Timo Moilanen

And Newton did not do this integral ,what changes nothing because I used the essential from the next integral I did

That is true the edges of disks are further away , and even if the mass is biggest there it does not compensate for the distance ^-2 Yes integral as in Swedish and I'm very confused by the antiderivative too . In online text they use integer ?

My paper starts with the integer and mainpoints how its done

Sorry I have complicated things. The relation between G an Ti is G*M1*M2/r^2 =Ti*p^2 *M1*M2/(p1^2*p2*2) , in witch case p1 is called l2 and p2 is l1 . this also changes const. Ti by 0.3% I do not have data on thousands of satellites and the mass of earth and other planets and sun I predict are not right The integer ( derivative ) show that Newton is right but I went one step further The M= gR^2/G went wrong , it apply only to homogen density spheres or randomly distributed many small "pieces" but not for densities organised in layers or shells

By slices I mean circular discs with radius y , and when getting rid of trigonometry as I tell in text it is very straight forward really. The integer being next to impossible I think is a mere excuse and lazines . But on the otherhand it contains no surprises and give 4/3*pi /d^2 . What yuo put in comes out . Its not a long story but typewriting it (no thanks). With little work it can be checked on net.cites , thats how I werified it and it is now written so it should adapt to those cites ( at least some of them) so you dont have to separately look up integer rules ( ones own memory is hardly a method of validating ) Not easily , but with measured gravitation from several heights it should give masses and thereby densities on different depths Best estimations on total mass is gotten from high as possible orbiting satellite "speeds" orbit times however one want to calculate centripetalforce vs. gravity "pull# Since I have no very exact estimation on constant Ti the best I can do is calculate Ti*M (MG) for earth that is (23,97) 24,0 *10^13 Nm2/kg for orbit 35786000m high

"A mass have two qualities size and weight " , size stays and weight depends on distance to where the other " source gravity" origins .Its own field Ti*M/ l^2 where l is meandistance to its own (M or V ..), Sorry p is the distance to () l is distance gravity and =p-2R^2/5p for a sphere The masses are not so straight forward as till now calculated . The example is on earth but sun is about twice heavier than till now assumed

Midpoint of gravity I mean where the longer away "slices of a sphere cause an equally strong force as the closer ones , And that is - 2R^2/5p for a sphere ( my later integer) divided by total gravity at distance p from sph. midpoint . Gravity qn average origins closer than sphere midpoint and therefore I calculate g= M*Ti /( p-R^2/5p)^2 = M*Ti /l^2 . The midpoint of sphere have no impact on gravity and shall be measured from its own mean value not that of the mass . g = Ti *M /(p-2R^2/5p)^2 p is distance to sphere midpoint , R sph. radius Ti new constant ( coz of different approach). p-2R^2/5p =l the ( shorter than p) distance for gravity (x/r^2) than mass (x/r) The p1 and p2 is for calculating the constant from a two sphere system ( Cavendish) All units in SI ( m, kg ......) The earth gravity on paper with the named quantities and "unnamed SI units

But since I do not agree on the planetary masses . Only thing so far that I'we been able to narrow down is the M*G for earth M*Ti for me to 24,0 *10^13 and that suggest I have too small value on my constant . I'm sure Newtons law is made to rather correct by adjusting planetary and other masses to "specifics" , but I still search for Cawendish experiment data to establish constant Ti and then compare to maybe at least ask a good WHY ! Sure Thank you there must be available couple hundred years of

I calculated it to over 32500 km . That and when I got M*Ti (M*G) =24,0 *10^13 tells me that my estimation on const.Ti = 2,95*10^-11 is too small . Sorry the satellite height answer vent to someone else Since my constant (Ti) still is to be "adjusted" and I disagree (somewhat) on mass of earth I calculated via the geostationary orbit that M*Ti ( MG used) should be 24,0 *10^13 for earth and that need by my meaning a very much bigger " iron core " since in my calc.model the core have less impact on surface gravity than outer layers

In g= M Ti /(p-2R^2/5p)^2 , M*Ti is equal to Newtonian G*M and l =p -2R^2/5p is for example in sun earth distance r= 1.5 *10^8 km l is very roughly 1200km smaller number ( 10^-6 difference) but distance same so Keplers laws G(m+M) do not make any difference . The big difference is that M/G to M/Ti is about 1:2 and why should g=TiM/l^2 be any different from g=GM/r^2 . They are so similar mathematically so I see no other way to determine Ti than using results from Cavendish experiment wirh spheres . Only spheres apply because the 2R^2/5p is not close enough on cylinders Since I have no near to exact value on my constant Ti i can only estimate that earth is close till what is assumed now but that need a sun ca. 2,3 times (1,5-2,5) more massive . to rest of stellar bodies is needed some 0,7 to 1,5 times more mass. The average density do not apply in my calc. Earths mass can be increased quite much by adding heavy elements in core with no dramatic impact on gravity at surface .As a "curiosity " the outer most dirt add to the g only by exact half its weight when measured from surface this can also be seen on spherical shell calculations when setting r=1 and R=1

For Me = 5,97 *10^24 it give 32500km = 28800km abowe surface but I'm not satisfied with input mass and simplification

