[allplanets-hollow] The Barrycenter

Scott comments in bold italics below

Scott,

I've been reading over your explanation for the tides. I understand
that the barrycenter is a critical point, but first of all I question, where
is the barrycenter? Where does it begin? I've repeated some interesting
comments by Cater in this regard at the bottom. Please comment.

But I'm not here looking for an excuse to put forward Cater's
theories, for once ( don't anybody fall out of their chair ). I have a

question: What are you saying when you say that an object, a

baseball, at the barrycenter would fall towards the sun with its

gravity cancelled out and fly off in a straight line and leave the

solar system forever? The sentence seems to be mixed up there,

and there seems to be a contradiction. Then you say " This means

that at any point in time, the portions of the Earth's surface that

are closer to the Sun than the barrycenter are prone to sheering
forces in that direction. They are a matter of no more than a

thousand miles or so closer to the sun than the perfect balance

point which is the barrycenter.

Dean, I was referring to the balance of forces. Any

object according to standard physics will have an inertia either

to remain in motion or to remain at rest. Any body that orbits

the Sun has a forward velocity of motion that if not combined

with the gravitational force from the Sun, would cause it to

fly in a straight line since there would be no force acting upon

**it to bend its flight or natural inertia to move in a straight line. **

In geometry, forces are measured as the combination of vectors

all acting on the same object. The barrycenter is that point

within the Earth where the Earth Moon **system orbits the Sun. **

I was sure everyone was aware that the Moon causes the Earth

to wobble in its orbit as the Earth/Moon system orbits the Sun.

The history of the barrycenter is that two astronomers

**were viewing a pair of planetoids or asteroids and they noticed **

from those observations that the time period of their

reappearance on far side of the Earth varied. They worked

it out that the Earth had a definite wobble in its course

around the sun. I don't have the exact particulars, but this

observation is actually in my eyes the source of a faulty

reckoning that the mass of the Moon is 1 part in roughly

81.3 compared to the Earth. If you take the mean orbital

radius of the Moon traveling around the Earth and divide

it by the measurement of the barrycenter you arrive nearly

exactly at the ratio of mass **I just mentioned. **

This value is divided into the accepted value for the

Earth's mass and you get the value of the moon's mass as

it is claimed to be. Y****ou arrive at exactly the one sixth

gravity that has always been claimed for the Moon surface

**drop rate or little g. **

To me this is bad science. Not only is the value of

gravity in the alleged Moon walk photographs nothing like

what a 1/6th gravity ought to show but before even considering

the question of which body provides what tidal motion on the

earth, we should have an accurate mass figure to begin with.

One time I asked a question of one of the AOL guru's

that are there to help students with science questions. I asked

what the effect would be on the barrycenter if the Moon were to

be moved further away from the Earth. One guy provided me

with lots of math and his conclusion was that since the Earth is

81.3 times the mass of the Moon, the barrycenter would always

be at the fulcrum or distance value of 1 / 81.3 times the mean

orbital distance that the Moon orbited the Earth by. He said

with great confidence that the barrycenter would move further

away from the center of the Earth.

One example that would support their fallacious argument

is that of a solid stick and the laws of leverage. That formula

is Weight times Leverage distance. In other words, this is the

reason one man once stated that if given a long enough pry bar

and a place to stand, he could lift the world. If the distance

between the Earth and Moon were to increase and they were

both balanced at the end of a long stick, then yes, the point of

fulcrum or barrycenter would move further away from the center

of the Earth, but that isn't the correct analogy for orbiting

planetary bodies even though it was exactly the same way this

AOL guru was approaching the question.

I find that his reasoning is false. In fact, the barrycenter

would move closer to the center of the planet (Earth), not away!

Two reasons make this clear to me. The first is that as bodies

orbit their primaries at ever greater distances, the velocity of

the orbiting planet is decreased. This causes the orbiting Moon

to take longer to complete one orbit. Secondly, and in

conjunction to the first point is that the further away one body

is from another, the less gravitational force there is between

the same two bodies. It is akin to an example I once told of

an adult grasping a 10 year old child and spinning the child

around himself in slow circles. Due to the weight of the child,

the adult has to lean in the opposite direction than where the

child is to balance **the tandem spin act they are both performing. **

If, on the other hand, an adult holds an object of lesser weight,

perhaps holding a smaller toddler, and very gently swings this

**burden **in circles, they are not required to lean near as much,

**if at all. **

In this case, the smaller child or object that is being swung

in slow lazy circles, is representative of the same Moon which

though containing the same mass, is at an increased distance.

This extra distance makes for less force exerted against the

Earth, which is the primary of the Moon's orbit and due both

to the inverse square law and Kepler's third law of planetary

**motion less force is exchanged between the two. **

It may further be stated that due to the conditions I have

just set out, the arc or circular shape of the orbit of increased

mean radius is slightly more straight than when the Moon was

orbiting closer to the Earth. If we consider it to be a question

of vector magnitudes, my point can be further proven as

sound reasoning:

The moon's mass or weight if you will is constant. If this

body is moving at 0.63 miles per second velocity in a straight

line, and you require the body to be continually turned slightly

so that it will orbit the Earth, a given force is required to make

it do so. In geometric terms this force can be assigned a rate

of acceleration which is the result of gravity inward, or a

centripetal force toward the center of its orbit. Anybody who

knows how to work with the geosynchronous orbit formula

knows that the further out one orbits, the slower that the orbiting

body is moving (Kepler's proven 3rd law of planetary motion)

and, according to the result of the formula, as distance increases,

the more the drop rate or acceleration slows. Another example

of all this is that it takes more force to alter the course of an

object propelled at high velocity ( force = mass times velocity )

than it does to alter the course of the same projectile propelled

at a slower velocity. There can be no doubt as to this reasoning!

