Timo Moilanen

Gravitation constant or not

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In my very thorough research on the gravity constant I made a connection between known constants ( G , c ; NA ) After a solid year of research on the gravitation constant , that I now say is 7.46209031*10-11 Nm2/kg2 , I have to question its real nature . It seem to be a mere modulus (multiplier) when modifying the qualities of kg*kg and mass .Here some short notes  picked from my work to this topic.

I'm very sorry for my still poor writing , and the shortness almost looses the ideas . On FB< Essence of gravitation > , hear from you.

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Why are you working in modern metric units?

 

Cavendish's apparatus was designed and largely built in the 1780s

 

The kilogramme was not intorduced until 1795 and has changed slightly several times since then.

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25 minutes ago, Timo Moilanen said:

After a solid year of research on the gravitation constant , that I now say is 7.46209031*10-11 Nm2/kg2

As this is different from the measured value, your hypothesis/guess is falsified. 

Bad luck. 

(I might make other comments but as you have posted images rather taking the time to write something, then I can't be bothered either. )

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Posted (edited)

It should be different , it's calculated on the principle that all dM parts have the same constant all time and Ti is calculated from that    Ftot . Traditionally the constant G is calculated from F where the from sides coming force component is left uncounted for ( it is not measurable ) , but by my meaning it must count at all distances same way as far away => cos (a)=1 The falsified suggestion is really offending 

Edited by Timo Moilanen
Calming

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!

Moderator Note

Last time you were asked to show how your idea could predict the orbit of geostationary orbits. 

That thread was closed due to your lack of engagement. Can you please start with the above challenge and then address the other points in that thread. Failure to do so will result in the closure of this thread and you not being allowed to reintroduce the topic. 

 

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Earth angular rotation speed w =2 pi rad /86164s  =7,2921*10-5/s  and the height over earth midpoint is r 

We have formula r=(GM/w^2)^(1/3)  for G I put in Ti 7.46 *10-11 Nm2/kg2but for earth mass I insist 5,344 *1024  kg 

r= (7.4972*1022 m)^(1/3) =42166415m =>  42166km- 6378km  =35788 km over sealevel 

My point is still that gravitation constant and earth mass is different from "customary belief" , so both must shange for satellite to stay in sky 

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6 hours ago, Timo Moilanen said:

It should be different

You have (as far as I can tell from your rambling description) the same equation with Ti substituted for G. The value of G is both measured and regularly used for a wide range of  applications (planetary motions, satellites, spacecraft, GPS, gravitational lending, and many experimental test of GR). 

As far as I can tell from description you have zero evidence that G is wrong and Ti is correct. 

1 hour ago, Timo Moilanen said:

My point is still that gravitation constant and earth mass is different from "customary belief" , so both must shange for satellite to stay in sky 

And the evidence for this is ... ?

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In measuring the value of Ti it differs from G because  I take into account the sideways parts of the dF " vectors "  that cancel out each other at close distances . At  longer distance all vectors come from same direction and the gravitation "work" at full efficiency =Ti  calculated from Ftot  .

All applications using gravity as a parameter are adjusted to G as " standardized value , and so are the " authoritative " science too . It's evident there must be consensus  about this ( thats why I renamed Ti) , but at some point I hope at least science stop multiplicating "the cosmos " with 0.89 , or dividing . I think it often is equalised ,that's no good excuse in the future .

The logic of measurement calculus G vs. Ti is the point , and only difference .

What comes to explaining , I'm a "foreigner ".

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6 minutes ago, Timo Moilanen said:

but at some point I hope at least science stop multiplicating "the cosmos " with 0.89 , or dividing

Please show an example where this is done. 

7 minutes ago, Timo Moilanen said:

In measuring the value of Ti

You haven't measured the value. Measurements show you are wrong. 

9 minutes ago, Timo Moilanen said:

All applications using gravity as a parameter are adjusted to G

How are they "adjusted"? The equations describing gravity were not changed after G was measured. 

