Proof that Gravitational Constant is not constant

Robittybob1 said:
Do you all agree that Gravitational constant may not be constant after all?
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Firstly, 'constants' which have dimensions (G has dimensions of $$m^{3}.kg^{-1}.s^{-2}$$) are not physically meaningful really. You should only consider dimensionless combinations, as they are independent of our arbitrary definitions of length and time etc. Secondly experiments are regularly done, including observations of distant galaxies, to see if dimensionless 'constants' such as the fine structure constant have changed in time. Extremely strong bounds are put on such variations already, far beyond anything you're going to get. Thirdly, the concept of varying constants is not something outside of theory. Coupling 'constants' are observed to vary with energy level, something the 2004 (if memory serves) Nobel Prize was given for. In fact, in string theory ALL 'constants' are actually dynamical quantities whose values are set by the particular solution to the equations of motion. Looking at time variation in such things could tell us about extra dimensions.

Robittybob1 said:
Is everyone happy now with the concept that gravitational radiation is like radiating away the system's gravitational attraction?
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No. Newtonian mechanics doesn't include gravitational radiation and therefore you cannot use an equation from Newtonian mechanics to link those concepts. Furthermore, a gravitational model which does include gravitational radiation, general relativity, explains the effect while G remains constant. Therefore there are other explanations, so your conclusion that the gravitational radiation is like radiating away the attraction is not necessarily true so asserting it as if it is is flawed.

Robittybob1 said:
(there could be a relativity aspect here for the energy associated with a moving mass goes up higher (E = MC^2/SQRT(1-v^2/c^2)) and plays a bigger part as the orbital velocity becomes more relativistic.
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You're showing you have no problem just throwing out equations you don't understand and talking about models you haven't any experience with. You're showing you're intellectually dishonest or lazy. Which one is it, are you doing all of this nonsense deliberately or do you not realise you're doing it?
 
AlphaNumeric said:
You're showing you have no problem just throwing out equations you don't understand and talking about models you haven't any experience with. You're showing you're intellectually dishonest or lazy. Which one is it, are you doing all of this nonsense deliberately or do you not realise you're doing it?
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Thanks once again for such an amazing reply. You show so much knowledge of the various aspects of science, things I know nothing about at all. Note I'm not even trying to use Newtonian equations to solve the problem, I was just using those formulas to show there is a major problem, and now I am trying with my limited knowledge to see how relativity might go some way to explain it. So I would say I'm not lazy or dishonest but just a person in a lower grade, and it is going to take time to get to the level (if ever) of RPenner, BruceP and you, AlphaNumeric.

I thought I had written down "(E = MC^2/SQRT(1-v^2/c^2))" correctly - maybe there is an error? Sorry.
(it seems a valid equation "Mass in special relativity" http://en.wikipedia.org/wiki/Mass_in_special_relativity)

The answer is in there somewhere linking mass, energy, and momentum, and now to put gravity into the mix as well!

There must be a connection for "gravitational radiation is linked to loss of orbital energy".
 
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Does anyone agree that the Newtonian analysis suggests that there is such a major discrepancy that just saying it is able to corrected by applying GR principles is insufficient.
A difference of 2.75 times in the orbital period is way more than can be accounted for by the relativistic velocity (1/1000 c) of the binary stars.
Does anyone agree with this?
 
Thinking about the consequences of what I am proposing. Will once the binary stars crashes into each other would the G ever return to the Universal G again? if so how was it going to do that or would it remain forever low? If that was the case would the black hole in the centre of our galaxy be affected by a lowered G as well?
The difference really is that the black-hole is not rushing around at a fraction of light speed. Hulse Taylor Binary will obviously end up as a rapidly rotating black-hole for the momentum of the system is going to be conserved (excepting that a degree of it will be radiated off a gravitational radiation.)
I wonder if Hulse and Taylor have thought about the ultimate fate of their binary?
 
The fact that the gravitational constant may have localities where it is less than 6.67384E-11 maybe hard to pick up. When one looks at the movement of the periastron maybe both formulas that account for the precession and gravitational radiation are still in a 1:1 PROPORTION regardless the actual value of G. So one can't tell what G is till one looks at the Newtonian orbital formulas. I keep on reading that the formulas are a "little out" for the Hulse Taylor Binary, but truthfully they aren't just a little out they are right out!
What was the radiation that Einstein predicted? Did Einstein ever question whether G could ever change?
 
