[totallackey]

They are basically dimensionless... 

Funny!

In response to Tome, you patently admit your whole routine is total shinola!

....

And neither do you demonstrate as to how the Sun's motion is insignificant to the orbital dynamics of the planets around it while traipses around the fictional Milky Way!!!

You have not a clue what you are talking about.

To your first ridiculous point: Newton's theory does not depend on the size of the objects, just their mass and the radius between them.  Even a complete idiot can see this:


f= Gm₁m₂/r²


That's the theory - where in there are the dimensions required?  They are not.  They have no bearing on the situation.

To your 2nd ridiculous point: The solar system is an inertial frame of reference WRT to the sun.  Uniform motion of the sun will not and can not affect the solar system as every body in it is also moving along with the sun at exactly the same speed in the direction of the sun's motion.  That's the explanation, there isn't any more one can say about it.  You don't understand that because you don't have the first clue about the mechanics of motion.  Why do you insist on displaying your lack of knowledge on this?

The onus is on you there Copernicus-lover...

Make me a CGI model of the heliocentric solar system, utilizing Newton, Kepler, and Einstein...

Not excuses!

Ahhh, I provided the model and it uses Newton, the onus is on you to prove it does not work.  I know you can't and I hate discussing things with people that proceed proudly from a position of profound ignorance, so I'll give you the last word ... our discussion is done until you provide some tangible proof.

So in your fictional construct, things that have mass must exist in a world without dimensions...

And you call my points silly?

WTF is the matter with you!?!?!

[AATW]

Great one Bill! Your main ammunition is an obvious typo, rather than any real or fundamental error. You even put it into your sig. Looks like a pretty desperate tactic.

It's certainly no match for your main ammunition of not being able to understand anything and then declaring yourself right.

We can all see how weak the argument is: Declaring that the moon is close to the earth and then crossing out the physics. High level stuff.

Yes, look at all the FE people queuing up to agree with you.
It's not even an argument, he's literally done the maths for you. Rene is getting himself in a muddle because he hasn't factored in the pull of the sun on the earth.

[Tom Bishop]

It's not even an argument, he's literally done the maths for you. Rene is getting himself in a muddle because he hasn't factored in the pull of the sun on the earth.

http://www.ralphrene.com/

My "Unproof of Newtonian Gravity" was one of my first. Using the Newtonian Formula, the listed gravitational constant and the accepted masses, and distances of Sun, Earth & Moon, I found that during New Moon, the Sun's attraction for the Moon Is almost 3 times that of the Earth. Therefore, once every month the Moon, obeying the stronger force, should cease orbiting the Earth and head toward the Sun. It doesn't!

Another physisist gas bag working for the National Science Foundation told me my Unproof of Gravity I was wrong because the Sun also pulls on the Earth. In other words you can no longer isolate forces in Physics. If this is the case every force table experiment must be discredited.

We can no longer isolate forces in physics?

Declaring that the moon and the earth are close, so it's practically the same thing, and then literally crossing out the physics equations that are working against you is not a valid solution to this.

[markjo]

It's not even an argument, he's literally done the maths for you. Rene is getting himself in a muddle because he hasn't factored in the pull of the sun on the earth.

http://www.ralphrene.com/

My "Unproof of Newtonian Gravity" was one of my first. Using the Newtonian Formula, the listed gravitational constant and the accepted masses, and distances of Sun, Earth & Moon, I found that during New Moon, the Sun's attraction for the Moon Is almost 3 times that of the Earth. Therefore, once every month the Moon, obeying the stronger force, should cease orbiting the Earth and head toward the Sun. It doesn't!

Another physisist gas bag working for the National Science Foundation told me my Unproof of Gravity I was wrong because the Sun also pulls on the Earth. In other words you can no longer isolate forces in Physics. If this is the case every force table experiment must be discredited.

We can no longer isolate forces in physics?

Declaring that the moon and the earth are close, so it's practically the same thing, and then literally crossing out the physics equations that are working against you is not a valid solution to this.

