Timo Moilanen Posted January 10, 2017 Share Posted January 10, 2017 (edited) 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 Edited January 10, 2017 by Timo Moilanen Link to comment Share on other sites More sharing options...
DrKrettin Posted January 10, 2017 Share Posted January 10, 2017 It's difficult to know what you are talking about. I guess you mean "law" and not "lav", and where is the attachment? Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 10, 2017 Author Share Posted January 10, 2017 Sorry computer denied sending attachment pages Link to comment Share on other sites More sharing options...
swansont Posted January 10, 2017 Share Posted January 10, 2017 Please use your law to find the orbital distance of a geostationary satellite, and explain why your method is easier or more accurate than Newton. 1 Link to comment Share on other sites More sharing options...
Sriman Dutta Posted January 10, 2017 Share Posted January 10, 2017 Explain all variables in your postulate. Link to comment Share on other sites More sharing options...
geordief Posted January 10, 2017 Share Posted January 10, 2017 How many posts are required to post attachments? Perhaps Timo Moilanen has not enough yet at only 6? Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 10, 2017 Author Share Posted January 10, 2017 s Link to comment Share on other sites More sharing options...
Strange Posted January 10, 2017 Share Posted January 10, 2017 I must conclude that gravity need to be calculated to the midpoint of gravity field instead of midpoint of mass . Well, it would be very easy for you to confirm this. You can calculate the orbital speed of a planet round the sun and compare it with what we measure. You could show how Kepler's laws are derived from your equation. You could, as someone else requested, calculate the orbital height of a geostationary satellite. Have you done any of these things to confirm your equation works? Link to comment Share on other sites More sharing options...
Sensei Posted January 10, 2017 Share Posted January 10, 2017 midpoint of gravity field instead of midpoint of mass . 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. Link to comment Share on other sites More sharing options...
swansont Posted January 10, 2017 Share Posted January 10, 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. But the force cancellation point is not the center of mass, in general, since one depends on r^2 and the other on r. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 11, 2017 Author Share Posted January 11, 2017 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" Please use your law to find the orbital distance of a geostationary satellite, and explain why your method is easier or more accurate than Newton. 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 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) It's difficult to know what you are talking about. I guess you mean "law" and not "lav", and where is the attachment? Sorry for my language skills Explain all variables in your postulate. 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 How many posts are required to post attachments? Perhaps Timo Moilanen has not enough yet at only 6? Sorry for that Link to comment Share on other sites More sharing options...
Strange Posted January 11, 2017 Share Posted January 11, 2017 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 . Until you can do that, you have nothing to discuss. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 11, 2017 Author Share Posted January 11, 2017 But the force cancellation point is not the center of mass, in general, since one depends on r^2 and the other on r. Until you can do that, you have nothing to discuss. That will not happen very soon as I think needed measurements do not exist yet Link to comment Share on other sites More sharing options...
Strange Posted January 11, 2017 Share Posted January 11, 2017 That will not happen very soon as I think needed measurements do not exist yet Perhaps you need to explain why you think these measurements do not exist. Then we can explain why you are wrong. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 11, 2017 Author Share Posted January 11, 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. 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 Link to comment Share on other sites More sharing options...
Strange Posted January 11, 2017 Share Posted January 11, 2017 Why can't you can use your new gravity "lav" (toilet) to calculate the mass of the Earth? I dare call the formula g= M*Ti /( p-R^2/5p )^2 a new gravity lav I don't know what g, M, Ti, p, and R represent but why can't you use this to calculate the mass of the Earth (is that M?) 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 . Sounds like you need to do an introductory course in calculus. https://en.wikipedia.org/wiki/Shell_theorem Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 11, 2017 Author Share Posted January 11, 2017 Perhaps you need to explain why you think these measurements do not exist. Then we can explain why you are wrong. 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 Why can't you can use your new gravity "lav" (toilet) to calculate the mass of the Earth? I don't know what g, M, Ti, p, and R represent but why can't you use this to calculate the mass of the Earth (is that M?) Sounds like you need to do an introductory course in calculus. https://en.wikipedia.org/wiki/Shell_theorem 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 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 Link to comment Share on other sites More sharing options...
Strange Posted January 11, 2017 Share Posted January 11, 2017 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 Your replies don't seem to have much connection to the question asked. Why can't you calculate the height of a geostationary satellite? Link to comment Share on other sites More sharing options...
swansont Posted January 11, 2017 Share Posted January 11, 2017 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 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. Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 11, 2017 Author Share Posted January 11, 2017 Please use your law to find the orbital distance of a geostationary satellite, and explain why your method is easier or more accurate than Newton. For Me = 5,97 *10^24 it give 32500km = 28800km abowe surface but I'm not satisfied with input mass and simplification Link to comment Share on other sites More sharing options...
Strange Posted January 11, 2017 Share Posted January 11, 2017 For Me = 5,97 *10^24 it give 32500km = 28800km abowe surface but I'm not satisfied with input mass and simplification 1. Please show how you calculated this. 2. It is wrong, therefore your model does not work. Link to comment Share on other sites More sharing options...
swansont Posted January 11, 2017 Share Posted January 11, 2017 For Me = 5,97 *10^24 it give 32500km = 28800km abowe surface but I'm not satisfied with input mass and simplification Show your work. Link to comment Share on other sites More sharing options...
Sensei Posted January 11, 2017 Share Posted January 11, 2017 (edited) 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 . 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? Edited January 11, 2017 by Sensei Link to comment Share on other sites More sharing options...
Timo Moilanen Posted January 12, 2017 Author Share Posted January 12, 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. 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 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? 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 Link to comment Share on other sites More sharing options...
Klaynos Posted January 12, 2017 Share Posted January 12, 2017 so I see no other way to determine Ti than using results from Cavendish experiment wirh spheres . ... ... Since I have no near to exact value on my constant Ti Why have you not done your first point here to solve your second? Wikipedia walks you through how to do this for Newtonian gravity do you should have no trouble for yours. Until you've got a reasonable value for Ti and have shown some reasonable correlation with what we can measure (orbital periods is a good start) at have nothing to discuss here. Insect it's worse than that because your single attempt at solving an orbital period problem got the wrong result, we should therefore dismiss your idea with no further comment. Link to comment Share on other sites More sharing options...
Recommended Posts