Proof that Gravitational Constant is not constant

Discussion in 'Pseudoscience Archive' started by Robittybob1, Oct 12, 2012.

1. Robittybob1BannedBanned

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http://www.members.shaw.ca/warmbeach/INDEX3.htm
" The Acceleration of earth's Gravity x earth orbit Time (exact lunar year) = the Velocity of Light.

(9.80175174 m/s2 x 30,585,600 s = 299,792,458 m/s)"

So if the Earth was a binary and falling into its companion, and hence speeding up, gravity would have to reduce to keep the product of orbit time and gravity, = speed of light?
Does this hold true for the binary stars as well! I think this is what was coming out of my macro, but sorry I'm not running it to completion today. I want to speed it up by going through the looping in 10 day lumps.

Looking into this a bit further, and it doesn't compute.
9.80665 m/s2 Times 3.15569e7 seconds = 309467473.4 m/sec => faster than light.

31556926 tropical year? Why choose 30,585,600?

OK it doesn't come out to the speed of light. But faster than the speed of light? Even though no one knows how fast gravity acts.
Weisberg and Taylor when analysing the HT binary use 31557600 seconds in a year.
When I get the computer fixed I'll use 309474338 m/sec as possible term to work with later.

Last edited: Nov 18, 2012

3. rpennerFully WiredValued Senior Member

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$24 = 2^3 \cdot 3 864 = 2^5 \cdot 3^3 G c^{-3} (m_p + m_c ) = \sqrt{\frac{P_b^5 (1 - e^2)^3 \dot{\omega}^3}{864 \pi^5} } G ( m_p + m_c ) = c^3 P_b^{\frac{5}{2}} (1 - e^2)^{\frac{3}{2}} \dot{\omega}^{\frac{3}{2}} 864^{-\frac{1}{2}} \pi^{-\frac{5}{2}} a = c \sqrt{\frac{P_b^3 \dot{\omega} (1 - e^2 )}{24 \pi^3}} = c P_b^{\frac{3}{2}} \dot{\omega}^{\frac{1}{2}} (1 - e^2 )^{\frac{1}{2}} 24^{-\frac{1}{2}} \pi^{-\frac{3}{2}} T = 2 \pi \sqrt{\frac{a^3}{G(m_p + m_c)}} \quad \quad \quad = 2 \pi a^{\frac{3}{2}}\left( G(m_p + m_c) \right)^{-\frac{1}{2}} \quad \quad \quad \quad \quad \quad = 2 \pi c^{\frac{3}{2}} P_b^{\frac{9}{4}} \dot{\omega}^{\frac{3}{4}} (1 - e^2 )^{\frac{3}{4}} 24^{-\frac{3}{4}} \pi^{-\frac{9}{4}} c^{-\frac{3}{2}} P_b^{-\frac{5}{4}} (1 - e^2)^{-\frac{3}{4}} \dot{\omega}^{-\frac{3}{4}} 864^{\frac{1}{4}} \pi^{\frac{5}{4}} \quad \quad \quad \quad \quad \quad \quad \quad \quad = 2^{1 - \frac{9}{4}+ \frac{5}{4}} \, 3^{-\frac{3}{4}+\frac{3}{4}} \, \pi^{1-\frac{9}{4}+\frac{5}{4}} \, c^{\frac{3}{2}-\frac{3}{2}} \, P_b^{\frac{9}{4}-\frac{5}{4}} \, \dot{\omega}^{\frac{3}{4}-\frac{3}{4}} \, (1 - e^2 )^{\frac{3}{4}-\frac{3}{4}} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = P_b$

Any other result would be wrong since a single model relates the observed values of $P_b$, $e$, and $\dot{\omega}$ to the model parameters of $a$ and $G(m_p + m_c)$.

5. Robittybob1BannedBanned

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So am I following you? By using the orbital period formula you prove that T = Pb the period of the binary is exactly as the predicted value of T? OK so why did I not get the same result? Thank you for your help, I'll look into over the next few days.

7. Robittybob1BannedBanned

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M1 = 1.4398 * 1.9891 * 10 ^ 30
M2 = 1.3886 * 1.9891 * 10 ^ 30
The above figures for the masses were the ones we have been using but in the paper quoted below the masses have been adjusted. Which masses are you using?

'Mp = 14414 +/- 0.0002 solar mass 'http://arxiv.org/pdf/astro-ph/0407149v1.pdf
'Mc = 13867 +/- 0.0002 solar mass 'http://arxiv.org/pdf/astro-ph/0407149v1.pdf

8. rpennerFully WiredValued Senior Member

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Who is "we" ?
None of the above. In all cases I have used algebra to replace $m_p$ and $m_c$ with algebraic formulas based on $P_b$, $e$, $\gamma$ and $\dot{\omega}$, so that I can estimate my errors.

For example, what is the definition of "year" for the quoted figure of $\dot{\omega}$. Astronomy is not your field or mine, but it appears that the standard for astronomers is not the tropical or sidereal year and not the Gregorian calendar approximation to a year, but the Julian approximation of 365.25 days of 24 hours of 3600 seconds.

Also, I would use a more recent estimate of solar mass that you use above.

