1

The gravitational constant

I will not repeat all the well-known historical facts about this constant it serves no

purpose. What I will show is that the gravitational constant must be determinable

from other known constants. The mainstream will receive this with protest; since

Albert Einstein’s work is respected and untouchable, not to forget that millions of

Dollars funding for new measurement methods of the constant of gravity are more

interesting than the simple truth ever will be. Newton’s gravity is thus degraded to

an approximation of Einstein’s work. However, I will forget he ever contributed and

must advise the reader to do the same, at least for a while. Not everything that

shines is gold.

I will only look at the “naked” theory as Newton delivered it. There can be no doubt

that Newton’s theory of gravity is working for every mass we look at, whether it’s

Earth, Jupiter, the Sun, or any other gravitating object in our universe. The only thing

that seems observed is that the greater the mass is, or the greater the speed of an

object, the more is the empirical result deviating from the theoretical. But this fact I

will push in front of this little investigation, first I will show some trivial facts which

seem to be of little interest for the physical society. I do not understand why,

because this is “evidence” I present you here, evidence Newton’s unrightfully

devaluated theory can deliver. No words shall be used to e- or devaluate special and

general relativity. Those theories I will leave for the mainstream to discuss. What I

will do is asking some irritating questions that show up from a scrutiny of Newton’s

theory.

However, if you have too, please try to reject it with rational comments and critics

and not by starting to say: “That is not right, because according to Einstein…” There

still is a great possibility that Einstein might have misunderstood a few things. A

possibility he uttered in his later years all by himself.

The first thing I will show is a comparison of the Chandrasekhar limit (about 1.44

solar mass) with the Newton limit. The Newton limit, rarely known, is a huge mass

derived from the fact that Iron is the last step in solar fusion and has some very

interesting relations to the mass limit of Chandrasekhar. There are different ways of

deriving the Newton limit, but I will here only show the most simple of them.

2

According to the basic equations of Newton’s gravity, every mass has a

characteristic set of data defining every aspect, like gravitational acceleration,

gravitational force, gravitational energy, surface-orbit velocity and characteristic

time, which is the time it takes to cross radius of the mass with the surface-orbit

velocity. As you see, I disregard the fact that Einstein stated that there are no

gravitational forces, but a, by the presence of mass, warped space. I allow myself to

doubt that very much! My presentation of the stuff will further make use of

numerical results and will show all 13 decimals, my calculator delivers. The mindless

claim that this is nonsense to do is the nonsensical part and I will not waste time

explaining why. It should be obvious for everybody using a minute or two thinking

about it. Using just three or four decimals in theoretical scrutinizes is able to hide

some important facts, so please beware of that thoughtless claim.

Newton’s gravity defines a characteristic time t0 in the following way

secmG

rt m

O⋅

=0

3

01.1

Surface-orbit velocity is given by

sec

m

r

mGv

m

O

⋅= 0

02.1

…so radius rm divided by vo gives the time t0

secmG

r

r

mG

r

v

rt m

m

mm

O⋅

=⋅

==0

3

00

03.1

But since mass is given by (G0 = 6, 67408(31) ⋅⋅⋅⋅10-11

according to NIST)

kgR

m Ch

3

4 3 ηπ ⋅⋅⋅= 04.1

...equation 1.03, the characteristic time t0, reduces to

secG

tηπ ⋅⋅⋅

=4

3

0

0 05.1

3

It is obvious now that only density ηηηη is determining the characteristic time of a

mass, which means that every element has its own characteristic time.

Let us now calculate this characteristic time for the Iron density:

secG

Fet 46035424687,67478604

3)(

0

0 =⋅⋅⋅

=π

06.1

Now we can find the Newton limit:

Radius of this mass is given of

mG

CcFetrN 21,202241054172

78604

3)(

0

2

0 =⋅⋅⋅

⋅=⋅=

π 07.1

...and this is the Newton limit, an Iron-mass of

kgG

CmN

38

3

0

6

10107622,7234481878604

3⋅=

⋅⋅⋅

⋅=

π 08.1

Or simply

kgG

ctmN

38

0

3

0 10107622,72344818 ⋅=⋅

= 09.1

The surface orbital velocity vo of this mass equals the speed of light c. The next

mass we look at is the Chandrasekhar limit. Mainstream physician have not (yet)

exactly defined this mass, there are given many different numbers, but the Danish

astrophysicist Jan Teuber †, gave a mass of mCh = 2,8656⋅⋅⋅⋅1030

kg as the

Chandrasekhar limit, it is this value I will use here as test-mass and compare with

the Newton limit.

