Dynamic Equilibrium of the Universe and Subsystems

[coldcreation]

triple post removed

[coldcreation]

Double post removed

[coldcreation]

It takes more energy to escape the solar system than it does to get to a Lagrange point in the solar system therefore your theory is proven false. No need for any atomic clocks.

 

Wrong. This proves that the concept has not yet been falsified. There is less curvature (gravity) in the direction of L1 (it is not a steeper slope as it would erroneously appear in a reduced dimension illustration. If you have difficulty visualizing the field potential I suggest you turn the illustration upside down: it is easier, it requires less energy, to get to L1 than to any other point at the same distance off the M1 - L1 line). The escape velocity should be less than 11.2 kilometers per second (~6.96 mi/s) toward L1 from the earth (not greater as you seem to have implied). I thought you would have understood this already (see my post to Will regarding tides etc.).

 

[Edited to add] On a reduced dimension cross section of the field, you see what looks like a deeper well at M1 and M2, a higher peak at L1, and thus a greater distance to climb out in order to attain L1. But remember, this is really a straight line towards L1. And since, the gravitational potential is less (equal to zero according to this hypothesis) the energy required to get to L1 should be less than expected. I don't know how to say that in more simple terms.

 

There is thus still a need for atomic clocks.

 

 

...Do you think NASA would not have noticed it took much more energy than predicted to get a spacecraft to L1?

 

Exactly the opposite. In case you missed it, in post 27 or 28 above, it was written that, unexpectedly, SOHO reached it’s halo orbit around the Sun-Earth L1 point six weeks ahead of schedule (on February 14, 1996), with enough fuel left over to sustain the orbital position for more than a decade—over twice the length of fuel-time anticipated preceding the launch (Martens 2001). Here is the full reference: Martens, P.C., 2001, Encyclopedia of Astronomy and Astrophysics, pp. 2448-2452 .

 

This would seem to imply (I detest conspiracy theories, so I won't state that this is the cause of SOHO's early arrival) that the gravitational well of the earth is more relaxed along the M1 - L1 line than expected according to the inverse square law, since the potential is weaker in that direction, i.e., the earths gravitational force is weaker towards L1.

 

The potential at L1 has been measured! It is not open for debate and it is not zero

 

Where is your full reference for that conjecture?

 

It is known that the gravitational field of the sun and earth cancel at L1, and that gravity is thus reduced to zero (that is why a test particle placed there 'feels' no acceleration). It is not known, however, the gravitational potential energy in comparison to the value attained at infinity. Indeed, I suspect the value is the same at L1 as it is at infinity, i.e., the zero gravitational potential energy value.

 

 

Instead of proposing a billion dollar mission to find something out - let me suggest wikipedia first.

 

Actually, the test can be accomplished by any future mission planned for L1 (or L2). Several of them are in the works (the Earth-Moon L1 would be perfect). All that is needed is for the mission to carry the atomic clocks, etc., park at L1 for the time it takes, then move on to a halo orbit (or the dark side of the moon) to continue the otherwise planned mission.

 

So the cost of testing the boundary condition of GR (inherent is nature) would be exceedingly low (several million at most on a joint mission). It is well worth it.

 

 

 

 

Something has only just begun.

 

 

 

 

 

CC

[Erasmus00]

But remember, this is really a straight line towards L1. And since, the gravitational potential is less (equal to zero according to this hypothesis) the energy required to get to L1 should be less than expected. I don't know how to say that in more simple terms.

 

You've made an error- you yourself have said the gravitational energy is equal to 0 at infinity (as well as lagrange points), which means that its a strictly non-positive quantity. 0 is larger than any negative number, so its actually harder to get to L1 in your theory, not easier.

 

I also reiterate that it breaks Einstein's equations. How, then, do you propose even defining a gravitational potential in your theory?

-Will

[coldcreation]

You've made an error- you yourself have said the gravitational energy is equal to 0 at infinity (as well as lagrange points), which means that its a strictly non-positive quantity. 0 is larger than any negative number, so its actually harder to get to L1 in your theory, not easier.