This theory says inderectly that the mass of sun must be about 2,3 times more and mass of earth happen to be somewhere close to now expected This is not a minor adjustment but rather a revolutionary approach what comes to the results . The math. thou is simple , only R diameter of sph. included to Newtons law Srry D/2

No it is a point where gravity seems on average to origin and is always inside the object .The near side create a stronger (bigger part of the field ) than the far side . The point depend on observation distance and is 2/5 R (radius of sphere)or smaller distance from masspoint ( my second integer) . l=p-2R^2/p and g = M* Ti /l^2

That will not happen very soon as I think needed measurements do not exist yet

Since the constant is not very exactly set jet and the mass of earth (and sun ) is questioned ( by me ) there is quite some way to go before comparing with measured facts . I needed gravity only as a parameter to quantum physics ( realized the old one is useless) and astronomers have equipment and know how to measure my saying in a "hearthbeat" I say that my method is the only right method but is not easier because a good knowlege of earths mass distribution is needed , and I really question the total mass of earth to begin with My method give (close to) near same distances but depend on loads of empiric data (earth mass distribution) Sorry for my language skills There are rally only M mass of the sphere, p distance to midpoint of sph. R radius of sph. and a new gravity constant Ti needed Sorry for that

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After thorough analysis of " arithmetic "( adding pieces on computer ) and then doing two " pen/paper " integers I must conclude that gravity need to be calculated to the midpoint of gravity field instead of midpoint of mass . And therefore I say that the method need to be as I briefly describe in attachment . I dare call the formula g= M*Ti /( p-R^2/5p )^2 a new gravity lav although it is made for only spheres so far ( and far away masses ) I ask you to read and consider my postulate because it is at least for me the only way of handling gravity so " ends meet" .The direction and distance from where the "weight "can be measured is by my meaning same as physically right .The new constant is needed and by small measures change the " universe " radically . Yours Timo Moilanen Please leave my name on copies and citates

Sorry again , meant vertical because we feel only the force downward not those pulling equally strongly to each side . Of the 13,5 m/s2 field only 9,8 is downward . And this means that in center the field is strongest (not absolutely true however rational) but have no direction or all directions are equally stong

The above mentioned lavs are sure accurate , but now we are adding up only the horizontal component , the down force (about integer cos(a/2) of total field ) a being 0 to 2pi around a continous cylinder "cable" .At surface distance being 67% of total in center 0 and far away about equal .The sphere further more have different mass points for crosection area and mass center point for each element . Simpler calculations with say 4-5 number accuracy can suggest the sqrt theory right as is it for the all over field , and a few section calculation can lead anywhere (inexact). Furthermore the difference is larger than 10^-4 only up to 1/10 of the radius 1,1 that .The smaller sphere in a Cavendish experiment is near enough and calculating earth mass (heterogen dencity) I came up with the r= 6369067 is equal point for inversce sqr and drops the mass of earth a stunning 0,12% (the mean radius 3671000 have little meaning in my calcs. but I'we not changed it)

Sorry I strugle with the language and . What I meant is that even the most perfect sphere can not be calculated as a point of mass . On the surface of the uniform density sphere g=mG/r2 is exact (not proven but assumed ) and by my calculations so close one get with a " computer" (15 digits ), but above the surface the simple equation for a mass point become inexact ( up to almost 0.3%) with a slightly stronger gravity field up to about 1,35 times the radius from center and then goes slightly under the simple value (18ppm at r*1.7).This "variation " is in the range of Cavendish type measurments for both equipment setup and measured differensies . I am sure this is not taken into account , and the mass of earth is said to be M= gr2/G and simple but laborious (computer so) math shows simply that for earth being "mixed density" it do not add up even on the surface .I could not find any precice details on done measurment , although there must be quite few (in last 100 years) . Could easily compare because the values ( I calculated ) depend only on the ratio R/r ( radius and distance) and G would be =F*r^2*k1*k2/(m1*m2) Sorry that is r/R distance per radius

While needing mass of sun for some speculative calc. I found that masses because of earths mass and G uncertainity is quite uncertain and merely agreed upon. Look't into it an ended up calculatin earth mass and found out ,by my meaning ,that earth mass by GM/r2 is certainly not correct .The simple formula work only on homogenous density spheres and gravitation only on far objects . To my surprise I found out that g near perfect spheres behave not "linear" to 1/r2 but does a miniscule swank and a high before at surface diving to center zero. I woul be very happy to get knowlege if this is taken into ackount when measuring G . So far I think earth (not homogenous density) 0,15% too small G.* M (only flat density so far) and G is by my understandin less than exact. Coul someone tell me if the anomalies from straight GM/r2 compared to my calcs. for a 3-dimensional sphere are taken in some way in account when measuring G . It was not surpricing on the otherhand to see that same calculated values apply to any perfect sphere for F(1/r2) /r any density and dimension .Te distance R *x is given multiples of any r and the used body in this table is 0,1524 m 158kg Pleace see charts and table Thank you for any interest It almost killed my 14 yo Dell summing up gravity of 36000 * 2pi (Guldin/Pappus) elements (could have used more than 200+ columns 19.6 g beräkn1.bmp Cavendish1.bmp Cavendish2.bmp