10 minutes ago, Timo Moilanen said:

but at some point I hope at least science stop multiplicating "the cosmos " with 0.89 , or dividing

Please show an example where this is done. 

10 minutes ago, Timo Moilanen said:

In measuring the value of Ti

You haven't measured the value. Measurements show you are wrong. 

10 minutes ago, Timo Moilanen said:

All applications using gravity as a parameter are adjusted to G

How are they "adjusted"? The equations describing gravity were not changed after G was measured. 

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All involved have agreed on a common (average ) value of G , and that is necessary to make "things work".

So far there have been no alternative at all . Now I introduce a coefficient that is unambiguous Ti= 1/2*c2 /NA /1000g/kg , all constants well known, and by my calculation also in signification exactly what G is supposed to stand for.

Here some more math to calculate Gravitation force and when needed Ti . I'm ready to prove my point expeimentally

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5 minutes ago, Timo Moilanen said:

All involved have agreed on a common (average ) value of G , and that is necessary to make "things work".

Exactly. That value is what is measured, it allows space craft to meet their destinations, allows the orbits of planets to be calculated. That is why it is the right value.

Replacing it with a different value would mean that satellites would fall out of the sky, space missions would fail and GPS would not work.

7 minutes ago, Timo Moilanen said:

Here some more math to calculate Gravitation force and when needed Ti .

As your math produces the wrong result, there is clearly an error somewhere.

Quote

I'm ready to prove my point expeimentally

Good. Let us know when you have some experimental data.

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2 hours ago, Timo Moilanen said:

All involved have agreed on a common (average ) value of G , and that is necessary to make "things work".

So far there have been no alternative at all . Now I introduce a coefficient that is unambiguous Ti= 1/2*c2 /NA /1000g/kg , all constants well known, and by my calculation also in signification exactly what G is supposed to stand for.

What are the units here? m2/(s2-mol)? How do you get Nm2/kg2 from that?

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I don't see a reply to my question in post#2.

 

This is an important point because you are working to quite a few significant figures in your calculations and it matters what values you take for measurements, since you are using converted values.

The conversion constants have (subtly) changed several times over the centuries.

 

To show that this is not an idle concern, here is a short anecdote from the second world war.

 

Once the Germans started bombing the English factories, the manufacture of spare parts was undertaken in America.
At that time, both England and America used pounds, inches etc.

When the spare parts were delivered they did not fit the machinery.

This was traced to a small difference in the size of an American inch v an English inch.
The Americans worked off English drawings, but used their micrometers to manufacture to perhaps a ten thou of an inch.

Today american inches are the same as english inches, but their gallons are still different.

 

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Would I be correct in saying that in all cases you think that 

GM = TiM'

Where G is the gravitational constant, M is mass, Ti is your constant and M' is your modified mass? 

 

 

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On 10/10/2017 at 0:59 PM, swansont said:

What are the units here? m2/(s2-mol)? How do you get Nm2/kg2 from that?

The units must stay as earlier to complete the formula , The mol-1 is not there  Ti=c^2*M'/2000NA ,M' = molar mass for protons and neutrons .  => (m/s)2*kg/mol/[(g/kg)*mol-1] . The kg to M' is" borrowed" ,units for Ti so far  (m2/s2 )/kg .1/2 Mc2   => in ( kg m2/s2) gives [E] =J=Nm and this by a "borrowed "kg  gives   Nm/kg. This is the potential energy between two masses(yet to come) at distance 1m  . but as we know this "spring" is not k*x=F (Hooke) but k/x^2=F  Since  Fgrav=k/x2 and UorE=k/x => [Ti]=Nm2/kg2.

here (the integral) require either to be negative , and i would prefer negative Ti to negative Energy 

On 10/10/2017 at 1:48 PM, Klaynos said:

Would I be correct in saying that in all cases you think that 

GM = TiM'

Where G is the gravitational constant, M is mass, Ti is your constant and M' is your modified mass? 