 

I also reiterate that it breaks Einstein's equations. How, then, do you propose even defining a gravitational potential in your theory?

-Will

 

I consider curvature, for convenience, to be nonzero and positive. So zero potential is less than any other value (similarly to the Kelvin temperature scale in concept, where zero is the lowest possible value). It is easier to get to L1, it does not require more energy than off the M2-L1-M1 line.

 

When you write "0 is larger than any negative number..." I assume you refer to gravitational energy being negative. I think the field structure, the concept of L1 points and gravity itself is easier to understand from the perspective of positive deviations from linearity. We say that gravity, curvature, or the intensity of the field potential diminishes with distance from a massive body until it attains zero. I had made that clear from the outset, or so I thought.

 

The concept still holds: it take less (not more) energy to travel along the M1 - L1 - M2 line than to travel anywhere else, because gravity (spacetime curvature) is diminished, less intense, relaxed, weaker, along that line.

 

I feel like I am highjacking this thread with a discussion on Lagrange points (and the gravitational potential energy at L1). So I will open a new thread where I will elaborate further.

 

 

 

CC

[HydrogenBond]

If we took the mass of the sun and spread it out to cover the entire solar system, its GR affect would be small. On the other hand, if we compress the same mass as tight as possible, the GR affect increases. What is preventing the GR from maxing out for the sun, like in the second scenario, are the other three forces and there physical limitations and energy output. Hypothetically, if we removed all the affects of the three other forces, the GR of the solar mass could increase, without adding any mass. The final gravity density is both mass and geometry dependant.

 

The GR of the universe that we see is a net affect, being prevented from reaching the theoretical GR max, because the other forces and the energy involved fluff geometry. This net affect is analogous to GR max minus the geometry fluff affect. The sun is one example. If we removed the affect of the other three forces, it could approach GR max for that mass. What is preventing this are the affect of other three forces. What we see is a net GR smaller than GR max because of the geometry limitations. The question is, if this minus aspect, which is keeping the universe matter fluffy to prevent the GR max geometry, decreases, so all the mass can approach the GR max, is there a number where the universe will begin to collapse?

 

The black holes are an example of GR max. So if all the matter of the universe was part of black holes, so we have universal GR max, would this cause the universe to collapse? This is the theoretical limit of 100% GR max. It may only require 90% GR max. Theoretically all we have to do is get rid of or lower the minus side. When fusion runs out of fuel, the minus side gets smaller allowing more GR. Sometimes this allows the GR net is to reach GR max, and form a black hole.

[modest]

The question is, if this minus aspect, which is keeping the universe matter fluffy to prevent the GR max geometry, decreases, so all the mass can approach the GR max, is there a number where the universe will begin to collapse?

 

 

I don't think so HydrogenBond. If the mass stays constant then the overall density stays constant (ignoring expansion for a moment). The curvature is determined by the density which shouldn't change over large enough distances even if densities become very high at very small distances.

 

In other words, cotton candy should weigh the same as the sugar that made it. Each crystal of sugar is like a black hole - high in density. The cotton candy is like the universe now. If we put the sugar in one container and the cotton candy in another container of equal size (the container representing the universe) then they both will have the same mass. The overall density in the container is therefore equal between them. It is that overall density that tells the universe to expand or collapse. Have I misunderstood?

 

-modest

[modest]

CC,

 

It honestly doesn't matter if you consider PE positive or negative. If you want to say it gets larger or smaller further from a mass. The only facts that need understood are these:

• It takes work to get from the surface of the earth to L1
  
• It takes more work to get from the surface of the earth out of the solar system
  
• We have been to L1 and measured the difference in PE from the surface to L1
  
• We have been to the edge of the solar system and measured the difference in PE there.
  