Just out of curiosity, do you or Ralph Rene ever take the earth's and moon's velocity into consideration when calculating their expected orbits?

Also, do you consider an n-body solution to be the only valid way to model the solar system?  If so, then why?

[BillO]

What did I tell you?  He has no understanding of even basic mathematical rules.

(A/A)*(B/C)=(B/C)*(D₁/D₂)


Since: A/A=1, therefore:

B/C=(B/C)*(D₁/D₂)

(B/C)*(C/B)=D₁/D₂


Since (B/C)*(C/B)=1, therefore:

D₁/D₂=1

D₁=D₂


Where A=GM/d², B=m_1, C=m₂, D₁=a₁, D₂=a₂

I'm betting this won't help either.

[Tom Bishop]

It's not even an argument, he's literally done the maths for you. Rene is getting himself in a muddle because he hasn't factored in the pull of the sun on the earth.

http://www.ralphrene.com/

My "Unproof of Newtonian Gravity" was one of my first. Using the Newtonian Formula, the listed gravitational constant and the accepted masses, and distances of Sun, Earth & Moon, I found that during New Moon, the Sun's attraction for the Moon Is almost 3 times that of the Earth. Therefore, once every month the Moon, obeying the stronger force, should cease orbiting the Earth and head toward the Sun. It doesn't!

Another physisist gas bag working for the National Science Foundation told me my Unproof of Gravity I was wrong because the Sun also pulls on the Earth. In other words you can no longer isolate forces in Physics. If this is the case every force table experiment must be discredited.

We can no longer isolate forces in physics?

Declaring that the moon and the earth are close, so it's practically the same thing, and then literally crossing out the physics equations that are working against you is not a valid solution to this.

Just out of curiosity, do you or Ralph Rene ever take the earth's and moon's velocity into consideration when calculating their expected orbits?

Also, do you consider an n-body solution to be the only valid way to model the solar system?  If so, then why?

As Rene says, at the point of New Moon the moon is experiencing the following forces:

Sun Force = 8.872 E13
Earth Force = 3.636 E13

Those were generated by Newton's Force = Mass / Distance^2. The distances between the earth and the moon, and the moon and the sun, are already accounted for in that equation. Declaring that the moon and earth is close, and are practically the same, so we should cross out the equations, is nonsensical. Newton's equation tells us how much force the moon is experiencing, with the distances already accounted for.

Yet we are supposed to believe that after the New Moon the moon will then move towards the earth?

I don't see how the speed of the moon has anything to do with it. A photon traveling at light speed would also move towards the greater force.

The n-body and three body solutions do isolate the forces; and this is why there are no Three Body Problem solutions that look anything like a heliocentric system. The forces are accounted for and the heliocentric system cannot exist.

[markjo]

As Rene says, at the point of New Moon the moon is experiencing the following forces:

Sun Force = 8.872 E13
Earth Force = 3.636 E13

Those were generated by Newton's Force = Mass / Distance^2. The distances between the earth and the moon, and the moon and the sun, are already accounted for in that equation. Declaring that the moon and earth is close, and are practically the same, so we should cross out the equations, is nonsensical. Newton's equation tells us how much force the moon is experiencing, with the distances already accounted for.

Yet we are supposed to believe that after the New Moon the moon will then move towards the earth?

How much force does the sun exert on the earth and how does that compare with the force that the sun exerts on the moon?

I don't see how the speed of the moon has anything to do with it. A photon traveling at light speed would also move towards the greater force.

Are you saying that the velocity of an object has nothing to do with its orbit around another object?

The n-body and three body solutions do isolate the forces; and this is why there are no Three Body Problem solutions that look anything like a heliocentric system. The forces are accounted for and the heliocentric system cannot exist.

Tom, are you insisting on an analytical solution to an n-body solar system or will a numerical solution suffice?  Granted, an analytical solution isn't possible, but numerical solutions are possible.