**Force is mass times acceleration or velocity. ** It is a foregone

**conclusion that the farther out a Moon orbits at, ** the less force

**is required upon the primary to maintain an orbit. ** So, I believe

that my logic in this is quite undeniable. The further out the

Moon orbits the earth, were a change in mean radius to occur,

the less force between both Earth and Moon would exist.

Reflecting my analogy, the rate of spin slows down and

since the forces that the adult has to balance against are lessened,

the less lean they are required to maintain to remain in balance

standing on a stationary spot**. This means that the barrycenter**

moves inward toward the center of the Earth since the influence

of a slower orbiting more remote object is less.

If they are all wet as to what would happen to the

barrycenter if the Moon's orbit grew in orbital circumference,

and I believe that I just proved that, then how can we trust in

the methodology that they used in determining the mass

**relationships between the Earth and the Moon? If we don'****t **

know that for certain, how is it possible for anyone to promote

**reliable theory pertaining to Lunar tides? This why I have **

**concentrated first on the Solar tides and I also don't rely on **

**force equations that to me must be in doubt being based on **

**faulty assumptions regarding the mass and thus the gravity **

**of the Moon. **

Now I am really confused. According to Cater's statement below,
this barrypoint is further away than 1,000 miles.
Cater's comments:

I read Mr. Cater's statement but it is not the barrycenter

he is referring to. He is talking about the neutral gravity

or gravi-spheres between the two bodies at which point the

gravitational forces between the Earth and Moon are in

perfect balance, or equivalence. The fact that Mr. Cater

**brings out is very correct in that the Moon mission **

rockets began to accelerate 44 or 45 thousand miles

away from the Moon rather than 22 thousand miles. The

only way this relates to the tides is the relationship to the

Lunar tide force which is really much stronger than the

1/6th surface gravity of Earth that is claimed. It cannot

be argued using establishment techniques that the Moon

has only 1/6th the surface gravity of earth since equal

gravi-spheres must reflect equal accelerations in both

directions or zero cm/sec^2 at that position.

Actually, and this is only a suspicion on my part,

I have long suspected that the equal gravi-spheres position

is actually even further away from the Moon than is stated

in the 45,000 mile figure. The reason for this belief is that

the space vehicle has been stated as travelling at about

2000 miles per hour at this point where acceleration began

to increase again after having slowed down escaping from

the Earth. This makes me curious. It seems to me that

there has never been a space vehicle that was perfectly at

rest in relationship between the the Earth and the Moon at

the equal gravi-spheres or neutral gravity position. All these

vehicles have been moving toward or away from the Moon

at velocity.

I wonder if a vehicle that WAS perfectly stopped at this

point and then nudged toward the Moon would not take a

considerable time to accelerate to the point where the dying

acceleration away from the Earth would be finally exceeded

by the accumulated ( 1/2 * rate * time ^2 ) acceleration formula

**for falling bodies. The question I have is whether or not the **

**space vehicle immediately benefits from crossing over the **

**equal gravi-sphere position or whether the rate of acceleration **

**must first surpass the vehicles current declining velocity in **

order for acceleration to commence. If equal gravi-spheres

were passed by space vehicles, would it not stand to reason

that **the natural **acceleration of the inverse square law would

**take **a while to exceed 2000 miles per hour or whatever the

velocity would **be at that point. ** I may be wrong about this

belief but I **don't believe it is **pivotal to the tide discussion.

****> " The anemic jumping feats of the astronauts under alleged,

ยทยทยท

On Fri, 16 Mar 2001 21:54:05 -0300 "Dean De Lucia" [email protected] writes:

one-sixth Earth gravity, as shown on the telecasts, represent

only part of the evidence of a high moon gravity. The consistent

reports of the point where the space ships entered the

gravitational influence of the moon indicated moon gravity
comparable to that of the Earth. If the Moon's surface gravity

were only one-sixth earth gravity, this point of entry, or the

point where the gravitational influence of the moon exceeds

that of Earth, would be approximately 22,000 miles from

the moon. This can easily be confirmed by elementary

mathematics and mechanics and will not be given here. The
distance will vary slightly percentagewise because the Moon's
distance from the Earth fluctuates. Since the advent of the

Apollo missions, the distance reported reported for this

point of entry has been consistently much greater than

22,000 miles. The distances claimed by various writers,

as well as the media, have varied from about 39,000 to

nearly 44,000 miles. This is, indeed, incredible since it

contradicts the consistent claims of a low moon gravity.

Interestingly enough, prior to the space program, this
distance was always given in the 20,000 to 22,000 range,
corresponding to one-sixth Earth gravity of the Moon. It

can be found in a number of earlier textbooks, including

Encyclopedia Britannica. Yet, the later editions of

Encyclopedia Britannica put this distance at about the 40,000

mile range. "

I will re-send Scott's explanation so that everyone can get

a look at it and refresh memories. It's important for the HET

to get to the bottom of gravity, and the behavior of tides

tells us a lot.

But if it ain't your bag, don't worry about it.

Dharma/Dean