 

These facts invalidate your idea if you turn your diagram upside down or not. If things are positive or negative or whatever. The fact is PE is understood and it has been measured. The difference in work between L1 and infinity is substantial. Voyager needed much more acceleration than SOHO. You can't get past that fact no matter how you draw a diagram or which way you hold it. Unless you want to completely redefine your PE to mean something it is not, you cannot say L1 has equal PE to infinity - saying that is wrong and you don't need atomic clocks to prove it.

 

By the way, usually doing work against gravity gives mass potential gravitational energy. Therefore moving higher in a gravity well means a higher accompanying PE. A book on a table has more PE than a book on the floor. If zero PE exists at infinity then all other PE is negative. Please look again closely at my graphs - they have accompanying numbers. That is helpful if you look at them.

 

-modest

[coldcreation]

CC

It honestly doesn't matter if you consider PE positive or negative.

 

That is my point. I consider gravity a positive displacement (deviation) from linearity.

 

 

We have been to L1 and measured the difference in PE from the surface

 

Your claim is false.

 

First, SOHO has never been to L1 (nor, hence, has it been docked there to carry out observations). It went directly (Edit: in a roundabout way :lol:) to its halo orbit and has been there ever since.

 

In order to verify the gravitational potential energy at L1 at least two things are needed to obtain accurate results. (a) The measuring devices (to test gravitational redshift/blueshift and time dilation) must be onboard, and (:) the craft must be situated at rest ON the line connecting the center of M1 and M2. Thus, the craft must be AT L1, not in orbital motion around it.

 

SOHO was not designed to measure gravitational potential energy and did NOT do so.

 

This is the reason you have not backed up your claim with a reference (recall I asked you for one above).

 

 

• It takes work to get from the surface of the earth to L1
  
• It takes more work to get from the surface of the earth out of the solar system
  
• We have been to L1 and measured the difference in PE from the surface to L1 [This is not factual. CC]
  
• We have been to the edge of the solar system and measured the difference in PE there.
  

 

These facts invalidate your idea.

 

So far the idea that gravitational potential energy is zero at L1 has not been invalidated.

 

Voyager needed much more acceleration than SOHO. You can't get past that fact no matter how you draw a diagram or which way you hold it. Unless you want to completely redefine your PE to mean something it is not, you cannot say L1 has equal PE to infinity - saying that is wrong and you don't need atomic clocks to prove it.

 

I am saying the same thing. It is easier to get to L1 when its gravitational potential energy is zero, than to get there when it is nonzero (positive), and so obviously is requires more energy to leave the solar system. [Edit: In fact, it requires more energy to leave the Earths gravitational well in all directions off the M1-L1-M2-L2 line].

 

...If zero PE exists at infinity then all other PE is negative.

 

Your choice of using the word "negative" is by convention only. I argue that if the zero of gravitational potential energy is the value obtained at infinity then all other values are positive (the value would thus increase toward massive bodies), again by convention.

 

 

 

CC

[modest]

We have been to L1 and measured the difference in PE from the surface

Your claim is false.

 

First, SOHO has never been to L1 (nor, hence, has it been docked there to carry out observations). It went directly to its halo orbit and has been there ever since.

 

In order to verify the gravitational potential energy at L1 at least two things are needed to obtain accurate results. (a) The measuring devices (to test gravitational redshift/blueshift and time dilation) must be onboard, and (b) the craft must be situated at rest ON the line connecting the center of M1 and M2. Thus, the craft must be AT L1, not in orbital motion around it.

 

SOHO was not designed to measure gravitational potential energy and did NOT do so.

I believe we can communicate sucessuflly. Please understand, I’m not out to prove you wrong. I’m just explaining as best I can what I understand things to be. Lagrange points are well-investigated by NASA. This is a rough orbit of WIND spacecraft all up-in a LaGrange point’s business.

