Strange Posted January 12, 2017 Posted January 12, 2017 In g= M Ti /(p-2R^2/5p)^2 , M*Ti is equal to Newtonian G*M If M*Ti = G*M, then Ti = G. Or do you need to take a class in basic arithmetic, as well as calculus? and l =p -2R^2/5p is for example in sun earth distance r= 1.5 *10^8 km There is no l in your equation. 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 As this data is available, why not use it? Or perhaps you think that the value of G was only measured once, by Cavendish using cylinders? You could start here, if you really don't know how G is measured: http://rsta.royalsocietypublishing.org/content/372/2026/20140022
Sensei Posted January 12, 2017 Posted January 12, 2017 (edited) 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 Heavier elements than Nickel are made only in minute amount during supernova explosion.. 7874 / 5513 = 1.43. Earth could be just 43% more heavier, if, and only if, it would be made of pure Iron. But it's not. Additionally, gravity on surface is not uniform, and can be disturbed even by existence of ores, minerals, etc. below the ground. And it's used during mineral exploration. https://en.wikipedia.org/wiki/Gravity_of_Earth https://en.wikipedia.org/wiki/Gravity_anomaly https://en.wikipedia.org/wiki/Exploration_geophysics https://en.wikipedia.org/wiki/Gravity_gradiometry Gravity anomaly map: . http://earthobservatory.nasa.gov/IOTD/view.php?id=3666 Edited January 12, 2017 by Sensei
Timo Moilanen Posted January 13, 2017 Author Posted January 13, 2017 Your replies don't seem to have much connection to the question asked. Why can't you calculate the height of a geostationary satellite? 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 Heavier elements than Nickel are made only in minute amount during supernova explosion.. 7874 / 5513 = 1.43. Earth could be just 43% more heavier, if, and only if, it would be made of pure Iron. But it's not. Additionally, gravity on surface is not uniform, and can be disturbed even by existence of ores, minerals, etc. below the ground. And it's used during mineral exploration. https://en.wikipedia.org/wiki/Gravity_of_Earth https://en.wikipedia.org/wiki/Gravity_anomaly https://en.wikipedia.org/wiki/Exploration_geophysics https://en.wikipedia.org/wiki/Gravity_gradiometry Gravity anomaly map: .PIA04652_lrg.jpg http://earthobservatory.nasa.gov/IOTD/view.php?id=3666 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 -2
Sensei Posted January 13, 2017 Posted January 13, 2017 If your calculations disagree with experimental data, your "law" will be dismissed. 1
Timo Moilanen Posted January 13, 2017 Author Posted January 13, 2017 You have to look at all of the calculations we can do with Newtonian gravity. These are sufficiently precise to say that the law works. That's what makes it the right method. Add to this that we have an exceedingly well-tested General theory of relativity that reduces to the Newtonian form. Given all of the data available that has been used to come up with the planetary masses and other values, you should be able to get whatever constants you need. 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 ! If your calculations disagree with experimental data, your "law" will be dismissed. Sure If M*Ti = G*M, then Ti = G. Or do you need to take a class in basic arithmetic, as well as calculus? There is no l in your equation. As this data is available, why not use it? Or perhaps you think that the value of G was only measured once, by Cavendish using cylinders? You could start here, if you really don't know how G is measured: http://rsta.royalsocietypublishing.org/content/372/2026/20140022 Thank you there must be available couple hundred years of
Klaynos Posted January 13, 2017 Posted January 13, 2017 Google scholar is good for these kinds of searches to find published data. https://scholar.google.co.uk/scholar?hl=en&as_sdt=0,5&q=measuring+the+gravitational+constant
swansont Posted January 13, 2017 Posted January 13, 2017 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 ! There is more astronomical data out there than masses. Do you disagree with orbital periods, too? How about distances?
Klaynos Posted January 13, 2017 Posted January 13, 2017 There is more astronomical data out there than masses. Do you disagree with orbital periods, too? How about distances? It'd probably break our understanding of the sun's fusion reactions as well.
imatfaal Posted January 13, 2017 Posted January 13, 2017 ! Moderator Note OK - Timo. I am going to insist upon seeing a worked example - your equation appears in your very first post and I am still not convinced you have even defined every variable. In your next post would you please repeat your equation with each individual variable elucidated (ie what is quantity it represents, units, etc) and a worked example - we insist on rigor in this forum and without this we may be forced to close down the thread.
Timo Moilanen Posted January 14, 2017 Author Posted January 14, 2017 What is "midpoint of gravity field"? How to calculate it? It is point where forces cancel each other between two objects? By "midpoint of mass" you probably meant center of mass, I suppose so. 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 . ! Moderator Note OK - Timo. I am going to insist upon seeing a worked example - your equation appears in your very first post and I am still not convinced you have even defined every variable. In your next post would you please repeat your equation with each individual variable elucidated (ie what is quantity it represents, units, etc) and a worked example - we insist on rigor in this forum and without this we may be forced to close down the thread. 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
Strange Posted January 14, 2017 Posted January 14, 2017 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 . Please show, in detail, how you derive this. What do you mean by "slices of a sphere"?