Radius of this Iron mass mCh with density ηηηη ==== 7860 kg/m3 is:

mckcm

r R

Ch

Ch 90719,443167738 849641,4782484578604

3 3

1

=⋅=⋅=

⋅⋅

⋅=

π 10.1

Comparing those two masses shows a remarkable relationship between them. No

doubt, that the Newton limit is a strange mass, no sun will ever reach almost 137

4

Million solar masses, but the active core of a galaxy, a quasar, can reach that size.

Because of many reasons, I will not even think the words “Black Hole”.

The orbital velocity at the surface of a mass is given by…

sec

m

r

mGv

m

o =⋅

= 0 11.1

…so the radius of every mass is now of cause defined by

mv

Gr

o

m ⋅=2

0 12.1

If now a mass reaches a value so big, that the gravitational force exceeds the

electrons force of repulsion, the mass is collapsing, according to present physics, to

become either a Neutron star or a so-called Black Hole, where all mass is contained

in a singularity. This is a bold allegation, since this would violate reason and the

theory claimed to predict Black Holes. How can you (or anybody) assign a density to

a hallucinated “entity” with no volume ? Since the speed of light is a limit, the

surface orbital velocity can only reach the speed of light. Otherwise, gravitational

energy of the collapsing mass will go against infinity, which obviously is impossible!

No valid argument can demand that (vo )2 →→→

→ ∞∞∞∞ and thus exceeds c2 and resulting in

a collapse-radius rCo →→→

→ 0. Black Holes are thus rejected as a wrong understanding of

gravity. Those considerations show clearly that the collapsed mass must have radius

mc

GrCo ⋅=

2

0 13.1

53532866152 456,==Co

ChE

r

rk 1.13.1

For the Chandrasekhar limit this means that collapse radius is

mk

cm

c

Gr

E

ChCo 9713301058,2127 849641,47824845

22

0 =⋅

=⋅= 14.1

Comparing now the radius before collapse with the collapse radius, we get

2

9713301058,2127

90719,443167738Ek

m

m= 1.14.1

5

This factor kE = 456, 35328661525 (see comparison, Table 1 below) is the

connection between the Newton limit and the Chandrasekhar limit… what does this

point at Gentlemen ?? Overcome your irrational fear and acknowledge that not

everything you once learned is true.

Data Newton-limit Chandrasekhar-limit Relation 1 Radius m RN= 2022410544172,21 m RCh= 443167738,90719 m kE=456,35328661525

2 Mass kg mN= 2,7234481810762⋅1038

mCh= 2,8656⋅1030

(kE)3

3 Gravity m/sec2

gN= 444397,98952565 gCh= 973,80253974259 kE

4 Vo m/sec Vo= 299792458 m/sec Vo= 656930,64297519 m/sec kE

5 t0 sec t0 = 674,60354246874 sec t0 = 674,60354246874 sec 1

6 tR=R/c sec tR = 674,60354246874 sec tR = 1,4782484584964 sec = kR kE

Table 1

The interesting part we see in row 6, tR = 1, 4782484584964 sec and the fact that

tR⋅⋅⋅⋅ kE = t0 this is a “package factor” for spherical atoms…. !! You may call it the

missing link to a gravitational theory that connects with atom physics.

The number 1, 4782484584964 needs first an adjustment, because it is a constant

with a very adventurous effect. Talking about mathematical beauty… (that is how a

dreamer works) is not something I want to compete with. Instead, I will present you

for a gravitational theory that covers everything necessary, as far as I can see.

This package factor is defined in a very special way and can only be proved 100 % if

it is possible to calculate the package density for spheres in a spherical container.

This was not possible for me with the mathematical tools available to me. This factor

is not and cannot be the Kepler conjecture, which is important to realize. Many

people argue that when the number of spheres goes against infinity, and thus the

size against zero… then the package factor will be the Kepler conjecture. This is NOT

true, since the Kepler conjecture is only valid for tetrahedral space. The “Barkholz

conjecture” is now that the package factor kR is the fourth power of the package

factor for circles in two dimensions and thus given by

7454783014446,114412

4

4

==

=

ππRk 15.1

"The power of simple models, the power of stripping things down — [Anderson]

taught me the power of reducing everything to the simplest possible idea and

following it to where it could go," Haldane said.