Thus
$\dot{\omega} = ( 2.337\underline{56683} \pm 0.00000277 )\times10^{-9} \, \textrm{s}^{-1} G c^{-3} m_p = ( 0.7100\underline{30} \pm 0.000096 ) \times 10^{-5} \, \textrm{s} m_p = \frac{ 2 c G c^{-3} m_p}{\frac{2 G M_{\odot}}{c^2}} M_{\odot} = ( 1.441\underline{5} \pm 0.0002 ) M_{\odot} m_p = ( 2.86\underline{65} \pm 0.0005 ) \times 10^{30} \, \textrm{kg}$

I might have gotten even better agreement with the paper if I redid the arithmetic in a high precision environment.

9. Robittybob1BannedBanned

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In the orbital period formula the two masses are added together, so even if one mass is overestimated it maybe balanced against an underestimation of its companion. Would you be able to define the Mp + Mc total with a higher degree of accuracy than just mass of the one star?

What is Mp + Mc in kilograms according to your calculations? (was that calculated using G as part of the equations?)

Is there a way of estimating the mass without having to resort to using a known G? For if G is a variable that changes due to gravitational radiation (as I have proposed), we (you and I) would need to isolate it somehow.

This mass is possibly based on its inertial mass, for if G varies there is no longer the same link between inertial mass and gravitational mass.

Last edited: Nov 19, 2012
10. Robittybob1BannedBanned

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I have attempted to go through and calculate T (or Pb as it is also called) and find that if I maximize and minimize the errors in the measurements I get Maximum and minimum periods and if I take the difference between them I get the following.
7.355449043 seconds TcMax - TcMin Using my previous masses.
7.356257787 seconds TpcMax - TpcMin Using the more recently quoted masses.

Now what I want to understand is since the period is known to an exceptional high degree (accurately known in order of 10^-12 seconds per orbit), so can that be used to define some of the less well known amounts?
For there is just not that amount of uncertainty in measuring the orbital period.

PS: Taylor and Weisberg make it clear they are using the Julian approximation of 365.25 days of 24 hours of 3600 seconds.

The orbital period is well defined.
Pi is defined to any degree of accuracy.
so all the variation must be be in the masses, the distance between them, or the gravitational constant. Can either of these two be define without referring to G?

Last edited: Nov 19, 2012
11. Robittybob1BannedBanned

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The orbital period of the Hulse Taylor Binary is well defined. (10^-12 seconds?)
Pi is defined to any degree of accuracy.
So all the inaccuracy must be be in the measurement of the masses, or the distance between them. Can either of these two be defined without referring to G? If they can it is possible to assume a variation of the gravitational constant is possible.

12. rpennerFully WiredValued Senior Member

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Incorrect. Most of the uncertainty in the sum of the masses (measured in the geometric units of seconds) comes from the relative uncertainty of $\dot{\omega}$ and $e$ which is further compounded by the uncertainty of $\gamma$ if you want to know the mass-ratios, and/or the uncertainty of $G$ if you wish to express the result in kilograms rather than solar masses.

13. Robittybob1BannedBanned

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Could you rephrase the last sentence please as I am having trouble understanding it. Particularly this bit "] if you want to know the mass-ratios, and/or the uncertainty of $G$ if you wish to express the result in kilograms rather than solar masses" Thanks.

Last edited: Nov 22, 2012
14. brucepValued Senior Member

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Unit kg or unit solar mass. Which is likely to have the smallest error bar when calculating G. For you it doesn't matter since G is an empirically measured coupling constant not a calculated coupling constant. For you it doesn't matter = your macro isn't capable of measuring the coupling constant G. Just like all these smart guys have been telling you from the start.

15. Robittybob1BannedBanned

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It is not the macro that does it but the application of the orbital period formula. If T and Pi are known to a high degree of accuracy the variation must come from an error in the determination of of G, (m1+m2) or a.
So we are continuing to look at which of these are measured and their errors to see if the explanation is that G in the binary has faded to some degree. The mechanism of paying back the negative energy debt seems to be a possibility as after each year it is possible to find values which totally balance the energy changes. Your argument has not won the day yet.

16. Robittybob1BannedBanned

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Learning about Keplerian orbital parameters and how they are determined... maybe I will be able to answer my own questions one day!

17. Robittybob1BannedBanned

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Getting over the stroke was my main problem last week and looks like this week too.

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19. Robittybob1BannedBanned

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My sight has improved, even though still not the best in the mornings. Having to type this with one eye closed to be able to see the keyboard. Can't wait till I am able to solve the issue.

20. Robittybob1BannedBanned

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Typing with both eyes open again. Now to resurrect the macro and see if G falls to zero in a black hole. I have a feeling it won't, but let the math tell the story.

21. Robittybob1BannedBanned

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You try and work a macro after a stroke! Seems to be too difficult.

Last edited: Dec 6, 2012
22. Robittybob1BannedBanned

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I read back over the last few pages of this thread, and was surprised to see some of the posters banned, but the memory of all my posts were still there. There doesn't seem to be anything wrong with my memory of the science here.

23. Robittybob1BannedBanned

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Doesn't the list of "Members who have read this thread" look like a sad bunch of names. You would wonder what is going on in the background? All those names who never post anything at all!
Tomorrow I'm going to revisit this thread and commence the macro again For us it will be the 12/12/12 what better date to do it on.