"If the force of gravity were reduced 50 percent, how would this change stars and the universe?
This is a complicated question because so many physical factors that describe stars and the universe are intertwined and rely on the value of 'G'.

Stars would get much larger in mass given the speeds of the particles that they contain. The core energy of these stars would be lower because there would be less gravitational potential energy per gram of mass to cause central temperatures to become very large. The stars would be more distended because radiation pressure would support the overlying stellar mass to much larger distances.

For some systems such as planets orbiting stars, or stars orbiting each other, the objects can become unbound and the parts can actually fly away because at a given distance, the orbital speed of the objects would now exceed the escape velocity for the system.

The Earth orbits the Sun at a speed that is just under the escape velocity from the Sun at this distance by a factor of about 40%. If the force of gravity was halved, its speed would be exactly the escape speed. In fact, any body orbiting in a circular orbit would become unbound if the force of gravity was reduced by a factor of two!"
http://www.astronomycafe.net/qadir/q2482.html

Well the above statement as far as I can see is not strictly true for there would still be gravity and the body moving apart would once again lose KE and this would limit their velocity and hence a new equilibrium would be established. But it would be wise for me to test the maths of this too.
Orbital energy stays the same but vary G and see what happens to "r" (the distance apart) of two massive bodies.
 
"Einstein's nemesis: di herculis

DI Herculis is an 8th-magnitude eclipsing binary about 2,000 light years from earth. These two young blue stars are very close -- only one fifth the distance from earth to our sun. They orbit about a common center of gravity every 10.55 days. So far, no problem!

The puzzle is that, as the two stars swing around one another, the axis of their orbit rotates or precesses too slowly. General relativity predicts a precession of 4.27°/century, but for DI Herculis the rate is only 1.05°/century. This does not sound like a figure large enough to get excited about, but it deeply troubles astronomers. D. Popper, an astronomer at UCLA, says:

"The observations are pretty clear. I don't think there's any question there's a discrepancy and, frankly, it is an important one and it's unresolved."
Accentuating the challenge to general relativity is the discovery that a second eclipsing binary, AC Camelopardalis, also violates general relativity in the same way. It seems that wherever gravitational fields are extremely strong and space-time, therefore, highly distorted, general relativity fails."
http://www.science-frontiers.com/sf103/sf103a04.htm

I see there maybe other star binaries I maybe able to test out whether lowering G solves the issues. Still early days. Keeping on searching ....
 
Robittybob1 said:
"If the force of gravity were reduced 50 percent, how would this change stars and the universe?
This is a complicated question because so many physical factors that describe stars and the universe are intertwined and rely on the value of 'G'.

Stars would get much larger in mass given the speeds of the particles that they contain. The core energy of these stars would be lower because there would be less gravitational potential energy per gram of mass to cause central temperatures to become very large. The stars would be more distended because radiation pressure would support the overlying stellar mass to much larger distances.

For some systems such as planets orbiting stars, or stars orbiting each other, the objects can become unbound and the parts can actually fly away because at a given distance, the orbital speed of the objects would now exceed the escape velocity for the system.

The Earth orbits the Sun at a speed that is just under the escape velocity from the Sun at this distance by a factor of about 40%. If the force of gravity was halved, its speed would be exactly the escape speed. In fact, any body orbiting in a circular orbit would become unbound if the force of gravity was reduced by a factor of two!"
http://www.astronomycafe.net/qadir/q2482.html

Well the above statement as far as I can see is not strictly true for there would still be gravity and the body moving apart would once again lose KE and this would limit their velocity and hence a new equilibrium would be established. But it would be wise for me to test the maths of this too.
Orbital energy stays the same but vary G and see what happens to "r" (the distance apart) of two massive bodies.
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Can anyone see where I'm going wrong with the following?

Orbital energy stays the same but vary G and see what happens to "r" (the distance apart) of two massive bodies.

Total orbital Energy (TE) = PE + KE
now the PE = negative whereas KE is positive but the absolute value of PE = 2 KE
So TE is going to be negative.