6

Or by the corrected gravitational theory

26

1

33

0

3

78604

3

⋅⋅⋅=

πGkR 16.1

The density for the standard particle is the density of Iron, since Iron is the last step

in solar fusion. This is an allegation I will now try to justify.

Consulting Table 1 tells us that the characteristic time of Iron is given by the factors

kR and kE …that means also the definition of the factor kR is justified.

sec 46874674,6035420 =⋅= ER kkt 17.1

This in itself is telling a great story… but mainstream physicists are not working like

that. It will take quite a while to digest this and finally laugh, as it did for me.

We go back to equation 1.05 and manipulate a little

⋅⋅⋅⋅=

⇔⋅⋅⋅

=

2

3

2

0

0

2

0

2

0

4

3

)(4

3

seckg

m

tG

secG

t

ηπ

ηπ 18.1

It must be clear for everybody (except for those dreaming that G0 is variable) that

⋅=

⋅⋅=⋅

3

2

0

2

0 5594,35770069244

3

m

seckg

Gt

πη 19.1

… necessarily must be constant too, so I expect naturally, since density defines t0,

that ηηηη is a constant. In other words, there IS a standard density, since

)(4

3 2

0

2

0 secG

tηπ ⋅⋅⋅

= 20.1

That means the gravitational constant IS determinable… we can calculate that

constant.

So… what is equation 1.20 suggesting gentlemen ?

7

We won’t go anywhere by discussing this from an Einstein perspective, I am a

hopeless guy that believes he found the philosopher's stone and not for a second

will I listen to voices that try to tell me what Einstein once said. I don’t care !

I have to show you how that works out. My suggestion for a package factor for

spherical atoms in a spherical container is given with equation 1.15 and leads to a

fully consistent gravitational theory. However, we have some problems we need to

solve in some way. Mother Nature has no units for distance and no units for mass.

You may protest now, but that is until now a not realized problem. Using meter and

kilogram does not induce any errors, but it closes the way to a further

understanding of gravity. The used equations are right and give the numerical right

answers, but are not able to show the way from gravity to atom physics. Let me

show you why:

All we can do with natures own parameter is defining the distance, or the radius of a

mass, by the time it takes the speed of light to cross the distance or radius.

Therefore, radius rm of a mass must be defined like this

)()()( metercxfsec

mcsectr rm ⋅=

⋅⋅⋅= 21.1

…or, in case of a homogenous mass, the Chandrasekhar limit, by a function of radius

rAtom and number Nam of atoms

( ) )()(

3/1meterNaxfrr mAtomm ⋅⋅=

22.1

The function f(x) will be called kR… that shows also the definition of mass, because

nature does not know about the human kilogram either, it only knows about the

number of fundamental mass-units, which we define in kg. For a planet like earth, it

is fine to know the average density and define the mass by volume and density, but

in case of an examination of border cases, like the homogenous Chandrasekhar limit,

we need very exact definitions in order to get an exact mathematical result.

Therefore, the consequence of equations 1.21 and 1.22 is that the mass-limit will be

defined this way

( )

)()(3

32

kgmr

cxfm atom

Atom

⋅⋅

= 23.1

8

If we now check for the radius of the Iron atom, we get when using equation 1.22

( )

( ) 3/1

3/1

)(

)()(

m

m

Atom

mAtomm

Naxf

rr

meterNaxfrr

⋅=

⇔⋅⋅=

24.1

f(x) is the package factor kR = 1,4782484584964

Radius of a 1 m3 sphere is

m99416203504908.04

3 3

1

=

⋅π

mk

mr

R

Fe

Fe

103

1

10392397388650,178604

3 −⋅=

⋅⋅⋅

⋅=

π 25.1

Testing with equation 1.25 and calculating the radius of a few elements gives those

results:

Element Official measured radius nm Calculated with package-factor kR

Lithium 0,152 0,15191

Sodium 0,185-0,186 0,18503

Iron 0,124-0,126 0,12397

Caesium 0,262-0,265 0,26552

Indium 0,162 0,161

Scandium 0,160 0,159

Vanadium 0,131 0,131

Table 2.

The deviation of the kR factor from my conjectured factor 4

144

π=Rk is just

1,0000358438923 !

It is of course not possible to get the right answers with a “foggy” definition like the

continuum definition of mass: volume multiplied by density. No matter how exact

volume and density are given, the resulting equations are the problem, they do not

allow to “open the door” to additional solutions.