[garygreen]

As Rene says, at the point of New Moon the moon is experiencing the following forces:

Sun Force = 8.872 E13
Earth Force = 3.636 E13

Those were generated by Newton's Force = Mass / Distance^2. The distances between the earth and the moon, and the moon and the sun, are already accounted for in that equation. Declaring that the moon and earth is close, and are practically the same, so we should cross out the equations, is nonsensical. Newton's equation tells us how much force the moon is experiencing, with the distances already accounted for.

a 0.5% difference in distance is a very small difference.  1 and 1.005 are both basically 1.

if you don't like that i cancelled out the distance terms, then just plug in the real values and do the calculation.  it's just algebra.

acceleration of the earth due to the sun = 0.0057 m/s^2

acceleration of the moon due to the sun = 0.0058 m/s^2

[BillO]

Sun Force = 8.872 E13
Earth Force = 3.636 E13

I wonder if the person that posted this realizes that an orbiting body is in free-fall and experiences zero net force?  No, I guess not - won't work on a fake earth.

If I'm not banned for life by tomorrow or the next day, I'll 'do the math' on this.

[HorstFue]

As Rene says, at the point of New Moon the moon is experiencing the following forces:

Sun Force = 8.872 E13
Earth Force = 3.636 E13

Those were generated by Newton's Force = Mass / Distance^2. The distances between the earth and the moon, and the moon and the sun, are already accounted for in that equation. Declaring that the moon and earth is close, and are practically the same, so we should cross out the equations, is nonsensical. Newton's equation tells us how much force the moon is experiencing, with the distances already accounted for.

Yet we are supposed to believe that after the New Moon the moon will then move towards the earth?

Gravity force of the sun doesn't matter. It's already canceled out by the centrifugal force.

Let's for a moment assume, earth does not attract the moon, e.g. assuming a moon being on the same orbit around the sun as earth, but a little behind, so that earth attraction is minimal. The moon would follow earth in a distance with exact the same speed. For the orbit mass does not matter, because both gravity and centrifugal force are proportional to the mass. So a small object can have the same orbit as a big object. So moon could have exact the same orbit as earth.

Now bring earth and moon back to their old "positions". At average both Earth and moon would still follow this orbit around the sun from above, but additionally orbit around themselves. The combined center of mass from earth and moon is still exactly on the orbit from above.

[BillO]

I promised this, so I'll deliver - even though I know it will be dismissed.  However the math is here for you to try yourself.

Fist let me point out someone is not doing their math right, and someone using that math is not even bothering to check it.  Because these numbers:

Sun Force = 8.872 E13
Earth Force = 3.636 E13

Are pure hokum.  Zetetic indeed.  See ** at the end.

Now to get on with it...

So, an object in a nearly circular orbit about a mass is neither accelerating toward the mass or away from it, but instead is traveling in more or less uniform velocity around the mass.  In other words is exist in a state where at any point on its orbit the force, and therefore the accelerations’ on it are in equilibrium.  There are two sources of acceleration, those from gravity and those that arise centrifugally form the orbit itself.  Here the task is to show that the accelerations on the moon are in equilibrium.  They must all add up to 0.  Mathematically stated: a_g-a_c=0, for each orbital instance.

Let’s use Mr. Rene's facts and figures.  According to him we are looking at moon’s orbit around the sun of

r = 1.498E011 m, and a sun’s mass of m=1.991E30 kg

And for the moon’s orbit around earth he gives the following:

r = 4.055E8 m, and an earth’s mass of 5.979E24 kg

He then goes on to calculate something he calls ‘relative force’, there is no such thing, but if we look carefully at the equation he uses:

1)   RF = MASS/DISTANCE², which we can see is similar to the equation for acceleration due to gravity:

2)   a = Gm/r², where G is the gravitational constant (6.67E-11), m is the mass of the object in kg, r the radius of the object’s orbit in meters, and a is the acceleration in m/s²

We are going to need that acceleration.  We can get that we can correct Rene’s ‘work’ substituting RF for  m/r² to get:

a_g = G*RF

Further, in order to determine the centrifugal acceleration we are also going to need to calculate the orbital velocity of each of these orbits.