Formula for TE = -GM1M2/2r = -G_2m1M2/2r_2 (conservation of Energy) so in fact to my surprise if G were to get smaller r or the semi major axis gets smaller too. Now that is surprising to me too. For I thought the masses would separate if G force got weaker but no they go closer to each other instead?

Is that really the case?

Energy stays the same but G declines (ascending order)
Gravitational Constant, ,Orbital radius
6.67384E-12, , 180401636.4
1.33477E-11, , 360803272.7
2.00215E-11, , 541204909.1
2.66954E-11, , 721606545.5
3.33692E-11, , 902008181.8
4.0043E-11 , , 1082409818
4.67169E-11, , 1262811455
5.33907E-11, , 1443213091
6.00646E-11, , 1623614727
6.67384E-11, , 1804016364

So that proves it but I still don't understand that at all!
So what would make the system fly apart as comment above suggested?
Is it twice the Kinetic energy? But where is that going to come from or is it increasing G rather than reducing it?

After testing higher G - No increasing G whilst keeping the total energy the same just made the "r" get larger. Whereas to me making the force of gravity stronger you would think things would orbit closer with the same amount of energy. But not according to the total orbital energy formula.
 
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Thinking about the reason for the apparent contradiction and the following is what I understand to be the reason for that.

The total orbital energy formula is looking at a mass or masses orbiting at a position and as if they had come in from infinity where their gravitational potential energy would have been originally zero. As they come in closer the GPE gets more negative meaning that they are in a bound state, and it will take that amount of energy to get them back out to infinity again.
So just by saying "if G was halved" you can't create energy simply out of nothing, so if an object had x amount of negative GPE you can NOT just overcome that by dialing up a new G.
If G is lowered as in the HTB system this adjustment is a slow process so things don't fly off into space but the new equilibrium is continually reestablished. The binary is orbiting a lot slower and closer in than it could if G was at the normal 6.67384E-11.

So the figures above represent the distances expected if G was lowered right from the start when the masses were at infinity. Then that makes sense, but it is senseless to talk of "halving the G force" without the mechanism to account for the energy (both the GPE and Kinetic Energy) for that mechanism is bound by the law of conservation of Energy.
Otherwise the Gravitational constant just won't change and it can't just happen in isolation.
 
From the amount that the HTB stars slows we might be able to estimate the speed of gravity. I'm not too certain if is possible, but give it a go!
The thought diagram and possible exchange is along the line for tidal acceleration of the Moon. There the force lines are ahead or behind the direct Centre Of Mass so there is a transfer of momentum which either increases or decreases the orbital energy. OK in the case of the Moon this momentum comes from the Earth's rotation, but in the case of the HTB the stars lose momentum (and energy??) Half goes to kinetic energy but the remainder is paid back into the negative GPE by lowering the strength of gravity. If that is the case there is no gravitational radiation as such.

http://www.talkorigins.org/faqs/moonrec.html (fig 3 shows the diagram I'm thinking of.)
 
Robittybob1 said:
Thinking about the reason for the apparent contradiction and the following is what I understand to be the reason for that.

The total orbital energy formula is looking at a mass or masses orbiting at a position and as if they had come in from infinity where their gravitational potential energy would have been originally zero. As they come in closer the GPE gets more negative meaning that they are in a bound state, and it will take that amount of energy to get them back out to infinity again.
So just by saying "if G was halved" you can't create energy simply out of nothing, so if an object had x amount of negative GPE you can NOT just overcome that by dialing up a new G.
If G is lowered as in the HTB system this adjustment is a slow process so things don't fly off into space but the new equilibrium is continually reestablished. The binary is orbiting a lot slower and closer in than it could if G was at the normal 6.67384E-11.

So the figures above represent the distances expected if G was lowered right from the start when the masses were at infinity. Then that makes sense, but it is senseless to talk of "halving the G force" without the mechanism to account for the energy (both the GPE and Kinetic Energy) for that mechanism is bound by the law of conservation of Energy.
Otherwise the Gravitational constant just won't change and it can't just happen in isolation.
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This macro gives the values of how much negative potential energy has been paid back.

Run this macro and see for yourself! 87% of the Energy debt has been paid back. Now I can see how binary black holes can dance and fling one out into free space. The energy debt had already been paid back! No debt > not bound any more.