 

 

But, this is a distraction. My only source needed is here or here. Which tells exactly how much potential is at L1. Actually, the second source has a nice pic I’d like to share with you:

 

 

The equation for gravitational potential energy must be [imath]PE = -GMm/r[/imath] in the LaGrange area as it has been measured by NASA (more on this to follow). [imath]PE = -GMm/r[/imath] can be simplified to [imath]PE = -GM/r[/imath] where one mass is negligible to the other. In this equation ‘GM’ is the Gravitational constant multiplied by the mass of either the sun or the earth. It is therefore a number which we know. ‘r’ is the radius or distance from the center of the sun or earth to the point L1 we are interested in.

 

Therefore with simple division we can solve for PE in one step. However, this will only solve for one body. How in heavens will we deal with both the sun and the earth? As it turns out (and is proofed on link above) all we need to do is add the potential of the two bodies to get total potential. Therefore with simple addition and division we can solve how much gravitational potential energy is at the earth/sun L1 point relative to the earth and sun. With this information we can make good scientific predictions. This sounds fun, so let’s do it:

 

Earth GM = [imath]3.986 \times 10^5[/imath]

Sun GM = [imath]1.32712 \times 10^{11}[/imath]

Earth r = [imath]1.53 \times 10^6[/imath]

Sun r = [imath]1.48 \times 10^8[/imath]

 

[math]PE =\left[ -\frac{GM_e}{r_e} \right] + \left[ -\frac{GM_s}{r_s} \right] [/math]

 

[math]PE = \left[ -\frac{ 3.986 \times 10^5 }{ 1.53 \times 10^6 } \right] + \left[ -\frac{1.32712 \times 10^{11}}{ 1.48 \times 10^8 } \right] =[/math]

 

[imath]-896.963\: kg\, km^2/s^2 = -8.96963 \times 10^8\: kg\, m^2/s^2 (joules) [/imath]

 

And we can do the same thing for the surface of the earth which I won’t show but is [imath]-9.49727 \times 10^8\: joules [/imath] Therefore the difference in potential gravitational energy between the surface of the earth and L1 by subtraction is 52.764 million jules. This is a prediction. We can investigate if we are correct here. The first thing to understand is that this value of potential energy is energy. If you look at the link I gave at the top of this post you will see how change in potential energy is equal to work. Therefore, we know how much work it takes to get to L1.

 

First step, let’s convert my value of PE to something more useful. We can convert kinetic energy to velocity by [imath]KE = 1/2mv^2[/imath]. We can use this to predict how fast a spacecraft that’s stationary at L1 will be when it falls back to earth. For this situation:

 

[math]PE = KE[/math] and

 

[math]velocity = \sqrt{2 \times KE}[/math].

 

[math]velocity = \sqrt{2 \times 5.2764 \times 10^7 J} = 10,272 m/s[/math].

 

Therefore we can predict that a spacecraft returning from L1 will enter our atmosphere between 10 and 11 kilometers per second. Understand CC, I make this prediction with only the mass of the earth and sun as my input variables. That’s it! So, is it correct? Has NASA proven the standard definition of PE at L1 right? Yes:

 

NASA chronology quarter 3 2001

The capsule successfully re-entered the atmosphere over Oregon at 11 km/s, but a wiring error resulted in the drogue parachute release mortar failing to fire at 33 km altitude.

The genesis probe returned from L1 with our predicted kinetic energy. I’m glad we have Newton and the scientists at NASA who understand all this.

 

Now that we know this method works let’s use it on your theory. All we need is the difference between PE at earth’s surface and the infinite point of earth and the sun. The infinite point is easy by -GM/r - it’s zero. We will say this value applies to both the infinite point and L1. We’ve already solved for the surface of the earth. We subtract the values to find the difference as before:

 

[math][-9.49727 \times 10^8] - [-0] = -9.49727 \times 10^8\: joules [/math]

 

converting to velocity:

 

[math]\sqrt{2 \times 9.49727 \times 10^8 J} = 43,583 m/s[/math]

 

Your theory predicts that spacecraft must be accelerated to 44 kilomoters per second to get to L1 and would return at the same rate. This is clearly not the case. This is what I mean when I say missions like SOHO and others have already measured the value of potential at lagrange points. Just the mere fact of getting them there requires NASA know just how much energy is needed. That energy is by definition potential energy.