3)   v = sqrt(Gm/r)

And the centrifugal acceleration, which is given by

4)   a_c = v²/r

So let’s begin.  We will first calculate the accelerations on the sun-moon instance using Rene’s numbers:

a_{g (sun-moon)} = 6.67E-11*1.991E30/(1.498E11)²

a_{g (sun-moon)} = 5.918E-3 m/s²

Now we’ll look at the centrifugal acceleration.  Using eq. 3) we get:

v_(sun-moon) = sqrt(6.67E-11*1.991E30/1.498E11)
v_(sun-moon) = 29774 m/s

Then from eq. 4) we get the centrifugal acceleration as:

a_{c (sun-moon)} = 29774²/1.498E11
a_{c (sun-moon)} = 5.918E-3 m/s²

Such that:

a_{g (sun-moon)}- a_{c (sun-moon)} = 5.918E-3 m/s²- 5.918E-3 m/s² = 0

Well, would you look at that!  No net acceleration and hence no net force.

Now let’s now look at the earth-moon instance:

a_{g (earth-moon)} = 6.67E-11*5.979E24/(4.055E08)²

a_{g (earth-moon)} = 2.425E-3 m/s²

Now we’ll look at the centrifugal acceleration as before:

v_(earth-moon) = sqrt(6.67E-11*5.979E24/4.055E08)
v_(earth-moon) = 991.7 m/s

And the centrifugal acceleration:

a_{c (earth-moon)} = 991.7²/4.055E08
a_{c (earth-moon)} = 2.425E-3 m/s²

Such that:
a_{g (earth-moon)}- a_{c (earth-moon)} = 2.425E-3 m/s²- 2.425E-3 m/s² = 0

My, oh my…

Just as a check on myself, let’s see if the ratio in the gravitational accelerations I calculated for the sun-moon and earth-moon match the ratio given by Rene.
5.918E-3 m/s²/ 2.425E-3 m/s² = 2.440

Conclusion, there is no net acceleration or force acting on the moon due to the earth or sun.

q.e.d.


**Now just for fun let me note something for those of you that think of Rene as a minor prophet for your FE hypothesis.    AS I stated before there is no ‘relative mass’ thingy calculated as m/r² and further,  he mixes his unit systems.  In the real world of physics, there are 3 unit systems usually used. cgs/Gaussian (centimeter-gram-second), mks/SI (meter-kilogram-second) and, rarely BE/fsg (foot-slug-second).  Your Mr. Rene used kg with km and ended up over estimating his ‘relative force’, whatever that is,  by a factor of 1 million!!!!  What a cac snámh!  And you trust this dude?  I will note that in the end he was taking a ratio and since this whole thing is linear in nature the ratio turned out to be correct and I admitted that before – however ignorance + 2 wrongs don’t make a right.  The real numbers:

Sun Force = 4.347E20 N
Earth Force = 1.781E20 N

Still a ratio of 2.440, but by doing the real math and producing the real numbers.  When you copy a cac snámh, you make yourself a cac snámh.  I thought zetetics were not supposed to copy anyone and do the research themselves?

[Tom Bishop]

So, an object in a nearly circular orbit about a mass is neither accelerating toward the mass or away from it, but instead is traveling in more or less uniform velocity around the mass.  In other words is exist in a state where at any point on its orbit the force, and therefore the accelerations’ on it are in equilibrium.  There are two sources of acceleration, those from gravity and those that arise centrifugally form the orbit itself.

...

Such that:

a_{g (sun-moon)}- a_{c (sun-moon)} = 5.918E-3 m/s²- 5.918E-3 m/s² = 0

...

Such that:
a_{g (earth-moon)}- a_{c (earth-moon)} = 2.425E-3 m/s²- 2.425E-3 m/s² = 0

You are just saying that because a two-body orbit is possible, where the inward pull is balanced by the outwards centrifugal pull, that orbits can exist in any situation, and we can just stack them on top of each other.

If the moon were perfectly orbiting the sun, with just those two bodies alone, the moon would have a net acceleration of 0 towards the sun. That is what you are saying, correct? But this says nothing about the pull of the moon towards the earth in that description.

