Timo Moilanen

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.

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

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

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/kg2 . 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

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

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 ".

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 m3 )^(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

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

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.

I have via research on the gravitation constant managed to couple Avogadros number ( as amount of neutons +protons ) to Newtons F=MG/r2 . It looks mush like gravitation would be in elementary particles and their features "before reaching relativistic circumstances " , But where did they" come from" url deleted

Sorry again . I'm too upset coz I published new formula and got banned from that site (Ursa). I wrote wrong and correct now g =Ti*M*(1/r^2 -3R^2/(14r^4)) and Ti =7,42*10^-11 Nm2/kg2 . I know my writing is bad but hardly a reason for banning me from a cite , so I suppose my work is stolen now . yours Timo Moilanen

As you begin (physics course) " for any uniform massdistribution " and this is very true for any "force" that is exactly proportional to d M/r^2 , and we probably can assume I do not think gravity is one of them , but should be calculated from cg or whatever midpoint gravity can be called and is situated dr= 2R^2/5r closer to observation height than sph. midpoint so dM/(r-2R^2/5r)^2 and that I'm going to explore further and it give almost impossible to " swallow " different perspective and after all maybe classic mechanics also is right ( Newton , Euler a.s.o) And I would be very interested in opinions of my integrals since they don't contradict any "modern belief " and should there for be harder to "declare" wrong . Newton would have loved the first (two parted) one

Now I have searched ( not very deep ) and did'n find the R1, R2 and r values that are essential for me There's more to dig Yes and before the parenthesis it's multiplied by 0 to be sure and that part of equation is gone ( =0) With R = 0,9999999 it make F=- GmM/ 2,00000001 but its not my equation so? The math is straight forward but as I expect you to mean the mass , so that is undone for any theorem

If you dont understand this short calc. I'm afraid I can't help from here EXACTLY what was the right answers again and how is the right way of finding them All measurements and most predictions from them will come out similar or very near , except ALL stellar MASSES change RADICALLY . So there really is only one minor adjustment

So you mean that this single site is where support /no support and thereby right/wrong is decided (by Strange)

You need to insert r=1 before reducing r away obviously F=-GmM/(4Rr^2)*{ 2R-(r^2-R^2)*[a.s.o.]} => F=-GmM/4R*{2R-0*[ ] } => F=-GmM /2

No because on shorter distances g = Ti M/(p-R^2/p)^2 must be right and that do not fit with G . So the only solution is that the "known" mass is much bigger . And G is supposed to be proven by the masses and the masses is calculated from G , how conwinient but in fact the same must also apply to Ti because we still have [kg] . The distance unit thou is[ R ] and distance therefor n*R =p

And who says the theory found no support ? It is anyhow without (magic ) and chaos, simply classical mechanics and some vectors and a reasonable simple logic . However the gravity particle look like , their qualities need to be summed up to form a gravity of "macro" physical size

And yet give different answer ?. One mainpoint is that now M*G = Ti *M

G measurements with newer methods and old theory will end up in G being 8,05*10^-11 Nm2/kg2 , when gap between masses goes to near 0 and R2 to small(free atoms ) .Cylinders give smaller ca. (8,05-6,67)/3+ 6,67 = 7,1 *10^-11

After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong

Just masses and constant G much wrong I should have said third and forth formula from end

There are for me useful "measurements to set const. Ti . Cavendish experiments with spheric masses .but I need details of the masses radiuses and r the distance . That might not seem very important to labs. who do experiments , otherwice than publicing them is like giving away their quite expensive job for free . G = Ti* f(R1,R2, r) as I see it

The result on small (laboratory) spheres will give same F(force) results as far as accuracy goes today , but " the my theory " will give a more "stable" constant not tending to change with increased mass for example as today. On the right scale (world) they will prove very different and to that I have not enough data or computing capacity and most of all astronomers skills and accumulated experience http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html Here you can insert R=1 and r=1 in the second from bottom and third from bottom formulas and get F= GmM/(2r^2) , That is absolutely my finding for the most outher shells arithmaticly done on computer . I can only say that Gauss had something little wrong or we interprete him too widely

The "my theory " will survive and prove right because physics do not need human acception . I do not see me allowed to keep this kind of "finding" to myself , but your "funktion" in this is for me more and more unclear