 

 

No M' is the molar mass and M is used for it ,This [M]=g/mol =kg/kmol , right from chemistry and GM=TiM   ,because M' is part of Ti

18 minutes ago, Timo Moilanen said:

The units must stay as earlier to complete the formula , The mol-1 is not there  Ti=c^2*M'/2000NA ,M' = molar mass for protons and neutrons .  => (m/s)2*kg/mol/[(g/kg)*mol-1] . The kg to M' is" borrowed" ,units for Ti so far  (m2/s2 )/kg .1/2 Mc2   => in ( kg m2/s2) gives [E] =J=Nm and this by a "borrowed "kg  gives   Nm/kg. This is the potential energy between two masses(yet to come) at distance 1m  . but as we know this "spring" is not k*x=F (Hooke) but k/x^2=F  Since  Fgrav=k/x2 and UorE=k/x => [Ti]=Nm2/kg2.

here (the integral) require either to be negative , and i would prefer negative Ti to negative Energy 

No M' is the molar mass and M is used for it ,This [M]=g/mol =kg/kmol , right from chemistry and GM=TiM   ,because M' is part of Ti

M' in this case = molar mass of (neutrons and protons )= 1g/mol = 1kg/1000mol=1kg/kmol

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Right, I was using M' as aX = bX where X is not the same quantity on both sides of the equations is frankly stupid. 

So TiM where M is he molar mass and Ti contains mass information? That must mean that Ti is body dependent and therefore not a constant. 

Your explanation is not clear to me. 

Where you calculated the geostationary height you just substituted Ti for G and a different value of M, let's call them M-1 and M-2. Let me ask my question again. 

Is GM-1 = TiM-2?

Where G is the gravitational constant, M-1 is mass, Ti is your constant and M-2 is your modified mass? 

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1 hour ago, Timo Moilanen said:

The units must stay as earlier to complete the formula , The mol-1 is not there  Ti=c^2*M'/2000NA ,M' = molar mass for protons and neutrons .  => (m/s)2*kg/mol/[(g/kg)*mol-1] . The kg to M' is" borrowed" ,units for Ti so far  (m2/s2 )/kg .1/2 Mc2   => in ( kg m2/s2) gives [E] =J=Nm and this by a "borrowed "kg  gives   Nm/kg. This is the potential energy between two masses(yet to come) at distance 1m  . but as we know this "spring" is not k*x=F (Hooke) but k/x^2=F  Since  Fgrav=k/x2 and UorE=k/x => [Ti]=Nm2/kg2.

So this formula, Ti= 1/2*c2 /NA /1000g/kg, is essentially meaningless. There's no underlying significance to it, since you have to put in a "modulus" to get the correct units (something you objected to in your OP)

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I still don't see a reply to my question in post#2

So I repeated that question in post #13, with additional amplification

So why is 13 unlucky for me since I still don't see a reply to my question?

 

Please note that answering legitimate questions is a requirement of this forum.

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On 10/8/2017 at 0:25 AM, Timo Moilanen said:

It should be different , it's calculated on the principle that all dM parts have the same constant all time and Ti is calculated from that    Ftot . Traditionally the constant G is calculated from F where the from sides coming force component is left uncounted for ( it is not measurable ) , but by my meaning it must count at all distances same way as far away => cos (a)=1 The falsified suggestion is really offending 

 

On 10/10/2017 at 3:42 AM, Timo Moilanen said:

In measuring the value of Ti it differs from G because  I take into account the sideways parts of the dF " vectors "  that cancel out each other at close distances . At  longer distance all vectors come from same direction and the gravitation "work" at full efficiency =Ti  calculated from Ftot  .

 

For homogenous spheres it makes no difference if integrated over the complete sphere, or simply calculated assuming r is at the centre point. The result is exactly the same, at any distance, whether using the basic formula or doing the calculus. The "inefficiencies" of the mass off the resultant vector line are exactly compensated by the "extra efficiencies" of the mass that is closer than r being a greater effect than the inefficiencies of the mass further away than r.