This is clearly shown by Table 1 row 6 (shown again next page):

9

Data Newton Chandrasekhar Relation 6 tR=R/c sec tR = 674,60354246874 sec tR = 1,4782484584964 sec = kR kE=456,35328661525

All we have to do now is accepting these very fantastic relations:

7454783014446,114412

4

4

==

=

ππRk 26.1

95398728781,6743

4144

13

4

0 =

⋅⋅

=

ππ

t 27.1

2938603004,4563

4144

12

4

=

⋅⋅

=

ππ

Ek 28.1

Provided those three equations above can be accepted as right, then for every

(homogenous) mass m with an arbitrary density ηηηη the following applies:

If )(Secc

rk m

R = and

m

E

r

mG

ck

⋅=

0

then

⋅⋅⋅⋅⋅=

⋅

⋅=

2

3

222

3

04

3

Seckg

m

kkmk

kcG

ERE

R

ηπ

It is obvious that the product 22

ER kk ⋅⋅η must be constant, but kR, kE and ηηηη are each

varying, depending on mass and density. Now since the product is a constant and

the gravitational constant is valid for every mass-density, kR and kE must be two

constants. kR as given above, is the package factor for spherical atoms in a spherical

container and kE is the factor given above as ratio between the speed of light and

surface orbital velocity of the mass-limit, and finally, ηηηη is the density, which must be

a constant too. Therefore, I suspect the AMU unit is representing this standard

density, but am afraid that the value of the AMU unit must be changed about

minus 0,274 %.

10

Why I think that kR is a package factor ?

We have that

93951137,455089222

0 =⋅= ER kkt 29.1

…and my conjecture is that package factor for spherical atoms in a spherical

container (the gravitating spherical mass limit) is 4

144

π, deviating from Table 1 only

1,0000358438923. A further part of my conjecture is that the factor kE is defined by

kR in the following way

2938603004,4563

1444

12

4

=

⋅⋅

=π

π

Ek 30.1

Table 1 gives for kE = 456,35328661525, which is deviating only 1,0001302369372

…the characteristic time t0 of the last step of gravitational fusion, Iron, is according

to my conjecture defined by

5398728782,6743

1444

13

4

0 =

⋅⋅

=π

π

t 31.1

…deviating from Table 1, t0 = 674,60354246874 sec only 1,0000943896619. As you

can see, the deviations are very small. The question is now to believe that this really

is the solution for a correction of Newton’s gravity. When I found this 14 years ago I

did not believe it, but I made the experiment because the deviations were so small.

I will now show the result of this small correction, which clearly points at that the

Chandrasekhar limit has to be understood as the mass limit of mass limits, the

supernova, there is but one supernova only. The observed different supernovae are

merely different because of different total mass. The brightest shining supernova is

the one very close to the Chandrasekhar limit and the less shining has a very big

mass, until we reach the variable stars with masses so big that the explosion of the

mass-limit only can blow up the star until gravity again takes over. That tells us that

the shorter the period the bigger the mass.

11

However, those thoughts are discussed at a later point in time, but not in this

article.

The secret is that the package factor is given by this:

7454783014446,1144

78604

34

26

1

33

3

==

⋅⋅⋅=

ππBa

RG

k 32.1

This allows determining the gravitational constant, which now is given by

2

311

2633

3

105676753430883,678604

3

seckg

m

kG

R

Ba⋅

⋅=⋅⋅⋅

= −

π 33.1

Yes… unbelievable but seems to be true. Using equation 1.18 to calculate G0 (now

called GBa) is giving

2

311

22105676753430883,6

78604

3

seckg

m

kkG

RE

Ba⋅

⋅=⋅⋅⋅⋅

= −

π 34.1

So and now you want to accuse me that I have constructed this in order to harvest

what you all dream about ?!

The factor kR for other densities is given for e.g. Lithium (534 kg/m3) by this

9296393870963,1534

7860144)(

26

1

4=

⋅=

πLithiumkR 35.1

That means a Lithium mass-limit should have radius

mcLithiumkR RLi 24111,491475887)( =⋅= 36.1

This is a Lithium mass of

kgR

m Li

Lithium

293

107276554401191,23

5344⋅=

⋅⋅⋅=

π 37.1

The collapsed radius of this mass would be

mmc

GcollapseR Lithium

Ba

Li 82280579342,197)(2

=⋅= 38.1

12

Giving a kE factor of

3

4)(5805901338,1578

)()(

12 π⋅⋅===

Lik

CollapseR

RLik R

Li

Li

E 39.1

The characteristic time of this mass is then

secLik

mG

RLit R

LithiumBa

Li 9046500815,25873

4)()(

133

0 =⋅⋅

=⋅

=π

40.1

You want to say: “It is a construction of yours.” …right ? Equations 1.26 to 1.28

seems to be valid in general… please do not come with the stereotypical answer that

I used a circular definition. Try to explain instead of… what is this I show here?