Sub GPE_V_GBPE()
'

' Macro recorded 25/10/2012 by Robittybob1

'V r( )=-GM*m/r
'Vr = -G * M1 * m2 / r

a = 992209000 'Measured Semi Major Axis in meters Weisberg and Taylor
'GB = 8.80128E-12 ' calculated GB from orbital period and SMA

'
Dim T As Double
Dim r As Double
Dim K As Double
Dim K2 As Double
Dim dr As Double
Dim Vr As Double 'GPE from infinity
Dim Vrb As Double 'GPE from infinity with lower G
Dim dr_dt As Double
Dim GB As Double
Dim c As Double
Dim M1 As Double
Dim M2 As Double

Sheets("Energy Debt Paid").Select
Range("A10").Select

G = 6.67384 * 10 ^ -11 'Gravitational constant measured in Solar System
GB = 8.80128E-12 ' calculated GB from orbital period and SMA
c = 299792458
M1 = 1.4398 * 1.9891 * 10 ^ 30
M2 = 1.3886 * 1.9891 * 10 ^ 30


'a = 992209000 'Measured Semi Major Axis in meters Weisberg and Taylor
r = 992209000

Vr = -G * M1 * M2 / r
Vrb = -GB * M1 * M2 / r
Range("G1") = Vr
Range("H1") = Vrb
Range("j1") = Vr - Vrb
'
Range("G1:j1").Select
'Selection.Insert Shift:=xlDown
End Sub
 
Robittybob1 said:
Take this article for instance: "Super-massive black hole is flung out of distant galaxy"

Read more: http://www.dailymail.co.uk/sciencet...-hole-flung-distant-galaxy.html#ixzz2AdPaF1F0

If two black holes orbiting can use an alteration in G to repay the GPE debt it is easier to see how they can be flung out of their gravitational well. Momentum and inertial mass remains the same even though G in the region is lessened.
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I think if the same poster posts 12 'posts in a row' in the same thread the thread should go to 'closed' due to lack of interest.
 
Neverfly said:
No, that wouldn't help.
Neither would closing the thread.
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No, I'm just slowly feeding the thread as I do a bit more research. Over the weekend I was looking into ways of using the data from the Hulse Taylor Binary to see if it could give a clue as to the speed of gravity. There is not that much available to give us a lead in on how to do this. Firstly I thought the movements would show a reason for the stars losing orbital energy but it ended up the other way around. Because gravity travels at the speed of light the stars would always be entering a region of higher gravitational field than they would if gravity was instantaneous, and hence they should be accelerated. So it was back to the drawing board!

This was confirmed from reading Wikipedia:
This passage in Wikipedia on "Speed of Gravity" talks of acceleration:
Laplace assumed that when an object like the Earth is moving around the Sun, the attraction of the Earth would not be toward the instantaneous position of the Sun, but toward where the Sun had been if its position was retarded using the relative velocity (this retardation actually does happen with the optical position of the Sun, and is called annual solar aberration). Putting the Sun immobile at the origin, when the Earth is moving in an orbit of radius R with velocity v presuming that the gravitational influence moves with velocity c, moves the Sun's true position ahead of its optical position, by an amount equal to vR/c, which is the travel time of gravity from the sun to the Earth times the relative velocity of the sun and the Earth. The pull of gravity (if it behaved like a wave, such as light) would then be always displaced in the direction of the Earth's velocity, so that the Earth would always be pulled toward the optical position of the Sun, rather than its actual position. This would cause a pull ahead of the Earth, which would cause the orbit of the Earth to spiral outward. Such an outspiral would be suppressed by an amount v/c compared to the force which keeps the Earth in orbit ;...
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http://en.wikipedia.org/wiki/Speed_of_gravity
 
brucep said:
I think if the same poster posts 12 'posts in a row' in the same thread the thread should go to 'closed' due to lack of interest.
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It might be worthwhile checking the number of times it is being read as well. If there were no readers I might as well go home. Can this be checked? OK Monday 5 PM 1083 views.
 
Neverfly said:
That much was clear. I'll admit that I'm just wondering when your research is going to lead you to the conclusion that you were in error...
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I have had to do that continually. You learn as you go. Your ideas are challenged and sometimes they seem right and other times you find you were confused. I like the exploration, and I don't know if I'll be found to be wrong, I'll just have to wait and see.
 
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