At greater distances these inefficiencies become less significant, but it is not relevant except when non spheres or varying densities are present since they otherwise cancel out at any distance.

Edited by J.C.MacSwell

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6 hours ago, Klaynos said:

That must mean that Ti is body dependent

G is body dependent  , Ti is not . As I write in my paper 2:nd page F = Ti *M1*M2 /r2  but   Ti = F * k1 * k2 * r2 /(M1 *M2 ) ,

And that is because a sphere is not  acting like a mass point when calculating Ti , but when measuring F and calculating Fm it is absolutely like a mass point. The slightly sideways dF  vectors and all different distances r + - R and cos(a) mean for example that Fm on surface on a even density sphere is M*Ti but  Ftot =3/2*M*Ti , k=1.5 

G is calculated directly from Fm  and measures is done when r is about 1.2 to1.3  R and k values k1 *k2 = some 1.12 

that should have been multiplied to G , but there wee no k values thought of . It is an oddity that F=M*Ti on a sphere , all other shapes and density divergence need much more calc. The "easy" calculus do'nt  go for Ftot . 

And for the easy  calculation , I have not heard anyone integrate V/r2 before or after I did it last December

 

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That doesn't appear to answer my question or it's not consistent with your answer finding geostationary altitude. 

  • Upvote 1

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52 minutes ago, Timo Moilanen said:

G is body dependent  , Ti is not . As I write in my paper 2:nd page F = Ti *M1*M2 /r2  but   Ti = F * k1 * k2 * r2 /(M1 *M2 ) ,

And that is because a sphere is not  acting like a mass point when calculating Ti , but when measuring F and calculating Fm it is absolutely like a mass point. The slightly sideways dF  vectors and all different distances r + - R and cos(a) mean for example that Fm on surface on a even density sphere is M*Ti but  Ftot =3/2*M*Ti , k=1.5 

The symmetry of the problem trivially shows that this is not the case. Any point with a sideways force has a matching point with the force in the opposite direction, present at the same r.

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Yes and at short distances these sideways "vector * sin(a)"cancel each other out , but they not disappear and on long distances they "work" in same radial direction (sin(a) =>0 ; cos(a)=>1)and make a sphere seem like a point mass.

And then there are the Sigma dM/r2 that is =3/2*M/r2 at r=R(surface) that I cal Ftot when multiplied by Ti

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23 hours ago, swansont said:

The symmetry of the problem trivially shows that this is not the case. Any point with a sideways force has a matching point with the force in the opposite direction, present at the same r.

I'm sorry I'm messing up . For objects the factor k is also needed when calculating Fm =TiM/(kr2 )  or F= Ti M1 M2 /(k1 k2 r2 )

k for spheres is Ftot  / Fm = 3p2 /2 - 3/4(p3 -p)*ln[(p+1)/(p-1)]  and this mean that spheres are not like point masses . 

This means Earth has an more  complex gravitation field than F/r2  and the shell distributed density " actually smoothen it some ,  but should by my meaning have made satellites on lower orbits dependent on empirical knowledge  , the geostationary ones are only 0.03% lower than would be for a point mass.

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2 hours ago, Timo Moilanen said:

I'm sorry I'm messing up . For objects the factor k is also needed when calculating Fm =TiM/(kr2 )  or F= Ti M1 M2 /(k1 k2 r2 )

k for spheres is Ftot  / Fm = 3p2 /2 - 3/4(p3 -p)*ln[(p+1)/(p-1)]  and this mean that spheres are not like point masses . 

This means Earth has an more  complex gravitation field than F/r2  and the shell distributed density " actually smoothen it some ,  but should by my meaning have made satellites on lower orbits dependent on empirical knowledge  , the geostationary ones are only 0.03% lower than would be for a point mass.

Why would a scalar matter when looking at whether vectors cancel?

Where is the asymmetry, required to have non-cancellation of "sideways" components?

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