Timo Moilanen Posted January 14, 2017 Author Posted January 14, 2017 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 "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 ..), "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 Then maybe you should learn how mass of Earth, and mass of Sun is calculated/estimated? p=m/V Average density is mass divided by volume. If you reverse equation you have: m=V*p V is volume of Earth, V = 4/3*PI*r^3 r - average radius of Earth r=6371000 m Then you just have to measure average density of object, to estimate its mass.. Density of water is 1000 kg/m^3 Density of CaCO3 is 2711 kg/m^3 Density of Iron is 7874 kg/m^3 But Earth is neither made of pure Iron, nor pure water, so average density must lie between these two values. And it's 5513 kg/m^3, slightly above the middle between 2711 and 7874 kg/m^3. ps. What is mass of Earth and Sun according to your calculations? 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
swansont Posted January 14, 2017 Posted January 14, 2017 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 You should be able to estimate the mass of the earth from that, right?
John Cuthber Posted January 14, 2017 Posted January 14, 2017 Then maybe you should learn how mass of Earth, and mass of Sun is calculated/estimated? p=m/V Average density is mass divided by volume. If you reverse equation you have: m=V*p V is volume of Earth, V = 4/3*PI*r^3 r - average radius of Earth r=6371000 m Then you just have to measure average density of object, to estimate its mass.. Density of water is 1000 kg/m^3 Density of CaCO3 is 2711 kg/m^3 Density of Iron is 7874 kg/m^3 But Earth is neither made of pure Iron, nor pure water, so average density must lie between these two values. And it's 5513 kg/m^3, slightly above the middle between 2711 and 7874 kg/m^3. ps. What is mass of Earth and Sun according to your calculations? You really need to get out of the habit of using spurious accuracy. Citing the density of iron to 4 figures is meaningless, since that's the density at low pressures and the stuff deep in the earth has a variable, and rather higher density near 10 g/ml. 1
Timo Moilanen Posted January 14, 2017 Author Posted January 14, 2017 Please show, in detail, how you derive this. What do you mean by "slices of a sphere"? 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 ) You should be able to estimate the mass of the earth from that, right? Not easily , but with measured gravitation from several heights it should give masses and thereby densities on different depths 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# 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
Strange Posted January 14, 2017 Posted January 14, 2017 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. That sounds as if it will give you the wrong answer. The mass at the outer edge of the disk will have less effect than the mass at the centre. This is why I asked you to show how you derive the equation. Please do that now. Someone called Newton did this properly years ago and got the right answer. You could learn from that. It might help you see where you have gone wrong. The integer being next to impossible I think is a mere excuse and lazines . I have no idea what you are talking about. 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# As there are thousands of satellites you can use (from the moon to geostationary satellites to the ISS) I don't understand why you do not have an exact value for Ti and the mass of the earth.
swansont Posted January 14, 2017 Posted January 14, 2017 As there are thousands of satellites you can use (from the moon to geostationary satellites to the ISS) I don't understand why you do not have an exact value for Ti and the mass of the earth. It's worse than that. Since we know that Newtonian physics works, you have an infinite number of solutions to use to do a parameter fit.
Timo Moilanen Posted January 15, 2017 Author Posted January 15, 2017 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 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% It's worse than that. Since we know that Newtonian physics works, you have an infinite number of solutions to use to do a parameter fit. I do not have data on thousands of satellites and the mass of earth and other planets and sun I predict are not right That sounds as if it will give you the wrong answer. The mass at the outer edge of the disk will have less effect than the mass at the centre. This is why I asked you to show how you derive the equation. Please do that now. Someone called Newton did this properly years ago and got the right answer. You could learn from that. It might help you see where you have gone wrong. I have no idea what you are talking about. As there are thousands of satellites you can use (from the moon to geostationary satellites to the ISS) I don't understand why you do not have an exact value for Ti and the mass of the earth. The integer ( derivative ) show that Newton is right but I went one step further 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
Strange Posted January 15, 2017 Posted January 15, 2017 That sounds as if it will give you the wrong answer. The mass at the outer edge of the disk will have less effect than the mass at the centre. This is why I asked you to show how you derive the equation. Please do that now. Still waiting for this...
Timo Moilanen Posted January 15, 2017 Author Posted January 15, 2017 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 My paper starts with the integer and mainpoints how its done
Strange Posted January 15, 2017 Posted January 15, 2017 My paper starts with the integer and mainpoints how its done Please explain it here (as required by the rules). p.s. I realise now when you say "integer" you mean "integral".
Timo Moilanen Posted January 15, 2017 Author Posted January 15, 2017 Still waiting for this... 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 Please explain it here (as required by the rules). p.s. I realise now when you say "integer" you mean "integral". Yes integral as in Swedish and I'm very confused by the antiderivative too . In online text they use integer ?
Strange Posted January 15, 2017 Posted January 15, 2017 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 So you will get the wrong answer. I expect this is the main reason your "theory" is wrong. There may be others. But as you have failed to explain anything, it is not clear. In online text they use integer ? I doubt it very much.
Timo Moilanen Posted January 15, 2017 Author Posted January 15, 2017 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 ? And Newton did not do this integral ,what changes nothing because I used the essential from the next integral I did
Strange Posted January 15, 2017 Posted January 15, 2017 And Newton did not do this integral Of course he did. Don't be so stupid. https://en.wikipedia.org/wiki/Shell_theorem As swansont points out you can test your equation by choosing a pair of spherical masses with any mass, any radius and any separation. Then compare your results with Newton. If your results are different then they are wrong.
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