 

So, why does your theory predict such high velocity? This is in fact the velocity we would expect if something at rest far outside the solar system fell into our sun/earth gravity well and hit the surface of the earth. We would expect it to enter at exactly this speed. By claiming the infinity point of PE is equal to the LaGrange point, you are saying the same about L1.

 

So, please, understand CC, we don't need atomic clocks to measure the value of PE at LaGrange points. We know GR and Newton are correct on this. It is not possible to say L1 has equal potential as infinity. It is proven otherwise by the fellas and ladies at NASA and the ESA who have to use PE every day to make spacecraft trajectories good. You and I are not so easily able to overrule them on this.

 

-modest

[HydrogenBond]

Maybe I am having a hard time explaining myself. The contraction of space-time and GR is not just mass dependant but also geometry dependant. In other words, in we took 5 solar masses and spread it out over the solar system and compared this to a black hole formed from the same mass, we have two different values of GR. For example, if we began with the black hole and somehow we could reverse it back to the expanded state, this will require the input of a lot of energy. These are not interchangeable states. The energy required to reverse the GR of the black hole reduces the level of the GR, to that of more expanded state.

 

A loose analogy of what I am trying to say is critical geometry with respect to radioactive materials. If we have a given mass of such a material, it can go critical or cause a chain reaction. If we cut it up into smaller pieces without changing the mass of material, and spread it out, we can create a geometry where it is non-critical. It doesn't even have to spread out real far. But if we reverse this geometry the situation changes.

 

One possible explanation has to do with the space-time reference that is created by the GR. From the point of view of the black hole, space-time is so contracted that the "edge" of the universe is in plain sight since it is within the infinite distance and time parameter. In the spread out reference, of the same amount of mass, the GR is lower so the reference overlap is different. Its impact is more localized and may not be enough to allow the critical GR geometry for reversal.

 

If something came out of a black hole and appeared somewhere else in the universe, that implies these two reference are overlapping at distances one may not expect from the zero reference. It doesn't have to travel via zero reference but uses the short cut reference that is more in line with the closer distances seen by the black hole. The affect of the other three forces, in terms of stars, is to fluff out the mass geometry so what the GR reference sees, isn't enough to close. Nature does this with the three forces of nature and their energy output.

[modest]

Maybe I am having a hard time explaining myself. The contraction of space-time and GR is not just mass dependant but also geometry dependant. In other words, in we took 5 solar masses and spread it out over the solar system and compared this to a black hole formed from the same mass, we have two different values of GR. For example, if we began with the black hole and somehow we could reverse it back to the expanded state, this will require the input of a lot of energy. These are not interchangeable states. The energy required to reverse the GR of the black hole reduces the level of the GR, to that of more expanded state.

 

I still think, even if the mass of the universe is entirely in the form of black holes that we can use the Friedmann / RW solution. If we accept the FLRW solution and its stipulation that the universe is homogeneous then I think we can answer your question. I, however, have a weak understanding of this - so we may need another opinion.

 

You are right that the evolution of the universe according to GR doesn’t depend on mass only. The factors I know of in the Friedmann equations are density (of energy or ‘mass’) and pressure (either positive via radiation or negative via the cosmological constant). In this framework I don’t think the scale factor would change or the evolution of the universe would be different if the mass/energy contribution were in the form of black holes. I certainly don’t think we would expect the geometry to change.

 

Your example of a nebula vs. an equally massive black hole is good I think. If you were one light year away from this nebula (or even closer) as it collapsed into a black hole you would not notice a different gravitational effect after the collapse. If you consider a sphere of one light year radius this observer and black hole make then the energy density inside is unchanged by the black hole. Three black holes over three light years is more homogeneous and 8 black holes over 8 light years is even more so. All topological differences between the nebula and black hole are less important the larger the scale. It's all a matter of scale. The earth is smoother than a marble - at scale.

