
>>>> a = v^2/r.   So this net radial acceleration vector (if unbalanced)
>>>> would
>>>> certainly do more than change the direction of the moon's velocity
>>>> vector.
>>>
>>> If something is moving and accelerating always (and only) in a
>>> direction perpendicular to its direction motion, then its velocity
>>> will only change in direction (not magnitude).
>>
>> Reading Tony gets one confused about physics, but that's not true.
>
> Yes it is true. His criteria that it's "accelerating always (and only)
> in a direction perpendicular to its direction motion" defines a circular
> orbit. Something in an elliptical orbit has an acceleration that is not
> always perpendicular to its direction of motion.

Oops. (I did say that reading Tony causes confusion.)

I thought for some seconds that perigee and apogee in elliptical orbits
were an exception - at those points the acceleration is instantaneously
perpendicular to the motion, but it's also at those points that the
change in magnitude is also instantaneously zero.

alias Ernest Major

[Ernest Major]

Yep. It's me that was confused.

--
alias Ernest Major

[Bill Rogers]

Tony, according to Newton a symmetric shell of matter exerts no gravitational force on its interior. If you think about that means that at the center of the earth you would be weightless. You'll eventually be able to tell us whether that's true or not, if you keep digging yourself into a progressively deeper hole.

[T Pagano]

On Sat, 21 Apr 2018 18:11:45 +0100, Bruce Stephens wrote:

> On 21/04/2018 17:07, T Pagano wrote:
>> 1. According to the standard explanation the acceleration vector is
>> perpendicular to the velocity vector.
>>
>> 2. However the problem with the standard explanation is that the
>> magnitude of that radial acceleration vector is certainly not 0:
>> a = v^2/r. So this net radial acceleration vector (if unbalanced)
>> would certainly do more than change the direction of the moon's
>> velocity vector.
>
> If something is moving and accelerating always (and only) in a direction
> perpendicular to its direction motion, then its velocity will only
> change in direction (not magnitude).
>

*STEPHENS' EXPLANATION OF THE ACCELERATION VECTOR ON A ROTATING BODY*

1. Let's assume for the sake of argument that I agree with Stephens'
explanation.
(a) That is, that the acceleration vector (perpendicular to the
velocity vector) on the Moon (in Stephens' example) has *NO OTHER EFFECT*
but to change the direction of the Moon's velocity vector.
(b) Its only effect is to keep the velocity vector tangential to
the circular orbit.
(c) And therefore the distance between moon and Earth does NOT
change.



***STEPHENS' ACCELERATION VECTOR EXPLANATION APPLIED TO THE EARTH**

2. Let's focus back to the Earth-Moon rotating system in the
Heliocentric Model and specifically back to the Earth.
(a) The Earth is rotating around the COM of the Earth-Moon
system with a tangential velocity vector and a perpendicular acceleration
vector.
(b) According to the Stephens' explanation this perpendicular
acceleration vector produces NO OTHER EFFECT but to change the direction
of the Earth's velocity vector. There is no change in the distance
between the Earth and Moon.
(C) Unfortunately this *conflicts* with Rogers' understanding of
the Differential Acceleration Theory of Tides.



**ROGERS' THEORY OF TIDES CONFLICTS WITH STEPHENS' ACCEL EXPLANATION**

3. Stephens' explanation of the radial acceleration vector on a body in
a circular orbit CONFLICTS with Rogers' understanding of the Differential
Acceleration Theory of Tides.

(a) According to Rogers' understanding of the Theory of Tides
(and from a frame of reference off of the Earth) there is a difference in
the magnitude of the acceleration vector (due to the moon's gravitational
force) on the far-side-ocean, Earth center and the near-side-ocean. A
rough depiction is shown below with the length of the acceleration vector
showing the relative difference in magnitude of the acceleration vectors.



far side ocean earth center near side ocean moon
acceleration acceleration acceleration

+ -> + ----> + --------> *
|
|
\|/ tangential velocity vector of earth


(b) CAUSE OF THE FAR-SIDE-TIDAL-BULGE: According to Rogers the
Earth is literally displacing towards the moon at a greater rate
(indicated by the differing acceleration vectors) than the far-side-ocean
causing the far-side-ocean to be literally "left behind" and hence bulge
out.