If the moon were perfectly orbiting the earth, with just those two bodies alone, the moon would have a net acceleration of 0 towards the earth. Again, that is what you are saying, correct? But this says nothing about the pull of the moon towards the sun in that description.

The pull of the moon towards the sun and the pull towards the earth is DIFFERENT.

We can't take those concepts of two-body orbits and combine them like magic to make a three body system. It doesn't work. The moon's pull from the sun is different than the moon's pull from the earth. That's the issue with the Three Body Problem: Two-Body orbits are possible, but stable orbits cannot be created with three or more bodies. It causes chaos.

If what you were saying were true, then it would be easy to predict the position of three bodies put into motion in that manner. They would have solved the Three Body Problem hundreds of years ago. They would have proclaimed "Stable orbits are easy! It's just two two-body orbits stacked on top of each other!" Your explanation is either an attempt at deception or a demonstration of ignoring the situation and searching really hard for excuses. The reader can decide which one to file you under.

[BillO]

I give up.

The math here has been done n-ways to Sunday and you have been provided with a program that provides a 3 body pure numerical solution (it will actually do up to 10).

I was told once by a very smart person that, while simple ignorance can be defeated through education, if it is willful there is nothing that can be done.

https://en.wiktionary.org/wiki/willful_ignorance

Noun

willful ignorance (uncountable)
1.   (idiomatic, law) A decision in bad faith to avoid becoming informed about something so as to avoid having to make undesirable decisions that such information might prompt.

Synonyms
•   (bad-faith decision to remain ignorant): vincible ignorance, willful blindness

Translations
 A decision in bad faith to avoid becoming informed

See also
•   Nelsonian knowledge

[markjo]

If what you were saying were true, then it would be easy to predict the position of three bodies put into motion in that manner.

But they can predict the positions of the 3 bodies in motion.  That's what makes things like gravitational assists and trips to other bodies in the solar system possible.

They would have solved the Three Body Problem hundreds of years ago. They would have proclaimed "Stable orbits are easy! It's just two two-body orbits stacked on top of each other!"

The moon has been shown to be moving away from the earth at the rate of about one inch per year.  An argument could be made that the moon's orbit around the earth doesn't qualify as stable.  It's also known that the orbits of the outer gas giant planets perturb each other periodically to the point where Jupiter either absorbed or ejected a number of bodies out of the solar system and Neptune and Uranus swapped orbits about 4 billion years ago.  I don't think that the solar system is quite as stable as you think that it should be.

[HorstFue]

...

You are just saying that because a two-body orbit is possible, where the inward pull is balanced by the outwards centrifugal pull, that orbits can exist in any situation, and we can just stack them on top of each other.

Ok, on a close inspection you are right, that's a 3 body problem, unsolvable.
But what would physicists do, when faced with an unsolvable problem? They make assumptions, approximations, that would transform the problem to a solvable one.
Distance to Sun is about 389 times the distance earth-moon. So the earth-moon combination is almost at the same distance to the Sun at any time. You could also say the combined earth-moon system is orbiting the Sun. 
I would assume, in this case you could "just stack them (the orbits) on top of each other".
At least for the very coarse question, if the moon will fall into the Sun or continue orbiting the earth.

The pull of the moon towards the sun and the pull towards the earth is DIFFERENT.

Yes,  I agree, but how DIFFERENT? My math. is not sufficient or this.
Can You, Tom give as a number? Or ask Ralph Renés for an answer.

Ah, sorry, later is not available right now. For not including centrifugal forces in his calculations, I just filed him under "nonsense".

[garygreen]

The pull of the moon towards the sun and the pull towards the earth is DIFFERENT.

acceleration of the earth due to the sun = 0.0057 m/s^2

acceleration of the moon due to the sun = 0.0058 m/s^2

that's a difference of 0.0001 m/s^2 when the forces are maximally different.

i still don't get what your point is.  astronomers have known about this fact at least since newton.  the moon's orbit is dynamic.  no one but you says that it's supposed to be perfectly stable.