 

Looking at the scale factor of the entire universe then would be properly modeled with the same density either with fluffy stars or black holes. So long as the overall mass/energy is the same - overall density will be the same. That density is the variable in FLRW to determine expansion which we could then say would be the same. After all, the one way a black hole can communicate with the universe is gravitationally.

 

What pops up in my mind is the contribution of radiation pressure. Stars and the like emit radiation while black holes (ignoring Hawking radiation) do not. Radiation has energy and therefore contributes to gravity. It does this in two ways - the first being through ordinary energy density (energy being equivalent to mass and all). The second is through pressure. Radiation has an associated pressure and pressure has an associated energy which in turn contributes to gravity. This contribution of radiation pressure to the field is ridiculously small relative to other factors today. It was an important factor in the initial instant of the big bang - but it gets less relevant with time.

 

The energy associated with a photon would still be there in the black hole. Therefore its contribution to gravity via energy density is accounted for just like a star’s photon. However, when the factors above (density and pressure) are combined it is necessary to say that radiation evolves with the scale factor differently than matter-mass. While ordinary mass evolves with the cube of the scale factor, radiation evolves with the forth power. This means that radiation looses it’s effect on gravity in an expanding universe faster than matter does. How this would play out if everything were locked up in black holes, I don’t know. New radiation would not be created so I guess the scale factor would be slightly different in the distant future. However, our universe is already matter dominated and if we include the cosmological constant as a factor - I think we can safely ignore everything I just said as even a slightly significant factor. In fact, when people solve the omega version of the Friedmann equation, I don't think they account for new radiation in omega-r. So, I think this is already safely ignored.

 

I really don’t see the overall geometry or scale factor changing if everything were in black holes. However, I really am a novice at this and perhaps someone else has insight I’m missing.

 

One possible explanation has to do with the space-time reference that is created by the GR. From the point of view of the black hole, space-time is so contracted that the "edge" of the universe is in plain sight since it is within the infinite distance and time parameter.

 

I don’t understand what you mean here. Someone in a black hole can see out. The event horizon really isn’t a boundary condition or the edge of anything. I’m not sure what you mean “infinite distance and time parameter”

 

The affect of the other three forces, in terms of stars, is to fluff out the mass geometry so what the GR reference sees, isn't enough to close. Nature does this with the three forces of nature and their energy output.

 

Energy is the thing that curves space time. Energetic matter is more massive than less-energetic matter where relativity is concerned. When a black hole swallows mass - it swallows the energy I think you’re talking about here. Its force of gravity is equivalent.

 

-modest

[coldcreation]

Your theory predicts that spacecraft must be accelerated to 44 kilomoters per second to get to L1 and would return at the same rate.

 

This is certainly not the case. As I have explained above there should be required slightly less energy (not more) to attain L1 (or L2) than required to obtain the same distance in any other direction.

 

I will obviously retract the claim that the value of gravitational potential energy at L1 is equal to the zero value obtained at infinity. My blunder.

 

However, the fact that gravity is cancelled at L1 and thus curvature is to zero (as we all agree) is the most important issue: to be discussed further.

 

Upon reconsideration, I have come to the conclusion that the value of potential energy at L1 is not the key issue. The core of the problem rests on the value of gravity (curvature) itself relative to potential energy of the system. So, though the value of potential energy at L1 is nonzero relative to the value of PE at infinity, the value of gravitational spacetime curvature is certainly zero relative to the system itself, it represents the local minima: a crucial fact. Though this fact is well-known and generally accepted, the importance of these gravity-free (curvature-free) surfaces (or points) - along with the structural modifications induced on the individual gravitational wells (surrounding each object in the system) in the regulation of stability - is not so well-known.

 

For all intents and purposes, the concept proposed for the mechanism responsible for mediating the equilibrium of gravitationally bounded systems, still holds, whether potential energy is zero or not at L1, since gravity is cancelled (curvature is reduced to zero) at key locations in the combined fields of massive bodies regardless, and that value does not change. That leads to two valid interpretations for observations (one of which will not require dark matter at galactic scales).