Let me show you a new property belonging to Newton’s theory, the gravitational

impulse:

The gravitational “eigenimpulse” of a mass pm is defined as follows

⋅=

⋅=⋅=

Sec

mkg

R

mGvmp

m

Ba

om

3

41.1

This "eigenimpulse” is the product of surface orbital velocity and mass and can of

cause never exceed mc, which is confirming the existence of a mass-limit very nicely.

This impulse is now given of this:

With the impulse given at the speed of light and using the package factor, we get…

CmR

mGNa

Rp Ch

Co

ChBa

Fe

Fe ⋅=⋅

=

⋅⋅

⋅⋅⋅=

33/1

4

4

4

0

144

3

78604

π

π 42.1

The gravitational constant is disappeared as you see, why do you think it

disappears?

I conclude further that the explosion of the supernova is NOT caused by some exotic

nuclear reaction, but merely by the fact that time stands still for the collapsed mass-

limit. That means the (imaginary) gravitational “eigenimpulse” becomes real.

13

The Chandrasekhar limit is now also given of

kgR

mCm

Fe

Fe

Ch

30

1233

433

103728659081538,278604144

3⋅=

⋅⋅⋅

⋅⋅⋅=

π 43.1

…or as already shown

)(101748659081538,2144 30

42

3

kgkG

Cm

EBa

Ch ⋅=⋅⋅

⋅=

π 44.1

The radius of the Iron atom is now

mm

r Fe

Fe

10

3/13

103712397240530,178601444

3 −⋅=

⋅⋅

⋅⋅=

π 45.1

…so the number of atoms in the mass limit is given by this

55

2

4

3

4

3

0 104580903881447,3144144

⋅=

⋅

=⋅

==

ππ FeFe

Ch

Fe

Fer

c

r

r

m

mNa 46.1

The gravitational acceleration of the mass limit is

22

0

sec90212936659,974

m

r

mGg

Ch

Ch

Ch =⋅

= 47.1

The gravitational acceleration of one Iron atom is given by

2

E

Fe

Fe

Fe

Fesec

m

k

r

R

mGg

16

242

0 107579543694420,5144 −

⋅==⋅

⋅⋅=

π 48.1

Now the total mass of NaCh=3,0903881447458⋅⋅⋅⋅1055

Iron atoms has a gravitational

acceleration of

2

3/15

4

42

0

sec90212936659,974

144

144 mNa

r

mGg Ch

Fe

Fe

Ch =

⋅

⋅

⋅

⋅⋅=

π

π 49.1

2

3/15

4

sec90212936659,974

144

mNagg ChFeCh =

⋅

⋅=

π 50.1

14

Very exact, isn’t it… and this is NOT manipulated by me! Try now to do this with

contemporary physics and let me know how and when you succeeded !

A true gravitational theory must be able to deal with the continuum definition of

mass as well as with the number of atoms and their radius. That is what I

demonstrate here !

However, we are not finish showing what this correction of Newton’s gravity is able

to do, but this will be in another article. Here I will just show a simple fact.

One last thing is the Lorentz factor, presented by the factor kE = 456, 2938603004:

The Lorentz-factor for the Chandrasekhar mass-limit is expressed by the factor kE

alone:

9850000024014,1

1

1

1

1

1

2

2

02

0

2

2

=

−

=

⋅

⋅−

=−

=

c

v

cr

mGk

kf

Ch

ChE

EL 51.1

This is NOT the right equation, as you will see at a later point in time. Equation 1.51

is here only given for pure demonstration of the Lorentz factor.

A further change of the theory of gravity will be shown later since the gravitational

Impulse shown above is NOT real for an object on the surface, or the gravitating

mass itself. This requires a complex theory and a negative gravitational constant and

is NOT considered here. However, that this seems to be overlooked shows me that

the praised “scientific method” is mentioned more often than seriously used !

What is your opinion about this !

Please read and test this carefully... it is not a sick idea, as David and David wanted

to convince me about. If it is judged wrong... then I must ask: “Why is it existing and

consistent ?”

Copyright 2002 Dec. 24. Berndt Barkholz