(c) So we have a conflict between Stephens' explanation about
the effect of the acceleration vector and Rogers' explanation in his
understanding of the theory of tides.

(1) According to Stephens the acceleration vector on the
Earth (perpendicular to the Earth's velocity vector) is ONLY causing the
direction of the Earth's velocity vector to remain tangential to the
circular orbit----IT HAS NOT OTHER EFFECT.

(2) According to Rogers' understanding of the theory of
tides the Earth is actually displacing towards the moon at a rate
determined by its acceleration vector. That rate is greater than the
rate of the far-side-ocean causing the far-side-ocean to bulge out.

[Bruce Stephens]

On 22/04/2018 15:57, T Pagano wrote:
> (1) According to Stephens the acceleration vector on the
> Earth (perpendicular to the Earth's velocity vector) is ONLY causing the
> direction of the Earth's velocity vector to remain tangential to the
> circular orbit----IT HAS NOT OTHER EFFECT.

I'm talking about a very idealised (simplified) system: one point moving
in a plane, in a circle around the origin at constant speed.

I was using this trivial system to show how one can *calculate* the
velocity and acceleration. (When I say "one can calculate", apparently
that doesn't include you since you've still failed to follow through
with the simple calculus.)

I'm just trying to argue that the acceleration (of this single point
moving in a circle at constant speed around the origin) is towards the
centre. That's all.

Given that you don't accept that, your intuitions about other possible
inconsistencies don't seem worth much.

[Dennis Feenstra]

" Is there ANY evidence that there is a net
acceleration on the Earth in the Earth-Moon Rotating System? ABSOLUTELY
NO"

You cannot have any orbit without acceleration.
You cannot have the moon orbit the earth with a barycenter at the center of the earth.
Hence, the Earth orbits the common barycenter.
Hence, the Earth undergoes acceleration.

[Don Cates]

You underestimate the depth of Tony's ignorance.

[jillery]

... and Tony's pride in his ignorance.

--
I disapprove of what you say, but I will defend to the death your right to say it.

Evelyn Beatrice Hall
Attributed to Voltaire

[Bob Casanova]

On Mon, 23 Apr 2018 08:52:44 -0500, the following appeared
in talk.origins, posted by Don Cates
<cate...@hotmail.com.invalid>:

Well, it seems one cannot *over*estimate that depth...

[gdgu...@gmail.com]

On Sunday, April 22, 2018 at 11:00:03 AM UTC-4, T Pagano wrote:

>
> (c) So we have a conflict between Stephens' explanation about
> the effect of the acceleration vector and Rogers' explanation in his
> understanding of the theory of tides.
>

We don't have a conflict, although I suppose I could see why you might think there is one.

> (1) According to Stephens the acceleration vector on the
> Earth (perpendicular to the Earth's velocity vector) is ONLY causing the
> direction of the Earth's velocity vector to remain tangential to the
> circular orbit----IT HAS NOT OTHER EFFECT.

That's an odd way to put it. The acceleration caused by the gravity between the Earth and Moon causes each body to *turn*, continuously as compared with the trajectories they would take absent that gravity. Looked at from far "north" in space. we'd see the Moon going around counter-clockwise - continually "turning left". If we were pretty observant, we'd see the Earth doing the same thing, but in a very small circle, smaller than the radius of the Earth.

That's acceleration. Toward the other body.

>
> (2) According to Rogers' understanding of the theory of
> tides the Earth is actually displacing towards the moon at a rate
> determined by its acceleration vector. That rate is greater than the
> rate of the far-side-ocean causing the far-side-ocean to bulge out.

And so it is, "displacing" more toward the Moon than it would have absent the Moon's gravity. But, as it is in a stable orbit, not getting closer to the Moon (to be more precise, the orbit is not a perfect circle; the distance actually does change in a cycle).

The gravity between the Earth and Moon causes each body to travel in an ellipse, at every moment "turning" toward the other body. The Moon side ocean turns a little more quickly than the body of the Earth. The far side ocean a little less quickly. Voila. Diurnal tides. There is no conflict.