 

In sum, it is not the potential energy at L1 that counts, it is the value of curvature (gravity) at Lagrange points that is most important. The problem is partly a geometric one, and partly a relative potential energy problem (not an absolute potential energy problem as I originally suspected).

 

Note: The gravitational redshift and time dilation test between L1 and Earth will still be required, but it will not be a test of the gravitational potential energy at L1. It will be required in order to e.g., test the geometric structure of the Earths gravitational field along the Earth-L1 (L2) line (and thus the inverse square law) in comparison with the structure of the field in all other directions. It may be shown that the field is not spherically symmetric, as is often erroneously assumed. Some of the other reasons the test should be performed are beyond the scope of this thread (e.g., those that point to the physical mechanism behind the gravitational interaction, behind lambda).

 

I shall, as stated above, open a new thread on this topic (and more), since the scope of the subject goes beyond the Dynamic Equilibrium of the Universe and Subsystems, and to delve into those other areas of contention here would inevitably cause the discussion to gravitate off-topic.

 

PS. Nice presentation above. As Hilton Ratcliffe once wrote with regards to my questioning: "The great value of this dialogue with you is that your questions compel me to get my ducks in a row, and most importantly, point to areas where I might be going wrong." I say the same to you modest. :hihi:

 

 

 

"The important thing is not to stop questioning.

Curiosity has its own reason for existing." Albert Einstein

 

 

[EDITED TO ADD:] For a further discussion of this topic and more see: The Physical Mechanism of Gravity - The Spatiotemporal Ground-State.

 

 

CC

[modest]

This is certainly not the case. As I have explained above there should be required slightly less energy (not more) to attain L1 (or L2) than required to obtain the same distance in any other direction.

 

I agree

 

I will obviously retract the claim that the value of gravitational potential energy at L1 is equal to the zero value obtained at infinity. My blunder.

 

That's very good

 

However, the fact that gravity is cancelled at L1 and thus curvature is to zero (as we all agree) is the most important issue: to be discussed further.

 

Agreed

 

In sum, it is not the potential energy at L1 that counts, it is the value of curvature (gravity) at Lagrange points that is most important. The problem is partly a geometric one, and partly a relative potential energy problem (not an absolute potential energy problem as I originally suspected).

 

Note: The gravitational redshift and time dilation test between L1 and Earth will still be required, but it will not be a test of the gravitational potential energy at L1. It will be required in order to e.g., test the geometric structure of the Earths gravitational field along the Earth-L1 (L2) line (and thus the inverse square law) in comparison with the structure of the field in all other directions. It may be shown that the field is not spherically symmetric, as is often erroneously assumed. Some of the other reasons the test should be performed are beyond the scope of this thread (e.g., those that point to the physical mechanism behind the gravitational interaction, behind lambda).

 

Ok, sure. I think what gives people trouble (or at least me) about LaGrange points is the normal Newtonian gravity/force or inverse square law is not by itself appropriate. LaGrange points being 3 body solutions, things are not as simple as the inverse square.

 

I shall, as stated above, open a new thread on this topic (and more), since the scope of the subject goes beyond the Dynamic Equilibrium of the Universe and Subsystems, and to delve into those other areas of contention here would inevitably cause the discussion to gravitate off-topic.

 

PS. Nice presentation above. As Hilton Ratcliffe once wrote with regards to my questioning: "The great value of this dialogue with you is that your questions compel me to get my ducks in a row, and most importantly, point to areas where I might be going wrong." I say the same to you modest. :eek_big:

 

Very good and thank you. I appreciate the nod because I spent about an hour trying to figure out why my numbers were coming up all crazy - which they were. It turned out I copied earth's GM wrong and that threw everything off, but it took a good amount of work to find the error. I'm glad it wasn't in vain.

 

-modest